Chapter 11Gases

2006 Prentice Hall

2

CHAPTER OUTLINE

Properties of Gases Pressure amp Its Measurement The Gas Laws Vapor Pressure amp Boiling Point Combined Gas Law Avogadrorsquos Law STP amp Molar Volume Ideal Gas Law Partial Pressures

3

PROPERTIESOF GASES

Gases are the least dense and most mobile of the three phases of matter

Particles of matter in the gas phase are spaced far apart from one another and move rapidly and collide with each other often

Gases occupy much greater space than the same amount of liquid or solid

This is because the gas particles are spaced apart from one another and are therefore compressible

Solid or liquid particles are spaced much closer and cannot be compressed

4

PROPERTIESOF GASES

Gases are characterized by four properties These are

1 Pressure (P)

2 Volume (V)

3 Temperature (T)

4 Amount (n)

5

KINETIC-MOLECULARTHEORY

Scientists use the kinetic-molecular theory (KMT) to describe the behavior of gases

The KMT consists of several postulates

1 Gases consist of small particles (atoms or molecules) that move randomly with rapid velocities

3 The distance between the particles is large compared to their size Therefore the volume occupied by gas molecules is small compared to the volume of the gas

2 Gas particles have little attraction for one another Therefore attractive forces between gas molecules can be ignored

6

KINETIC-MOLECULARTHEORY

5 The average kinetic energy of gas molecules is directly proportional to the absolute temperature (Kelvin)

Animation

4 Gas particles move in straight lines and collide with each other and the container frequently The force of collision of the gas particles with the walls of the container causes pressure

2

CHAPTER OUTLINE

Properties of Gases Pressure amp Its Measurement The Gas Laws Vapor Pressure amp Boiling Point Combined Gas Law Avogadrorsquos Law STP amp Molar Volume Ideal Gas Law Partial Pressures

3

PROPERTIESOF GASES

Gases are the least dense and most mobile of the three phases of matter

Particles of matter in the gas phase are spaced far apart from one another and move rapidly and collide with each other often

Gases occupy much greater space than the same amount of liquid or solid

This is because the gas particles are spaced apart from one another and are therefore compressible

Solid or liquid particles are spaced much closer and cannot be compressed

4

PROPERTIESOF GASES

Gases are characterized by four properties These are

1 Pressure (P)

2 Volume (V)

3 Temperature (T)

4 Amount (n)

5

KINETIC-MOLECULARTHEORY

Scientists use the kinetic-molecular theory (KMT) to describe the behavior of gases

The KMT consists of several postulates

1 Gases consist of small particles (atoms or molecules) that move randomly with rapid velocities

3 The distance between the particles is large compared to their size Therefore the volume occupied by gas molecules is small compared to the volume of the gas

2 Gas particles have little attraction for one another Therefore attractive forces between gas molecules can be ignored

6

KINETIC-MOLECULARTHEORY

5 The average kinetic energy of gas molecules is directly proportional to the absolute temperature (Kelvin)

Animation

4 Gas particles move in straight lines and collide with each other and the container frequently The force of collision of the gas particles with the walls of the container causes pressure

3

PROPERTIESOF GASES

Gases are the least dense and most mobile of the three phases of matter

Particles of matter in the gas phase are spaced far apart from one another and move rapidly and collide with each other often

Gases occupy much greater space than the same amount of liquid or solid

This is because the gas particles are spaced apart from one another and are therefore compressible

Solid or liquid particles are spaced much closer and cannot be compressed

4

PROPERTIESOF GASES

Gases are characterized by four properties These are

1 Pressure (P)

2 Volume (V)

3 Temperature (T)

4 Amount (n)

5

KINETIC-MOLECULARTHEORY

Scientists use the kinetic-molecular theory (KMT) to describe the behavior of gases

The KMT consists of several postulates

1 Gases consist of small particles (atoms or molecules) that move randomly with rapid velocities

3 The distance between the particles is large compared to their size Therefore the volume occupied by gas molecules is small compared to the volume of the gas

2 Gas particles have little attraction for one another Therefore attractive forces between gas molecules can be ignored

6

KINETIC-MOLECULARTHEORY

5 The average kinetic energy of gas molecules is directly proportional to the absolute temperature (Kelvin)

Animation

4 Gas particles move in straight lines and collide with each other and the container frequently The force of collision of the gas particles with the walls of the container causes pressure

4

PROPERTIESOF GASES

Gases are characterized by four properties These are

1 Pressure (P)

2 Volume (V)

3 Temperature (T)

4 Amount (n)

5

KINETIC-MOLECULARTHEORY

Scientists use the kinetic-molecular theory (KMT) to describe the behavior of gases

The KMT consists of several postulates

1 Gases consist of small particles (atoms or molecules) that move randomly with rapid velocities

3 The distance between the particles is large compared to their size Therefore the volume occupied by gas molecules is small compared to the volume of the gas

2 Gas particles have little attraction for one another Therefore attractive forces between gas molecules can be ignored

6

KINETIC-MOLECULARTHEORY

5 The average kinetic energy of gas molecules is directly proportional to the absolute temperature (Kelvin)

Animation

4 Gas particles move in straight lines and collide with each other and the container frequently The force of collision of the gas particles with the walls of the container causes pressure

5

KINETIC-MOLECULARTHEORY

Scientists use the kinetic-molecular theory (KMT) to describe the behavior of gases

The KMT consists of several postulates

1 Gases consist of small particles (atoms or molecules) that move randomly with rapid velocities

3 The distance between the particles is large compared to their size Therefore the volume occupied by gas molecules is small compared to the volume of the gas

2 Gas particles have little attraction for one another Therefore attractive forces between gas molecules can be ignored

6

KINETIC-MOLECULARTHEORY

5 The average kinetic energy of gas molecules is directly proportional to the absolute temperature (Kelvin)

Animation

4 Gas particles move in straight lines and collide with each other and the container frequently The force of collision of the gas particles with the walls of the container causes pressure

6

KINETIC-MOLECULARTHEORY

5 The average kinetic energy of gas molecules is directly proportional to the absolute temperature (Kelvin)

Animation

4 Gas particles move in straight lines and collide with each other and the container frequently The force of collision of the gas particles with the walls of the container causes pressure

Tros Introductory Chemistry Chapter

7

Kinetic Molecular Theory

8

PRESSURE ampITS MEASUREMENT

Pressure is the result of collision of gas particles with the sides of the container Pressure is defined as the force per unit area

Pressure is measured in units of atmosphere (atm) or mmHg or torr The SI unit of pressure is pascal (Pa) or kilopascal (kPa)

1 atm = 760 mmHg = 101325 kPa

1 mmHg = 1 torr

Tros Introductory Chemistry Chapter

9

Common Units of Pressure

Unit Average Air Pressure at Sea Level

pascal (Pa) 101325

kilopascal (kPa) 101325

atmosphere (atm) 1 (exactly)

millimeters of mercury (mmHg) 760 (exactly)

inches of mercury (inHg) 2992

torr (torr) 760 (exactly)

pounds per square inch (psi lbsin2) 147

10

PRESSURE ampITS MEASUREMENT

Atmospheric pressure can be measured with the use of a barometer

Mercury is used in a barometer due to its high density

At sea level the mercury stands at 760 mm above its base

Tros Introductory Chemistry Chapter

11

Atmospheric Pressure amp Altitude

bull the higher up in the atmosphere you go the lower the atmospheric pressure is around youat the surface the atmospheric pressure is

147 psi but at 10000 ft is is only 100 psibull rapid changes in atmospheric pressure

may cause your ears to ldquopoprdquo due to an imbalance in pressure on either side of your ear drum

Tros Introductory Chemistry Chapter

12

Pressure Imbalance in Ear

If there is a differencein pressure acrossthe eardrum membranethe membrane will bepushed out ndash what we commonly call a ldquopopped eardrumrdquo

13

Example 1

The atmospheric pressure at Walnut CA is 740 mmHg Calculate this pressure in torr and atm

1 mmHg = 1 torr

740 mmHg = 740 torr

740 mmHg atm

x mmHg760

1= 0974 atm

14

Example 2

The barometer at a location reads 112 atm Calculate the pressure in mmHg and torr

1 mmHg = 1 torr

851 mmHg = 851 torr

112 atm mmHg

x atm1760

= 851 mmHg

15

PRESSURE amp MOLES OF A GAS

The pressure of a gas is directly proportional to the number of particles (moles) present

The greater the moles of the gas the greater the pressure

16

BOYLErsquoS LAW

The relationship of pressure and volume in gases is called Boylersquos Law

At constant temperature the volume of a fixed amount of gas is inversely proportional to its pressure

As pressure increases the volume of the gas decreases

Pressure and volume of a

gas are inversely

proportional

17

BOYLErsquoS LAW

Boylersquos Law can be mathematically expressed as

P1 x V1 = P2 x V2

P and V under initial

condition

P and V under final

condition

18

Example 1

A sample of gas has a volume of 12-L and a pressure of 4500 mmHg What is the volume of the gas when the pressure is reduced to 750 mmHg

1 12

2

P VV =

P

V2 = 72 L

P1= 4500 mmHg

P2= 750 mmHg

V1 = 12 L

V2 =

P1 x V1 = P2 x V2

(4500 mmHg)(12 L)=

(750 mmHg)

Pressure decreases Volume

increases

19

Example 2

A sample of hydrogen gas occupies 40 L at 650 mmHg What volume would it occupy at 20 atm

1 12

2

P VV =

P

V2 = 17 L

P1= 650 mmHg

P2= 20 atm

V1 = 40 L

V2 =

P1 x V1 = P2 x V2

(650 mmHg)(40 L)=

(1520 mmHg)Pressures must be in the same

unit

P2 = 20 atm = 1520 mmHgVolume decreases

Pressure increases

20

CHARLESrsquoS LAW

The relationship of temperature and volume in gases is called Charlesrsquo Law

At constant pressure the volume of a fixed amount of gas is directly proportional to its absolute temperature

As temperature increases the volume of the gas increases

Temperature and volume of a gas

are directly proportional

21

CHARLESrsquoS LAW

Charlesrsquo Law can be mathematically expressed as

V and T under initial

condition

1 2

1 2

V V=

T T

V and T under final condition

T must be in units of K

22

Example 1

A 20-L sample of a gas is cooled from 298K to 278 K at constant pressure What is the new volume of the gas

22 1

1

TV = V x

T

V2 = 19 L

V1 = 20 L

V2 =

T1 = 298 K

T2 = 278 K

278 K= 20 L x

298 K

Volume decreases1 2

1 2

V V=

T T

Temperature decreases

23

Example 2

If 200 L of oxygen is cooled from 100ordmC to 0ordmC what is the new volume

22 1

1

TV = V x

T

V2 = 146 L

V1 = 200 L

V2 =

T1 = 100ordmC

T2 = 0ordmC

273 K= 200 L x

373 K

T1= 100ordmC + 273 = 373 K

T2= 0ordmC + 273 = 273 K

Temperatures must be in

KVolume

decreasesTemperature

decreases

24

GAY-LUSSACrsquoSLAW

The relationship of temperature and pressure in gases is called Gay-Lussacrsquos Law

At constant volume the pressure of a fixed amount of gas is directly proportional to its absolute temperature

As temperature increases the

pressure of the gas increases

Temperature and pressure of a gas

are directly proportional

25

GAY-LUSSACrsquoSLAW

Gay-Lussacrsquos Law can be mathematically expressed as

P and T under initial

condition

1 2

1 2

P P=

T T

P and T under final conditionT must be in

units of K

26

Example 1

An aerosol spray can has a pressure of 40 atm at 25C What pressure will the can have if it is placed in a fire and reaches temperature of 400C

P1 = 40 atm

P2 =

T1 = 25ordmC

T2 = 400ordmC

T1= 25ordmC + 273 = 298 K

T2= 400ordmC + 273 = 673 K

Temperatures must be in K

27

Example 1

An aerosol spray can has a pressure of 40 atm at 25C What pressure will the can have if it is placed in a fire and reaches temperature of 400C

22 1

1

TP = P x

T

P2 = 90 atm

P1 = 40 atm

P2 =

T1 = 298 K

T2 = 673 K

673 K= 40 L x

298 K

Pressure increases

1 2

1 2

P P=

T T

Temperature increases

28

Example 2

A cylinder of gas with a volume of 150-L and a pressure of 965 mmHg is stored at a temperature of 55C To what temperature must the cylinder be cooled to reach a pressure of 850 mmHg

22 1

1

PT = T x

P

T2 = 289 K

P1 = 965 mmHg

P2 = 850 mmHg

T1 = 55ordmC

T2 =

850 mmHg= 328 K x

965 mmHg

T1= 55ordmC + 273 = 328 K

T2 = 16ordmC

Temperature decreases

Pressure decreases

29

VAPOR PRESSUREamp BOILING POINT

In an open container liquid molecules at the surface that possess sufficient energy can break away from the surface and become gas particles or vapor

In a closed container these gas particles can accumulate and create pressure called vapor pressure

Vapor pressure is defined as the pressure above a liquid at a given temperature Vapor pressure varies with each liquid and increases with temperature

30

VAPOR PRESSUREamp BOILING POINT

Listed below is the vapor pressure of water at various temperatures

31

VAPOR PRESSUREamp BOILING POINT

A liquid reaches its boiling point when its vapor pressure becomes equal to the external pressure (atmospheric pressure)

For example at sea level water reaches its boiling point at 100C since its vapor pressure is 760 mmHg at this temperature

32

VAPOR PRESSUREamp BOILING POINT

At higher altitudes where atmospheric pressure is lower water reaches boiling point at temperatures lower than 100C

For example in Denver where atmospheric pressure is 630 mmHg water boils at 95C since its vapor pressure is 630 mmHg at this temperature

33

COMBINEDGAS LAW

All pressure-volume-temperature relationships can be combined into a single relationship called the combined gas law

1 1 2 2

1 2

P V P V=

T T

Initial condition

This law is useful for studying the effect of changes in two variables

Final condition

34

COMBINEDGAS LAW

The individual gas laws studied previously are embodied in the combined gas law

1 1 2 2

1 2

P V P V=

T T

Boylersquos LawCharlesrsquos

LawGay-Lussacrsquos

Law

35

Example 1

A 250-mL sample of gas has a pressure of 400 atm at a temperature of 10C What is the volume of the gas at a pressure of 100 atm and a temperature of 18C

1 22 1

2 1

P TV = V x x

P T

V2 = 103 mL

P1= 400 atm V1= 250 mL T1= 283 K P2= 100 atm V2= T2 = 291 K

2

400 atm 291 KV = 250 mL x x

100 atm 283 K

1 1 2 2

1 2

P V P V=

T TBoth pressure

and temp change

Must use combined

gas law

36

AVOGADROrsquoSLAW

The relationship of moles and volume in gases is called Avogadrorsquos Law

At constant temperature and pressure the volume of a fixed amount of gas is directly proportional to the number of moles

As number of moles increases

the volume of the gas

increases

Number of moles and volume of a gas are directly

proportional

Tros Introductory Chemistry Chapter

37

Avogadrorsquos Law

38

AVOGADROrsquoSLAW

As a result of Avogadrorsquos Law equal volumes of different gases at the same temp and pressure contain equal number of moles (molecules)

2 Tanks of gas of equal volume at the same T amp P contain the same number of molecules

39

AVOGADROrsquoSLAW

For example

2 H2 (g) + 1 O2 (g) 2 H2O (g)

2 molecules 1 molecule 2 molecules

2 moles 1 mole 2 moles

2 Liters 1 Liter 2 Liters

Avogadrorsquos Law also allows chemists to relate volumes and moles of a gas in a chemical reaction

This relationship is only valid for

gaseous substances

40

Example 1

A sample of helium gas with a mass of 180 g occupies 16 Liters at a particular temperature and pressure What mass of oxygen would occupy16 L at the same temperature and pressure

1 mol= 180 g x

400 gmol of He = 450 mol

mol of O2 = mol of He = 450 mol

Same Temp amp Pressure

mass of O2

320 g= 450 mol x

1 mol= 144 g

41

Example 2

How many Liters of NH3 can be produced from reaction of 18 L of H2 with excess N2 as shown below

18 L H23

2

L NHx

L H= 12 L NH3

N2 (g) + 3 H2 (g) 2 NH3 (g)

32

42

STP amp MOLAR VOLUME

To better understand the factors that affect gas behavior a set of standard conditions have been chosen for use and are referred to as Standard Temperature and Pressure (STP)

STP =

1 atm (760 mmHg)

0ordmC (273 K)

43

STP amp MOLAR VOLUME

At STP conditions one mole of any gas is observed to occupy a volume of 224 L

V = 224 L

Molar Volume at

STP

44

Example 1

If 200 L of a gas at STP has a mass of 323 g what is the molar mass of the gas

mol200 L x

Lmol of gas =

2241

= 00893 mol

g

molMolar mass =

323 g=

00893 mol= 362 gmol Molar

volume at STP

45

Example 2

A sample of gas has a volume of 250 L at 730 mmHg and 20C What is the volume of this gas at STP

1 22 1

2 1

P TV = V x x

P T

V2 = 224 L

P1= 730 mmHg V1= 250 L T1= 293 K P2= 760 mmHg V2= T2 = 273 K

2

730 mmHg 273 KV = 250 L x x

760 mmHg 293 K

46

IDEAL GAS LAW

Combining all the laws that describe the behavior of gases one can obtain a useful relationship that relates the volume of a gas to the temperature pressure and number of moles

n R TV =

P

Universal gas constant

R = 00821 L atmmol K

47

IDEAL GAS LAW

This relationship is called the Ideal Gas Law and commonly written as

P V = n R T

Pressure in atm Volume

in Liters

Number of moles

Temp in K

48

Example 1

A sample of H2 gas has a volume of 856 L at a temp of 0C and pressure of 15 atm Calculate the moles of gas present

PVn =

RT

= 057 mol

P = 15 atm V = 856 L T = 273 K n = R = 00821

PV = nRT

(15 atm)(856 L)=

L atm(00821 )(273 K)

mol K

49

Example 2

What volume does 400 g of N2 gas occupy at 10C and 750 mmHg

P =750 mmHg V = T = 283 K n = not given m = 400 g R = 00821

mol of N2 =1 mol

400 g x280 g

= 143 mol

P = 750 mmHg1 atm

x760 mmHg

Moles must be calculated

from mass

Pressure must be

converted to atm

P = 0987 atm

50

Example 2

What volume does 400 g of N2 gas occupy at 10C and 750 mmHg

nRTV =

P

V = 337 L

P =0987 atm V = T = 283 K n = 143 R = 00821

L atm(143 mol)(00821 )(283 K)

mol KV = 0987 atm

Tros Introductory Chemistry Chapter

51

Ideal vs Real Gases

bull Real gases often do not behave like ideal gases at high pressure or low temperature

bull Ideal gas laws assume1) no attractions between gas molecules2) gas molecules do not take up space based on the Kinetic-Molecular Theory

bull at low temperatures and high pressures these assumptions are not valid

Tros Introductory Chemistry Chapter

52

Ideal vs Real Gases

53

PARTIAL PRESSURES

Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases

Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture

In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas

54

DALTONrsquoSLAW

In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture

This is called Daltonrsquos law of partial pressures

Total pressure of a gas mixture

Ptotal = P1 + P2 + P3 +

Sum of the partial pressures of the gases in the mixture

=

55

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = PHe + PAr

Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture

= 60 atm

56

PARTIAL PRESSURES

The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture

For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container

57

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = 60 atm

The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture

Twice as many moles of Ar

compared to He

PAr = 2 PHe

58

Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank

Ptotal = Poxygen + Phelium

Poxygen = 1140 mmHg x

1 atm

760 mmHg= 150 atm

Phelium

= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm

59

Example 2

A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture

Ptotal = Poxygen + Pnitrogen

Pnitrogen = 2 Poxygen

Twice as many moles of N2

compared to O2

Pnitrogen = 23 (Ptotal ) = 20 atm

Poxygen = 13 (Ptotal ) = 10 atm

60

THE END

• Slide 1
• CHAPTER OUTLINE
• PROPERTIES OF GASES
• PROPERTIES OF GASES (2)
• KINETIC-MOLECULAR THEORY
• KINETIC-MOLECULAR THEORY (2)
• Kinetic Molecular Theory
• PRESSURE amp ITS MEASUREMENT
• Common Units of Pressure
• PRESSURE amp ITS MEASUREMENT (2)
• Atmospheric Pressure amp Altitude
• Pressure Imbalance in Ear
• Example 1
• Example 2
• PRESSURE amp MOLES OF A GAS
• BOYLErsquoS LAW
• BOYLErsquoS LAW (2)
• Example 1 (2)
• Example 2 (2)
• CHARLESrsquoS LAW
• CHARLESrsquoS LAW (2)
• Example 1 (3)
• Example 2 (3)
• GAY-LUSSACrsquoS LAW
• GAY-LUSSACrsquoS LAW (2)
• Example 1 (4)
• Example 1 (5)
• Example 2 (4)
• VAPOR PRESSURE amp BOILING POINT
• VAPOR PRESSURE amp BOILING POINT (2)
• VAPOR PRESSURE amp BOILING POINT (3)
• VAPOR PRESSURE amp BOILING POINT (4)
• COMBINED GAS LAW
• COMBINED GAS LAW (2)
• Example 1 (6)
• AVOGADROrsquoS LAW
• Avogadrorsquos Law
• AVOGADROrsquoS LAW (2)
• AVOGADROrsquoS LAW (3)
• Example 1 (7)
• Example 2 (5)
• STP amp MOLAR VOLUME
• STP amp MOLAR VOLUME (2)
• Example 1 (8)
• Example 2 (6)
• IDEAL GAS LAW
• IDEAL GAS LAW (2)
• Example 1 (9)
• Example 2 (7)
• Example 2 (8)
• Ideal vs Real Gases
• Ideal vs Real Gases (2)
• PARTIAL PRESSURES
• DALTONrsquoS LAW
• PARTIAL PRESSURES (2)
• PARTIAL PRESSURES (3)
• PARTIAL PRESSURES (4)
• Example 1 (10)
• Example 2 (9)
• Slide 60

Tros Introductory Chemistry Chapter

9

Common Units of Pressure

Unit Average Air Pressure at Sea Level

pascal (Pa) 101325

kilopascal (kPa) 101325

atmosphere (atm) 1 (exactly)

millimeters of mercury (mmHg) 760 (exactly)

inches of mercury (inHg) 2992

torr (torr) 760 (exactly)

pounds per square inch (psi lbsin2) 147

10

PRESSURE ampITS MEASUREMENT

Atmospheric pressure can be measured with the use of a barometer

Mercury is used in a barometer due to its high density

At sea level the mercury stands at 760 mm above its base

Tros Introductory Chemistry Chapter

11

Atmospheric Pressure amp Altitude

bull the higher up in the atmosphere you go the lower the atmospheric pressure is around youat the surface the atmospheric pressure is

147 psi but at 10000 ft is is only 100 psibull rapid changes in atmospheric pressure

may cause your ears to ldquopoprdquo due to an imbalance in pressure on either side of your ear drum

Tros Introductory Chemistry Chapter

12

Pressure Imbalance in Ear

If there is a differencein pressure acrossthe eardrum membranethe membrane will bepushed out ndash what we commonly call a ldquopopped eardrumrdquo

13

Example 1

The atmospheric pressure at Walnut CA is 740 mmHg Calculate this pressure in torr and atm

1 mmHg = 1 torr

740 mmHg = 740 torr

740 mmHg atm

x mmHg760

1= 0974 atm

14

Example 2

The barometer at a location reads 112 atm Calculate the pressure in mmHg and torr

1 mmHg = 1 torr

851 mmHg = 851 torr

112 atm mmHg

x atm1760

= 851 mmHg

15

PRESSURE amp MOLES OF A GAS

The pressure of a gas is directly proportional to the number of particles (moles) present

The greater the moles of the gas the greater the pressure

16

BOYLErsquoS LAW

The relationship of pressure and volume in gases is called Boylersquos Law

At constant temperature the volume of a fixed amount of gas is inversely proportional to its pressure

As pressure increases the volume of the gas decreases

Pressure and volume of a

gas are inversely

proportional

17

BOYLErsquoS LAW

Boylersquos Law can be mathematically expressed as

P1 x V1 = P2 x V2

P and V under initial

condition

P and V under final

condition

18

Example 1

A sample of gas has a volume of 12-L and a pressure of 4500 mmHg What is the volume of the gas when the pressure is reduced to 750 mmHg

1 12

2

P VV =

P

V2 = 72 L

P1= 4500 mmHg

P2= 750 mmHg

V1 = 12 L

V2 =

P1 x V1 = P2 x V2

(4500 mmHg)(12 L)=

(750 mmHg)

Pressure decreases Volume

increases

19

Example 2

A sample of hydrogen gas occupies 40 L at 650 mmHg What volume would it occupy at 20 atm

1 12

2

P VV =

P

V2 = 17 L

P1= 650 mmHg

P2= 20 atm

V1 = 40 L

V2 =

P1 x V1 = P2 x V2

(650 mmHg)(40 L)=

(1520 mmHg)Pressures must be in the same

unit

P2 = 20 atm = 1520 mmHgVolume decreases

Pressure increases

20

CHARLESrsquoS LAW

The relationship of temperature and volume in gases is called Charlesrsquo Law

At constant pressure the volume of a fixed amount of gas is directly proportional to its absolute temperature

As temperature increases the volume of the gas increases

Temperature and volume of a gas

are directly proportional

21

CHARLESrsquoS LAW

Charlesrsquo Law can be mathematically expressed as

V and T under initial

condition

1 2

1 2

V V=

T T

V and T under final condition

T must be in units of K

22

Example 1

A 20-L sample of a gas is cooled from 298K to 278 K at constant pressure What is the new volume of the gas

22 1

1

TV = V x

T

V2 = 19 L

V1 = 20 L

V2 =

T1 = 298 K

T2 = 278 K

278 K= 20 L x

298 K

Volume decreases1 2

1 2

V V=

T T

Temperature decreases

23

Example 2

If 200 L of oxygen is cooled from 100ordmC to 0ordmC what is the new volume

22 1

1

TV = V x

T

V2 = 146 L

V1 = 200 L

V2 =

T1 = 100ordmC

T2 = 0ordmC

273 K= 200 L x

373 K

T1= 100ordmC + 273 = 373 K

T2= 0ordmC + 273 = 273 K

Temperatures must be in

KVolume

decreasesTemperature

decreases

24

GAY-LUSSACrsquoSLAW

The relationship of temperature and pressure in gases is called Gay-Lussacrsquos Law

At constant volume the pressure of a fixed amount of gas is directly proportional to its absolute temperature

As temperature increases the

pressure of the gas increases

Temperature and pressure of a gas

are directly proportional

25

GAY-LUSSACrsquoSLAW

Gay-Lussacrsquos Law can be mathematically expressed as

P and T under initial

condition

1 2

1 2

P P=

T T

P and T under final conditionT must be in

units of K

26

Example 1

An aerosol spray can has a pressure of 40 atm at 25C What pressure will the can have if it is placed in a fire and reaches temperature of 400C

P1 = 40 atm

P2 =

T1 = 25ordmC

T2 = 400ordmC

T1= 25ordmC + 273 = 298 K

T2= 400ordmC + 273 = 673 K

Temperatures must be in K

27

Example 1

An aerosol spray can has a pressure of 40 atm at 25C What pressure will the can have if it is placed in a fire and reaches temperature of 400C

22 1

1

TP = P x

T

P2 = 90 atm

P1 = 40 atm

P2 =

T1 = 298 K

T2 = 673 K

673 K= 40 L x

298 K

Pressure increases

1 2

1 2

P P=

T T

Temperature increases

28

Example 2

A cylinder of gas with a volume of 150-L and a pressure of 965 mmHg is stored at a temperature of 55C To what temperature must the cylinder be cooled to reach a pressure of 850 mmHg

22 1

1

PT = T x

P

T2 = 289 K

P1 = 965 mmHg

P2 = 850 mmHg

T1 = 55ordmC

T2 =

850 mmHg= 328 K x

965 mmHg

T1= 55ordmC + 273 = 328 K

T2 = 16ordmC

Temperature decreases

Pressure decreases

29

VAPOR PRESSUREamp BOILING POINT

In an open container liquid molecules at the surface that possess sufficient energy can break away from the surface and become gas particles or vapor

In a closed container these gas particles can accumulate and create pressure called vapor pressure

Vapor pressure is defined as the pressure above a liquid at a given temperature Vapor pressure varies with each liquid and increases with temperature

30

VAPOR PRESSUREamp BOILING POINT

Listed below is the vapor pressure of water at various temperatures

31

VAPOR PRESSUREamp BOILING POINT

A liquid reaches its boiling point when its vapor pressure becomes equal to the external pressure (atmospheric pressure)

For example at sea level water reaches its boiling point at 100C since its vapor pressure is 760 mmHg at this temperature

32

VAPOR PRESSUREamp BOILING POINT

At higher altitudes where atmospheric pressure is lower water reaches boiling point at temperatures lower than 100C

For example in Denver where atmospheric pressure is 630 mmHg water boils at 95C since its vapor pressure is 630 mmHg at this temperature

33

COMBINEDGAS LAW

All pressure-volume-temperature relationships can be combined into a single relationship called the combined gas law

1 1 2 2

1 2

P V P V=

T T

Initial condition

This law is useful for studying the effect of changes in two variables

Final condition

34

COMBINEDGAS LAW

The individual gas laws studied previously are embodied in the combined gas law

1 1 2 2

1 2

P V P V=

T T

Boylersquos LawCharlesrsquos

LawGay-Lussacrsquos

Law

35

Example 1

A 250-mL sample of gas has a pressure of 400 atm at a temperature of 10C What is the volume of the gas at a pressure of 100 atm and a temperature of 18C

1 22 1

2 1

P TV = V x x

P T

V2 = 103 mL

P1= 400 atm V1= 250 mL T1= 283 K P2= 100 atm V2= T2 = 291 K

2

400 atm 291 KV = 250 mL x x

100 atm 283 K

1 1 2 2

1 2

P V P V=

T TBoth pressure

and temp change

Must use combined

gas law

36

AVOGADROrsquoSLAW

The relationship of moles and volume in gases is called Avogadrorsquos Law

At constant temperature and pressure the volume of a fixed amount of gas is directly proportional to the number of moles

As number of moles increases

the volume of the gas

increases

Number of moles and volume of a gas are directly

proportional

Tros Introductory Chemistry Chapter

37

Avogadrorsquos Law

38

AVOGADROrsquoSLAW

As a result of Avogadrorsquos Law equal volumes of different gases at the same temp and pressure contain equal number of moles (molecules)

2 Tanks of gas of equal volume at the same T amp P contain the same number of molecules

39

AVOGADROrsquoSLAW

For example

2 H2 (g) + 1 O2 (g) 2 H2O (g)

2 molecules 1 molecule 2 molecules

2 moles 1 mole 2 moles

2 Liters 1 Liter 2 Liters

Avogadrorsquos Law also allows chemists to relate volumes and moles of a gas in a chemical reaction

This relationship is only valid for

gaseous substances

40

Example 1

A sample of helium gas with a mass of 180 g occupies 16 Liters at a particular temperature and pressure What mass of oxygen would occupy16 L at the same temperature and pressure

1 mol= 180 g x

400 gmol of He = 450 mol

mol of O2 = mol of He = 450 mol

Same Temp amp Pressure

mass of O2

320 g= 450 mol x

1 mol= 144 g

41

Example 2

How many Liters of NH3 can be produced from reaction of 18 L of H2 with excess N2 as shown below

18 L H23

2

L NHx

L H= 12 L NH3

N2 (g) + 3 H2 (g) 2 NH3 (g)

32

42

STP amp MOLAR VOLUME

To better understand the factors that affect gas behavior a set of standard conditions have been chosen for use and are referred to as Standard Temperature and Pressure (STP)

STP =

1 atm (760 mmHg)

0ordmC (273 K)

43

STP amp MOLAR VOLUME

At STP conditions one mole of any gas is observed to occupy a volume of 224 L

V = 224 L

Molar Volume at

STP

44

Example 1

If 200 L of a gas at STP has a mass of 323 g what is the molar mass of the gas

mol200 L x

Lmol of gas =

2241

= 00893 mol

g

molMolar mass =

323 g=

00893 mol= 362 gmol Molar

volume at STP

45

Example 2

A sample of gas has a volume of 250 L at 730 mmHg and 20C What is the volume of this gas at STP

1 22 1

2 1

P TV = V x x

P T

V2 = 224 L

P1= 730 mmHg V1= 250 L T1= 293 K P2= 760 mmHg V2= T2 = 273 K

2

730 mmHg 273 KV = 250 L x x

760 mmHg 293 K

46

IDEAL GAS LAW

Combining all the laws that describe the behavior of gases one can obtain a useful relationship that relates the volume of a gas to the temperature pressure and number of moles

n R TV =

P

Universal gas constant

R = 00821 L atmmol K

47

IDEAL GAS LAW

This relationship is called the Ideal Gas Law and commonly written as

P V = n R T

Pressure in atm Volume

in Liters

Number of moles

Temp in K

48

Example 1

A sample of H2 gas has a volume of 856 L at a temp of 0C and pressure of 15 atm Calculate the moles of gas present

PVn =

RT

= 057 mol

P = 15 atm V = 856 L T = 273 K n = R = 00821

PV = nRT

(15 atm)(856 L)=

L atm(00821 )(273 K)

mol K

49

Example 2

What volume does 400 g of N2 gas occupy at 10C and 750 mmHg

P =750 mmHg V = T = 283 K n = not given m = 400 g R = 00821

mol of N2 =1 mol

400 g x280 g

= 143 mol

P = 750 mmHg1 atm

x760 mmHg

Moles must be calculated

from mass

Pressure must be

converted to atm

P = 0987 atm

50

Example 2

What volume does 400 g of N2 gas occupy at 10C and 750 mmHg

nRTV =

P

V = 337 L

P =0987 atm V = T = 283 K n = 143 R = 00821

L atm(143 mol)(00821 )(283 K)

mol KV = 0987 atm

Tros Introductory Chemistry Chapter

51

Ideal vs Real Gases

bull Real gases often do not behave like ideal gases at high pressure or low temperature

bull Ideal gas laws assume1) no attractions between gas molecules2) gas molecules do not take up space based on the Kinetic-Molecular Theory

bull at low temperatures and high pressures these assumptions are not valid

Tros Introductory Chemistry Chapter

52

Ideal vs Real Gases

53

PARTIAL PRESSURES

Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases

Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture

In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas

54

DALTONrsquoSLAW

In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture

This is called Daltonrsquos law of partial pressures

Total pressure of a gas mixture

Ptotal = P1 + P2 + P3 +

Sum of the partial pressures of the gases in the mixture

=

55

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = PHe + PAr

Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture

= 60 atm

56

PARTIAL PRESSURES

The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture

For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container

57

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = 60 atm

The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture

Twice as many moles of Ar

compared to He

PAr = 2 PHe

58

Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank

Ptotal = Poxygen + Phelium

Poxygen = 1140 mmHg x

1 atm

760 mmHg= 150 atm

Phelium

= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm

59

Example 2

A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture

Ptotal = Poxygen + Pnitrogen

Pnitrogen = 2 Poxygen

Twice as many moles of N2

compared to O2

Pnitrogen = 23 (Ptotal ) = 20 atm

Poxygen = 13 (Ptotal ) = 10 atm

60

THE END

• Slide 1
• CHAPTER OUTLINE
• PROPERTIES OF GASES
• PROPERTIES OF GASES (2)
• KINETIC-MOLECULAR THEORY
• KINETIC-MOLECULAR THEORY (2)
• Kinetic Molecular Theory
• PRESSURE amp ITS MEASUREMENT
• Common Units of Pressure
• PRESSURE amp ITS MEASUREMENT (2)
• Atmospheric Pressure amp Altitude
• Pressure Imbalance in Ear
• Example 1
• Example 2
• PRESSURE amp MOLES OF A GAS
• BOYLErsquoS LAW
• BOYLErsquoS LAW (2)
• Example 1 (2)
• Example 2 (2)
• CHARLESrsquoS LAW
• CHARLESrsquoS LAW (2)
• Example 1 (3)
• Example 2 (3)
• GAY-LUSSACrsquoS LAW
• GAY-LUSSACrsquoS LAW (2)
• Example 1 (4)
• Example 1 (5)
• Example 2 (4)
• VAPOR PRESSURE amp BOILING POINT
• VAPOR PRESSURE amp BOILING POINT (2)
• VAPOR PRESSURE amp BOILING POINT (3)
• VAPOR PRESSURE amp BOILING POINT (4)
• COMBINED GAS LAW
• COMBINED GAS LAW (2)
• Example 1 (6)
• AVOGADROrsquoS LAW
• Avogadrorsquos Law
• AVOGADROrsquoS LAW (2)
• AVOGADROrsquoS LAW (3)
• Example 1 (7)
• Example 2 (5)
• STP amp MOLAR VOLUME
• STP amp MOLAR VOLUME (2)
• Example 1 (8)
• Example 2 (6)
• IDEAL GAS LAW
• IDEAL GAS LAW (2)
• Example 1 (9)
• Example 2 (7)
• Example 2 (8)
• Ideal vs Real Gases
• Ideal vs Real Gases (2)
• PARTIAL PRESSURES
• DALTONrsquoS LAW
• PARTIAL PRESSURES (2)
• PARTIAL PRESSURES (3)
• PARTIAL PRESSURES (4)
• Example 1 (10)
• Example 2 (9)
• Slide 60

10

PRESSURE ampITS MEASUREMENT

Atmospheric pressure can be measured with the use of a barometer

Mercury is used in a barometer due to its high density

At sea level the mercury stands at 760 mm above its base

Tros Introductory Chemistry Chapter

11

Atmospheric Pressure amp Altitude

bull the higher up in the atmosphere you go the lower the atmospheric pressure is around youat the surface the atmospheric pressure is

147 psi but at 10000 ft is is only 100 psibull rapid changes in atmospheric pressure

may cause your ears to ldquopoprdquo due to an imbalance in pressure on either side of your ear drum

Tros Introductory Chemistry Chapter

12

Pressure Imbalance in Ear

If there is a differencein pressure acrossthe eardrum membranethe membrane will bepushed out ndash what we commonly call a ldquopopped eardrumrdquo

13

Example 1

The atmospheric pressure at Walnut CA is 740 mmHg Calculate this pressure in torr and atm

1 mmHg = 1 torr

740 mmHg = 740 torr

740 mmHg atm

x mmHg760

1= 0974 atm

14

Example 2

The barometer at a location reads 112 atm Calculate the pressure in mmHg and torr

1 mmHg = 1 torr

851 mmHg = 851 torr

112 atm mmHg

x atm1760

= 851 mmHg

15

PRESSURE amp MOLES OF A GAS

The pressure of a gas is directly proportional to the number of particles (moles) present

The greater the moles of the gas the greater the pressure

16

BOYLErsquoS LAW

The relationship of pressure and volume in gases is called Boylersquos Law

At constant temperature the volume of a fixed amount of gas is inversely proportional to its pressure

As pressure increases the volume of the gas decreases

Pressure and volume of a

gas are inversely

proportional

17

BOYLErsquoS LAW

Boylersquos Law can be mathematically expressed as

P1 x V1 = P2 x V2

P and V under initial

condition

P and V under final

condition

18

Example 1

A sample of gas has a volume of 12-L and a pressure of 4500 mmHg What is the volume of the gas when the pressure is reduced to 750 mmHg

1 12

2

P VV =

P

V2 = 72 L

P1= 4500 mmHg

P2= 750 mmHg

V1 = 12 L

V2 =

P1 x V1 = P2 x V2

(4500 mmHg)(12 L)=

(750 mmHg)

Pressure decreases Volume

increases

19

Example 2

A sample of hydrogen gas occupies 40 L at 650 mmHg What volume would it occupy at 20 atm

1 12

2

P VV =

P

V2 = 17 L

P1= 650 mmHg

P2= 20 atm

V1 = 40 L

V2 =

P1 x V1 = P2 x V2

(650 mmHg)(40 L)=

(1520 mmHg)Pressures must be in the same

unit

P2 = 20 atm = 1520 mmHgVolume decreases

Pressure increases

20

CHARLESrsquoS LAW

The relationship of temperature and volume in gases is called Charlesrsquo Law

At constant pressure the volume of a fixed amount of gas is directly proportional to its absolute temperature

As temperature increases the volume of the gas increases

Temperature and volume of a gas

are directly proportional

21

CHARLESrsquoS LAW

Charlesrsquo Law can be mathematically expressed as

V and T under initial

condition

1 2

1 2

V V=

T T

V and T under final condition

T must be in units of K

22

Example 1

A 20-L sample of a gas is cooled from 298K to 278 K at constant pressure What is the new volume of the gas

22 1

1

TV = V x

T

V2 = 19 L

V1 = 20 L

V2 =

T1 = 298 K

T2 = 278 K

278 K= 20 L x

298 K

Volume decreases1 2

1 2

V V=

T T

Temperature decreases

23

Example 2

If 200 L of oxygen is cooled from 100ordmC to 0ordmC what is the new volume

22 1

1

TV = V x

T

V2 = 146 L

V1 = 200 L

V2 =

T1 = 100ordmC

T2 = 0ordmC

273 K= 200 L x

373 K

T1= 100ordmC + 273 = 373 K

T2= 0ordmC + 273 = 273 K

Temperatures must be in

KVolume

decreasesTemperature

decreases

24

GAY-LUSSACrsquoSLAW

The relationship of temperature and pressure in gases is called Gay-Lussacrsquos Law

At constant volume the pressure of a fixed amount of gas is directly proportional to its absolute temperature

As temperature increases the

pressure of the gas increases

Temperature and pressure of a gas

are directly proportional

25

GAY-LUSSACrsquoSLAW

Gay-Lussacrsquos Law can be mathematically expressed as

P and T under initial

condition

1 2

1 2

P P=

T T

P and T under final conditionT must be in

units of K

26

Example 1

An aerosol spray can has a pressure of 40 atm at 25C What pressure will the can have if it is placed in a fire and reaches temperature of 400C

P1 = 40 atm

P2 =

T1 = 25ordmC

T2 = 400ordmC

T1= 25ordmC + 273 = 298 K

T2= 400ordmC + 273 = 673 K

Temperatures must be in K

27

Example 1

An aerosol spray can has a pressure of 40 atm at 25C What pressure will the can have if it is placed in a fire and reaches temperature of 400C

22 1

1

TP = P x

T

P2 = 90 atm

P1 = 40 atm

P2 =

T1 = 298 K

T2 = 673 K

673 K= 40 L x

298 K

Pressure increases

1 2

1 2

P P=

T T

Temperature increases

28

Example 2

A cylinder of gas with a volume of 150-L and a pressure of 965 mmHg is stored at a temperature of 55C To what temperature must the cylinder be cooled to reach a pressure of 850 mmHg

22 1

1

PT = T x

P

T2 = 289 K

P1 = 965 mmHg

P2 = 850 mmHg

T1 = 55ordmC

T2 =

850 mmHg= 328 K x

965 mmHg

T1= 55ordmC + 273 = 328 K

T2 = 16ordmC

Temperature decreases

Pressure decreases

29

VAPOR PRESSUREamp BOILING POINT

In an open container liquid molecules at the surface that possess sufficient energy can break away from the surface and become gas particles or vapor

In a closed container these gas particles can accumulate and create pressure called vapor pressure

Vapor pressure is defined as the pressure above a liquid at a given temperature Vapor pressure varies with each liquid and increases with temperature

30

VAPOR PRESSUREamp BOILING POINT

Listed below is the vapor pressure of water at various temperatures

31

VAPOR PRESSUREamp BOILING POINT

A liquid reaches its boiling point when its vapor pressure becomes equal to the external pressure (atmospheric pressure)

For example at sea level water reaches its boiling point at 100C since its vapor pressure is 760 mmHg at this temperature

32

VAPOR PRESSUREamp BOILING POINT

At higher altitudes where atmospheric pressure is lower water reaches boiling point at temperatures lower than 100C

For example in Denver where atmospheric pressure is 630 mmHg water boils at 95C since its vapor pressure is 630 mmHg at this temperature

33

COMBINEDGAS LAW

All pressure-volume-temperature relationships can be combined into a single relationship called the combined gas law

1 1 2 2

1 2

P V P V=

T T

Initial condition

This law is useful for studying the effect of changes in two variables

Final condition

34

COMBINEDGAS LAW

The individual gas laws studied previously are embodied in the combined gas law

1 1 2 2

1 2

P V P V=

T T

Boylersquos LawCharlesrsquos

LawGay-Lussacrsquos

Law

35

Example 1

A 250-mL sample of gas has a pressure of 400 atm at a temperature of 10C What is the volume of the gas at a pressure of 100 atm and a temperature of 18C

1 22 1

2 1

P TV = V x x

P T

V2 = 103 mL

P1= 400 atm V1= 250 mL T1= 283 K P2= 100 atm V2= T2 = 291 K

2

400 atm 291 KV = 250 mL x x

100 atm 283 K

1 1 2 2

1 2

P V P V=

T TBoth pressure

and temp change

Must use combined

gas law

36

AVOGADROrsquoSLAW

The relationship of moles and volume in gases is called Avogadrorsquos Law

At constant temperature and pressure the volume of a fixed amount of gas is directly proportional to the number of moles

As number of moles increases

the volume of the gas

increases

Number of moles and volume of a gas are directly

proportional

Tros Introductory Chemistry Chapter

37

Avogadrorsquos Law

38

AVOGADROrsquoSLAW

As a result of Avogadrorsquos Law equal volumes of different gases at the same temp and pressure contain equal number of moles (molecules)

2 Tanks of gas of equal volume at the same T amp P contain the same number of molecules

39

AVOGADROrsquoSLAW

For example

2 H2 (g) + 1 O2 (g) 2 H2O (g)

2 molecules 1 molecule 2 molecules

2 moles 1 mole 2 moles

2 Liters 1 Liter 2 Liters

Avogadrorsquos Law also allows chemists to relate volumes and moles of a gas in a chemical reaction

This relationship is only valid for

gaseous substances

40

Example 1

A sample of helium gas with a mass of 180 g occupies 16 Liters at a particular temperature and pressure What mass of oxygen would occupy16 L at the same temperature and pressure

1 mol= 180 g x

400 gmol of He = 450 mol

mol of O2 = mol of He = 450 mol

Same Temp amp Pressure

mass of O2

320 g= 450 mol x

1 mol= 144 g

41

Example 2

How many Liters of NH3 can be produced from reaction of 18 L of H2 with excess N2 as shown below

18 L H23

2

L NHx

L H= 12 L NH3

N2 (g) + 3 H2 (g) 2 NH3 (g)

32

42

STP amp MOLAR VOLUME

To better understand the factors that affect gas behavior a set of standard conditions have been chosen for use and are referred to as Standard Temperature and Pressure (STP)

STP =

1 atm (760 mmHg)

0ordmC (273 K)

43

STP amp MOLAR VOLUME

At STP conditions one mole of any gas is observed to occupy a volume of 224 L

V = 224 L

Molar Volume at

STP

44

Example 1

If 200 L of a gas at STP has a mass of 323 g what is the molar mass of the gas

mol200 L x

Lmol of gas =

2241

= 00893 mol

g

molMolar mass =

323 g=

00893 mol= 362 gmol Molar

volume at STP

45

Example 2

A sample of gas has a volume of 250 L at 730 mmHg and 20C What is the volume of this gas at STP

1 22 1

2 1

P TV = V x x

P T

V2 = 224 L

P1= 730 mmHg V1= 250 L T1= 293 K P2= 760 mmHg V2= T2 = 273 K

2

730 mmHg 273 KV = 250 L x x

760 mmHg 293 K

46

IDEAL GAS LAW

Combining all the laws that describe the behavior of gases one can obtain a useful relationship that relates the volume of a gas to the temperature pressure and number of moles

n R TV =

P

Universal gas constant

R = 00821 L atmmol K

47

IDEAL GAS LAW

This relationship is called the Ideal Gas Law and commonly written as

P V = n R T

Pressure in atm Volume

in Liters

Number of moles

Temp in K

48

Example 1

A sample of H2 gas has a volume of 856 L at a temp of 0C and pressure of 15 atm Calculate the moles of gas present

PVn =

RT

= 057 mol

P = 15 atm V = 856 L T = 273 K n = R = 00821

PV = nRT

(15 atm)(856 L)=

L atm(00821 )(273 K)

mol K

49

Example 2

What volume does 400 g of N2 gas occupy at 10C and 750 mmHg

P =750 mmHg V = T = 283 K n = not given m = 400 g R = 00821

mol of N2 =1 mol

400 g x280 g

= 143 mol

P = 750 mmHg1 atm

x760 mmHg

Moles must be calculated

from mass

Pressure must be

converted to atm

P = 0987 atm

50

Example 2

What volume does 400 g of N2 gas occupy at 10C and 750 mmHg

nRTV =

P

V = 337 L

P =0987 atm V = T = 283 K n = 143 R = 00821

L atm(143 mol)(00821 )(283 K)

mol KV = 0987 atm

Tros Introductory Chemistry Chapter

51

Ideal vs Real Gases

bull Real gases often do not behave like ideal gases at high pressure or low temperature

bull Ideal gas laws assume1) no attractions between gas molecules2) gas molecules do not take up space based on the Kinetic-Molecular Theory

bull at low temperatures and high pressures these assumptions are not valid

Tros Introductory Chemistry Chapter

52

Ideal vs Real Gases

53

PARTIAL PRESSURES

Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases

Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture

In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas

54

DALTONrsquoSLAW

In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture

This is called Daltonrsquos law of partial pressures

Total pressure of a gas mixture

Ptotal = P1 + P2 + P3 +

Sum of the partial pressures of the gases in the mixture

=

55

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = PHe + PAr

Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture

= 60 atm

56

PARTIAL PRESSURES

The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture

For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container

57

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = 60 atm

The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture

Twice as many moles of Ar

compared to He

PAr = 2 PHe

58

Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank

Ptotal = Poxygen + Phelium

Poxygen = 1140 mmHg x

1 atm

760 mmHg= 150 atm

Phelium

= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm

59

Example 2

A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture

Ptotal = Poxygen + Pnitrogen

Pnitrogen = 2 Poxygen

Twice as many moles of N2

compared to O2

Pnitrogen = 23 (Ptotal ) = 20 atm

Poxygen = 13 (Ptotal ) = 10 atm

60

THE END

• Slide 1
• CHAPTER OUTLINE
• PROPERTIES OF GASES
• PROPERTIES OF GASES (2)
• KINETIC-MOLECULAR THEORY
• KINETIC-MOLECULAR THEORY (2)
• Kinetic Molecular Theory
• PRESSURE amp ITS MEASUREMENT
• Common Units of Pressure
• PRESSURE amp ITS MEASUREMENT (2)
• Atmospheric Pressure amp Altitude
• Pressure Imbalance in Ear
• Example 1
• Example 2
• PRESSURE amp MOLES OF A GAS
• BOYLErsquoS LAW
• BOYLErsquoS LAW (2)
• Example 1 (2)
• Example 2 (2)
• CHARLESrsquoS LAW
• CHARLESrsquoS LAW (2)
• Example 1 (3)
• Example 2 (3)
• GAY-LUSSACrsquoS LAW
• GAY-LUSSACrsquoS LAW (2)
• Example 1 (4)
• Example 1 (5)
• Example 2 (4)
• VAPOR PRESSURE amp BOILING POINT
• VAPOR PRESSURE amp BOILING POINT (2)
• VAPOR PRESSURE amp BOILING POINT (3)
• VAPOR PRESSURE amp BOILING POINT (4)
• COMBINED GAS LAW
• COMBINED GAS LAW (2)
• Example 1 (6)
• AVOGADROrsquoS LAW
• Avogadrorsquos Law
• AVOGADROrsquoS LAW (2)
• AVOGADROrsquoS LAW (3)
• Example 1 (7)
• Example 2 (5)
• STP amp MOLAR VOLUME
• STP amp MOLAR VOLUME (2)
• Example 1 (8)
• Example 2 (6)
• IDEAL GAS LAW
• IDEAL GAS LAW (2)
• Example 1 (9)
• Example 2 (7)
• Example 2 (8)
• Ideal vs Real Gases
• Ideal vs Real Gases (2)
• PARTIAL PRESSURES
• DALTONrsquoS LAW
• PARTIAL PRESSURES (2)
• PARTIAL PRESSURES (3)
• PARTIAL PRESSURES (4)
• Example 1 (10)
• Example 2 (9)
• Slide 60

Tros Introductory Chemistry Chapter

11

Atmospheric Pressure amp Altitude

bull the higher up in the atmosphere you go the lower the atmospheric pressure is around youat the surface the atmospheric pressure is

147 psi but at 10000 ft is is only 100 psibull rapid changes in atmospheric pressure

may cause your ears to ldquopoprdquo due to an imbalance in pressure on either side of your ear drum

Tros Introductory Chemistry Chapter

12

Pressure Imbalance in Ear

If there is a differencein pressure acrossthe eardrum membranethe membrane will bepushed out ndash what we commonly call a ldquopopped eardrumrdquo

13

Example 1

The atmospheric pressure at Walnut CA is 740 mmHg Calculate this pressure in torr and atm

1 mmHg = 1 torr

740 mmHg = 740 torr

740 mmHg atm

x mmHg760

1= 0974 atm

14

Example 2

The barometer at a location reads 112 atm Calculate the pressure in mmHg and torr

1 mmHg = 1 torr

851 mmHg = 851 torr

112 atm mmHg

x atm1760

= 851 mmHg

15

PRESSURE amp MOLES OF A GAS

The pressure of a gas is directly proportional to the number of particles (moles) present

The greater the moles of the gas the greater the pressure

16

BOYLErsquoS LAW

The relationship of pressure and volume in gases is called Boylersquos Law

At constant temperature the volume of a fixed amount of gas is inversely proportional to its pressure

As pressure increases the volume of the gas decreases

Pressure and volume of a

gas are inversely

proportional

17

BOYLErsquoS LAW

Boylersquos Law can be mathematically expressed as

P1 x V1 = P2 x V2

P and V under initial

condition

P and V under final

condition

18

Example 1

A sample of gas has a volume of 12-L and a pressure of 4500 mmHg What is the volume of the gas when the pressure is reduced to 750 mmHg

1 12

2

P VV =

P

V2 = 72 L

P1= 4500 mmHg

P2= 750 mmHg

V1 = 12 L

V2 =

P1 x V1 = P2 x V2

(4500 mmHg)(12 L)=

(750 mmHg)

Pressure decreases Volume

increases

19

Example 2

A sample of hydrogen gas occupies 40 L at 650 mmHg What volume would it occupy at 20 atm

1 12

2

P VV =

P

V2 = 17 L

P1= 650 mmHg

P2= 20 atm

V1 = 40 L

V2 =

P1 x V1 = P2 x V2

(650 mmHg)(40 L)=

(1520 mmHg)Pressures must be in the same

unit

P2 = 20 atm = 1520 mmHgVolume decreases

Pressure increases

20

CHARLESrsquoS LAW

The relationship of temperature and volume in gases is called Charlesrsquo Law

At constant pressure the volume of a fixed amount of gas is directly proportional to its absolute temperature

As temperature increases the volume of the gas increases

Temperature and volume of a gas

are directly proportional

21

CHARLESrsquoS LAW

Charlesrsquo Law can be mathematically expressed as

V and T under initial

condition

1 2

1 2

V V=

T T

V and T under final condition

T must be in units of K

22

Example 1

A 20-L sample of a gas is cooled from 298K to 278 K at constant pressure What is the new volume of the gas

22 1

1

TV = V x

T

V2 = 19 L

V1 = 20 L

V2 =

T1 = 298 K

T2 = 278 K

278 K= 20 L x

298 K

Volume decreases1 2

1 2

V V=

T T

Temperature decreases

23

Example 2

If 200 L of oxygen is cooled from 100ordmC to 0ordmC what is the new volume

22 1

1

TV = V x

T

V2 = 146 L

V1 = 200 L

V2 =

T1 = 100ordmC

T2 = 0ordmC

273 K= 200 L x

373 K

T1= 100ordmC + 273 = 373 K

T2= 0ordmC + 273 = 273 K

Temperatures must be in

KVolume

decreasesTemperature

decreases

24

GAY-LUSSACrsquoSLAW

The relationship of temperature and pressure in gases is called Gay-Lussacrsquos Law

At constant volume the pressure of a fixed amount of gas is directly proportional to its absolute temperature

As temperature increases the

pressure of the gas increases

Temperature and pressure of a gas

are directly proportional

25

GAY-LUSSACrsquoSLAW

Gay-Lussacrsquos Law can be mathematically expressed as

P and T under initial

condition

1 2

1 2

P P=

T T

P and T under final conditionT must be in

units of K

26

Example 1

An aerosol spray can has a pressure of 40 atm at 25C What pressure will the can have if it is placed in a fire and reaches temperature of 400C

P1 = 40 atm

P2 =

T1 = 25ordmC

T2 = 400ordmC

T1= 25ordmC + 273 = 298 K

T2= 400ordmC + 273 = 673 K

Temperatures must be in K

27

Example 1

An aerosol spray can has a pressure of 40 atm at 25C What pressure will the can have if it is placed in a fire and reaches temperature of 400C

22 1

1

TP = P x

T

P2 = 90 atm

P1 = 40 atm

P2 =

T1 = 298 K

T2 = 673 K

673 K= 40 L x

298 K

Pressure increases

1 2

1 2

P P=

T T

Temperature increases

28

Example 2

A cylinder of gas with a volume of 150-L and a pressure of 965 mmHg is stored at a temperature of 55C To what temperature must the cylinder be cooled to reach a pressure of 850 mmHg

22 1

1

PT = T x

P

T2 = 289 K

P1 = 965 mmHg

P2 = 850 mmHg

T1 = 55ordmC

T2 =

850 mmHg= 328 K x

965 mmHg

T1= 55ordmC + 273 = 328 K

T2 = 16ordmC

Temperature decreases

Pressure decreases

29

VAPOR PRESSUREamp BOILING POINT

In an open container liquid molecules at the surface that possess sufficient energy can break away from the surface and become gas particles or vapor

In a closed container these gas particles can accumulate and create pressure called vapor pressure

Vapor pressure is defined as the pressure above a liquid at a given temperature Vapor pressure varies with each liquid and increases with temperature

30

VAPOR PRESSUREamp BOILING POINT

Listed below is the vapor pressure of water at various temperatures

31

VAPOR PRESSUREamp BOILING POINT

A liquid reaches its boiling point when its vapor pressure becomes equal to the external pressure (atmospheric pressure)

For example at sea level water reaches its boiling point at 100C since its vapor pressure is 760 mmHg at this temperature

32

VAPOR PRESSUREamp BOILING POINT

At higher altitudes where atmospheric pressure is lower water reaches boiling point at temperatures lower than 100C

For example in Denver where atmospheric pressure is 630 mmHg water boils at 95C since its vapor pressure is 630 mmHg at this temperature

33

COMBINEDGAS LAW

All pressure-volume-temperature relationships can be combined into a single relationship called the combined gas law

1 1 2 2

1 2

P V P V=

T T

Initial condition

This law is useful for studying the effect of changes in two variables

Final condition

34

COMBINEDGAS LAW

The individual gas laws studied previously are embodied in the combined gas law

1 1 2 2

1 2

P V P V=

T T

Boylersquos LawCharlesrsquos

LawGay-Lussacrsquos

Law

35

Example 1

A 250-mL sample of gas has a pressure of 400 atm at a temperature of 10C What is the volume of the gas at a pressure of 100 atm and a temperature of 18C

1 22 1

2 1

P TV = V x x

P T

V2 = 103 mL

P1= 400 atm V1= 250 mL T1= 283 K P2= 100 atm V2= T2 = 291 K

2

400 atm 291 KV = 250 mL x x

100 atm 283 K

1 1 2 2

1 2

P V P V=

T TBoth pressure

and temp change

Must use combined

gas law

36

AVOGADROrsquoSLAW

The relationship of moles and volume in gases is called Avogadrorsquos Law

At constant temperature and pressure the volume of a fixed amount of gas is directly proportional to the number of moles

As number of moles increases

the volume of the gas

increases

Number of moles and volume of a gas are directly

proportional

Tros Introductory Chemistry Chapter

37

Avogadrorsquos Law

38

AVOGADROrsquoSLAW

As a result of Avogadrorsquos Law equal volumes of different gases at the same temp and pressure contain equal number of moles (molecules)

2 Tanks of gas of equal volume at the same T amp P contain the same number of molecules

39

AVOGADROrsquoSLAW

For example

2 H2 (g) + 1 O2 (g) 2 H2O (g)

2 molecules 1 molecule 2 molecules

2 moles 1 mole 2 moles

2 Liters 1 Liter 2 Liters

Avogadrorsquos Law also allows chemists to relate volumes and moles of a gas in a chemical reaction

This relationship is only valid for

gaseous substances

40

Example 1

A sample of helium gas with a mass of 180 g occupies 16 Liters at a particular temperature and pressure What mass of oxygen would occupy16 L at the same temperature and pressure

1 mol= 180 g x

400 gmol of He = 450 mol

mol of O2 = mol of He = 450 mol

Same Temp amp Pressure

mass of O2

320 g= 450 mol x

1 mol= 144 g

41

Example 2

How many Liters of NH3 can be produced from reaction of 18 L of H2 with excess N2 as shown below

18 L H23

2

L NHx

L H= 12 L NH3

N2 (g) + 3 H2 (g) 2 NH3 (g)

32

42

STP amp MOLAR VOLUME

To better understand the factors that affect gas behavior a set of standard conditions have been chosen for use and are referred to as Standard Temperature and Pressure (STP)

STP =

1 atm (760 mmHg)

0ordmC (273 K)

43

STP amp MOLAR VOLUME

At STP conditions one mole of any gas is observed to occupy a volume of 224 L

V = 224 L

Molar Volume at

STP

44

Example 1

If 200 L of a gas at STP has a mass of 323 g what is the molar mass of the gas

mol200 L x

Lmol of gas =

2241

= 00893 mol

g

molMolar mass =

323 g=

00893 mol= 362 gmol Molar

volume at STP

45

Example 2

A sample of gas has a volume of 250 L at 730 mmHg and 20C What is the volume of this gas at STP

1 22 1

2 1

P TV = V x x

P T

V2 = 224 L

P1= 730 mmHg V1= 250 L T1= 293 K P2= 760 mmHg V2= T2 = 273 K

2

730 mmHg 273 KV = 250 L x x

760 mmHg 293 K

46

IDEAL GAS LAW

Combining all the laws that describe the behavior of gases one can obtain a useful relationship that relates the volume of a gas to the temperature pressure and number of moles

n R TV =

P

Universal gas constant

R = 00821 L atmmol K

47

IDEAL GAS LAW

This relationship is called the Ideal Gas Law and commonly written as

P V = n R T

Pressure in atm Volume

in Liters

Number of moles

Temp in K

48

Example 1

A sample of H2 gas has a volume of 856 L at a temp of 0C and pressure of 15 atm Calculate the moles of gas present

PVn =

RT

= 057 mol

P = 15 atm V = 856 L T = 273 K n = R = 00821

PV = nRT

(15 atm)(856 L)=

L atm(00821 )(273 K)

mol K

49

Example 2

What volume does 400 g of N2 gas occupy at 10C and 750 mmHg

P =750 mmHg V = T = 283 K n = not given m = 400 g R = 00821

mol of N2 =1 mol

400 g x280 g

= 143 mol

P = 750 mmHg1 atm

x760 mmHg

Moles must be calculated

from mass

Pressure must be

converted to atm

P = 0987 atm

50

Example 2

What volume does 400 g of N2 gas occupy at 10C and 750 mmHg

nRTV =

P

V = 337 L

P =0987 atm V = T = 283 K n = 143 R = 00821

L atm(143 mol)(00821 )(283 K)

mol KV = 0987 atm

Tros Introductory Chemistry Chapter

51

Ideal vs Real Gases

bull Real gases often do not behave like ideal gases at high pressure or low temperature

bull Ideal gas laws assume1) no attractions between gas molecules2) gas molecules do not take up space based on the Kinetic-Molecular Theory

bull at low temperatures and high pressures these assumptions are not valid

Tros Introductory Chemistry Chapter

52

Ideal vs Real Gases

53

PARTIAL PRESSURES

Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases

Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture

In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas

54

DALTONrsquoSLAW

In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture

This is called Daltonrsquos law of partial pressures

Total pressure of a gas mixture

Ptotal = P1 + P2 + P3 +

Sum of the partial pressures of the gases in the mixture

=

55

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = PHe + PAr

Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture

= 60 atm

56

PARTIAL PRESSURES

The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture

For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container

57

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = 60 atm

The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture

Twice as many moles of Ar

compared to He

PAr = 2 PHe

58

Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank

Ptotal = Poxygen + Phelium

Poxygen = 1140 mmHg x

1 atm

760 mmHg= 150 atm

Phelium

= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm

59

Example 2

A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture

Ptotal = Poxygen + Pnitrogen

Pnitrogen = 2 Poxygen

Twice as many moles of N2

compared to O2

Pnitrogen = 23 (Ptotal ) = 20 atm

Poxygen = 13 (Ptotal ) = 10 atm

60

THE END

• Slide 1
• CHAPTER OUTLINE
• PROPERTIES OF GASES
• PROPERTIES OF GASES (2)
• KINETIC-MOLECULAR THEORY
• KINETIC-MOLECULAR THEORY (2)
• Kinetic Molecular Theory
• PRESSURE amp ITS MEASUREMENT
• Common Units of Pressure
• PRESSURE amp ITS MEASUREMENT (2)
• Atmospheric Pressure amp Altitude
• Pressure Imbalance in Ear
• Example 1
• Example 2
• PRESSURE amp MOLES OF A GAS
• BOYLErsquoS LAW
• BOYLErsquoS LAW (2)
• Example 1 (2)
• Example 2 (2)
• CHARLESrsquoS LAW
• CHARLESrsquoS LAW (2)
• Example 1 (3)
• Example 2 (3)
• GAY-LUSSACrsquoS LAW
• GAY-LUSSACrsquoS LAW (2)
• Example 1 (4)
• Example 1 (5)
• Example 2 (4)
• VAPOR PRESSURE amp BOILING POINT
• VAPOR PRESSURE amp BOILING POINT (2)
• VAPOR PRESSURE amp BOILING POINT (3)
• VAPOR PRESSURE amp BOILING POINT (4)
• COMBINED GAS LAW
• COMBINED GAS LAW (2)
• Example 1 (6)
• AVOGADROrsquoS LAW
• Avogadrorsquos Law
• AVOGADROrsquoS LAW (2)
• AVOGADROrsquoS LAW (3)
• Example 1 (7)
• Example 2 (5)
• STP amp MOLAR VOLUME
• STP amp MOLAR VOLUME (2)
• Example 1 (8)
• Example 2 (6)
• IDEAL GAS LAW
• IDEAL GAS LAW (2)
• Example 1 (9)
• Example 2 (7)
• Example 2 (8)
• Ideal vs Real Gases
• Ideal vs Real Gases (2)
• PARTIAL PRESSURES
• DALTONrsquoS LAW
• PARTIAL PRESSURES (2)
• PARTIAL PRESSURES (3)
• PARTIAL PRESSURES (4)
• Example 1 (10)
• Example 2 (9)
• Slide 60

Tros Introductory Chemistry Chapter

12

Pressure Imbalance in Ear

If there is a differencein pressure acrossthe eardrum membranethe membrane will bepushed out ndash what we commonly call a ldquopopped eardrumrdquo

13

Example 1

The atmospheric pressure at Walnut CA is 740 mmHg Calculate this pressure in torr and atm

1 mmHg = 1 torr

740 mmHg = 740 torr

740 mmHg atm

x mmHg760

1= 0974 atm

14

Example 2

The barometer at a location reads 112 atm Calculate the pressure in mmHg and torr

1 mmHg = 1 torr

851 mmHg = 851 torr

112 atm mmHg

x atm1760

= 851 mmHg

15

PRESSURE amp MOLES OF A GAS

The pressure of a gas is directly proportional to the number of particles (moles) present

The greater the moles of the gas the greater the pressure

16

BOYLErsquoS LAW

The relationship of pressure and volume in gases is called Boylersquos Law

At constant temperature the volume of a fixed amount of gas is inversely proportional to its pressure

As pressure increases the volume of the gas decreases

Pressure and volume of a

gas are inversely

proportional

17

BOYLErsquoS LAW

Boylersquos Law can be mathematically expressed as

P1 x V1 = P2 x V2

P and V under initial

condition

P and V under final

condition

18

Example 1

A sample of gas has a volume of 12-L and a pressure of 4500 mmHg What is the volume of the gas when the pressure is reduced to 750 mmHg

1 12

2

P VV =

P

V2 = 72 L

P1= 4500 mmHg

P2= 750 mmHg

V1 = 12 L

V2 =

P1 x V1 = P2 x V2

(4500 mmHg)(12 L)=

(750 mmHg)

Pressure decreases Volume

increases

19

Example 2

A sample of hydrogen gas occupies 40 L at 650 mmHg What volume would it occupy at 20 atm

1 12

2

P VV =

P

V2 = 17 L

P1= 650 mmHg

P2= 20 atm

V1 = 40 L

V2 =

P1 x V1 = P2 x V2

(650 mmHg)(40 L)=

(1520 mmHg)Pressures must be in the same

unit

P2 = 20 atm = 1520 mmHgVolume decreases

Pressure increases

20

CHARLESrsquoS LAW

The relationship of temperature and volume in gases is called Charlesrsquo Law

At constant pressure the volume of a fixed amount of gas is directly proportional to its absolute temperature

As temperature increases the volume of the gas increases

Temperature and volume of a gas

are directly proportional

21

CHARLESrsquoS LAW

Charlesrsquo Law can be mathematically expressed as

V and T under initial

condition

1 2

1 2

V V=

T T

V and T under final condition

T must be in units of K

22

Example 1

A 20-L sample of a gas is cooled from 298K to 278 K at constant pressure What is the new volume of the gas

22 1

1

TV = V x

T

V2 = 19 L

V1 = 20 L

V2 =

T1 = 298 K

T2 = 278 K

278 K= 20 L x

298 K

Volume decreases1 2

1 2

V V=

T T

Temperature decreases

23

Example 2

If 200 L of oxygen is cooled from 100ordmC to 0ordmC what is the new volume

22 1

1

TV = V x

T

V2 = 146 L

V1 = 200 L

V2 =

T1 = 100ordmC

T2 = 0ordmC

273 K= 200 L x

373 K

T1= 100ordmC + 273 = 373 K

T2= 0ordmC + 273 = 273 K

Temperatures must be in

KVolume

decreasesTemperature

decreases

24

GAY-LUSSACrsquoSLAW

The relationship of temperature and pressure in gases is called Gay-Lussacrsquos Law

At constant volume the pressure of a fixed amount of gas is directly proportional to its absolute temperature

As temperature increases the

pressure of the gas increases

Temperature and pressure of a gas

are directly proportional

25

GAY-LUSSACrsquoSLAW

Gay-Lussacrsquos Law can be mathematically expressed as

P and T under initial

condition

1 2

1 2

P P=

T T

P and T under final conditionT must be in

units of K

26

Example 1

An aerosol spray can has a pressure of 40 atm at 25C What pressure will the can have if it is placed in a fire and reaches temperature of 400C

P1 = 40 atm

P2 =

T1 = 25ordmC

T2 = 400ordmC

T1= 25ordmC + 273 = 298 K

T2= 400ordmC + 273 = 673 K

Temperatures must be in K

27

Example 1

An aerosol spray can has a pressure of 40 atm at 25C What pressure will the can have if it is placed in a fire and reaches temperature of 400C

22 1

1

TP = P x

T

P2 = 90 atm

P1 = 40 atm

P2 =

T1 = 298 K

T2 = 673 K

673 K= 40 L x

298 K

Pressure increases

1 2

1 2

P P=

T T

Temperature increases

28

Example 2

A cylinder of gas with a volume of 150-L and a pressure of 965 mmHg is stored at a temperature of 55C To what temperature must the cylinder be cooled to reach a pressure of 850 mmHg

22 1

1

PT = T x

P

T2 = 289 K

P1 = 965 mmHg

P2 = 850 mmHg

T1 = 55ordmC

T2 =

850 mmHg= 328 K x

965 mmHg

T1= 55ordmC + 273 = 328 K

T2 = 16ordmC

Temperature decreases

Pressure decreases

29

VAPOR PRESSUREamp BOILING POINT

In an open container liquid molecules at the surface that possess sufficient energy can break away from the surface and become gas particles or vapor

In a closed container these gas particles can accumulate and create pressure called vapor pressure

Vapor pressure is defined as the pressure above a liquid at a given temperature Vapor pressure varies with each liquid and increases with temperature

30

VAPOR PRESSUREamp BOILING POINT

Listed below is the vapor pressure of water at various temperatures

31

VAPOR PRESSUREamp BOILING POINT

A liquid reaches its boiling point when its vapor pressure becomes equal to the external pressure (atmospheric pressure)

For example at sea level water reaches its boiling point at 100C since its vapor pressure is 760 mmHg at this temperature

32

VAPOR PRESSUREamp BOILING POINT

At higher altitudes where atmospheric pressure is lower water reaches boiling point at temperatures lower than 100C

For example in Denver where atmospheric pressure is 630 mmHg water boils at 95C since its vapor pressure is 630 mmHg at this temperature

33

COMBINEDGAS LAW

All pressure-volume-temperature relationships can be combined into a single relationship called the combined gas law

1 1 2 2

1 2

P V P V=

T T

Initial condition

This law is useful for studying the effect of changes in two variables

Final condition

34

COMBINEDGAS LAW

The individual gas laws studied previously are embodied in the combined gas law

1 1 2 2

1 2

P V P V=

T T

Boylersquos LawCharlesrsquos

LawGay-Lussacrsquos

Law

35

Example 1

A 250-mL sample of gas has a pressure of 400 atm at a temperature of 10C What is the volume of the gas at a pressure of 100 atm and a temperature of 18C

1 22 1

2 1

P TV = V x x

P T

V2 = 103 mL

P1= 400 atm V1= 250 mL T1= 283 K P2= 100 atm V2= T2 = 291 K

2

400 atm 291 KV = 250 mL x x

100 atm 283 K

1 1 2 2

1 2

P V P V=

T TBoth pressure

and temp change

Must use combined

gas law

36

AVOGADROrsquoSLAW

The relationship of moles and volume in gases is called Avogadrorsquos Law

At constant temperature and pressure the volume of a fixed amount of gas is directly proportional to the number of moles

As number of moles increases

the volume of the gas

increases

Number of moles and volume of a gas are directly

proportional

Tros Introductory Chemistry Chapter

37

Avogadrorsquos Law

38

AVOGADROrsquoSLAW

As a result of Avogadrorsquos Law equal volumes of different gases at the same temp and pressure contain equal number of moles (molecules)

2 Tanks of gas of equal volume at the same T amp P contain the same number of molecules

39

AVOGADROrsquoSLAW

For example

2 H2 (g) + 1 O2 (g) 2 H2O (g)

2 molecules 1 molecule 2 molecules

2 moles 1 mole 2 moles

2 Liters 1 Liter 2 Liters

Avogadrorsquos Law also allows chemists to relate volumes and moles of a gas in a chemical reaction

This relationship is only valid for

gaseous substances

40

Example 1

A sample of helium gas with a mass of 180 g occupies 16 Liters at a particular temperature and pressure What mass of oxygen would occupy16 L at the same temperature and pressure

1 mol= 180 g x

400 gmol of He = 450 mol

mol of O2 = mol of He = 450 mol

Same Temp amp Pressure

mass of O2

320 g= 450 mol x

1 mol= 144 g

41

Example 2

How many Liters of NH3 can be produced from reaction of 18 L of H2 with excess N2 as shown below

18 L H23

2

L NHx

L H= 12 L NH3

N2 (g) + 3 H2 (g) 2 NH3 (g)

32

42

STP amp MOLAR VOLUME

To better understand the factors that affect gas behavior a set of standard conditions have been chosen for use and are referred to as Standard Temperature and Pressure (STP)

STP =

1 atm (760 mmHg)

0ordmC (273 K)

43

STP amp MOLAR VOLUME

At STP conditions one mole of any gas is observed to occupy a volume of 224 L

V = 224 L

Molar Volume at

STP

44

Example 1

If 200 L of a gas at STP has a mass of 323 g what is the molar mass of the gas

mol200 L x

Lmol of gas =

2241

= 00893 mol

g

molMolar mass =

323 g=

00893 mol= 362 gmol Molar

volume at STP

45

Example 2

A sample of gas has a volume of 250 L at 730 mmHg and 20C What is the volume of this gas at STP

1 22 1

2 1

P TV = V x x

P T

V2 = 224 L

P1= 730 mmHg V1= 250 L T1= 293 K P2= 760 mmHg V2= T2 = 273 K

2

730 mmHg 273 KV = 250 L x x

760 mmHg 293 K

46

IDEAL GAS LAW

Combining all the laws that describe the behavior of gases one can obtain a useful relationship that relates the volume of a gas to the temperature pressure and number of moles

n R TV =

P

Universal gas constant

R = 00821 L atmmol K

47

IDEAL GAS LAW

This relationship is called the Ideal Gas Law and commonly written as

P V = n R T

Pressure in atm Volume

in Liters

Number of moles

Temp in K

48

Example 1

A sample of H2 gas has a volume of 856 L at a temp of 0C and pressure of 15 atm Calculate the moles of gas present

PVn =

RT

= 057 mol

P = 15 atm V = 856 L T = 273 K n = R = 00821

PV = nRT

(15 atm)(856 L)=

L atm(00821 )(273 K)

mol K

49

Example 2

What volume does 400 g of N2 gas occupy at 10C and 750 mmHg

P =750 mmHg V = T = 283 K n = not given m = 400 g R = 00821

mol of N2 =1 mol

400 g x280 g

= 143 mol

P = 750 mmHg1 atm

x760 mmHg

Moles must be calculated

from mass

Pressure must be

converted to atm

P = 0987 atm

50

Example 2

What volume does 400 g of N2 gas occupy at 10C and 750 mmHg

nRTV =

P

V = 337 L

P =0987 atm V = T = 283 K n = 143 R = 00821

L atm(143 mol)(00821 )(283 K)

mol KV = 0987 atm

Tros Introductory Chemistry Chapter

51

Ideal vs Real Gases

bull Real gases often do not behave like ideal gases at high pressure or low temperature

bull Ideal gas laws assume1) no attractions between gas molecules2) gas molecules do not take up space based on the Kinetic-Molecular Theory

bull at low temperatures and high pressures these assumptions are not valid

Tros Introductory Chemistry Chapter

52

Ideal vs Real Gases

53

PARTIAL PRESSURES

Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases

Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture

In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas

54

DALTONrsquoSLAW

In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture

This is called Daltonrsquos law of partial pressures

Total pressure of a gas mixture

Ptotal = P1 + P2 + P3 +

Sum of the partial pressures of the gases in the mixture

=

55

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = PHe + PAr

Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture

= 60 atm

56

PARTIAL PRESSURES

The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture

For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container

57

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = 60 atm

The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture

Twice as many moles of Ar

compared to He

PAr = 2 PHe

58

Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank

Ptotal = Poxygen + Phelium

Poxygen = 1140 mmHg x

1 atm

760 mmHg= 150 atm

Phelium

= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm

59

Example 2

A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture

Ptotal = Poxygen + Pnitrogen

Pnitrogen = 2 Poxygen

Twice as many moles of N2

compared to O2

Pnitrogen = 23 (Ptotal ) = 20 atm

Poxygen = 13 (Ptotal ) = 10 atm

60

THE END

• Slide 1
• CHAPTER OUTLINE
• PROPERTIES OF GASES
• PROPERTIES OF GASES (2)
• KINETIC-MOLECULAR THEORY
• KINETIC-MOLECULAR THEORY (2)
• Kinetic Molecular Theory
• PRESSURE amp ITS MEASUREMENT
• Common Units of Pressure
• PRESSURE amp ITS MEASUREMENT (2)
• Atmospheric Pressure amp Altitude
• Pressure Imbalance in Ear
• Example 1
• Example 2
• PRESSURE amp MOLES OF A GAS
• BOYLErsquoS LAW
• BOYLErsquoS LAW (2)
• Example 1 (2)
• Example 2 (2)
• CHARLESrsquoS LAW
• CHARLESrsquoS LAW (2)
• Example 1 (3)
• Example 2 (3)
• GAY-LUSSACrsquoS LAW
• GAY-LUSSACrsquoS LAW (2)
• Example 1 (4)
• Example 1 (5)
• Example 2 (4)
• VAPOR PRESSURE amp BOILING POINT
• VAPOR PRESSURE amp BOILING POINT (2)
• VAPOR PRESSURE amp BOILING POINT (3)
• VAPOR PRESSURE amp BOILING POINT (4)
• COMBINED GAS LAW
• COMBINED GAS LAW (2)
• Example 1 (6)
• AVOGADROrsquoS LAW
• Avogadrorsquos Law
• AVOGADROrsquoS LAW (2)
• AVOGADROrsquoS LAW (3)
• Example 1 (7)
• Example 2 (5)
• STP amp MOLAR VOLUME
• STP amp MOLAR VOLUME (2)
• Example 1 (8)
• Example 2 (6)
• IDEAL GAS LAW
• IDEAL GAS LAW (2)
• Example 1 (9)
• Example 2 (7)
• Example 2 (8)
• Ideal vs Real Gases
• Ideal vs Real Gases (2)
• PARTIAL PRESSURES
• DALTONrsquoS LAW
• PARTIAL PRESSURES (2)
• PARTIAL PRESSURES (3)
• PARTIAL PRESSURES (4)
• Example 1 (10)
• Example 2 (9)
• Slide 60

13

Example 1

The atmospheric pressure at Walnut CA is 740 mmHg Calculate this pressure in torr and atm

1 mmHg = 1 torr

740 mmHg = 740 torr

740 mmHg atm

x mmHg760

1= 0974 atm

14

Example 2

The barometer at a location reads 112 atm Calculate the pressure in mmHg and torr

1 mmHg = 1 torr

851 mmHg = 851 torr

112 atm mmHg

x atm1760

= 851 mmHg

15

PRESSURE amp MOLES OF A GAS

The pressure of a gas is directly proportional to the number of particles (moles) present

The greater the moles of the gas the greater the pressure

16

BOYLErsquoS LAW

The relationship of pressure and volume in gases is called Boylersquos Law

At constant temperature the volume of a fixed amount of gas is inversely proportional to its pressure

As pressure increases the volume of the gas decreases

Pressure and volume of a

gas are inversely

proportional

17

BOYLErsquoS LAW

Boylersquos Law can be mathematically expressed as

P1 x V1 = P2 x V2

P and V under initial

condition

P and V under final

condition

18

Example 1

A sample of gas has a volume of 12-L and a pressure of 4500 mmHg What is the volume of the gas when the pressure is reduced to 750 mmHg

1 12

2

P VV =

P

V2 = 72 L

P1= 4500 mmHg

P2= 750 mmHg

V1 = 12 L

V2 =

P1 x V1 = P2 x V2

(4500 mmHg)(12 L)=

(750 mmHg)

Pressure decreases Volume

increases

19

Example 2

A sample of hydrogen gas occupies 40 L at 650 mmHg What volume would it occupy at 20 atm

1 12

2

P VV =

P

V2 = 17 L

P1= 650 mmHg

P2= 20 atm

V1 = 40 L

V2 =

P1 x V1 = P2 x V2

(650 mmHg)(40 L)=

(1520 mmHg)Pressures must be in the same

unit

P2 = 20 atm = 1520 mmHgVolume decreases

Pressure increases

20

CHARLESrsquoS LAW

The relationship of temperature and volume in gases is called Charlesrsquo Law

At constant pressure the volume of a fixed amount of gas is directly proportional to its absolute temperature

As temperature increases the volume of the gas increases

Temperature and volume of a gas

are directly proportional

21

CHARLESrsquoS LAW

Charlesrsquo Law can be mathematically expressed as

V and T under initial

condition

1 2

1 2

V V=

T T

V and T under final condition

T must be in units of K

22

Example 1

A 20-L sample of a gas is cooled from 298K to 278 K at constant pressure What is the new volume of the gas

22 1

1

TV = V x

T

V2 = 19 L

V1 = 20 L

V2 =

T1 = 298 K

T2 = 278 K

278 K= 20 L x

298 K

Volume decreases1 2

1 2

V V=

T T

Temperature decreases

23

Example 2

If 200 L of oxygen is cooled from 100ordmC to 0ordmC what is the new volume

22 1

1

TV = V x

T

V2 = 146 L

V1 = 200 L

V2 =

T1 = 100ordmC

T2 = 0ordmC

273 K= 200 L x

373 K

T1= 100ordmC + 273 = 373 K

T2= 0ordmC + 273 = 273 K

Temperatures must be in

KVolume

decreasesTemperature

decreases

24

GAY-LUSSACrsquoSLAW

The relationship of temperature and pressure in gases is called Gay-Lussacrsquos Law

At constant volume the pressure of a fixed amount of gas is directly proportional to its absolute temperature

As temperature increases the

pressure of the gas increases

Temperature and pressure of a gas

are directly proportional

25

GAY-LUSSACrsquoSLAW

Gay-Lussacrsquos Law can be mathematically expressed as

P and T under initial

condition

1 2

1 2

P P=

T T

P and T under final conditionT must be in

units of K

26

Example 1

An aerosol spray can has a pressure of 40 atm at 25C What pressure will the can have if it is placed in a fire and reaches temperature of 400C

P1 = 40 atm

P2 =

T1 = 25ordmC

T2 = 400ordmC

T1= 25ordmC + 273 = 298 K

T2= 400ordmC + 273 = 673 K

Temperatures must be in K

27

Example 1

An aerosol spray can has a pressure of 40 atm at 25C What pressure will the can have if it is placed in a fire and reaches temperature of 400C

22 1

1

TP = P x

T

P2 = 90 atm

P1 = 40 atm

P2 =

T1 = 298 K

T2 = 673 K

673 K= 40 L x

298 K

Pressure increases

1 2

1 2

P P=

T T

Temperature increases

28

Example 2

A cylinder of gas with a volume of 150-L and a pressure of 965 mmHg is stored at a temperature of 55C To what temperature must the cylinder be cooled to reach a pressure of 850 mmHg

22 1

1

PT = T x

P

T2 = 289 K

P1 = 965 mmHg

P2 = 850 mmHg

T1 = 55ordmC

T2 =

850 mmHg= 328 K x

965 mmHg

T1= 55ordmC + 273 = 328 K

T2 = 16ordmC

Temperature decreases

Pressure decreases

29

VAPOR PRESSUREamp BOILING POINT

In an open container liquid molecules at the surface that possess sufficient energy can break away from the surface and become gas particles or vapor

In a closed container these gas particles can accumulate and create pressure called vapor pressure

Vapor pressure is defined as the pressure above a liquid at a given temperature Vapor pressure varies with each liquid and increases with temperature

30

VAPOR PRESSUREamp BOILING POINT

Listed below is the vapor pressure of water at various temperatures

31

VAPOR PRESSUREamp BOILING POINT

A liquid reaches its boiling point when its vapor pressure becomes equal to the external pressure (atmospheric pressure)

For example at sea level water reaches its boiling point at 100C since its vapor pressure is 760 mmHg at this temperature

32

VAPOR PRESSUREamp BOILING POINT

At higher altitudes where atmospheric pressure is lower water reaches boiling point at temperatures lower than 100C

For example in Denver where atmospheric pressure is 630 mmHg water boils at 95C since its vapor pressure is 630 mmHg at this temperature

33

COMBINEDGAS LAW

All pressure-volume-temperature relationships can be combined into a single relationship called the combined gas law

1 1 2 2

1 2

P V P V=

T T

Initial condition

This law is useful for studying the effect of changes in two variables

Final condition

34

COMBINEDGAS LAW

The individual gas laws studied previously are embodied in the combined gas law

1 1 2 2

1 2

P V P V=

T T

Boylersquos LawCharlesrsquos

LawGay-Lussacrsquos

Law

35

Example 1

A 250-mL sample of gas has a pressure of 400 atm at a temperature of 10C What is the volume of the gas at a pressure of 100 atm and a temperature of 18C

1 22 1

2 1

P TV = V x x

P T

V2 = 103 mL

P1= 400 atm V1= 250 mL T1= 283 K P2= 100 atm V2= T2 = 291 K

2

400 atm 291 KV = 250 mL x x

100 atm 283 K

1 1 2 2

1 2

P V P V=

T TBoth pressure

and temp change

Must use combined

gas law

36

AVOGADROrsquoSLAW

The relationship of moles and volume in gases is called Avogadrorsquos Law

At constant temperature and pressure the volume of a fixed amount of gas is directly proportional to the number of moles

As number of moles increases

the volume of the gas

increases

Number of moles and volume of a gas are directly

proportional

Tros Introductory Chemistry Chapter

37

Avogadrorsquos Law

38

AVOGADROrsquoSLAW

As a result of Avogadrorsquos Law equal volumes of different gases at the same temp and pressure contain equal number of moles (molecules)

2 Tanks of gas of equal volume at the same T amp P contain the same number of molecules

39

AVOGADROrsquoSLAW

For example

2 H2 (g) + 1 O2 (g) 2 H2O (g)

2 molecules 1 molecule 2 molecules

2 moles 1 mole 2 moles

2 Liters 1 Liter 2 Liters

Avogadrorsquos Law also allows chemists to relate volumes and moles of a gas in a chemical reaction

This relationship is only valid for

gaseous substances

40

Example 1

A sample of helium gas with a mass of 180 g occupies 16 Liters at a particular temperature and pressure What mass of oxygen would occupy16 L at the same temperature and pressure

1 mol= 180 g x

400 gmol of He = 450 mol

mol of O2 = mol of He = 450 mol

Same Temp amp Pressure

mass of O2

320 g= 450 mol x

1 mol= 144 g

41

Example 2

How many Liters of NH3 can be produced from reaction of 18 L of H2 with excess N2 as shown below

18 L H23

2

L NHx

L H= 12 L NH3

N2 (g) + 3 H2 (g) 2 NH3 (g)

32

42

STP amp MOLAR VOLUME

To better understand the factors that affect gas behavior a set of standard conditions have been chosen for use and are referred to as Standard Temperature and Pressure (STP)

STP =

1 atm (760 mmHg)

0ordmC (273 K)

43

STP amp MOLAR VOLUME

At STP conditions one mole of any gas is observed to occupy a volume of 224 L

V = 224 L

Molar Volume at

STP

44

Example 1

If 200 L of a gas at STP has a mass of 323 g what is the molar mass of the gas

mol200 L x

Lmol of gas =

2241

= 00893 mol

g

molMolar mass =

323 g=

00893 mol= 362 gmol Molar

volume at STP

45

Example 2

A sample of gas has a volume of 250 L at 730 mmHg and 20C What is the volume of this gas at STP

1 22 1

2 1

P TV = V x x

P T

V2 = 224 L

P1= 730 mmHg V1= 250 L T1= 293 K P2= 760 mmHg V2= T2 = 273 K

2

730 mmHg 273 KV = 250 L x x

760 mmHg 293 K

46

IDEAL GAS LAW

Combining all the laws that describe the behavior of gases one can obtain a useful relationship that relates the volume of a gas to the temperature pressure and number of moles

n R TV =

P

Universal gas constant

R = 00821 L atmmol K

47

IDEAL GAS LAW

This relationship is called the Ideal Gas Law and commonly written as

P V = n R T

Pressure in atm Volume

in Liters

Number of moles

Temp in K

48

Example 1

A sample of H2 gas has a volume of 856 L at a temp of 0C and pressure of 15 atm Calculate the moles of gas present

PVn =

RT

= 057 mol

P = 15 atm V = 856 L T = 273 K n = R = 00821

PV = nRT

(15 atm)(856 L)=

L atm(00821 )(273 K)

mol K

49

Example 2

What volume does 400 g of N2 gas occupy at 10C and 750 mmHg

P =750 mmHg V = T = 283 K n = not given m = 400 g R = 00821

mol of N2 =1 mol

400 g x280 g

= 143 mol

P = 750 mmHg1 atm

x760 mmHg

Moles must be calculated

from mass

Pressure must be

converted to atm

P = 0987 atm

50

Example 2

What volume does 400 g of N2 gas occupy at 10C and 750 mmHg

nRTV =

P

V = 337 L

P =0987 atm V = T = 283 K n = 143 R = 00821

L atm(143 mol)(00821 )(283 K)

mol KV = 0987 atm

Tros Introductory Chemistry Chapter

51

Ideal vs Real Gases

bull Real gases often do not behave like ideal gases at high pressure or low temperature

bull Ideal gas laws assume1) no attractions between gas molecules2) gas molecules do not take up space based on the Kinetic-Molecular Theory

bull at low temperatures and high pressures these assumptions are not valid

Tros Introductory Chemistry Chapter

52

Ideal vs Real Gases

53

PARTIAL PRESSURES

Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases

Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture

In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas

54

DALTONrsquoSLAW

In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture

This is called Daltonrsquos law of partial pressures

Total pressure of a gas mixture

Ptotal = P1 + P2 + P3 +

Sum of the partial pressures of the gases in the mixture

=

55

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = PHe + PAr

Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture

= 60 atm

56

PARTIAL PRESSURES

The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture

For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container

57

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = 60 atm

The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture

Twice as many moles of Ar

compared to He

PAr = 2 PHe

58

Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank

Ptotal = Poxygen + Phelium

Poxygen = 1140 mmHg x

1 atm

760 mmHg= 150 atm

Phelium

= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm

59

Example 2

A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture

Ptotal = Poxygen + Pnitrogen

Pnitrogen = 2 Poxygen

Twice as many moles of N2

compared to O2

Pnitrogen = 23 (Ptotal ) = 20 atm

Poxygen = 13 (Ptotal ) = 10 atm

60

THE END

• Slide 1
• CHAPTER OUTLINE
• PROPERTIES OF GASES
• PROPERTIES OF GASES (2)
• KINETIC-MOLECULAR THEORY
• KINETIC-MOLECULAR THEORY (2)
• Kinetic Molecular Theory
• PRESSURE amp ITS MEASUREMENT
• Common Units of Pressure
• PRESSURE amp ITS MEASUREMENT (2)
• Atmospheric Pressure amp Altitude
• Pressure Imbalance in Ear
• Example 1
• Example 2
• PRESSURE amp MOLES OF A GAS
• BOYLErsquoS LAW
• BOYLErsquoS LAW (2)
• Example 1 (2)
• Example 2 (2)
• CHARLESrsquoS LAW
• CHARLESrsquoS LAW (2)
• Example 1 (3)
• Example 2 (3)
• GAY-LUSSACrsquoS LAW
• GAY-LUSSACrsquoS LAW (2)
• Example 1 (4)
• Example 1 (5)
• Example 2 (4)
• VAPOR PRESSURE amp BOILING POINT
• VAPOR PRESSURE amp BOILING POINT (2)
• VAPOR PRESSURE amp BOILING POINT (3)
• VAPOR PRESSURE amp BOILING POINT (4)
• COMBINED GAS LAW
• COMBINED GAS LAW (2)
• Example 1 (6)
• AVOGADROrsquoS LAW
• Avogadrorsquos Law
• AVOGADROrsquoS LAW (2)
• AVOGADROrsquoS LAW (3)
• Example 1 (7)
• Example 2 (5)
• STP amp MOLAR VOLUME
• STP amp MOLAR VOLUME (2)
• Example 1 (8)
• Example 2 (6)
• IDEAL GAS LAW
• IDEAL GAS LAW (2)
• Example 1 (9)
• Example 2 (7)
• Example 2 (8)
• Ideal vs Real Gases
• Ideal vs Real Gases (2)
• PARTIAL PRESSURES
• DALTONrsquoS LAW
• PARTIAL PRESSURES (2)
• PARTIAL PRESSURES (3)
• PARTIAL PRESSURES (4)
• Example 1 (10)
• Example 2 (9)
• Slide 60

14

Example 2

The barometer at a location reads 112 atm Calculate the pressure in mmHg and torr

1 mmHg = 1 torr

851 mmHg = 851 torr

112 atm mmHg

x atm1760

= 851 mmHg

15

PRESSURE amp MOLES OF A GAS

The pressure of a gas is directly proportional to the number of particles (moles) present

The greater the moles of the gas the greater the pressure

16

BOYLErsquoS LAW

The relationship of pressure and volume in gases is called Boylersquos Law

At constant temperature the volume of a fixed amount of gas is inversely proportional to its pressure

As pressure increases the volume of the gas decreases

Pressure and volume of a

gas are inversely

proportional

17

BOYLErsquoS LAW

Boylersquos Law can be mathematically expressed as

P1 x V1 = P2 x V2

P and V under initial

condition

P and V under final

condition

18

Example 1

A sample of gas has a volume of 12-L and a pressure of 4500 mmHg What is the volume of the gas when the pressure is reduced to 750 mmHg

1 12

2

P VV =

P

V2 = 72 L

P1= 4500 mmHg

P2= 750 mmHg

V1 = 12 L

V2 =

P1 x V1 = P2 x V2

(4500 mmHg)(12 L)=

(750 mmHg)

Pressure decreases Volume

increases

19

Example 2

A sample of hydrogen gas occupies 40 L at 650 mmHg What volume would it occupy at 20 atm

1 12

2

P VV =

P

V2 = 17 L

P1= 650 mmHg

P2= 20 atm

V1 = 40 L

V2 =

P1 x V1 = P2 x V2

(650 mmHg)(40 L)=

(1520 mmHg)Pressures must be in the same

unit

P2 = 20 atm = 1520 mmHgVolume decreases

Pressure increases

20

CHARLESrsquoS LAW

The relationship of temperature and volume in gases is called Charlesrsquo Law

At constant pressure the volume of a fixed amount of gas is directly proportional to its absolute temperature

As temperature increases the volume of the gas increases

Temperature and volume of a gas

are directly proportional

21

CHARLESrsquoS LAW

Charlesrsquo Law can be mathematically expressed as

V and T under initial

condition

1 2

1 2

V V=

T T

V and T under final condition

T must be in units of K

22

Example 1

A 20-L sample of a gas is cooled from 298K to 278 K at constant pressure What is the new volume of the gas

22 1

1

TV = V x

T

V2 = 19 L

V1 = 20 L

V2 =

T1 = 298 K

T2 = 278 K

278 K= 20 L x

298 K

Volume decreases1 2

1 2

V V=

T T

Temperature decreases

23

Example 2

If 200 L of oxygen is cooled from 100ordmC to 0ordmC what is the new volume

22 1

1

TV = V x

T

V2 = 146 L

V1 = 200 L

V2 =

T1 = 100ordmC

T2 = 0ordmC

273 K= 200 L x

373 K

T1= 100ordmC + 273 = 373 K

T2= 0ordmC + 273 = 273 K

Temperatures must be in

KVolume

decreasesTemperature

decreases

24

GAY-LUSSACrsquoSLAW

The relationship of temperature and pressure in gases is called Gay-Lussacrsquos Law

At constant volume the pressure of a fixed amount of gas is directly proportional to its absolute temperature

As temperature increases the

pressure of the gas increases

Temperature and pressure of a gas

are directly proportional

25

GAY-LUSSACrsquoSLAW

Gay-Lussacrsquos Law can be mathematically expressed as

P and T under initial

condition

1 2

1 2

P P=

T T

P and T under final conditionT must be in

units of K

26

Example 1

An aerosol spray can has a pressure of 40 atm at 25C What pressure will the can have if it is placed in a fire and reaches temperature of 400C

P1 = 40 atm

P2 =

T1 = 25ordmC

T2 = 400ordmC

T1= 25ordmC + 273 = 298 K

T2= 400ordmC + 273 = 673 K

Temperatures must be in K

27

Example 1

An aerosol spray can has a pressure of 40 atm at 25C What pressure will the can have if it is placed in a fire and reaches temperature of 400C

22 1

1

TP = P x

T

P2 = 90 atm

P1 = 40 atm

P2 =

T1 = 298 K

T2 = 673 K

673 K= 40 L x

298 K

Pressure increases

1 2

1 2

P P=

T T

Temperature increases

28

Example 2

A cylinder of gas with a volume of 150-L and a pressure of 965 mmHg is stored at a temperature of 55C To what temperature must the cylinder be cooled to reach a pressure of 850 mmHg

22 1

1

PT = T x

P

T2 = 289 K

P1 = 965 mmHg

P2 = 850 mmHg

T1 = 55ordmC

T2 =

850 mmHg= 328 K x

965 mmHg

T1= 55ordmC + 273 = 328 K

T2 = 16ordmC

Temperature decreases

Pressure decreases

29

VAPOR PRESSUREamp BOILING POINT

In an open container liquid molecules at the surface that possess sufficient energy can break away from the surface and become gas particles or vapor

In a closed container these gas particles can accumulate and create pressure called vapor pressure

Vapor pressure is defined as the pressure above a liquid at a given temperature Vapor pressure varies with each liquid and increases with temperature

30

VAPOR PRESSUREamp BOILING POINT

Listed below is the vapor pressure of water at various temperatures

31

VAPOR PRESSUREamp BOILING POINT

A liquid reaches its boiling point when its vapor pressure becomes equal to the external pressure (atmospheric pressure)

For example at sea level water reaches its boiling point at 100C since its vapor pressure is 760 mmHg at this temperature

32

VAPOR PRESSUREamp BOILING POINT

At higher altitudes where atmospheric pressure is lower water reaches boiling point at temperatures lower than 100C

For example in Denver where atmospheric pressure is 630 mmHg water boils at 95C since its vapor pressure is 630 mmHg at this temperature

33

COMBINEDGAS LAW

All pressure-volume-temperature relationships can be combined into a single relationship called the combined gas law

1 1 2 2

1 2

P V P V=

T T

Initial condition

This law is useful for studying the effect of changes in two variables

Final condition

34

COMBINEDGAS LAW

The individual gas laws studied previously are embodied in the combined gas law

1 1 2 2

1 2

P V P V=

T T

Boylersquos LawCharlesrsquos

LawGay-Lussacrsquos

Law

35

Example 1

A 250-mL sample of gas has a pressure of 400 atm at a temperature of 10C What is the volume of the gas at a pressure of 100 atm and a temperature of 18C

1 22 1

2 1

P TV = V x x

P T

V2 = 103 mL

P1= 400 atm V1= 250 mL T1= 283 K P2= 100 atm V2= T2 = 291 K

2

400 atm 291 KV = 250 mL x x

100 atm 283 K

1 1 2 2

1 2

P V P V=

T TBoth pressure

and temp change

Must use combined

gas law

36

AVOGADROrsquoSLAW

The relationship of moles and volume in gases is called Avogadrorsquos Law

At constant temperature and pressure the volume of a fixed amount of gas is directly proportional to the number of moles

As number of moles increases

the volume of the gas

increases

Number of moles and volume of a gas are directly

proportional

Tros Introductory Chemistry Chapter

37

Avogadrorsquos Law

38

AVOGADROrsquoSLAW

As a result of Avogadrorsquos Law equal volumes of different gases at the same temp and pressure contain equal number of moles (molecules)

2 Tanks of gas of equal volume at the same T amp P contain the same number of molecules

39

AVOGADROrsquoSLAW

For example

2 H2 (g) + 1 O2 (g) 2 H2O (g)

2 molecules 1 molecule 2 molecules

2 moles 1 mole 2 moles

2 Liters 1 Liter 2 Liters

Avogadrorsquos Law also allows chemists to relate volumes and moles of a gas in a chemical reaction

This relationship is only valid for

gaseous substances

40

Example 1

A sample of helium gas with a mass of 180 g occupies 16 Liters at a particular temperature and pressure What mass of oxygen would occupy16 L at the same temperature and pressure

1 mol= 180 g x

400 gmol of He = 450 mol

mol of O2 = mol of He = 450 mol

Same Temp amp Pressure

mass of O2

320 g= 450 mol x

1 mol= 144 g

41

Example 2

How many Liters of NH3 can be produced from reaction of 18 L of H2 with excess N2 as shown below

18 L H23

2

L NHx

L H= 12 L NH3

N2 (g) + 3 H2 (g) 2 NH3 (g)

32

42

STP amp MOLAR VOLUME

To better understand the factors that affect gas behavior a set of standard conditions have been chosen for use and are referred to as Standard Temperature and Pressure (STP)

STP =

1 atm (760 mmHg)

0ordmC (273 K)

43

STP amp MOLAR VOLUME

At STP conditions one mole of any gas is observed to occupy a volume of 224 L

V = 224 L

Molar Volume at

STP

44

Example 1

If 200 L of a gas at STP has a mass of 323 g what is the molar mass of the gas

mol200 L x

Lmol of gas =

2241

= 00893 mol

g

molMolar mass =

323 g=

00893 mol= 362 gmol Molar

volume at STP

45

Example 2

A sample of gas has a volume of 250 L at 730 mmHg and 20C What is the volume of this gas at STP

1 22 1

2 1

P TV = V x x

P T

V2 = 224 L

P1= 730 mmHg V1= 250 L T1= 293 K P2= 760 mmHg V2= T2 = 273 K

2

730 mmHg 273 KV = 250 L x x

760 mmHg 293 K

46

IDEAL GAS LAW

Combining all the laws that describe the behavior of gases one can obtain a useful relationship that relates the volume of a gas to the temperature pressure and number of moles

n R TV =

P

Universal gas constant

R = 00821 L atmmol K

47

IDEAL GAS LAW

This relationship is called the Ideal Gas Law and commonly written as

P V = n R T

Pressure in atm Volume

in Liters

Number of moles

Temp in K

48

Example 1

A sample of H2 gas has a volume of 856 L at a temp of 0C and pressure of 15 atm Calculate the moles of gas present

PVn =

RT

= 057 mol

P = 15 atm V = 856 L T = 273 K n = R = 00821

PV = nRT

(15 atm)(856 L)=

L atm(00821 )(273 K)

mol K

49

Example 2

What volume does 400 g of N2 gas occupy at 10C and 750 mmHg

P =750 mmHg V = T = 283 K n = not given m = 400 g R = 00821

mol of N2 =1 mol

400 g x280 g

= 143 mol

P = 750 mmHg1 atm

x760 mmHg

Moles must be calculated

from mass

Pressure must be

converted to atm

P = 0987 atm

50

Example 2

What volume does 400 g of N2 gas occupy at 10C and 750 mmHg

nRTV =

P

V = 337 L

P =0987 atm V = T = 283 K n = 143 R = 00821

L atm(143 mol)(00821 )(283 K)

mol KV = 0987 atm

Tros Introductory Chemistry Chapter

51

Ideal vs Real Gases

bull Real gases often do not behave like ideal gases at high pressure or low temperature

bull Ideal gas laws assume1) no attractions between gas molecules2) gas molecules do not take up space based on the Kinetic-Molecular Theory

bull at low temperatures and high pressures these assumptions are not valid

Tros Introductory Chemistry Chapter

52

Ideal vs Real Gases

53

PARTIAL PRESSURES

Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases

Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture

In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas

54

DALTONrsquoSLAW

In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture

This is called Daltonrsquos law of partial pressures

Total pressure of a gas mixture

Ptotal = P1 + P2 + P3 +

Sum of the partial pressures of the gases in the mixture

=

55

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = PHe + PAr

Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture

= 60 atm

56

PARTIAL PRESSURES

The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture

For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container

57

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = 60 atm

The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture

Twice as many moles of Ar

compared to He

PAr = 2 PHe

58

Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank

Ptotal = Poxygen + Phelium

Poxygen = 1140 mmHg x

1 atm

760 mmHg= 150 atm

Phelium

= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm

59

Example 2

A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture

Ptotal = Poxygen + Pnitrogen

Pnitrogen = 2 Poxygen

Twice as many moles of N2

compared to O2

Pnitrogen = 23 (Ptotal ) = 20 atm

Poxygen = 13 (Ptotal ) = 10 atm

60

THE END

• Slide 1
• CHAPTER OUTLINE
• PROPERTIES OF GASES
• PROPERTIES OF GASES (2)
• KINETIC-MOLECULAR THEORY
• KINETIC-MOLECULAR THEORY (2)
• Kinetic Molecular Theory
• PRESSURE amp ITS MEASUREMENT
• Common Units of Pressure
• PRESSURE amp ITS MEASUREMENT (2)
• Atmospheric Pressure amp Altitude
• Pressure Imbalance in Ear
• Example 1
• Example 2
• PRESSURE amp MOLES OF A GAS
• BOYLErsquoS LAW
• BOYLErsquoS LAW (2)
• Example 1 (2)
• Example 2 (2)
• CHARLESrsquoS LAW
• CHARLESrsquoS LAW (2)
• Example 1 (3)
• Example 2 (3)
• GAY-LUSSACrsquoS LAW
• GAY-LUSSACrsquoS LAW (2)
• Example 1 (4)
• Example 1 (5)
• Example 2 (4)
• VAPOR PRESSURE amp BOILING POINT
• VAPOR PRESSURE amp BOILING POINT (2)
• VAPOR PRESSURE amp BOILING POINT (3)
• VAPOR PRESSURE amp BOILING POINT (4)
• COMBINED GAS LAW
• COMBINED GAS LAW (2)
• Example 1 (6)
• AVOGADROrsquoS LAW
• Avogadrorsquos Law
• AVOGADROrsquoS LAW (2)
• AVOGADROrsquoS LAW (3)
• Example 1 (7)
• Example 2 (5)
• STP amp MOLAR VOLUME
• STP amp MOLAR VOLUME (2)
• Example 1 (8)
• Example 2 (6)
• IDEAL GAS LAW
• IDEAL GAS LAW (2)
• Example 1 (9)
• Example 2 (7)
• Example 2 (8)
• Ideal vs Real Gases
• Ideal vs Real Gases (2)
• PARTIAL PRESSURES
• DALTONrsquoS LAW
• PARTIAL PRESSURES (2)
• PARTIAL PRESSURES (3)
• PARTIAL PRESSURES (4)
• Example 1 (10)
• Example 2 (9)
• Slide 60

15

PRESSURE amp MOLES OF A GAS

The pressure of a gas is directly proportional to the number of particles (moles) present

The greater the moles of the gas the greater the pressure

16

BOYLErsquoS LAW

The relationship of pressure and volume in gases is called Boylersquos Law

At constant temperature the volume of a fixed amount of gas is inversely proportional to its pressure

As pressure increases the volume of the gas decreases

Pressure and volume of a

gas are inversely

proportional

17

BOYLErsquoS LAW

Boylersquos Law can be mathematically expressed as

P1 x V1 = P2 x V2

P and V under initial

condition

P and V under final

condition

18

Example 1

A sample of gas has a volume of 12-L and a pressure of 4500 mmHg What is the volume of the gas when the pressure is reduced to 750 mmHg

1 12

2

P VV =

P

V2 = 72 L

P1= 4500 mmHg

P2= 750 mmHg

V1 = 12 L

V2 =

P1 x V1 = P2 x V2

(4500 mmHg)(12 L)=

(750 mmHg)

Pressure decreases Volume

increases

19

Example 2

A sample of hydrogen gas occupies 40 L at 650 mmHg What volume would it occupy at 20 atm

1 12

2

P VV =

P

V2 = 17 L

P1= 650 mmHg

P2= 20 atm

V1 = 40 L

V2 =

P1 x V1 = P2 x V2

(650 mmHg)(40 L)=

(1520 mmHg)Pressures must be in the same

unit

P2 = 20 atm = 1520 mmHgVolume decreases

Pressure increases

20

CHARLESrsquoS LAW

The relationship of temperature and volume in gases is called Charlesrsquo Law

At constant pressure the volume of a fixed amount of gas is directly proportional to its absolute temperature

As temperature increases the volume of the gas increases

Temperature and volume of a gas

are directly proportional

21

CHARLESrsquoS LAW

Charlesrsquo Law can be mathematically expressed as

V and T under initial

condition

1 2

1 2

V V=

T T

V and T under final condition

T must be in units of K

22

Example 1

A 20-L sample of a gas is cooled from 298K to 278 K at constant pressure What is the new volume of the gas

22 1

1

TV = V x

T

V2 = 19 L

V1 = 20 L

V2 =

T1 = 298 K

T2 = 278 K

278 K= 20 L x

298 K

Volume decreases1 2

1 2

V V=

T T

Temperature decreases

23

Example 2

If 200 L of oxygen is cooled from 100ordmC to 0ordmC what is the new volume

22 1

1

TV = V x

T

V2 = 146 L

V1 = 200 L

V2 =

T1 = 100ordmC

T2 = 0ordmC

273 K= 200 L x

373 K

T1= 100ordmC + 273 = 373 K

T2= 0ordmC + 273 = 273 K

Temperatures must be in

KVolume

decreasesTemperature

decreases

24

GAY-LUSSACrsquoSLAW

The relationship of temperature and pressure in gases is called Gay-Lussacrsquos Law

At constant volume the pressure of a fixed amount of gas is directly proportional to its absolute temperature

As temperature increases the

pressure of the gas increases

Temperature and pressure of a gas

are directly proportional

25

GAY-LUSSACrsquoSLAW

Gay-Lussacrsquos Law can be mathematically expressed as

P and T under initial

condition

1 2

1 2

P P=

T T

P and T under final conditionT must be in

units of K

26

Example 1

An aerosol spray can has a pressure of 40 atm at 25C What pressure will the can have if it is placed in a fire and reaches temperature of 400C

P1 = 40 atm

P2 =

T1 = 25ordmC

T2 = 400ordmC

T1= 25ordmC + 273 = 298 K

T2= 400ordmC + 273 = 673 K

Temperatures must be in K

27

Example 1

An aerosol spray can has a pressure of 40 atm at 25C What pressure will the can have if it is placed in a fire and reaches temperature of 400C

22 1

1

TP = P x

T

P2 = 90 atm

P1 = 40 atm

P2 =

T1 = 298 K

T2 = 673 K

673 K= 40 L x

298 K

Pressure increases

1 2

1 2

P P=

T T

Temperature increases

28

Example 2

A cylinder of gas with a volume of 150-L and a pressure of 965 mmHg is stored at a temperature of 55C To what temperature must the cylinder be cooled to reach a pressure of 850 mmHg

22 1

1

PT = T x

P

T2 = 289 K

P1 = 965 mmHg

P2 = 850 mmHg

T1 = 55ordmC

T2 =

850 mmHg= 328 K x

965 mmHg

T1= 55ordmC + 273 = 328 K

T2 = 16ordmC

Temperature decreases

Pressure decreases

29

VAPOR PRESSUREamp BOILING POINT

In an open container liquid molecules at the surface that possess sufficient energy can break away from the surface and become gas particles or vapor

In a closed container these gas particles can accumulate and create pressure called vapor pressure

Vapor pressure is defined as the pressure above a liquid at a given temperature Vapor pressure varies with each liquid and increases with temperature

30

VAPOR PRESSUREamp BOILING POINT

Listed below is the vapor pressure of water at various temperatures

31

VAPOR PRESSUREamp BOILING POINT

A liquid reaches its boiling point when its vapor pressure becomes equal to the external pressure (atmospheric pressure)

For example at sea level water reaches its boiling point at 100C since its vapor pressure is 760 mmHg at this temperature

32

VAPOR PRESSUREamp BOILING POINT

At higher altitudes where atmospheric pressure is lower water reaches boiling point at temperatures lower than 100C

For example in Denver where atmospheric pressure is 630 mmHg water boils at 95C since its vapor pressure is 630 mmHg at this temperature

33

COMBINEDGAS LAW

All pressure-volume-temperature relationships can be combined into a single relationship called the combined gas law

1 1 2 2

1 2

P V P V=

T T

Initial condition

This law is useful for studying the effect of changes in two variables

Final condition

34

COMBINEDGAS LAW

The individual gas laws studied previously are embodied in the combined gas law

1 1 2 2

1 2

P V P V=

T T

Boylersquos LawCharlesrsquos

LawGay-Lussacrsquos

Law

35

Example 1

A 250-mL sample of gas has a pressure of 400 atm at a temperature of 10C What is the volume of the gas at a pressure of 100 atm and a temperature of 18C

1 22 1

2 1

P TV = V x x

P T

V2 = 103 mL

P1= 400 atm V1= 250 mL T1= 283 K P2= 100 atm V2= T2 = 291 K

2

400 atm 291 KV = 250 mL x x

100 atm 283 K

1 1 2 2

1 2

P V P V=

T TBoth pressure

and temp change

Must use combined

gas law

36

AVOGADROrsquoSLAW

The relationship of moles and volume in gases is called Avogadrorsquos Law

At constant temperature and pressure the volume of a fixed amount of gas is directly proportional to the number of moles

As number of moles increases

the volume of the gas

increases

Number of moles and volume of a gas are directly

proportional

Tros Introductory Chemistry Chapter

37

Avogadrorsquos Law

38

AVOGADROrsquoSLAW

As a result of Avogadrorsquos Law equal volumes of different gases at the same temp and pressure contain equal number of moles (molecules)

2 Tanks of gas of equal volume at the same T amp P contain the same number of molecules

39

AVOGADROrsquoSLAW

For example

2 H2 (g) + 1 O2 (g) 2 H2O (g)

2 molecules 1 molecule 2 molecules

2 moles 1 mole 2 moles

2 Liters 1 Liter 2 Liters

Avogadrorsquos Law also allows chemists to relate volumes and moles of a gas in a chemical reaction

This relationship is only valid for

gaseous substances

40

Example 1

A sample of helium gas with a mass of 180 g occupies 16 Liters at a particular temperature and pressure What mass of oxygen would occupy16 L at the same temperature and pressure

1 mol= 180 g x

400 gmol of He = 450 mol

mol of O2 = mol of He = 450 mol

Same Temp amp Pressure

mass of O2

320 g= 450 mol x

1 mol= 144 g

41

Example 2

How many Liters of NH3 can be produced from reaction of 18 L of H2 with excess N2 as shown below

18 L H23

2

L NHx

L H= 12 L NH3

N2 (g) + 3 H2 (g) 2 NH3 (g)

32

42

STP amp MOLAR VOLUME

To better understand the factors that affect gas behavior a set of standard conditions have been chosen for use and are referred to as Standard Temperature and Pressure (STP)

STP =

1 atm (760 mmHg)

0ordmC (273 K)

43

STP amp MOLAR VOLUME

At STP conditions one mole of any gas is observed to occupy a volume of 224 L

V = 224 L

Molar Volume at

STP

44

Example 1

If 200 L of a gas at STP has a mass of 323 g what is the molar mass of the gas

mol200 L x

Lmol of gas =

2241

= 00893 mol

g

molMolar mass =

323 g=

00893 mol= 362 gmol Molar

volume at STP

45

Example 2

A sample of gas has a volume of 250 L at 730 mmHg and 20C What is the volume of this gas at STP

1 22 1

2 1

P TV = V x x

P T

V2 = 224 L

P1= 730 mmHg V1= 250 L T1= 293 K P2= 760 mmHg V2= T2 = 273 K

2

730 mmHg 273 KV = 250 L x x

760 mmHg 293 K

46

IDEAL GAS LAW

Combining all the laws that describe the behavior of gases one can obtain a useful relationship that relates the volume of a gas to the temperature pressure and number of moles

n R TV =

P

Universal gas constant

R = 00821 L atmmol K

47

IDEAL GAS LAW

This relationship is called the Ideal Gas Law and commonly written as

P V = n R T

Pressure in atm Volume

in Liters

Number of moles

Temp in K

48

Example 1

A sample of H2 gas has a volume of 856 L at a temp of 0C and pressure of 15 atm Calculate the moles of gas present

PVn =

RT

= 057 mol

P = 15 atm V = 856 L T = 273 K n = R = 00821

PV = nRT

(15 atm)(856 L)=

L atm(00821 )(273 K)

mol K

49

Example 2

What volume does 400 g of N2 gas occupy at 10C and 750 mmHg

P =750 mmHg V = T = 283 K n = not given m = 400 g R = 00821

mol of N2 =1 mol

400 g x280 g

= 143 mol

P = 750 mmHg1 atm

x760 mmHg

Moles must be calculated

from mass

Pressure must be

converted to atm

P = 0987 atm

50

Example 2

What volume does 400 g of N2 gas occupy at 10C and 750 mmHg

nRTV =

P

V = 337 L

P =0987 atm V = T = 283 K n = 143 R = 00821

L atm(143 mol)(00821 )(283 K)

mol KV = 0987 atm

Tros Introductory Chemistry Chapter

51

Ideal vs Real Gases

bull Real gases often do not behave like ideal gases at high pressure or low temperature

bull Ideal gas laws assume1) no attractions between gas molecules2) gas molecules do not take up space based on the Kinetic-Molecular Theory

bull at low temperatures and high pressures these assumptions are not valid

Tros Introductory Chemistry Chapter

52

Ideal vs Real Gases

53

PARTIAL PRESSURES

Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases

Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture

In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas

54

DALTONrsquoSLAW

In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture

This is called Daltonrsquos law of partial pressures

Total pressure of a gas mixture

Ptotal = P1 + P2 + P3 +

Sum of the partial pressures of the gases in the mixture

=

55

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = PHe + PAr

Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture

= 60 atm

56

PARTIAL PRESSURES

The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture

For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container

57

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = 60 atm

The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture

Twice as many moles of Ar

compared to He

PAr = 2 PHe

58

Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank

Ptotal = Poxygen + Phelium

Poxygen = 1140 mmHg x

1 atm

760 mmHg= 150 atm

Phelium

= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm

59

Example 2

A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture

Ptotal = Poxygen + Pnitrogen

Pnitrogen = 2 Poxygen

Twice as many moles of N2

compared to O2

Pnitrogen = 23 (Ptotal ) = 20 atm

Poxygen = 13 (Ptotal ) = 10 atm

60

THE END

• Slide 1
• CHAPTER OUTLINE
• PROPERTIES OF GASES
• PROPERTIES OF GASES (2)
• KINETIC-MOLECULAR THEORY
• KINETIC-MOLECULAR THEORY (2)
• Kinetic Molecular Theory
• PRESSURE amp ITS MEASUREMENT
• Common Units of Pressure
• PRESSURE amp ITS MEASUREMENT (2)
• Atmospheric Pressure amp Altitude
• Pressure Imbalance in Ear
• Example 1
• Example 2
• PRESSURE amp MOLES OF A GAS
• BOYLErsquoS LAW
• BOYLErsquoS LAW (2)
• Example 1 (2)
• Example 2 (2)
• CHARLESrsquoS LAW
• CHARLESrsquoS LAW (2)
• Example 1 (3)
• Example 2 (3)
• GAY-LUSSACrsquoS LAW
• GAY-LUSSACrsquoS LAW (2)
• Example 1 (4)
• Example 1 (5)
• Example 2 (4)
• VAPOR PRESSURE amp BOILING POINT
• VAPOR PRESSURE amp BOILING POINT (2)
• VAPOR PRESSURE amp BOILING POINT (3)
• VAPOR PRESSURE amp BOILING POINT (4)
• COMBINED GAS LAW
• COMBINED GAS LAW (2)
• Example 1 (6)
• AVOGADROrsquoS LAW
• Avogadrorsquos Law
• AVOGADROrsquoS LAW (2)
• AVOGADROrsquoS LAW (3)
• Example 1 (7)
• Example 2 (5)
• STP amp MOLAR VOLUME
• STP amp MOLAR VOLUME (2)
• Example 1 (8)
• Example 2 (6)
• IDEAL GAS LAW
• IDEAL GAS LAW (2)
• Example 1 (9)
• Example 2 (7)
• Example 2 (8)
• Ideal vs Real Gases
• Ideal vs Real Gases (2)
• PARTIAL PRESSURES
• DALTONrsquoS LAW
• PARTIAL PRESSURES (2)
• PARTIAL PRESSURES (3)
• PARTIAL PRESSURES (4)
• Example 1 (10)
• Example 2 (9)
• Slide 60

16

BOYLErsquoS LAW

The relationship of pressure and volume in gases is called Boylersquos Law

At constant temperature the volume of a fixed amount of gas is inversely proportional to its pressure

As pressure increases the volume of the gas decreases

Pressure and volume of a

gas are inversely

proportional

17

BOYLErsquoS LAW

Boylersquos Law can be mathematically expressed as

P1 x V1 = P2 x V2

P and V under initial

condition

P and V under final

condition

18

Example 1

A sample of gas has a volume of 12-L and a pressure of 4500 mmHg What is the volume of the gas when the pressure is reduced to 750 mmHg

1 12

2

P VV =

P

V2 = 72 L

P1= 4500 mmHg

P2= 750 mmHg

V1 = 12 L

V2 =

P1 x V1 = P2 x V2

(4500 mmHg)(12 L)=

(750 mmHg)

Pressure decreases Volume

increases

19

Example 2

A sample of hydrogen gas occupies 40 L at 650 mmHg What volume would it occupy at 20 atm

1 12

2

P VV =

P

V2 = 17 L

P1= 650 mmHg

P2= 20 atm

V1 = 40 L

V2 =

P1 x V1 = P2 x V2

(650 mmHg)(40 L)=

(1520 mmHg)Pressures must be in the same

unit

P2 = 20 atm = 1520 mmHgVolume decreases

Pressure increases

20

CHARLESrsquoS LAW

The relationship of temperature and volume in gases is called Charlesrsquo Law

At constant pressure the volume of a fixed amount of gas is directly proportional to its absolute temperature

As temperature increases the volume of the gas increases

Temperature and volume of a gas

are directly proportional

21

CHARLESrsquoS LAW

Charlesrsquo Law can be mathematically expressed as

V and T under initial

condition

1 2

1 2

V V=

T T

V and T under final condition

T must be in units of K

22

Example 1

A 20-L sample of a gas is cooled from 298K to 278 K at constant pressure What is the new volume of the gas

22 1

1

TV = V x

T

V2 = 19 L

V1 = 20 L

V2 =

T1 = 298 K

T2 = 278 K

278 K= 20 L x

298 K

Volume decreases1 2

1 2

V V=

T T

Temperature decreases

23

Example 2

If 200 L of oxygen is cooled from 100ordmC to 0ordmC what is the new volume

22 1

1

TV = V x

T

V2 = 146 L

V1 = 200 L

V2 =

T1 = 100ordmC

T2 = 0ordmC

273 K= 200 L x

373 K

T1= 100ordmC + 273 = 373 K

T2= 0ordmC + 273 = 273 K

Temperatures must be in

KVolume

decreasesTemperature

decreases

24

GAY-LUSSACrsquoSLAW

The relationship of temperature and pressure in gases is called Gay-Lussacrsquos Law

At constant volume the pressure of a fixed amount of gas is directly proportional to its absolute temperature

As temperature increases the

pressure of the gas increases

Temperature and pressure of a gas

are directly proportional

25

GAY-LUSSACrsquoSLAW

Gay-Lussacrsquos Law can be mathematically expressed as

P and T under initial

condition

1 2

1 2

P P=

T T

P and T under final conditionT must be in

units of K

26

Example 1

An aerosol spray can has a pressure of 40 atm at 25C What pressure will the can have if it is placed in a fire and reaches temperature of 400C

P1 = 40 atm

P2 =

T1 = 25ordmC

T2 = 400ordmC

T1= 25ordmC + 273 = 298 K

T2= 400ordmC + 273 = 673 K

Temperatures must be in K

27

Example 1

An aerosol spray can has a pressure of 40 atm at 25C What pressure will the can have if it is placed in a fire and reaches temperature of 400C

22 1

1

TP = P x

T

P2 = 90 atm

P1 = 40 atm

P2 =

T1 = 298 K

T2 = 673 K

673 K= 40 L x

298 K

Pressure increases

1 2

1 2

P P=

T T

Temperature increases

28

Example 2

A cylinder of gas with a volume of 150-L and a pressure of 965 mmHg is stored at a temperature of 55C To what temperature must the cylinder be cooled to reach a pressure of 850 mmHg

22 1

1

PT = T x

P

T2 = 289 K

P1 = 965 mmHg

P2 = 850 mmHg

T1 = 55ordmC

T2 =

850 mmHg= 328 K x

965 mmHg

T1= 55ordmC + 273 = 328 K

T2 = 16ordmC

Temperature decreases

Pressure decreases

29

VAPOR PRESSUREamp BOILING POINT

In an open container liquid molecules at the surface that possess sufficient energy can break away from the surface and become gas particles or vapor

In a closed container these gas particles can accumulate and create pressure called vapor pressure

Vapor pressure is defined as the pressure above a liquid at a given temperature Vapor pressure varies with each liquid and increases with temperature

30

VAPOR PRESSUREamp BOILING POINT

Listed below is the vapor pressure of water at various temperatures

31

VAPOR PRESSUREamp BOILING POINT

A liquid reaches its boiling point when its vapor pressure becomes equal to the external pressure (atmospheric pressure)

For example at sea level water reaches its boiling point at 100C since its vapor pressure is 760 mmHg at this temperature

32

VAPOR PRESSUREamp BOILING POINT

At higher altitudes where atmospheric pressure is lower water reaches boiling point at temperatures lower than 100C

For example in Denver where atmospheric pressure is 630 mmHg water boils at 95C since its vapor pressure is 630 mmHg at this temperature

33

COMBINEDGAS LAW

All pressure-volume-temperature relationships can be combined into a single relationship called the combined gas law

1 1 2 2

1 2

P V P V=

T T

Initial condition

This law is useful for studying the effect of changes in two variables

Final condition

34

COMBINEDGAS LAW

The individual gas laws studied previously are embodied in the combined gas law

1 1 2 2

1 2

P V P V=

T T

Boylersquos LawCharlesrsquos

LawGay-Lussacrsquos

Law

35

Example 1

A 250-mL sample of gas has a pressure of 400 atm at a temperature of 10C What is the volume of the gas at a pressure of 100 atm and a temperature of 18C

1 22 1

2 1

P TV = V x x

P T

V2 = 103 mL

P1= 400 atm V1= 250 mL T1= 283 K P2= 100 atm V2= T2 = 291 K

2

400 atm 291 KV = 250 mL x x

100 atm 283 K

1 1 2 2

1 2

P V P V=

T TBoth pressure

and temp change

Must use combined

gas law

36

AVOGADROrsquoSLAW

The relationship of moles and volume in gases is called Avogadrorsquos Law

At constant temperature and pressure the volume of a fixed amount of gas is directly proportional to the number of moles

As number of moles increases

the volume of the gas

increases

Number of moles and volume of a gas are directly

proportional

Tros Introductory Chemistry Chapter

37

Avogadrorsquos Law

38

AVOGADROrsquoSLAW

As a result of Avogadrorsquos Law equal volumes of different gases at the same temp and pressure contain equal number of moles (molecules)

2 Tanks of gas of equal volume at the same T amp P contain the same number of molecules

39

AVOGADROrsquoSLAW

For example

2 H2 (g) + 1 O2 (g) 2 H2O (g)

2 molecules 1 molecule 2 molecules

2 moles 1 mole 2 moles

2 Liters 1 Liter 2 Liters

Avogadrorsquos Law also allows chemists to relate volumes and moles of a gas in a chemical reaction

This relationship is only valid for

gaseous substances

40

Example 1

A sample of helium gas with a mass of 180 g occupies 16 Liters at a particular temperature and pressure What mass of oxygen would occupy16 L at the same temperature and pressure

1 mol= 180 g x

400 gmol of He = 450 mol

mol of O2 = mol of He = 450 mol

Same Temp amp Pressure

mass of O2

320 g= 450 mol x

1 mol= 144 g

41

Example 2

How many Liters of NH3 can be produced from reaction of 18 L of H2 with excess N2 as shown below

18 L H23

2

L NHx

L H= 12 L NH3

N2 (g) + 3 H2 (g) 2 NH3 (g)

32

42

STP amp MOLAR VOLUME

To better understand the factors that affect gas behavior a set of standard conditions have been chosen for use and are referred to as Standard Temperature and Pressure (STP)

STP =

1 atm (760 mmHg)

0ordmC (273 K)

43

STP amp MOLAR VOLUME

At STP conditions one mole of any gas is observed to occupy a volume of 224 L

V = 224 L

Molar Volume at

STP

44

Example 1

If 200 L of a gas at STP has a mass of 323 g what is the molar mass of the gas

mol200 L x

Lmol of gas =

2241

= 00893 mol

g

molMolar mass =

323 g=

00893 mol= 362 gmol Molar

volume at STP

45

Example 2

A sample of gas has a volume of 250 L at 730 mmHg and 20C What is the volume of this gas at STP

1 22 1

2 1

P TV = V x x

P T

V2 = 224 L

P1= 730 mmHg V1= 250 L T1= 293 K P2= 760 mmHg V2= T2 = 273 K

2

730 mmHg 273 KV = 250 L x x

760 mmHg 293 K

46

IDEAL GAS LAW

Combining all the laws that describe the behavior of gases one can obtain a useful relationship that relates the volume of a gas to the temperature pressure and number of moles

n R TV =

P

Universal gas constant

R = 00821 L atmmol K

47

IDEAL GAS LAW

This relationship is called the Ideal Gas Law and commonly written as

P V = n R T

Pressure in atm Volume

in Liters

Number of moles

Temp in K

48

Example 1

A sample of H2 gas has a volume of 856 L at a temp of 0C and pressure of 15 atm Calculate the moles of gas present

PVn =

RT

= 057 mol

P = 15 atm V = 856 L T = 273 K n = R = 00821

PV = nRT

(15 atm)(856 L)=

L atm(00821 )(273 K)

mol K

49

Example 2

What volume does 400 g of N2 gas occupy at 10C and 750 mmHg

P =750 mmHg V = T = 283 K n = not given m = 400 g R = 00821

mol of N2 =1 mol

400 g x280 g

= 143 mol

P = 750 mmHg1 atm

x760 mmHg

Moles must be calculated

from mass

Pressure must be

converted to atm

P = 0987 atm

50

Example 2

What volume does 400 g of N2 gas occupy at 10C and 750 mmHg

nRTV =

P

V = 337 L

P =0987 atm V = T = 283 K n = 143 R = 00821

L atm(143 mol)(00821 )(283 K)

mol KV = 0987 atm

Tros Introductory Chemistry Chapter

51

Ideal vs Real Gases

bull Real gases often do not behave like ideal gases at high pressure or low temperature

bull Ideal gas laws assume1) no attractions between gas molecules2) gas molecules do not take up space based on the Kinetic-Molecular Theory

bull at low temperatures and high pressures these assumptions are not valid

Tros Introductory Chemistry Chapter

52

Ideal vs Real Gases

53

PARTIAL PRESSURES

Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases

Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture

In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas

54

DALTONrsquoSLAW

In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture

This is called Daltonrsquos law of partial pressures

Total pressure of a gas mixture

Ptotal = P1 + P2 + P3 +

Sum of the partial pressures of the gases in the mixture

=

55

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = PHe + PAr

Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture

= 60 atm

56

PARTIAL PRESSURES

The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture

For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container

57

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = 60 atm

The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture

Twice as many moles of Ar

compared to He

PAr = 2 PHe

58

Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank

Ptotal = Poxygen + Phelium

Poxygen = 1140 mmHg x

1 atm

760 mmHg= 150 atm

Phelium

= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm

59

Example 2

A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture

Ptotal = Poxygen + Pnitrogen

Pnitrogen = 2 Poxygen

Twice as many moles of N2

compared to O2

Pnitrogen = 23 (Ptotal ) = 20 atm

Poxygen = 13 (Ptotal ) = 10 atm

60

THE END

• Slide 1
• CHAPTER OUTLINE
• PROPERTIES OF GASES
• PROPERTIES OF GASES (2)
• KINETIC-MOLECULAR THEORY
• KINETIC-MOLECULAR THEORY (2)
• Kinetic Molecular Theory
• PRESSURE amp ITS MEASUREMENT
• Common Units of Pressure
• PRESSURE amp ITS MEASUREMENT (2)
• Atmospheric Pressure amp Altitude
• Pressure Imbalance in Ear
• Example 1
• Example 2
• PRESSURE amp MOLES OF A GAS
• BOYLErsquoS LAW
• BOYLErsquoS LAW (2)
• Example 1 (2)
• Example 2 (2)
• CHARLESrsquoS LAW
• CHARLESrsquoS LAW (2)
• Example 1 (3)
• Example 2 (3)
• GAY-LUSSACrsquoS LAW
• GAY-LUSSACrsquoS LAW (2)
• Example 1 (4)
• Example 1 (5)
• Example 2 (4)
• VAPOR PRESSURE amp BOILING POINT
• VAPOR PRESSURE amp BOILING POINT (2)
• VAPOR PRESSURE amp BOILING POINT (3)
• VAPOR PRESSURE amp BOILING POINT (4)
• COMBINED GAS LAW
• COMBINED GAS LAW (2)
• Example 1 (6)
• AVOGADROrsquoS LAW
• Avogadrorsquos Law
• AVOGADROrsquoS LAW (2)
• AVOGADROrsquoS LAW (3)
• Example 1 (7)
• Example 2 (5)
• STP amp MOLAR VOLUME
• STP amp MOLAR VOLUME (2)
• Example 1 (8)
• Example 2 (6)
• IDEAL GAS LAW
• IDEAL GAS LAW (2)
• Example 1 (9)
• Example 2 (7)
• Example 2 (8)
• Ideal vs Real Gases
• Ideal vs Real Gases (2)
• PARTIAL PRESSURES
• DALTONrsquoS LAW
• PARTIAL PRESSURES (2)
• PARTIAL PRESSURES (3)
• PARTIAL PRESSURES (4)
• Example 1 (10)
• Example 2 (9)
• Slide 60

17

BOYLErsquoS LAW

Boylersquos Law can be mathematically expressed as

P1 x V1 = P2 x V2

P and V under initial

condition

P and V under final

condition

18

Example 1

A sample of gas has a volume of 12-L and a pressure of 4500 mmHg What is the volume of the gas when the pressure is reduced to 750 mmHg

1 12

2

P VV =

P

V2 = 72 L

P1= 4500 mmHg

P2= 750 mmHg

V1 = 12 L

V2 =

P1 x V1 = P2 x V2

(4500 mmHg)(12 L)=

(750 mmHg)

Pressure decreases Volume

increases

19

Example 2

A sample of hydrogen gas occupies 40 L at 650 mmHg What volume would it occupy at 20 atm

1 12

2

P VV =

P

V2 = 17 L

P1= 650 mmHg

P2= 20 atm

V1 = 40 L

V2 =

P1 x V1 = P2 x V2

(650 mmHg)(40 L)=

(1520 mmHg)Pressures must be in the same

unit

P2 = 20 atm = 1520 mmHgVolume decreases

Pressure increases

20

CHARLESrsquoS LAW

The relationship of temperature and volume in gases is called Charlesrsquo Law

At constant pressure the volume of a fixed amount of gas is directly proportional to its absolute temperature

As temperature increases the volume of the gas increases

Temperature and volume of a gas

are directly proportional

21

CHARLESrsquoS LAW

Charlesrsquo Law can be mathematically expressed as

V and T under initial

condition

1 2

1 2

V V=

T T

V and T under final condition

T must be in units of K

22

Example 1

A 20-L sample of a gas is cooled from 298K to 278 K at constant pressure What is the new volume of the gas

22 1

1

TV = V x

T

V2 = 19 L

V1 = 20 L

V2 =

T1 = 298 K

T2 = 278 K

278 K= 20 L x

298 K

Volume decreases1 2

1 2

V V=

T T

Temperature decreases

23

Example 2

If 200 L of oxygen is cooled from 100ordmC to 0ordmC what is the new volume

22 1

1

TV = V x

T

V2 = 146 L

V1 = 200 L

V2 =

T1 = 100ordmC

T2 = 0ordmC

273 K= 200 L x

373 K

T1= 100ordmC + 273 = 373 K

T2= 0ordmC + 273 = 273 K

Temperatures must be in

KVolume

decreasesTemperature

decreases

24

GAY-LUSSACrsquoSLAW

The relationship of temperature and pressure in gases is called Gay-Lussacrsquos Law

At constant volume the pressure of a fixed amount of gas is directly proportional to its absolute temperature

As temperature increases the

pressure of the gas increases

Temperature and pressure of a gas

are directly proportional

25

GAY-LUSSACrsquoSLAW

Gay-Lussacrsquos Law can be mathematically expressed as

P and T under initial

condition

1 2

1 2

P P=

T T

P and T under final conditionT must be in

units of K

26

Example 1

An aerosol spray can has a pressure of 40 atm at 25C What pressure will the can have if it is placed in a fire and reaches temperature of 400C

P1 = 40 atm

P2 =

T1 = 25ordmC

T2 = 400ordmC

T1= 25ordmC + 273 = 298 K

T2= 400ordmC + 273 = 673 K

Temperatures must be in K

27

Example 1

An aerosol spray can has a pressure of 40 atm at 25C What pressure will the can have if it is placed in a fire and reaches temperature of 400C

22 1

1

TP = P x

T

P2 = 90 atm

P1 = 40 atm

P2 =

T1 = 298 K

T2 = 673 K

673 K= 40 L x

298 K

Pressure increases

1 2

1 2

P P=

T T

Temperature increases

28

Example 2

A cylinder of gas with a volume of 150-L and a pressure of 965 mmHg is stored at a temperature of 55C To what temperature must the cylinder be cooled to reach a pressure of 850 mmHg

22 1

1

PT = T x

P

T2 = 289 K

P1 = 965 mmHg

P2 = 850 mmHg

T1 = 55ordmC

T2 =

850 mmHg= 328 K x

965 mmHg

T1= 55ordmC + 273 = 328 K

T2 = 16ordmC

Temperature decreases

Pressure decreases

29

VAPOR PRESSUREamp BOILING POINT

In an open container liquid molecules at the surface that possess sufficient energy can break away from the surface and become gas particles or vapor

In a closed container these gas particles can accumulate and create pressure called vapor pressure

Vapor pressure is defined as the pressure above a liquid at a given temperature Vapor pressure varies with each liquid and increases with temperature

30

VAPOR PRESSUREamp BOILING POINT

Listed below is the vapor pressure of water at various temperatures

31

VAPOR PRESSUREamp BOILING POINT

A liquid reaches its boiling point when its vapor pressure becomes equal to the external pressure (atmospheric pressure)

For example at sea level water reaches its boiling point at 100C since its vapor pressure is 760 mmHg at this temperature

32

VAPOR PRESSUREamp BOILING POINT

At higher altitudes where atmospheric pressure is lower water reaches boiling point at temperatures lower than 100C

For example in Denver where atmospheric pressure is 630 mmHg water boils at 95C since its vapor pressure is 630 mmHg at this temperature

33

COMBINEDGAS LAW

All pressure-volume-temperature relationships can be combined into a single relationship called the combined gas law

1 1 2 2

1 2

P V P V=

T T

Initial condition

This law is useful for studying the effect of changes in two variables

Final condition

34

COMBINEDGAS LAW

The individual gas laws studied previously are embodied in the combined gas law

1 1 2 2

1 2

P V P V=

T T

Boylersquos LawCharlesrsquos

LawGay-Lussacrsquos

Law

35

Example 1

A 250-mL sample of gas has a pressure of 400 atm at a temperature of 10C What is the volume of the gas at a pressure of 100 atm and a temperature of 18C

1 22 1

2 1

P TV = V x x

P T

V2 = 103 mL

P1= 400 atm V1= 250 mL T1= 283 K P2= 100 atm V2= T2 = 291 K

2

400 atm 291 KV = 250 mL x x

100 atm 283 K

1 1 2 2

1 2

P V P V=

T TBoth pressure

and temp change

Must use combined

gas law

36

AVOGADROrsquoSLAW

The relationship of moles and volume in gases is called Avogadrorsquos Law

At constant temperature and pressure the volume of a fixed amount of gas is directly proportional to the number of moles

As number of moles increases

the volume of the gas

increases

Number of moles and volume of a gas are directly

proportional

Tros Introductory Chemistry Chapter

37

Avogadrorsquos Law

38

AVOGADROrsquoSLAW

As a result of Avogadrorsquos Law equal volumes of different gases at the same temp and pressure contain equal number of moles (molecules)

2 Tanks of gas of equal volume at the same T amp P contain the same number of molecules

39

AVOGADROrsquoSLAW

For example

2 H2 (g) + 1 O2 (g) 2 H2O (g)

2 molecules 1 molecule 2 molecules

2 moles 1 mole 2 moles

2 Liters 1 Liter 2 Liters

Avogadrorsquos Law also allows chemists to relate volumes and moles of a gas in a chemical reaction

This relationship is only valid for

gaseous substances

40

Example 1

A sample of helium gas with a mass of 180 g occupies 16 Liters at a particular temperature and pressure What mass of oxygen would occupy16 L at the same temperature and pressure

1 mol= 180 g x

400 gmol of He = 450 mol

mol of O2 = mol of He = 450 mol

Same Temp amp Pressure

mass of O2

320 g= 450 mol x

1 mol= 144 g

41

Example 2

How many Liters of NH3 can be produced from reaction of 18 L of H2 with excess N2 as shown below

18 L H23

2

L NHx

L H= 12 L NH3

N2 (g) + 3 H2 (g) 2 NH3 (g)

32

42

STP amp MOLAR VOLUME

To better understand the factors that affect gas behavior a set of standard conditions have been chosen for use and are referred to as Standard Temperature and Pressure (STP)

STP =

1 atm (760 mmHg)

0ordmC (273 K)

43

STP amp MOLAR VOLUME

At STP conditions one mole of any gas is observed to occupy a volume of 224 L

V = 224 L

Molar Volume at

STP

44

Example 1

If 200 L of a gas at STP has a mass of 323 g what is the molar mass of the gas

mol200 L x

Lmol of gas =

2241

= 00893 mol

g

molMolar mass =

323 g=

00893 mol= 362 gmol Molar

volume at STP

45

Example 2

A sample of gas has a volume of 250 L at 730 mmHg and 20C What is the volume of this gas at STP

1 22 1

2 1

P TV = V x x

P T

V2 = 224 L

P1= 730 mmHg V1= 250 L T1= 293 K P2= 760 mmHg V2= T2 = 273 K

2

730 mmHg 273 KV = 250 L x x

760 mmHg 293 K

46

IDEAL GAS LAW

Combining all the laws that describe the behavior of gases one can obtain a useful relationship that relates the volume of a gas to the temperature pressure and number of moles

n R TV =

P

Universal gas constant

R = 00821 L atmmol K

47

IDEAL GAS LAW

This relationship is called the Ideal Gas Law and commonly written as

P V = n R T

Pressure in atm Volume

in Liters

Number of moles

Temp in K

48

Example 1

A sample of H2 gas has a volume of 856 L at a temp of 0C and pressure of 15 atm Calculate the moles of gas present

PVn =

RT

= 057 mol

P = 15 atm V = 856 L T = 273 K n = R = 00821

PV = nRT

(15 atm)(856 L)=

L atm(00821 )(273 K)

mol K

49

Example 2

What volume does 400 g of N2 gas occupy at 10C and 750 mmHg

P =750 mmHg V = T = 283 K n = not given m = 400 g R = 00821

mol of N2 =1 mol

400 g x280 g

= 143 mol

P = 750 mmHg1 atm

x760 mmHg

Moles must be calculated

from mass

Pressure must be

converted to atm

P = 0987 atm

50

Example 2

What volume does 400 g of N2 gas occupy at 10C and 750 mmHg

nRTV =

P

V = 337 L

P =0987 atm V = T = 283 K n = 143 R = 00821

L atm(143 mol)(00821 )(283 K)

mol KV = 0987 atm

Tros Introductory Chemistry Chapter

51

Ideal vs Real Gases

bull Real gases often do not behave like ideal gases at high pressure or low temperature

bull Ideal gas laws assume1) no attractions between gas molecules2) gas molecules do not take up space based on the Kinetic-Molecular Theory

bull at low temperatures and high pressures these assumptions are not valid

Tros Introductory Chemistry Chapter

52

Ideal vs Real Gases

53

PARTIAL PRESSURES

Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases

Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture

In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas

54

DALTONrsquoSLAW

In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture

This is called Daltonrsquos law of partial pressures

Total pressure of a gas mixture

Ptotal = P1 + P2 + P3 +

Sum of the partial pressures of the gases in the mixture

=

55

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = PHe + PAr

Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture

= 60 atm

56

PARTIAL PRESSURES

The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture

For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container

57

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = 60 atm

The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture

Twice as many moles of Ar

compared to He

PAr = 2 PHe

58

Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank

Ptotal = Poxygen + Phelium

Poxygen = 1140 mmHg x

1 atm

760 mmHg= 150 atm

Phelium

= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm

59

Example 2

A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture

Ptotal = Poxygen + Pnitrogen

Pnitrogen = 2 Poxygen

Twice as many moles of N2

compared to O2

Pnitrogen = 23 (Ptotal ) = 20 atm

Poxygen = 13 (Ptotal ) = 10 atm

60

THE END

• Slide 1
• CHAPTER OUTLINE
• PROPERTIES OF GASES
• PROPERTIES OF GASES (2)
• KINETIC-MOLECULAR THEORY
• KINETIC-MOLECULAR THEORY (2)
• Kinetic Molecular Theory
• PRESSURE amp ITS MEASUREMENT
• Common Units of Pressure
• PRESSURE amp ITS MEASUREMENT (2)
• Atmospheric Pressure amp Altitude
• Pressure Imbalance in Ear
• Example 1
• Example 2
• PRESSURE amp MOLES OF A GAS
• BOYLErsquoS LAW
• BOYLErsquoS LAW (2)
• Example 1 (2)
• Example 2 (2)
• CHARLESrsquoS LAW
• CHARLESrsquoS LAW (2)
• Example 1 (3)
• Example 2 (3)
• GAY-LUSSACrsquoS LAW
• GAY-LUSSACrsquoS LAW (2)
• Example 1 (4)
• Example 1 (5)
• Example 2 (4)
• VAPOR PRESSURE amp BOILING POINT
• VAPOR PRESSURE amp BOILING POINT (2)
• VAPOR PRESSURE amp BOILING POINT (3)
• VAPOR PRESSURE amp BOILING POINT (4)
• COMBINED GAS LAW
• COMBINED GAS LAW (2)
• Example 1 (6)
• AVOGADROrsquoS LAW
• Avogadrorsquos Law
• AVOGADROrsquoS LAW (2)
• AVOGADROrsquoS LAW (3)
• Example 1 (7)
• Example 2 (5)
• STP amp MOLAR VOLUME
• STP amp MOLAR VOLUME (2)
• Example 1 (8)
• Example 2 (6)
• IDEAL GAS LAW
• IDEAL GAS LAW (2)
• Example 1 (9)
• Example 2 (7)
• Example 2 (8)
• Ideal vs Real Gases
• Ideal vs Real Gases (2)
• PARTIAL PRESSURES
• DALTONrsquoS LAW
• PARTIAL PRESSURES (2)
• PARTIAL PRESSURES (3)
• PARTIAL PRESSURES (4)
• Example 1 (10)
• Example 2 (9)
• Slide 60

18

Example 1

A sample of gas has a volume of 12-L and a pressure of 4500 mmHg What is the volume of the gas when the pressure is reduced to 750 mmHg

1 12

2

P VV =

P

V2 = 72 L

P1= 4500 mmHg

P2= 750 mmHg

V1 = 12 L

V2 =

P1 x V1 = P2 x V2

(4500 mmHg)(12 L)=

(750 mmHg)

Pressure decreases Volume

increases

19

Example 2

A sample of hydrogen gas occupies 40 L at 650 mmHg What volume would it occupy at 20 atm

1 12

2

P VV =

P

V2 = 17 L

P1= 650 mmHg

P2= 20 atm

V1 = 40 L

V2 =

P1 x V1 = P2 x V2

(650 mmHg)(40 L)=

(1520 mmHg)Pressures must be in the same

unit

P2 = 20 atm = 1520 mmHgVolume decreases

Pressure increases

20

CHARLESrsquoS LAW

The relationship of temperature and volume in gases is called Charlesrsquo Law

At constant pressure the volume of a fixed amount of gas is directly proportional to its absolute temperature

As temperature increases the volume of the gas increases

Temperature and volume of a gas

are directly proportional

21

CHARLESrsquoS LAW

Charlesrsquo Law can be mathematically expressed as

V and T under initial

condition

1 2

1 2

V V=

T T

V and T under final condition

T must be in units of K

22

Example 1

A 20-L sample of a gas is cooled from 298K to 278 K at constant pressure What is the new volume of the gas

22 1

1

TV = V x

T

V2 = 19 L

V1 = 20 L

V2 =

T1 = 298 K

T2 = 278 K

278 K= 20 L x

298 K

Volume decreases1 2

1 2

V V=

T T

Temperature decreases

23

Example 2

If 200 L of oxygen is cooled from 100ordmC to 0ordmC what is the new volume

22 1

1

TV = V x

T

V2 = 146 L

V1 = 200 L

V2 =

T1 = 100ordmC

T2 = 0ordmC

273 K= 200 L x

373 K

T1= 100ordmC + 273 = 373 K

T2= 0ordmC + 273 = 273 K

Temperatures must be in

KVolume

decreasesTemperature

decreases

24

GAY-LUSSACrsquoSLAW

The relationship of temperature and pressure in gases is called Gay-Lussacrsquos Law

At constant volume the pressure of a fixed amount of gas is directly proportional to its absolute temperature

As temperature increases the

pressure of the gas increases

Temperature and pressure of a gas

are directly proportional

25

GAY-LUSSACrsquoSLAW

Gay-Lussacrsquos Law can be mathematically expressed as

P and T under initial

condition

1 2

1 2

P P=

T T

P and T under final conditionT must be in

units of K

26

Example 1

An aerosol spray can has a pressure of 40 atm at 25C What pressure will the can have if it is placed in a fire and reaches temperature of 400C

P1 = 40 atm

P2 =

T1 = 25ordmC

T2 = 400ordmC

T1= 25ordmC + 273 = 298 K

T2= 400ordmC + 273 = 673 K

Temperatures must be in K

27

Example 1

An aerosol spray can has a pressure of 40 atm at 25C What pressure will the can have if it is placed in a fire and reaches temperature of 400C

22 1

1

TP = P x

T

P2 = 90 atm

P1 = 40 atm

P2 =

T1 = 298 K

T2 = 673 K

673 K= 40 L x

298 K

Pressure increases

1 2

1 2

P P=

T T

Temperature increases

28

Example 2

A cylinder of gas with a volume of 150-L and a pressure of 965 mmHg is stored at a temperature of 55C To what temperature must the cylinder be cooled to reach a pressure of 850 mmHg

22 1

1

PT = T x

P

T2 = 289 K

P1 = 965 mmHg

P2 = 850 mmHg

T1 = 55ordmC

T2 =

850 mmHg= 328 K x

965 mmHg

T1= 55ordmC + 273 = 328 K

T2 = 16ordmC

Temperature decreases

Pressure decreases

29

VAPOR PRESSUREamp BOILING POINT

In an open container liquid molecules at the surface that possess sufficient energy can break away from the surface and become gas particles or vapor

In a closed container these gas particles can accumulate and create pressure called vapor pressure

Vapor pressure is defined as the pressure above a liquid at a given temperature Vapor pressure varies with each liquid and increases with temperature

30

VAPOR PRESSUREamp BOILING POINT

Listed below is the vapor pressure of water at various temperatures

31

VAPOR PRESSUREamp BOILING POINT

A liquid reaches its boiling point when its vapor pressure becomes equal to the external pressure (atmospheric pressure)

For example at sea level water reaches its boiling point at 100C since its vapor pressure is 760 mmHg at this temperature

32

VAPOR PRESSUREamp BOILING POINT

At higher altitudes where atmospheric pressure is lower water reaches boiling point at temperatures lower than 100C

For example in Denver where atmospheric pressure is 630 mmHg water boils at 95C since its vapor pressure is 630 mmHg at this temperature

33

COMBINEDGAS LAW

All pressure-volume-temperature relationships can be combined into a single relationship called the combined gas law

1 1 2 2

1 2

P V P V=

T T

Initial condition

This law is useful for studying the effect of changes in two variables

Final condition

34

COMBINEDGAS LAW

The individual gas laws studied previously are embodied in the combined gas law

1 1 2 2

1 2

P V P V=

T T

Boylersquos LawCharlesrsquos

LawGay-Lussacrsquos

Law

35

Example 1

A 250-mL sample of gas has a pressure of 400 atm at a temperature of 10C What is the volume of the gas at a pressure of 100 atm and a temperature of 18C

1 22 1

2 1

P TV = V x x

P T

V2 = 103 mL

P1= 400 atm V1= 250 mL T1= 283 K P2= 100 atm V2= T2 = 291 K

2

400 atm 291 KV = 250 mL x x

100 atm 283 K

1 1 2 2

1 2

P V P V=

T TBoth pressure

and temp change

Must use combined

gas law

36

AVOGADROrsquoSLAW

The relationship of moles and volume in gases is called Avogadrorsquos Law

At constant temperature and pressure the volume of a fixed amount of gas is directly proportional to the number of moles

As number of moles increases

the volume of the gas

increases

Number of moles and volume of a gas are directly

proportional

Tros Introductory Chemistry Chapter

37

Avogadrorsquos Law

38

AVOGADROrsquoSLAW

As a result of Avogadrorsquos Law equal volumes of different gases at the same temp and pressure contain equal number of moles (molecules)

2 Tanks of gas of equal volume at the same T amp P contain the same number of molecules

39

AVOGADROrsquoSLAW

For example

2 H2 (g) + 1 O2 (g) 2 H2O (g)

2 molecules 1 molecule 2 molecules

2 moles 1 mole 2 moles

2 Liters 1 Liter 2 Liters

Avogadrorsquos Law also allows chemists to relate volumes and moles of a gas in a chemical reaction

This relationship is only valid for

gaseous substances

40

Example 1

A sample of helium gas with a mass of 180 g occupies 16 Liters at a particular temperature and pressure What mass of oxygen would occupy16 L at the same temperature and pressure

1 mol= 180 g x

400 gmol of He = 450 mol

mol of O2 = mol of He = 450 mol

Same Temp amp Pressure

mass of O2

320 g= 450 mol x

1 mol= 144 g

41

Example 2

How many Liters of NH3 can be produced from reaction of 18 L of H2 with excess N2 as shown below

18 L H23

2

L NHx

L H= 12 L NH3

N2 (g) + 3 H2 (g) 2 NH3 (g)

32

42

STP amp MOLAR VOLUME

To better understand the factors that affect gas behavior a set of standard conditions have been chosen for use and are referred to as Standard Temperature and Pressure (STP)

STP =

1 atm (760 mmHg)

0ordmC (273 K)

43

STP amp MOLAR VOLUME

At STP conditions one mole of any gas is observed to occupy a volume of 224 L

V = 224 L

Molar Volume at

STP

44

Example 1

If 200 L of a gas at STP has a mass of 323 g what is the molar mass of the gas

mol200 L x

Lmol of gas =

2241

= 00893 mol

g

molMolar mass =

323 g=

00893 mol= 362 gmol Molar

volume at STP

45

Example 2

A sample of gas has a volume of 250 L at 730 mmHg and 20C What is the volume of this gas at STP

1 22 1

2 1

P TV = V x x

P T

V2 = 224 L

P1= 730 mmHg V1= 250 L T1= 293 K P2= 760 mmHg V2= T2 = 273 K

2

730 mmHg 273 KV = 250 L x x

760 mmHg 293 K

46

IDEAL GAS LAW

Combining all the laws that describe the behavior of gases one can obtain a useful relationship that relates the volume of a gas to the temperature pressure and number of moles

n R TV =

P

Universal gas constant

R = 00821 L atmmol K

47

IDEAL GAS LAW

This relationship is called the Ideal Gas Law and commonly written as

P V = n R T

Pressure in atm Volume

in Liters

Number of moles

Temp in K

48

Example 1

A sample of H2 gas has a volume of 856 L at a temp of 0C and pressure of 15 atm Calculate the moles of gas present

PVn =

RT

= 057 mol

P = 15 atm V = 856 L T = 273 K n = R = 00821

PV = nRT

(15 atm)(856 L)=

L atm(00821 )(273 K)

mol K

49

Example 2

What volume does 400 g of N2 gas occupy at 10C and 750 mmHg

P =750 mmHg V = T = 283 K n = not given m = 400 g R = 00821

mol of N2 =1 mol

400 g x280 g

= 143 mol

P = 750 mmHg1 atm

x760 mmHg

Moles must be calculated

from mass

Pressure must be

converted to atm

P = 0987 atm

50

Example 2

What volume does 400 g of N2 gas occupy at 10C and 750 mmHg

nRTV =

P

V = 337 L

P =0987 atm V = T = 283 K n = 143 R = 00821

L atm(143 mol)(00821 )(283 K)

mol KV = 0987 atm

Tros Introductory Chemistry Chapter

51

Ideal vs Real Gases

bull Real gases often do not behave like ideal gases at high pressure or low temperature

bull Ideal gas laws assume1) no attractions between gas molecules2) gas molecules do not take up space based on the Kinetic-Molecular Theory

bull at low temperatures and high pressures these assumptions are not valid

Tros Introductory Chemistry Chapter

52

Ideal vs Real Gases

53

PARTIAL PRESSURES

Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases

Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture

In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas

54

DALTONrsquoSLAW

In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture

This is called Daltonrsquos law of partial pressures

Total pressure of a gas mixture

Ptotal = P1 + P2 + P3 +

Sum of the partial pressures of the gases in the mixture

=

55

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = PHe + PAr

Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture

= 60 atm

56

PARTIAL PRESSURES

The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture

For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container

57

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = 60 atm

The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture

Twice as many moles of Ar

compared to He

PAr = 2 PHe

58

Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank

Ptotal = Poxygen + Phelium

Poxygen = 1140 mmHg x

1 atm

760 mmHg= 150 atm

Phelium

= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm

59

Example 2

A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture

Ptotal = Poxygen + Pnitrogen

Pnitrogen = 2 Poxygen

Twice as many moles of N2

compared to O2

Pnitrogen = 23 (Ptotal ) = 20 atm

Poxygen = 13 (Ptotal ) = 10 atm

60

THE END

• Slide 1
• CHAPTER OUTLINE
• PROPERTIES OF GASES
• PROPERTIES OF GASES (2)
• KINETIC-MOLECULAR THEORY
• KINETIC-MOLECULAR THEORY (2)
• Kinetic Molecular Theory
• PRESSURE amp ITS MEASUREMENT
• Common Units of Pressure
• PRESSURE amp ITS MEASUREMENT (2)
• Atmospheric Pressure amp Altitude
• Pressure Imbalance in Ear
• Example 1
• Example 2
• PRESSURE amp MOLES OF A GAS
• BOYLErsquoS LAW
• BOYLErsquoS LAW (2)
• Example 1 (2)
• Example 2 (2)
• CHARLESrsquoS LAW
• CHARLESrsquoS LAW (2)
• Example 1 (3)
• Example 2 (3)
• GAY-LUSSACrsquoS LAW
• GAY-LUSSACrsquoS LAW (2)
• Example 1 (4)
• Example 1 (5)
• Example 2 (4)
• VAPOR PRESSURE amp BOILING POINT
• VAPOR PRESSURE amp BOILING POINT (2)
• VAPOR PRESSURE amp BOILING POINT (3)
• VAPOR PRESSURE amp BOILING POINT (4)
• COMBINED GAS LAW
• COMBINED GAS LAW (2)
• Example 1 (6)
• AVOGADROrsquoS LAW
• Avogadrorsquos Law
• AVOGADROrsquoS LAW (2)
• AVOGADROrsquoS LAW (3)
• Example 1 (7)
• Example 2 (5)
• STP amp MOLAR VOLUME
• STP amp MOLAR VOLUME (2)
• Example 1 (8)
• Example 2 (6)
• IDEAL GAS LAW
• IDEAL GAS LAW (2)
• Example 1 (9)
• Example 2 (7)
• Example 2 (8)
• Ideal vs Real Gases
• Ideal vs Real Gases (2)
• PARTIAL PRESSURES
• DALTONrsquoS LAW
• PARTIAL PRESSURES (2)
• PARTIAL PRESSURES (3)
• PARTIAL PRESSURES (4)
• Example 1 (10)
• Example 2 (9)
• Slide 60

19

Example 2

A sample of hydrogen gas occupies 40 L at 650 mmHg What volume would it occupy at 20 atm

1 12

2

P VV =

P

V2 = 17 L

P1= 650 mmHg

P2= 20 atm

V1 = 40 L

V2 =

P1 x V1 = P2 x V2

(650 mmHg)(40 L)=

(1520 mmHg)Pressures must be in the same

unit

P2 = 20 atm = 1520 mmHgVolume decreases

Pressure increases

20

CHARLESrsquoS LAW

The relationship of temperature and volume in gases is called Charlesrsquo Law

At constant pressure the volume of a fixed amount of gas is directly proportional to its absolute temperature

As temperature increases the volume of the gas increases

Temperature and volume of a gas

are directly proportional

21

CHARLESrsquoS LAW

Charlesrsquo Law can be mathematically expressed as

V and T under initial

condition

1 2

1 2

V V=

T T

V and T under final condition

T must be in units of K

22

Example 1

A 20-L sample of a gas is cooled from 298K to 278 K at constant pressure What is the new volume of the gas

22 1

1

TV = V x

T

V2 = 19 L

V1 = 20 L

V2 =

T1 = 298 K

T2 = 278 K

278 K= 20 L x

298 K

Volume decreases1 2

1 2

V V=

T T

Temperature decreases

23

Example 2

If 200 L of oxygen is cooled from 100ordmC to 0ordmC what is the new volume

22 1

1

TV = V x

T

V2 = 146 L

V1 = 200 L

V2 =

T1 = 100ordmC

T2 = 0ordmC

273 K= 200 L x

373 K

T1= 100ordmC + 273 = 373 K

T2= 0ordmC + 273 = 273 K

Temperatures must be in

KVolume

decreasesTemperature

decreases

24

GAY-LUSSACrsquoSLAW

The relationship of temperature and pressure in gases is called Gay-Lussacrsquos Law

At constant volume the pressure of a fixed amount of gas is directly proportional to its absolute temperature

As temperature increases the

pressure of the gas increases

Temperature and pressure of a gas

are directly proportional

25

GAY-LUSSACrsquoSLAW

Gay-Lussacrsquos Law can be mathematically expressed as

P and T under initial

condition

1 2

1 2

P P=

T T

P and T under final conditionT must be in

units of K

26

Example 1

An aerosol spray can has a pressure of 40 atm at 25C What pressure will the can have if it is placed in a fire and reaches temperature of 400C

P1 = 40 atm

P2 =

T1 = 25ordmC

T2 = 400ordmC

T1= 25ordmC + 273 = 298 K

T2= 400ordmC + 273 = 673 K

Temperatures must be in K

27

Example 1

An aerosol spray can has a pressure of 40 atm at 25C What pressure will the can have if it is placed in a fire and reaches temperature of 400C

22 1

1

TP = P x

T

P2 = 90 atm

P1 = 40 atm

P2 =

T1 = 298 K

T2 = 673 K

673 K= 40 L x

298 K

Pressure increases

1 2

1 2

P P=

T T

Temperature increases

28

Example 2

A cylinder of gas with a volume of 150-L and a pressure of 965 mmHg is stored at a temperature of 55C To what temperature must the cylinder be cooled to reach a pressure of 850 mmHg

22 1

1

PT = T x

P

T2 = 289 K

P1 = 965 mmHg

P2 = 850 mmHg

T1 = 55ordmC

T2 =

850 mmHg= 328 K x

965 mmHg

T1= 55ordmC + 273 = 328 K

T2 = 16ordmC

Temperature decreases

Pressure decreases

29

VAPOR PRESSUREamp BOILING POINT

In an open container liquid molecules at the surface that possess sufficient energy can break away from the surface and become gas particles or vapor

In a closed container these gas particles can accumulate and create pressure called vapor pressure

Vapor pressure is defined as the pressure above a liquid at a given temperature Vapor pressure varies with each liquid and increases with temperature

30

VAPOR PRESSUREamp BOILING POINT

Listed below is the vapor pressure of water at various temperatures

31

VAPOR PRESSUREamp BOILING POINT

A liquid reaches its boiling point when its vapor pressure becomes equal to the external pressure (atmospheric pressure)

For example at sea level water reaches its boiling point at 100C since its vapor pressure is 760 mmHg at this temperature

32

VAPOR PRESSUREamp BOILING POINT

At higher altitudes where atmospheric pressure is lower water reaches boiling point at temperatures lower than 100C

For example in Denver where atmospheric pressure is 630 mmHg water boils at 95C since its vapor pressure is 630 mmHg at this temperature

33

COMBINEDGAS LAW

All pressure-volume-temperature relationships can be combined into a single relationship called the combined gas law

1 1 2 2

1 2

P V P V=

T T

Initial condition

This law is useful for studying the effect of changes in two variables

Final condition

34

COMBINEDGAS LAW

The individual gas laws studied previously are embodied in the combined gas law

1 1 2 2

1 2

P V P V=

T T

Boylersquos LawCharlesrsquos

LawGay-Lussacrsquos

Law

35

Example 1

A 250-mL sample of gas has a pressure of 400 atm at a temperature of 10C What is the volume of the gas at a pressure of 100 atm and a temperature of 18C

1 22 1

2 1

P TV = V x x

P T

V2 = 103 mL

P1= 400 atm V1= 250 mL T1= 283 K P2= 100 atm V2= T2 = 291 K

2

400 atm 291 KV = 250 mL x x

100 atm 283 K

1 1 2 2

1 2

P V P V=

T TBoth pressure

and temp change

Must use combined

gas law

36

AVOGADROrsquoSLAW

The relationship of moles and volume in gases is called Avogadrorsquos Law

At constant temperature and pressure the volume of a fixed amount of gas is directly proportional to the number of moles

As number of moles increases

the volume of the gas

increases

Number of moles and volume of a gas are directly

proportional

Tros Introductory Chemistry Chapter

37

Avogadrorsquos Law

38

AVOGADROrsquoSLAW

As a result of Avogadrorsquos Law equal volumes of different gases at the same temp and pressure contain equal number of moles (molecules)

2 Tanks of gas of equal volume at the same T amp P contain the same number of molecules

39

AVOGADROrsquoSLAW

For example

2 H2 (g) + 1 O2 (g) 2 H2O (g)

2 molecules 1 molecule 2 molecules

2 moles 1 mole 2 moles

2 Liters 1 Liter 2 Liters

Avogadrorsquos Law also allows chemists to relate volumes and moles of a gas in a chemical reaction

This relationship is only valid for

gaseous substances

40

Example 1

A sample of helium gas with a mass of 180 g occupies 16 Liters at a particular temperature and pressure What mass of oxygen would occupy16 L at the same temperature and pressure

1 mol= 180 g x

400 gmol of He = 450 mol

mol of O2 = mol of He = 450 mol

Same Temp amp Pressure

mass of O2

320 g= 450 mol x

1 mol= 144 g

41

Example 2

How many Liters of NH3 can be produced from reaction of 18 L of H2 with excess N2 as shown below

18 L H23

2

L NHx

L H= 12 L NH3

N2 (g) + 3 H2 (g) 2 NH3 (g)

32

42

STP amp MOLAR VOLUME

To better understand the factors that affect gas behavior a set of standard conditions have been chosen for use and are referred to as Standard Temperature and Pressure (STP)

STP =

1 atm (760 mmHg)

0ordmC (273 K)

43

STP amp MOLAR VOLUME

At STP conditions one mole of any gas is observed to occupy a volume of 224 L

V = 224 L

Molar Volume at

STP

44

Example 1

If 200 L of a gas at STP has a mass of 323 g what is the molar mass of the gas

mol200 L x

Lmol of gas =

2241

= 00893 mol

g

molMolar mass =

323 g=

00893 mol= 362 gmol Molar

volume at STP

45

Example 2

A sample of gas has a volume of 250 L at 730 mmHg and 20C What is the volume of this gas at STP

1 22 1

2 1

P TV = V x x

P T

V2 = 224 L

P1= 730 mmHg V1= 250 L T1= 293 K P2= 760 mmHg V2= T2 = 273 K

2

730 mmHg 273 KV = 250 L x x

760 mmHg 293 K

46

IDEAL GAS LAW

Combining all the laws that describe the behavior of gases one can obtain a useful relationship that relates the volume of a gas to the temperature pressure and number of moles

n R TV =

P

Universal gas constant

R = 00821 L atmmol K

47

IDEAL GAS LAW

This relationship is called the Ideal Gas Law and commonly written as

P V = n R T

Pressure in atm Volume

in Liters

Number of moles

Temp in K

48

Example 1

A sample of H2 gas has a volume of 856 L at a temp of 0C and pressure of 15 atm Calculate the moles of gas present

PVn =

RT

= 057 mol

P = 15 atm V = 856 L T = 273 K n = R = 00821

PV = nRT

(15 atm)(856 L)=

L atm(00821 )(273 K)

mol K

49

Example 2

What volume does 400 g of N2 gas occupy at 10C and 750 mmHg

P =750 mmHg V = T = 283 K n = not given m = 400 g R = 00821

mol of N2 =1 mol

400 g x280 g

= 143 mol

P = 750 mmHg1 atm

x760 mmHg

Moles must be calculated

from mass

Pressure must be

converted to atm

P = 0987 atm

50

Example 2

What volume does 400 g of N2 gas occupy at 10C and 750 mmHg

nRTV =

P

V = 337 L

P =0987 atm V = T = 283 K n = 143 R = 00821

L atm(143 mol)(00821 )(283 K)

mol KV = 0987 atm

Tros Introductory Chemistry Chapter

51

Ideal vs Real Gases

bull Real gases often do not behave like ideal gases at high pressure or low temperature

bull Ideal gas laws assume1) no attractions between gas molecules2) gas molecules do not take up space based on the Kinetic-Molecular Theory

bull at low temperatures and high pressures these assumptions are not valid

Tros Introductory Chemistry Chapter

52

Ideal vs Real Gases

53

PARTIAL PRESSURES

Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases

Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture

In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas

54

DALTONrsquoSLAW

In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture

This is called Daltonrsquos law of partial pressures

Total pressure of a gas mixture

Ptotal = P1 + P2 + P3 +

Sum of the partial pressures of the gases in the mixture

=

55

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = PHe + PAr

Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture

= 60 atm

56

PARTIAL PRESSURES

The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture

For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container

57

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = 60 atm

The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture

Twice as many moles of Ar

compared to He

PAr = 2 PHe

58

Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank

Ptotal = Poxygen + Phelium

Poxygen = 1140 mmHg x

1 atm

760 mmHg= 150 atm

Phelium

= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm

59

Example 2

A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture

Ptotal = Poxygen + Pnitrogen

Pnitrogen = 2 Poxygen

Twice as many moles of N2

compared to O2

Pnitrogen = 23 (Ptotal ) = 20 atm

Poxygen = 13 (Ptotal ) = 10 atm

60

THE END

• Slide 1
• CHAPTER OUTLINE
• PROPERTIES OF GASES
• PROPERTIES OF GASES (2)
• KINETIC-MOLECULAR THEORY
• KINETIC-MOLECULAR THEORY (2)
• Kinetic Molecular Theory
• PRESSURE amp ITS MEASUREMENT
• Common Units of Pressure
• PRESSURE amp ITS MEASUREMENT (2)
• Atmospheric Pressure amp Altitude
• Pressure Imbalance in Ear
• Example 1
• Example 2
• PRESSURE amp MOLES OF A GAS
• BOYLErsquoS LAW
• BOYLErsquoS LAW (2)
• Example 1 (2)
• Example 2 (2)
• CHARLESrsquoS LAW
• CHARLESrsquoS LAW (2)
• Example 1 (3)
• Example 2 (3)
• GAY-LUSSACrsquoS LAW
• GAY-LUSSACrsquoS LAW (2)
• Example 1 (4)
• Example 1 (5)
• Example 2 (4)
• VAPOR PRESSURE amp BOILING POINT
• VAPOR PRESSURE amp BOILING POINT (2)
• VAPOR PRESSURE amp BOILING POINT (3)
• VAPOR PRESSURE amp BOILING POINT (4)
• COMBINED GAS LAW
• COMBINED GAS LAW (2)
• Example 1 (6)
• AVOGADROrsquoS LAW
• Avogadrorsquos Law
• AVOGADROrsquoS LAW (2)
• AVOGADROrsquoS LAW (3)
• Example 1 (7)
• Example 2 (5)
• STP amp MOLAR VOLUME
• STP amp MOLAR VOLUME (2)
• Example 1 (8)
• Example 2 (6)
• IDEAL GAS LAW
• IDEAL GAS LAW (2)
• Example 1 (9)
• Example 2 (7)
• Example 2 (8)
• Ideal vs Real Gases
• Ideal vs Real Gases (2)
• PARTIAL PRESSURES
• DALTONrsquoS LAW
• PARTIAL PRESSURES (2)
• PARTIAL PRESSURES (3)
• PARTIAL PRESSURES (4)
• Example 1 (10)
• Example 2 (9)
• Slide 60

20

CHARLESrsquoS LAW

The relationship of temperature and volume in gases is called Charlesrsquo Law

At constant pressure the volume of a fixed amount of gas is directly proportional to its absolute temperature

As temperature increases the volume of the gas increases

Temperature and volume of a gas

are directly proportional

21

CHARLESrsquoS LAW

Charlesrsquo Law can be mathematically expressed as

V and T under initial

condition

1 2

1 2

V V=

T T

V and T under final condition

T must be in units of K

22

Example 1

A 20-L sample of a gas is cooled from 298K to 278 K at constant pressure What is the new volume of the gas

22 1

1

TV = V x

T

V2 = 19 L

V1 = 20 L

V2 =

T1 = 298 K

T2 = 278 K

278 K= 20 L x

298 K

Volume decreases1 2

1 2

V V=

T T

Temperature decreases

23

Example 2

If 200 L of oxygen is cooled from 100ordmC to 0ordmC what is the new volume

22 1

1

TV = V x

T

V2 = 146 L

V1 = 200 L

V2 =

T1 = 100ordmC

T2 = 0ordmC

273 K= 200 L x

373 K

T1= 100ordmC + 273 = 373 K

T2= 0ordmC + 273 = 273 K

Temperatures must be in

KVolume

decreasesTemperature

decreases

24

GAY-LUSSACrsquoSLAW

The relationship of temperature and pressure in gases is called Gay-Lussacrsquos Law

At constant volume the pressure of a fixed amount of gas is directly proportional to its absolute temperature

As temperature increases the

pressure of the gas increases

Temperature and pressure of a gas

are directly proportional

25

GAY-LUSSACrsquoSLAW

Gay-Lussacrsquos Law can be mathematically expressed as

P and T under initial

condition

1 2

1 2

P P=

T T

P and T under final conditionT must be in

units of K

26

Example 1

An aerosol spray can has a pressure of 40 atm at 25C What pressure will the can have if it is placed in a fire and reaches temperature of 400C

P1 = 40 atm

P2 =

T1 = 25ordmC

T2 = 400ordmC

T1= 25ordmC + 273 = 298 K

T2= 400ordmC + 273 = 673 K

Temperatures must be in K

27

Example 1

An aerosol spray can has a pressure of 40 atm at 25C What pressure will the can have if it is placed in a fire and reaches temperature of 400C

22 1

1

TP = P x

T

P2 = 90 atm

P1 = 40 atm

P2 =

T1 = 298 K

T2 = 673 K

673 K= 40 L x

298 K

Pressure increases

1 2

1 2

P P=

T T

Temperature increases

28

Example 2

A cylinder of gas with a volume of 150-L and a pressure of 965 mmHg is stored at a temperature of 55C To what temperature must the cylinder be cooled to reach a pressure of 850 mmHg

22 1

1

PT = T x

P

T2 = 289 K

P1 = 965 mmHg

P2 = 850 mmHg

T1 = 55ordmC

T2 =

850 mmHg= 328 K x

965 mmHg

T1= 55ordmC + 273 = 328 K

T2 = 16ordmC

Temperature decreases

Pressure decreases

29

VAPOR PRESSUREamp BOILING POINT

In an open container liquid molecules at the surface that possess sufficient energy can break away from the surface and become gas particles or vapor

In a closed container these gas particles can accumulate and create pressure called vapor pressure

Vapor pressure is defined as the pressure above a liquid at a given temperature Vapor pressure varies with each liquid and increases with temperature

30

VAPOR PRESSUREamp BOILING POINT

Listed below is the vapor pressure of water at various temperatures

31

VAPOR PRESSUREamp BOILING POINT

A liquid reaches its boiling point when its vapor pressure becomes equal to the external pressure (atmospheric pressure)

For example at sea level water reaches its boiling point at 100C since its vapor pressure is 760 mmHg at this temperature

32

VAPOR PRESSUREamp BOILING POINT

At higher altitudes where atmospheric pressure is lower water reaches boiling point at temperatures lower than 100C

For example in Denver where atmospheric pressure is 630 mmHg water boils at 95C since its vapor pressure is 630 mmHg at this temperature

33

COMBINEDGAS LAW

All pressure-volume-temperature relationships can be combined into a single relationship called the combined gas law

1 1 2 2

1 2

P V P V=

T T

Initial condition

This law is useful for studying the effect of changes in two variables

Final condition

34

COMBINEDGAS LAW

The individual gas laws studied previously are embodied in the combined gas law

1 1 2 2

1 2

P V P V=

T T

Boylersquos LawCharlesrsquos

LawGay-Lussacrsquos

Law

35

Example 1

A 250-mL sample of gas has a pressure of 400 atm at a temperature of 10C What is the volume of the gas at a pressure of 100 atm and a temperature of 18C

1 22 1

2 1

P TV = V x x

P T

V2 = 103 mL

P1= 400 atm V1= 250 mL T1= 283 K P2= 100 atm V2= T2 = 291 K

2

400 atm 291 KV = 250 mL x x

100 atm 283 K

1 1 2 2

1 2

P V P V=

T TBoth pressure

and temp change

Must use combined

gas law

36

AVOGADROrsquoSLAW

The relationship of moles and volume in gases is called Avogadrorsquos Law

At constant temperature and pressure the volume of a fixed amount of gas is directly proportional to the number of moles

As number of moles increases

the volume of the gas

increases

Number of moles and volume of a gas are directly

proportional

Tros Introductory Chemistry Chapter

37

Avogadrorsquos Law

38

AVOGADROrsquoSLAW

As a result of Avogadrorsquos Law equal volumes of different gases at the same temp and pressure contain equal number of moles (molecules)

2 Tanks of gas of equal volume at the same T amp P contain the same number of molecules

39

AVOGADROrsquoSLAW

For example

2 H2 (g) + 1 O2 (g) 2 H2O (g)

2 molecules 1 molecule 2 molecules

2 moles 1 mole 2 moles

2 Liters 1 Liter 2 Liters

Avogadrorsquos Law also allows chemists to relate volumes and moles of a gas in a chemical reaction

This relationship is only valid for

gaseous substances

40

Example 1

A sample of helium gas with a mass of 180 g occupies 16 Liters at a particular temperature and pressure What mass of oxygen would occupy16 L at the same temperature and pressure

1 mol= 180 g x

400 gmol of He = 450 mol

mol of O2 = mol of He = 450 mol

Same Temp amp Pressure

mass of O2

320 g= 450 mol x

1 mol= 144 g

41

Example 2

How many Liters of NH3 can be produced from reaction of 18 L of H2 with excess N2 as shown below

18 L H23

2

L NHx

L H= 12 L NH3

N2 (g) + 3 H2 (g) 2 NH3 (g)

32

42

STP amp MOLAR VOLUME

To better understand the factors that affect gas behavior a set of standard conditions have been chosen for use and are referred to as Standard Temperature and Pressure (STP)

STP =

1 atm (760 mmHg)

0ordmC (273 K)

43

STP amp MOLAR VOLUME

At STP conditions one mole of any gas is observed to occupy a volume of 224 L

V = 224 L

Molar Volume at

STP

44

Example 1

If 200 L of a gas at STP has a mass of 323 g what is the molar mass of the gas

mol200 L x

Lmol of gas =

2241

= 00893 mol

g

molMolar mass =

323 g=

00893 mol= 362 gmol Molar

volume at STP

45

Example 2

A sample of gas has a volume of 250 L at 730 mmHg and 20C What is the volume of this gas at STP

1 22 1

2 1

P TV = V x x

P T

V2 = 224 L

P1= 730 mmHg V1= 250 L T1= 293 K P2= 760 mmHg V2= T2 = 273 K

2

730 mmHg 273 KV = 250 L x x

760 mmHg 293 K

46

IDEAL GAS LAW

Combining all the laws that describe the behavior of gases one can obtain a useful relationship that relates the volume of a gas to the temperature pressure and number of moles

n R TV =

P

Universal gas constant

R = 00821 L atmmol K

47

IDEAL GAS LAW

This relationship is called the Ideal Gas Law and commonly written as

P V = n R T

Pressure in atm Volume

in Liters

Number of moles

Temp in K

48

Example 1

A sample of H2 gas has a volume of 856 L at a temp of 0C and pressure of 15 atm Calculate the moles of gas present

PVn =

RT

= 057 mol

P = 15 atm V = 856 L T = 273 K n = R = 00821

PV = nRT

(15 atm)(856 L)=

L atm(00821 )(273 K)

mol K

49

Example 2

What volume does 400 g of N2 gas occupy at 10C and 750 mmHg

P =750 mmHg V = T = 283 K n = not given m = 400 g R = 00821

mol of N2 =1 mol

400 g x280 g

= 143 mol

P = 750 mmHg1 atm

x760 mmHg

Moles must be calculated

from mass

Pressure must be

converted to atm

P = 0987 atm

50

Example 2

What volume does 400 g of N2 gas occupy at 10C and 750 mmHg

nRTV =

P

V = 337 L

P =0987 atm V = T = 283 K n = 143 R = 00821

L atm(143 mol)(00821 )(283 K)

mol KV = 0987 atm

Tros Introductory Chemistry Chapter

51

Ideal vs Real Gases

bull Real gases often do not behave like ideal gases at high pressure or low temperature

bull Ideal gas laws assume1) no attractions between gas molecules2) gas molecules do not take up space based on the Kinetic-Molecular Theory

bull at low temperatures and high pressures these assumptions are not valid

Tros Introductory Chemistry Chapter

52

Ideal vs Real Gases

53

PARTIAL PRESSURES

Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases

Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture

In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas

54

DALTONrsquoSLAW

In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture

This is called Daltonrsquos law of partial pressures

Total pressure of a gas mixture

Ptotal = P1 + P2 + P3 +

Sum of the partial pressures of the gases in the mixture

=

55

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = PHe + PAr

Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture

= 60 atm

56

PARTIAL PRESSURES

The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture

For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container

57

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = 60 atm

The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture

Twice as many moles of Ar

compared to He

PAr = 2 PHe

58

Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank

Ptotal = Poxygen + Phelium

Poxygen = 1140 mmHg x

1 atm

760 mmHg= 150 atm

Phelium

= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm

59

Example 2

A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture

Ptotal = Poxygen + Pnitrogen

Pnitrogen = 2 Poxygen

Twice as many moles of N2

compared to O2

Pnitrogen = 23 (Ptotal ) = 20 atm

Poxygen = 13 (Ptotal ) = 10 atm

60

THE END

• Slide 1
• CHAPTER OUTLINE
• PROPERTIES OF GASES
• PROPERTIES OF GASES (2)
• KINETIC-MOLECULAR THEORY
• KINETIC-MOLECULAR THEORY (2)
• Kinetic Molecular Theory
• PRESSURE amp ITS MEASUREMENT
• Common Units of Pressure
• PRESSURE amp ITS MEASUREMENT (2)
• Atmospheric Pressure amp Altitude
• Pressure Imbalance in Ear
• Example 1
• Example 2
• PRESSURE amp MOLES OF A GAS
• BOYLErsquoS LAW
• BOYLErsquoS LAW (2)
• Example 1 (2)
• Example 2 (2)
• CHARLESrsquoS LAW
• CHARLESrsquoS LAW (2)
• Example 1 (3)
• Example 2 (3)
• GAY-LUSSACrsquoS LAW
• GAY-LUSSACrsquoS LAW (2)
• Example 1 (4)
• Example 1 (5)
• Example 2 (4)
• VAPOR PRESSURE amp BOILING POINT
• VAPOR PRESSURE amp BOILING POINT (2)
• VAPOR PRESSURE amp BOILING POINT (3)
• VAPOR PRESSURE amp BOILING POINT (4)
• COMBINED GAS LAW
• COMBINED GAS LAW (2)
• Example 1 (6)
• AVOGADROrsquoS LAW
• Avogadrorsquos Law
• AVOGADROrsquoS LAW (2)
• AVOGADROrsquoS LAW (3)
• Example 1 (7)
• Example 2 (5)
• STP amp MOLAR VOLUME
• STP amp MOLAR VOLUME (2)
• Example 1 (8)
• Example 2 (6)
• IDEAL GAS LAW
• IDEAL GAS LAW (2)
• Example 1 (9)
• Example 2 (7)
• Example 2 (8)
• Ideal vs Real Gases
• Ideal vs Real Gases (2)
• PARTIAL PRESSURES
• DALTONrsquoS LAW
• PARTIAL PRESSURES (2)
• PARTIAL PRESSURES (3)
• PARTIAL PRESSURES (4)
• Example 1 (10)
• Example 2 (9)
• Slide 60

21

CHARLESrsquoS LAW

Charlesrsquo Law can be mathematically expressed as

V and T under initial

condition

1 2

1 2

V V=

T T

V and T under final condition

T must be in units of K

22

Example 1

A 20-L sample of a gas is cooled from 298K to 278 K at constant pressure What is the new volume of the gas

22 1

1

TV = V x

T

V2 = 19 L

V1 = 20 L

V2 =

T1 = 298 K

T2 = 278 K

278 K= 20 L x

298 K

Volume decreases1 2

1 2

V V=

T T

Temperature decreases

23

Example 2

If 200 L of oxygen is cooled from 100ordmC to 0ordmC what is the new volume

22 1

1

TV = V x

T

V2 = 146 L

V1 = 200 L

V2 =

T1 = 100ordmC

T2 = 0ordmC

273 K= 200 L x

373 K

T1= 100ordmC + 273 = 373 K

T2= 0ordmC + 273 = 273 K

Temperatures must be in

KVolume

decreasesTemperature

decreases

24

GAY-LUSSACrsquoSLAW

The relationship of temperature and pressure in gases is called Gay-Lussacrsquos Law

At constant volume the pressure of a fixed amount of gas is directly proportional to its absolute temperature

As temperature increases the

pressure of the gas increases

Temperature and pressure of a gas

are directly proportional

25

GAY-LUSSACrsquoSLAW

Gay-Lussacrsquos Law can be mathematically expressed as

P and T under initial

condition

1 2

1 2

P P=

T T

P and T under final conditionT must be in

units of K

26

Example 1

An aerosol spray can has a pressure of 40 atm at 25C What pressure will the can have if it is placed in a fire and reaches temperature of 400C

P1 = 40 atm

P2 =

T1 = 25ordmC

T2 = 400ordmC

T1= 25ordmC + 273 = 298 K

T2= 400ordmC + 273 = 673 K

Temperatures must be in K

27

Example 1

An aerosol spray can has a pressure of 40 atm at 25C What pressure will the can have if it is placed in a fire and reaches temperature of 400C

22 1

1

TP = P x

T

P2 = 90 atm

P1 = 40 atm

P2 =

T1 = 298 K

T2 = 673 K

673 K= 40 L x

298 K

Pressure increases

1 2

1 2

P P=

T T

Temperature increases

28

Example 2

A cylinder of gas with a volume of 150-L and a pressure of 965 mmHg is stored at a temperature of 55C To what temperature must the cylinder be cooled to reach a pressure of 850 mmHg

22 1

1

PT = T x

P

T2 = 289 K

P1 = 965 mmHg

P2 = 850 mmHg

T1 = 55ordmC

T2 =

850 mmHg= 328 K x

965 mmHg

T1= 55ordmC + 273 = 328 K

T2 = 16ordmC

Temperature decreases

Pressure decreases

29

VAPOR PRESSUREamp BOILING POINT

In an open container liquid molecules at the surface that possess sufficient energy can break away from the surface and become gas particles or vapor

In a closed container these gas particles can accumulate and create pressure called vapor pressure

Vapor pressure is defined as the pressure above a liquid at a given temperature Vapor pressure varies with each liquid and increases with temperature

30

VAPOR PRESSUREamp BOILING POINT

Listed below is the vapor pressure of water at various temperatures

31

VAPOR PRESSUREamp BOILING POINT

A liquid reaches its boiling point when its vapor pressure becomes equal to the external pressure (atmospheric pressure)

For example at sea level water reaches its boiling point at 100C since its vapor pressure is 760 mmHg at this temperature

32

VAPOR PRESSUREamp BOILING POINT

At higher altitudes where atmospheric pressure is lower water reaches boiling point at temperatures lower than 100C

For example in Denver where atmospheric pressure is 630 mmHg water boils at 95C since its vapor pressure is 630 mmHg at this temperature

33

COMBINEDGAS LAW

All pressure-volume-temperature relationships can be combined into a single relationship called the combined gas law

1 1 2 2

1 2

P V P V=

T T

Initial condition

This law is useful for studying the effect of changes in two variables

Final condition

34

COMBINEDGAS LAW

The individual gas laws studied previously are embodied in the combined gas law

1 1 2 2

1 2

P V P V=

T T

Boylersquos LawCharlesrsquos

LawGay-Lussacrsquos

Law

35

Example 1

A 250-mL sample of gas has a pressure of 400 atm at a temperature of 10C What is the volume of the gas at a pressure of 100 atm and a temperature of 18C

1 22 1

2 1

P TV = V x x

P T

V2 = 103 mL

P1= 400 atm V1= 250 mL T1= 283 K P2= 100 atm V2= T2 = 291 K

2

400 atm 291 KV = 250 mL x x

100 atm 283 K

1 1 2 2

1 2

P V P V=

T TBoth pressure

and temp change

Must use combined

gas law

36

AVOGADROrsquoSLAW

The relationship of moles and volume in gases is called Avogadrorsquos Law

At constant temperature and pressure the volume of a fixed amount of gas is directly proportional to the number of moles

As number of moles increases

the volume of the gas

increases

Number of moles and volume of a gas are directly

proportional

Tros Introductory Chemistry Chapter

37

Avogadrorsquos Law

38

AVOGADROrsquoSLAW

As a result of Avogadrorsquos Law equal volumes of different gases at the same temp and pressure contain equal number of moles (molecules)

2 Tanks of gas of equal volume at the same T amp P contain the same number of molecules

39

AVOGADROrsquoSLAW

For example

2 H2 (g) + 1 O2 (g) 2 H2O (g)

2 molecules 1 molecule 2 molecules

2 moles 1 mole 2 moles

2 Liters 1 Liter 2 Liters

Avogadrorsquos Law also allows chemists to relate volumes and moles of a gas in a chemical reaction

This relationship is only valid for

gaseous substances

40

Example 1

A sample of helium gas with a mass of 180 g occupies 16 Liters at a particular temperature and pressure What mass of oxygen would occupy16 L at the same temperature and pressure

1 mol= 180 g x

400 gmol of He = 450 mol

mol of O2 = mol of He = 450 mol

Same Temp amp Pressure

mass of O2

320 g= 450 mol x

1 mol= 144 g

41

Example 2

How many Liters of NH3 can be produced from reaction of 18 L of H2 with excess N2 as shown below

18 L H23

2

L NHx

L H= 12 L NH3

N2 (g) + 3 H2 (g) 2 NH3 (g)

32

42

STP amp MOLAR VOLUME

To better understand the factors that affect gas behavior a set of standard conditions have been chosen for use and are referred to as Standard Temperature and Pressure (STP)

STP =

1 atm (760 mmHg)

0ordmC (273 K)

43

STP amp MOLAR VOLUME

At STP conditions one mole of any gas is observed to occupy a volume of 224 L

V = 224 L

Molar Volume at

STP

44

Example 1

If 200 L of a gas at STP has a mass of 323 g what is the molar mass of the gas

mol200 L x

Lmol of gas =

2241

= 00893 mol

g

molMolar mass =

323 g=

00893 mol= 362 gmol Molar

volume at STP

45

Example 2

A sample of gas has a volume of 250 L at 730 mmHg and 20C What is the volume of this gas at STP

1 22 1

2 1

P TV = V x x

P T

V2 = 224 L

P1= 730 mmHg V1= 250 L T1= 293 K P2= 760 mmHg V2= T2 = 273 K

2

730 mmHg 273 KV = 250 L x x

760 mmHg 293 K

46

IDEAL GAS LAW

Combining all the laws that describe the behavior of gases one can obtain a useful relationship that relates the volume of a gas to the temperature pressure and number of moles

n R TV =

P

Universal gas constant

R = 00821 L atmmol K

47

IDEAL GAS LAW

This relationship is called the Ideal Gas Law and commonly written as

P V = n R T

Pressure in atm Volume

in Liters

Number of moles

Temp in K

48

Example 1

A sample of H2 gas has a volume of 856 L at a temp of 0C and pressure of 15 atm Calculate the moles of gas present

PVn =

RT

= 057 mol

P = 15 atm V = 856 L T = 273 K n = R = 00821

PV = nRT

(15 atm)(856 L)=

L atm(00821 )(273 K)

mol K

49

Example 2

What volume does 400 g of N2 gas occupy at 10C and 750 mmHg

P =750 mmHg V = T = 283 K n = not given m = 400 g R = 00821

mol of N2 =1 mol

400 g x280 g

= 143 mol

P = 750 mmHg1 atm

x760 mmHg

Moles must be calculated

from mass

Pressure must be

converted to atm

P = 0987 atm

50

Example 2

What volume does 400 g of N2 gas occupy at 10C and 750 mmHg

nRTV =

P

V = 337 L

P =0987 atm V = T = 283 K n = 143 R = 00821

L atm(143 mol)(00821 )(283 K)

mol KV = 0987 atm

Tros Introductory Chemistry Chapter

51

Ideal vs Real Gases

bull Real gases often do not behave like ideal gases at high pressure or low temperature

bull Ideal gas laws assume1) no attractions between gas molecules2) gas molecules do not take up space based on the Kinetic-Molecular Theory

bull at low temperatures and high pressures these assumptions are not valid

Tros Introductory Chemistry Chapter

52

Ideal vs Real Gases

53

PARTIAL PRESSURES

Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases

Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture

In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas

54

DALTONrsquoSLAW

In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture

This is called Daltonrsquos law of partial pressures

Total pressure of a gas mixture

Ptotal = P1 + P2 + P3 +

Sum of the partial pressures of the gases in the mixture

=

55

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = PHe + PAr

Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture

= 60 atm

56

PARTIAL PRESSURES

The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture

For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container

57

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = 60 atm

The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture

Twice as many moles of Ar

compared to He

PAr = 2 PHe

58

Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank

Ptotal = Poxygen + Phelium

Poxygen = 1140 mmHg x

1 atm

760 mmHg= 150 atm

Phelium

= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm

59

Example 2

A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture

Ptotal = Poxygen + Pnitrogen

Pnitrogen = 2 Poxygen

Twice as many moles of N2

compared to O2

Pnitrogen = 23 (Ptotal ) = 20 atm

Poxygen = 13 (Ptotal ) = 10 atm

60

THE END

• Slide 1
• CHAPTER OUTLINE
• PROPERTIES OF GASES
• PROPERTIES OF GASES (2)
• KINETIC-MOLECULAR THEORY
• KINETIC-MOLECULAR THEORY (2)
• Kinetic Molecular Theory
• PRESSURE amp ITS MEASUREMENT
• Common Units of Pressure
• PRESSURE amp ITS MEASUREMENT (2)
• Atmospheric Pressure amp Altitude
• Pressure Imbalance in Ear
• Example 1
• Example 2
• PRESSURE amp MOLES OF A GAS
• BOYLErsquoS LAW
• BOYLErsquoS LAW (2)
• Example 1 (2)
• Example 2 (2)
• CHARLESrsquoS LAW
• CHARLESrsquoS LAW (2)
• Example 1 (3)
• Example 2 (3)
• GAY-LUSSACrsquoS LAW
• GAY-LUSSACrsquoS LAW (2)
• Example 1 (4)
• Example 1 (5)
• Example 2 (4)
• VAPOR PRESSURE amp BOILING POINT
• VAPOR PRESSURE amp BOILING POINT (2)
• VAPOR PRESSURE amp BOILING POINT (3)
• VAPOR PRESSURE amp BOILING POINT (4)
• COMBINED GAS LAW
• COMBINED GAS LAW (2)
• Example 1 (6)
• AVOGADROrsquoS LAW
• Avogadrorsquos Law
• AVOGADROrsquoS LAW (2)
• AVOGADROrsquoS LAW (3)
• Example 1 (7)
• Example 2 (5)
• STP amp MOLAR VOLUME
• STP amp MOLAR VOLUME (2)
• Example 1 (8)
• Example 2 (6)
• IDEAL GAS LAW
• IDEAL GAS LAW (2)
• Example 1 (9)
• Example 2 (7)
• Example 2 (8)
• Ideal vs Real Gases
• Ideal vs Real Gases (2)
• PARTIAL PRESSURES
• DALTONrsquoS LAW
• PARTIAL PRESSURES (2)
• PARTIAL PRESSURES (3)
• PARTIAL PRESSURES (4)
• Example 1 (10)
• Example 2 (9)
• Slide 60

22

Example 1

A 20-L sample of a gas is cooled from 298K to 278 K at constant pressure What is the new volume of the gas

22 1

1

TV = V x

T

V2 = 19 L

V1 = 20 L

V2 =

T1 = 298 K

T2 = 278 K

278 K= 20 L x

298 K

Volume decreases1 2

1 2

V V=

T T

Temperature decreases

23

Example 2

If 200 L of oxygen is cooled from 100ordmC to 0ordmC what is the new volume

22 1

1

TV = V x

T

V2 = 146 L

V1 = 200 L

V2 =

T1 = 100ordmC

T2 = 0ordmC

273 K= 200 L x

373 K

T1= 100ordmC + 273 = 373 K

T2= 0ordmC + 273 = 273 K

Temperatures must be in

KVolume

decreasesTemperature

decreases

24

GAY-LUSSACrsquoSLAW

The relationship of temperature and pressure in gases is called Gay-Lussacrsquos Law

At constant volume the pressure of a fixed amount of gas is directly proportional to its absolute temperature

As temperature increases the

pressure of the gas increases

Temperature and pressure of a gas

are directly proportional

25

GAY-LUSSACrsquoSLAW

Gay-Lussacrsquos Law can be mathematically expressed as

P and T under initial

condition

1 2

1 2

P P=

T T

P and T under final conditionT must be in

units of K

26

Example 1

An aerosol spray can has a pressure of 40 atm at 25C What pressure will the can have if it is placed in a fire and reaches temperature of 400C

P1 = 40 atm

P2 =

T1 = 25ordmC

T2 = 400ordmC

T1= 25ordmC + 273 = 298 K

T2= 400ordmC + 273 = 673 K

Temperatures must be in K

27

Example 1

An aerosol spray can has a pressure of 40 atm at 25C What pressure will the can have if it is placed in a fire and reaches temperature of 400C

22 1

1

TP = P x

T

P2 = 90 atm

P1 = 40 atm

P2 =

T1 = 298 K

T2 = 673 K

673 K= 40 L x

298 K

Pressure increases

1 2

1 2

P P=

T T

Temperature increases

28

Example 2

A cylinder of gas with a volume of 150-L and a pressure of 965 mmHg is stored at a temperature of 55C To what temperature must the cylinder be cooled to reach a pressure of 850 mmHg

22 1

1

PT = T x

P

T2 = 289 K

P1 = 965 mmHg

P2 = 850 mmHg

T1 = 55ordmC

T2 =

850 mmHg= 328 K x

965 mmHg

T1= 55ordmC + 273 = 328 K

T2 = 16ordmC

Temperature decreases

Pressure decreases

29

VAPOR PRESSUREamp BOILING POINT

In an open container liquid molecules at the surface that possess sufficient energy can break away from the surface and become gas particles or vapor

In a closed container these gas particles can accumulate and create pressure called vapor pressure

Vapor pressure is defined as the pressure above a liquid at a given temperature Vapor pressure varies with each liquid and increases with temperature

30

VAPOR PRESSUREamp BOILING POINT

Listed below is the vapor pressure of water at various temperatures

31

VAPOR PRESSUREamp BOILING POINT

A liquid reaches its boiling point when its vapor pressure becomes equal to the external pressure (atmospheric pressure)

For example at sea level water reaches its boiling point at 100C since its vapor pressure is 760 mmHg at this temperature

32

VAPOR PRESSUREamp BOILING POINT

At higher altitudes where atmospheric pressure is lower water reaches boiling point at temperatures lower than 100C

For example in Denver where atmospheric pressure is 630 mmHg water boils at 95C since its vapor pressure is 630 mmHg at this temperature

33

COMBINEDGAS LAW

All pressure-volume-temperature relationships can be combined into a single relationship called the combined gas law

1 1 2 2

1 2

P V P V=

T T

Initial condition

This law is useful for studying the effect of changes in two variables

Final condition

34

COMBINEDGAS LAW

The individual gas laws studied previously are embodied in the combined gas law

1 1 2 2

1 2

P V P V=

T T

Boylersquos LawCharlesrsquos

LawGay-Lussacrsquos

Law

35

Example 1

A 250-mL sample of gas has a pressure of 400 atm at a temperature of 10C What is the volume of the gas at a pressure of 100 atm and a temperature of 18C

1 22 1

2 1

P TV = V x x

P T

V2 = 103 mL

P1= 400 atm V1= 250 mL T1= 283 K P2= 100 atm V2= T2 = 291 K

2

400 atm 291 KV = 250 mL x x

100 atm 283 K

1 1 2 2

1 2

P V P V=

T TBoth pressure

and temp change

Must use combined

gas law

36

AVOGADROrsquoSLAW

The relationship of moles and volume in gases is called Avogadrorsquos Law

At constant temperature and pressure the volume of a fixed amount of gas is directly proportional to the number of moles

As number of moles increases

the volume of the gas

increases

Number of moles and volume of a gas are directly

proportional

Tros Introductory Chemistry Chapter

37

Avogadrorsquos Law

38

AVOGADROrsquoSLAW

As a result of Avogadrorsquos Law equal volumes of different gases at the same temp and pressure contain equal number of moles (molecules)

2 Tanks of gas of equal volume at the same T amp P contain the same number of molecules

39

AVOGADROrsquoSLAW

For example

2 H2 (g) + 1 O2 (g) 2 H2O (g)

2 molecules 1 molecule 2 molecules

2 moles 1 mole 2 moles

2 Liters 1 Liter 2 Liters

Avogadrorsquos Law also allows chemists to relate volumes and moles of a gas in a chemical reaction

This relationship is only valid for

gaseous substances

40

Example 1

A sample of helium gas with a mass of 180 g occupies 16 Liters at a particular temperature and pressure What mass of oxygen would occupy16 L at the same temperature and pressure

1 mol= 180 g x

400 gmol of He = 450 mol

mol of O2 = mol of He = 450 mol

Same Temp amp Pressure

mass of O2

320 g= 450 mol x

1 mol= 144 g

41

Example 2

How many Liters of NH3 can be produced from reaction of 18 L of H2 with excess N2 as shown below

18 L H23

2

L NHx

L H= 12 L NH3

N2 (g) + 3 H2 (g) 2 NH3 (g)

32

42

STP amp MOLAR VOLUME

To better understand the factors that affect gas behavior a set of standard conditions have been chosen for use and are referred to as Standard Temperature and Pressure (STP)

STP =

1 atm (760 mmHg)

0ordmC (273 K)

43

STP amp MOLAR VOLUME

At STP conditions one mole of any gas is observed to occupy a volume of 224 L

V = 224 L

Molar Volume at

STP

44

Example 1

If 200 L of a gas at STP has a mass of 323 g what is the molar mass of the gas

mol200 L x

Lmol of gas =

2241

= 00893 mol

g

molMolar mass =

323 g=

00893 mol= 362 gmol Molar

volume at STP

45

Example 2

A sample of gas has a volume of 250 L at 730 mmHg and 20C What is the volume of this gas at STP

1 22 1

2 1

P TV = V x x

P T

V2 = 224 L

P1= 730 mmHg V1= 250 L T1= 293 K P2= 760 mmHg V2= T2 = 273 K

2

730 mmHg 273 KV = 250 L x x

760 mmHg 293 K

46

IDEAL GAS LAW

Combining all the laws that describe the behavior of gases one can obtain a useful relationship that relates the volume of a gas to the temperature pressure and number of moles

n R TV =

P

Universal gas constant

R = 00821 L atmmol K

47

IDEAL GAS LAW

This relationship is called the Ideal Gas Law and commonly written as

P V = n R T

Pressure in atm Volume

in Liters

Number of moles

Temp in K

48

Example 1

A sample of H2 gas has a volume of 856 L at a temp of 0C and pressure of 15 atm Calculate the moles of gas present

PVn =

RT

= 057 mol

P = 15 atm V = 856 L T = 273 K n = R = 00821

PV = nRT

(15 atm)(856 L)=

L atm(00821 )(273 K)

mol K

49

Example 2

What volume does 400 g of N2 gas occupy at 10C and 750 mmHg

P =750 mmHg V = T = 283 K n = not given m = 400 g R = 00821

mol of N2 =1 mol

400 g x280 g

= 143 mol

P = 750 mmHg1 atm

x760 mmHg

Moles must be calculated

from mass

Pressure must be

converted to atm

P = 0987 atm

50

Example 2

What volume does 400 g of N2 gas occupy at 10C and 750 mmHg

nRTV =

P

V = 337 L

P =0987 atm V = T = 283 K n = 143 R = 00821

L atm(143 mol)(00821 )(283 K)

mol KV = 0987 atm

Tros Introductory Chemistry Chapter

51

Ideal vs Real Gases

bull Real gases often do not behave like ideal gases at high pressure or low temperature

bull Ideal gas laws assume1) no attractions between gas molecules2) gas molecules do not take up space based on the Kinetic-Molecular Theory

bull at low temperatures and high pressures these assumptions are not valid

Tros Introductory Chemistry Chapter

52

Ideal vs Real Gases

53

PARTIAL PRESSURES

Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases

Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture

In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas

54

DALTONrsquoSLAW

In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture

This is called Daltonrsquos law of partial pressures

Total pressure of a gas mixture

Ptotal = P1 + P2 + P3 +

Sum of the partial pressures of the gases in the mixture

=

55

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = PHe + PAr

Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture

= 60 atm

56

PARTIAL PRESSURES

The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture

For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container

57

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = 60 atm

The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture

Twice as many moles of Ar

compared to He

PAr = 2 PHe

58

Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank

Ptotal = Poxygen + Phelium

Poxygen = 1140 mmHg x

1 atm

760 mmHg= 150 atm

Phelium

= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm

59

Example 2

A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture

Ptotal = Poxygen + Pnitrogen

Pnitrogen = 2 Poxygen

Twice as many moles of N2

compared to O2

Pnitrogen = 23 (Ptotal ) = 20 atm

Poxygen = 13 (Ptotal ) = 10 atm

60

THE END

• Slide 1
• CHAPTER OUTLINE
• PROPERTIES OF GASES
• PROPERTIES OF GASES (2)
• KINETIC-MOLECULAR THEORY
• KINETIC-MOLECULAR THEORY (2)
• Kinetic Molecular Theory
• PRESSURE amp ITS MEASUREMENT
• Common Units of Pressure
• PRESSURE amp ITS MEASUREMENT (2)
• Atmospheric Pressure amp Altitude
• Pressure Imbalance in Ear
• Example 1
• Example 2
• PRESSURE amp MOLES OF A GAS
• BOYLErsquoS LAW
• BOYLErsquoS LAW (2)
• Example 1 (2)
• Example 2 (2)
• CHARLESrsquoS LAW
• CHARLESrsquoS LAW (2)
• Example 1 (3)
• Example 2 (3)
• GAY-LUSSACrsquoS LAW
• GAY-LUSSACrsquoS LAW (2)
• Example 1 (4)
• Example 1 (5)
• Example 2 (4)
• VAPOR PRESSURE amp BOILING POINT
• VAPOR PRESSURE amp BOILING POINT (2)
• VAPOR PRESSURE amp BOILING POINT (3)
• VAPOR PRESSURE amp BOILING POINT (4)
• COMBINED GAS LAW
• COMBINED GAS LAW (2)
• Example 1 (6)
• AVOGADROrsquoS LAW
• Avogadrorsquos Law
• AVOGADROrsquoS LAW (2)
• AVOGADROrsquoS LAW (3)
• Example 1 (7)
• Example 2 (5)
• STP amp MOLAR VOLUME
• STP amp MOLAR VOLUME (2)
• Example 1 (8)
• Example 2 (6)
• IDEAL GAS LAW
• IDEAL GAS LAW (2)
• Example 1 (9)
• Example 2 (7)
• Example 2 (8)
• Ideal vs Real Gases
• Ideal vs Real Gases (2)
• PARTIAL PRESSURES
• DALTONrsquoS LAW
• PARTIAL PRESSURES (2)
• PARTIAL PRESSURES (3)
• PARTIAL PRESSURES (4)
• Example 1 (10)
• Example 2 (9)
• Slide 60

23

Example 2

If 200 L of oxygen is cooled from 100ordmC to 0ordmC what is the new volume

22 1

1

TV = V x

T

V2 = 146 L

V1 = 200 L

V2 =

T1 = 100ordmC

T2 = 0ordmC

273 K= 200 L x

373 K

T1= 100ordmC + 273 = 373 K

T2= 0ordmC + 273 = 273 K

Temperatures must be in

KVolume

decreasesTemperature

decreases

24

GAY-LUSSACrsquoSLAW

The relationship of temperature and pressure in gases is called Gay-Lussacrsquos Law

At constant volume the pressure of a fixed amount of gas is directly proportional to its absolute temperature

As temperature increases the

pressure of the gas increases

Temperature and pressure of a gas

are directly proportional

25

GAY-LUSSACrsquoSLAW

Gay-Lussacrsquos Law can be mathematically expressed as

P and T under initial

condition

1 2

1 2

P P=

T T

P and T under final conditionT must be in

units of K

26

Example 1

An aerosol spray can has a pressure of 40 atm at 25C What pressure will the can have if it is placed in a fire and reaches temperature of 400C

P1 = 40 atm

P2 =

T1 = 25ordmC

T2 = 400ordmC

T1= 25ordmC + 273 = 298 K

T2= 400ordmC + 273 = 673 K

Temperatures must be in K

27

Example 1

An aerosol spray can has a pressure of 40 atm at 25C What pressure will the can have if it is placed in a fire and reaches temperature of 400C

22 1

1

TP = P x

T

P2 = 90 atm

P1 = 40 atm

P2 =

T1 = 298 K

T2 = 673 K

673 K= 40 L x

298 K

Pressure increases

1 2

1 2

P P=

T T

Temperature increases

28

Example 2

A cylinder of gas with a volume of 150-L and a pressure of 965 mmHg is stored at a temperature of 55C To what temperature must the cylinder be cooled to reach a pressure of 850 mmHg

22 1

1

PT = T x

P

T2 = 289 K

P1 = 965 mmHg

P2 = 850 mmHg

T1 = 55ordmC

T2 =

850 mmHg= 328 K x

965 mmHg

T1= 55ordmC + 273 = 328 K

T2 = 16ordmC

Temperature decreases

Pressure decreases

29

VAPOR PRESSUREamp BOILING POINT

In an open container liquid molecules at the surface that possess sufficient energy can break away from the surface and become gas particles or vapor

In a closed container these gas particles can accumulate and create pressure called vapor pressure

Vapor pressure is defined as the pressure above a liquid at a given temperature Vapor pressure varies with each liquid and increases with temperature

30

VAPOR PRESSUREamp BOILING POINT

Listed below is the vapor pressure of water at various temperatures

31

VAPOR PRESSUREamp BOILING POINT

A liquid reaches its boiling point when its vapor pressure becomes equal to the external pressure (atmospheric pressure)

For example at sea level water reaches its boiling point at 100C since its vapor pressure is 760 mmHg at this temperature

32

VAPOR PRESSUREamp BOILING POINT

At higher altitudes where atmospheric pressure is lower water reaches boiling point at temperatures lower than 100C

For example in Denver where atmospheric pressure is 630 mmHg water boils at 95C since its vapor pressure is 630 mmHg at this temperature

33

COMBINEDGAS LAW

All pressure-volume-temperature relationships can be combined into a single relationship called the combined gas law

1 1 2 2

1 2

P V P V=

T T

Initial condition

This law is useful for studying the effect of changes in two variables

Final condition

34

COMBINEDGAS LAW

The individual gas laws studied previously are embodied in the combined gas law

1 1 2 2

1 2

P V P V=

T T

Boylersquos LawCharlesrsquos

LawGay-Lussacrsquos

Law

35

Example 1

A 250-mL sample of gas has a pressure of 400 atm at a temperature of 10C What is the volume of the gas at a pressure of 100 atm and a temperature of 18C

1 22 1

2 1

P TV = V x x

P T

V2 = 103 mL

P1= 400 atm V1= 250 mL T1= 283 K P2= 100 atm V2= T2 = 291 K

2

400 atm 291 KV = 250 mL x x

100 atm 283 K

1 1 2 2

1 2

P V P V=

T TBoth pressure

and temp change

Must use combined

gas law

36

AVOGADROrsquoSLAW

The relationship of moles and volume in gases is called Avogadrorsquos Law

At constant temperature and pressure the volume of a fixed amount of gas is directly proportional to the number of moles

As number of moles increases

the volume of the gas

increases

Number of moles and volume of a gas are directly

proportional

Tros Introductory Chemistry Chapter

37

Avogadrorsquos Law

38

AVOGADROrsquoSLAW

As a result of Avogadrorsquos Law equal volumes of different gases at the same temp and pressure contain equal number of moles (molecules)

2 Tanks of gas of equal volume at the same T amp P contain the same number of molecules

39

AVOGADROrsquoSLAW

For example

2 H2 (g) + 1 O2 (g) 2 H2O (g)

2 molecules 1 molecule 2 molecules

2 moles 1 mole 2 moles

2 Liters 1 Liter 2 Liters

Avogadrorsquos Law also allows chemists to relate volumes and moles of a gas in a chemical reaction

This relationship is only valid for

gaseous substances

40

Example 1

A sample of helium gas with a mass of 180 g occupies 16 Liters at a particular temperature and pressure What mass of oxygen would occupy16 L at the same temperature and pressure

1 mol= 180 g x

400 gmol of He = 450 mol

mol of O2 = mol of He = 450 mol

Same Temp amp Pressure

mass of O2

320 g= 450 mol x

1 mol= 144 g

41

Example 2

How many Liters of NH3 can be produced from reaction of 18 L of H2 with excess N2 as shown below

18 L H23

2

L NHx

L H= 12 L NH3

N2 (g) + 3 H2 (g) 2 NH3 (g)

32

42

STP amp MOLAR VOLUME

To better understand the factors that affect gas behavior a set of standard conditions have been chosen for use and are referred to as Standard Temperature and Pressure (STP)

STP =

1 atm (760 mmHg)

0ordmC (273 K)

43

STP amp MOLAR VOLUME

At STP conditions one mole of any gas is observed to occupy a volume of 224 L

V = 224 L

Molar Volume at

STP

44

Example 1

If 200 L of a gas at STP has a mass of 323 g what is the molar mass of the gas

mol200 L x

Lmol of gas =

2241

= 00893 mol

g

molMolar mass =

323 g=

00893 mol= 362 gmol Molar

volume at STP

45

Example 2

A sample of gas has a volume of 250 L at 730 mmHg and 20C What is the volume of this gas at STP

1 22 1

2 1

P TV = V x x

P T

V2 = 224 L

P1= 730 mmHg V1= 250 L T1= 293 K P2= 760 mmHg V2= T2 = 273 K

2

730 mmHg 273 KV = 250 L x x

760 mmHg 293 K

46

IDEAL GAS LAW

Combining all the laws that describe the behavior of gases one can obtain a useful relationship that relates the volume of a gas to the temperature pressure and number of moles

n R TV =

P

Universal gas constant

R = 00821 L atmmol K

47

IDEAL GAS LAW

This relationship is called the Ideal Gas Law and commonly written as

P V = n R T

Pressure in atm Volume

in Liters

Number of moles

Temp in K

48

Example 1

A sample of H2 gas has a volume of 856 L at a temp of 0C and pressure of 15 atm Calculate the moles of gas present

PVn =

RT

= 057 mol

P = 15 atm V = 856 L T = 273 K n = R = 00821

PV = nRT

(15 atm)(856 L)=

L atm(00821 )(273 K)

mol K

49

Example 2

What volume does 400 g of N2 gas occupy at 10C and 750 mmHg

P =750 mmHg V = T = 283 K n = not given m = 400 g R = 00821

mol of N2 =1 mol

400 g x280 g

= 143 mol

P = 750 mmHg1 atm

x760 mmHg

Moles must be calculated

from mass

Pressure must be

converted to atm

P = 0987 atm

50

Example 2

What volume does 400 g of N2 gas occupy at 10C and 750 mmHg

nRTV =

P

V = 337 L

P =0987 atm V = T = 283 K n = 143 R = 00821

L atm(143 mol)(00821 )(283 K)

mol KV = 0987 atm

Tros Introductory Chemistry Chapter

51

Ideal vs Real Gases

bull Real gases often do not behave like ideal gases at high pressure or low temperature

bull Ideal gas laws assume1) no attractions between gas molecules2) gas molecules do not take up space based on the Kinetic-Molecular Theory

bull at low temperatures and high pressures these assumptions are not valid

Tros Introductory Chemistry Chapter

52

Ideal vs Real Gases

53

PARTIAL PRESSURES

Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases

Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture

In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas

54

DALTONrsquoSLAW

In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture

This is called Daltonrsquos law of partial pressures

Total pressure of a gas mixture

Ptotal = P1 + P2 + P3 +

Sum of the partial pressures of the gases in the mixture

=

55

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = PHe + PAr

Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture

= 60 atm

56

PARTIAL PRESSURES

The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture

For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container

57

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = 60 atm

The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture

Twice as many moles of Ar

compared to He

PAr = 2 PHe

58

Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank

Ptotal = Poxygen + Phelium

Poxygen = 1140 mmHg x

1 atm

760 mmHg= 150 atm

Phelium

= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm

59

Example 2

A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture

Ptotal = Poxygen + Pnitrogen

Pnitrogen = 2 Poxygen

Twice as many moles of N2

compared to O2

Pnitrogen = 23 (Ptotal ) = 20 atm

Poxygen = 13 (Ptotal ) = 10 atm

60

THE END

• Slide 1
• CHAPTER OUTLINE
• PROPERTIES OF GASES
• PROPERTIES OF GASES (2)
• KINETIC-MOLECULAR THEORY
• KINETIC-MOLECULAR THEORY (2)
• Kinetic Molecular Theory
• PRESSURE amp ITS MEASUREMENT
• Common Units of Pressure
• PRESSURE amp ITS MEASUREMENT (2)
• Atmospheric Pressure amp Altitude
• Pressure Imbalance in Ear
• Example 1
• Example 2
• PRESSURE amp MOLES OF A GAS
• BOYLErsquoS LAW
• BOYLErsquoS LAW (2)
• Example 1 (2)
• Example 2 (2)
• CHARLESrsquoS LAW
• CHARLESrsquoS LAW (2)
• Example 1 (3)
• Example 2 (3)
• GAY-LUSSACrsquoS LAW
• GAY-LUSSACrsquoS LAW (2)
• Example 1 (4)
• Example 1 (5)
• Example 2 (4)
• VAPOR PRESSURE amp BOILING POINT
• VAPOR PRESSURE amp BOILING POINT (2)
• VAPOR PRESSURE amp BOILING POINT (3)
• VAPOR PRESSURE amp BOILING POINT (4)
• COMBINED GAS LAW
• COMBINED GAS LAW (2)
• Example 1 (6)
• AVOGADROrsquoS LAW
• Avogadrorsquos Law
• AVOGADROrsquoS LAW (2)
• AVOGADROrsquoS LAW (3)
• Example 1 (7)
• Example 2 (5)
• STP amp MOLAR VOLUME
• STP amp MOLAR VOLUME (2)
• Example 1 (8)
• Example 2 (6)
• IDEAL GAS LAW
• IDEAL GAS LAW (2)
• Example 1 (9)
• Example 2 (7)
• Example 2 (8)
• Ideal vs Real Gases
• Ideal vs Real Gases (2)
• PARTIAL PRESSURES
• DALTONrsquoS LAW
• PARTIAL PRESSURES (2)
• PARTIAL PRESSURES (3)
• PARTIAL PRESSURES (4)
• Example 1 (10)
• Example 2 (9)
• Slide 60

24

GAY-LUSSACrsquoSLAW

The relationship of temperature and pressure in gases is called Gay-Lussacrsquos Law

At constant volume the pressure of a fixed amount of gas is directly proportional to its absolute temperature

As temperature increases the

pressure of the gas increases

Temperature and pressure of a gas

are directly proportional

25

GAY-LUSSACrsquoSLAW

Gay-Lussacrsquos Law can be mathematically expressed as

P and T under initial

condition

1 2

1 2

P P=

T T

P and T under final conditionT must be in

units of K

26

Example 1

An aerosol spray can has a pressure of 40 atm at 25C What pressure will the can have if it is placed in a fire and reaches temperature of 400C

P1 = 40 atm

P2 =

T1 = 25ordmC

T2 = 400ordmC

T1= 25ordmC + 273 = 298 K

T2= 400ordmC + 273 = 673 K

Temperatures must be in K

27

Example 1

An aerosol spray can has a pressure of 40 atm at 25C What pressure will the can have if it is placed in a fire and reaches temperature of 400C

22 1

1

TP = P x

T

P2 = 90 atm

P1 = 40 atm

P2 =

T1 = 298 K

T2 = 673 K

673 K= 40 L x

298 K

Pressure increases

1 2

1 2

P P=

T T

Temperature increases

28

Example 2

A cylinder of gas with a volume of 150-L and a pressure of 965 mmHg is stored at a temperature of 55C To what temperature must the cylinder be cooled to reach a pressure of 850 mmHg

22 1

1

PT = T x

P

T2 = 289 K

P1 = 965 mmHg

P2 = 850 mmHg

T1 = 55ordmC

T2 =

850 mmHg= 328 K x

965 mmHg

T1= 55ordmC + 273 = 328 K

T2 = 16ordmC

Temperature decreases

Pressure decreases

29

VAPOR PRESSUREamp BOILING POINT

In an open container liquid molecules at the surface that possess sufficient energy can break away from the surface and become gas particles or vapor

In a closed container these gas particles can accumulate and create pressure called vapor pressure

Vapor pressure is defined as the pressure above a liquid at a given temperature Vapor pressure varies with each liquid and increases with temperature

30

VAPOR PRESSUREamp BOILING POINT

Listed below is the vapor pressure of water at various temperatures

31

VAPOR PRESSUREamp BOILING POINT

A liquid reaches its boiling point when its vapor pressure becomes equal to the external pressure (atmospheric pressure)

For example at sea level water reaches its boiling point at 100C since its vapor pressure is 760 mmHg at this temperature

32

VAPOR PRESSUREamp BOILING POINT

At higher altitudes where atmospheric pressure is lower water reaches boiling point at temperatures lower than 100C

For example in Denver where atmospheric pressure is 630 mmHg water boils at 95C since its vapor pressure is 630 mmHg at this temperature

33

COMBINEDGAS LAW

All pressure-volume-temperature relationships can be combined into a single relationship called the combined gas law

1 1 2 2

1 2

P V P V=

T T

Initial condition

This law is useful for studying the effect of changes in two variables

Final condition

34

COMBINEDGAS LAW

The individual gas laws studied previously are embodied in the combined gas law

1 1 2 2

1 2

P V P V=

T T

Boylersquos LawCharlesrsquos

LawGay-Lussacrsquos

Law

35

Example 1

A 250-mL sample of gas has a pressure of 400 atm at a temperature of 10C What is the volume of the gas at a pressure of 100 atm and a temperature of 18C

1 22 1

2 1

P TV = V x x

P T

V2 = 103 mL

P1= 400 atm V1= 250 mL T1= 283 K P2= 100 atm V2= T2 = 291 K

2

400 atm 291 KV = 250 mL x x

100 atm 283 K

1 1 2 2

1 2

P V P V=

T TBoth pressure

and temp change

Must use combined

gas law

36

AVOGADROrsquoSLAW

The relationship of moles and volume in gases is called Avogadrorsquos Law

At constant temperature and pressure the volume of a fixed amount of gas is directly proportional to the number of moles

As number of moles increases

the volume of the gas

increases

Number of moles and volume of a gas are directly

proportional

Tros Introductory Chemistry Chapter

37

Avogadrorsquos Law

38

AVOGADROrsquoSLAW

As a result of Avogadrorsquos Law equal volumes of different gases at the same temp and pressure contain equal number of moles (molecules)

2 Tanks of gas of equal volume at the same T amp P contain the same number of molecules

39

AVOGADROrsquoSLAW

For example

2 H2 (g) + 1 O2 (g) 2 H2O (g)

2 molecules 1 molecule 2 molecules

2 moles 1 mole 2 moles

2 Liters 1 Liter 2 Liters

Avogadrorsquos Law also allows chemists to relate volumes and moles of a gas in a chemical reaction

This relationship is only valid for

gaseous substances

40

Example 1

A sample of helium gas with a mass of 180 g occupies 16 Liters at a particular temperature and pressure What mass of oxygen would occupy16 L at the same temperature and pressure

1 mol= 180 g x

400 gmol of He = 450 mol

mol of O2 = mol of He = 450 mol

Same Temp amp Pressure

mass of O2

320 g= 450 mol x

1 mol= 144 g

41

Example 2

How many Liters of NH3 can be produced from reaction of 18 L of H2 with excess N2 as shown below

18 L H23

2

L NHx

L H= 12 L NH3

N2 (g) + 3 H2 (g) 2 NH3 (g)

32

42

STP amp MOLAR VOLUME

To better understand the factors that affect gas behavior a set of standard conditions have been chosen for use and are referred to as Standard Temperature and Pressure (STP)

STP =

1 atm (760 mmHg)

0ordmC (273 K)

43

STP amp MOLAR VOLUME

At STP conditions one mole of any gas is observed to occupy a volume of 224 L

V = 224 L

Molar Volume at

STP

44

Example 1

If 200 L of a gas at STP has a mass of 323 g what is the molar mass of the gas

mol200 L x

Lmol of gas =

2241

= 00893 mol

g

molMolar mass =

323 g=

00893 mol= 362 gmol Molar

volume at STP

45

Example 2

A sample of gas has a volume of 250 L at 730 mmHg and 20C What is the volume of this gas at STP

1 22 1

2 1

P TV = V x x

P T

V2 = 224 L

P1= 730 mmHg V1= 250 L T1= 293 K P2= 760 mmHg V2= T2 = 273 K

2

730 mmHg 273 KV = 250 L x x

760 mmHg 293 K

46

IDEAL GAS LAW

Combining all the laws that describe the behavior of gases one can obtain a useful relationship that relates the volume of a gas to the temperature pressure and number of moles

n R TV =

P

Universal gas constant

R = 00821 L atmmol K

47

IDEAL GAS LAW

This relationship is called the Ideal Gas Law and commonly written as

P V = n R T

Pressure in atm Volume

in Liters

Number of moles

Temp in K

48

Example 1

A sample of H2 gas has a volume of 856 L at a temp of 0C and pressure of 15 atm Calculate the moles of gas present

PVn =

RT

= 057 mol

P = 15 atm V = 856 L T = 273 K n = R = 00821

PV = nRT

(15 atm)(856 L)=

L atm(00821 )(273 K)

mol K

49

Example 2

What volume does 400 g of N2 gas occupy at 10C and 750 mmHg

P =750 mmHg V = T = 283 K n = not given m = 400 g R = 00821

mol of N2 =1 mol

400 g x280 g

= 143 mol

P = 750 mmHg1 atm

x760 mmHg

Moles must be calculated

from mass

Pressure must be

converted to atm

P = 0987 atm

50

Example 2

What volume does 400 g of N2 gas occupy at 10C and 750 mmHg

nRTV =

P

V = 337 L

P =0987 atm V = T = 283 K n = 143 R = 00821

L atm(143 mol)(00821 )(283 K)

mol KV = 0987 atm

Tros Introductory Chemistry Chapter

51

Ideal vs Real Gases

bull Real gases often do not behave like ideal gases at high pressure or low temperature

bull Ideal gas laws assume1) no attractions between gas molecules2) gas molecules do not take up space based on the Kinetic-Molecular Theory

bull at low temperatures and high pressures these assumptions are not valid

Tros Introductory Chemistry Chapter

52

Ideal vs Real Gases

53

PARTIAL PRESSURES

Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases

Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture

In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas

54

DALTONrsquoSLAW

In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture

This is called Daltonrsquos law of partial pressures

Total pressure of a gas mixture

Ptotal = P1 + P2 + P3 +

Sum of the partial pressures of the gases in the mixture

=

55

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = PHe + PAr

Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture

= 60 atm

56

PARTIAL PRESSURES

The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture

For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container

57

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = 60 atm

The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture

Twice as many moles of Ar

compared to He

PAr = 2 PHe

58

Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank

Ptotal = Poxygen + Phelium

Poxygen = 1140 mmHg x

1 atm

760 mmHg= 150 atm

Phelium

= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm

59

Example 2

A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture

Ptotal = Poxygen + Pnitrogen

Pnitrogen = 2 Poxygen

Twice as many moles of N2

compared to O2

Pnitrogen = 23 (Ptotal ) = 20 atm

Poxygen = 13 (Ptotal ) = 10 atm

60

THE END

• Slide 1
• CHAPTER OUTLINE
• PROPERTIES OF GASES
• PROPERTIES OF GASES (2)
• KINETIC-MOLECULAR THEORY
• KINETIC-MOLECULAR THEORY (2)
• Kinetic Molecular Theory
• PRESSURE amp ITS MEASUREMENT
• Common Units of Pressure
• PRESSURE amp ITS MEASUREMENT (2)
• Atmospheric Pressure amp Altitude
• Pressure Imbalance in Ear
• Example 1
• Example 2
• PRESSURE amp MOLES OF A GAS
• BOYLErsquoS LAW
• BOYLErsquoS LAW (2)
• Example 1 (2)
• Example 2 (2)
• CHARLESrsquoS LAW
• CHARLESrsquoS LAW (2)
• Example 1 (3)
• Example 2 (3)
• GAY-LUSSACrsquoS LAW
• GAY-LUSSACrsquoS LAW (2)
• Example 1 (4)
• Example 1 (5)
• Example 2 (4)
• VAPOR PRESSURE amp BOILING POINT
• VAPOR PRESSURE amp BOILING POINT (2)
• VAPOR PRESSURE amp BOILING POINT (3)
• VAPOR PRESSURE amp BOILING POINT (4)
• COMBINED GAS LAW
• COMBINED GAS LAW (2)
• Example 1 (6)
• AVOGADROrsquoS LAW
• Avogadrorsquos Law
• AVOGADROrsquoS LAW (2)
• AVOGADROrsquoS LAW (3)
• Example 1 (7)
• Example 2 (5)
• STP amp MOLAR VOLUME
• STP amp MOLAR VOLUME (2)
• Example 1 (8)
• Example 2 (6)
• IDEAL GAS LAW
• IDEAL GAS LAW (2)
• Example 1 (9)
• Example 2 (7)
• Example 2 (8)
• Ideal vs Real Gases
• Ideal vs Real Gases (2)
• PARTIAL PRESSURES
• DALTONrsquoS LAW
• PARTIAL PRESSURES (2)
• PARTIAL PRESSURES (3)
• PARTIAL PRESSURES (4)
• Example 1 (10)
• Example 2 (9)
• Slide 60

25

GAY-LUSSACrsquoSLAW

Gay-Lussacrsquos Law can be mathematically expressed as

P and T under initial

condition

1 2

1 2

P P=

T T

P and T under final conditionT must be in

units of K

26

Example 1

An aerosol spray can has a pressure of 40 atm at 25C What pressure will the can have if it is placed in a fire and reaches temperature of 400C

P1 = 40 atm

P2 =

T1 = 25ordmC

T2 = 400ordmC

T1= 25ordmC + 273 = 298 K

T2= 400ordmC + 273 = 673 K

Temperatures must be in K

27

Example 1

An aerosol spray can has a pressure of 40 atm at 25C What pressure will the can have if it is placed in a fire and reaches temperature of 400C

22 1

1

TP = P x

T

P2 = 90 atm

P1 = 40 atm

P2 =

T1 = 298 K

T2 = 673 K

673 K= 40 L x

298 K

Pressure increases

1 2

1 2

P P=

T T

Temperature increases

28

Example 2

A cylinder of gas with a volume of 150-L and a pressure of 965 mmHg is stored at a temperature of 55C To what temperature must the cylinder be cooled to reach a pressure of 850 mmHg

22 1

1

PT = T x

P

T2 = 289 K

P1 = 965 mmHg

P2 = 850 mmHg

T1 = 55ordmC

T2 =

850 mmHg= 328 K x

965 mmHg

T1= 55ordmC + 273 = 328 K

T2 = 16ordmC

Temperature decreases

Pressure decreases

29

VAPOR PRESSUREamp BOILING POINT

In an open container liquid molecules at the surface that possess sufficient energy can break away from the surface and become gas particles or vapor

In a closed container these gas particles can accumulate and create pressure called vapor pressure

Vapor pressure is defined as the pressure above a liquid at a given temperature Vapor pressure varies with each liquid and increases with temperature

30

VAPOR PRESSUREamp BOILING POINT

Listed below is the vapor pressure of water at various temperatures

31

VAPOR PRESSUREamp BOILING POINT

A liquid reaches its boiling point when its vapor pressure becomes equal to the external pressure (atmospheric pressure)

For example at sea level water reaches its boiling point at 100C since its vapor pressure is 760 mmHg at this temperature

32

VAPOR PRESSUREamp BOILING POINT

At higher altitudes where atmospheric pressure is lower water reaches boiling point at temperatures lower than 100C

For example in Denver where atmospheric pressure is 630 mmHg water boils at 95C since its vapor pressure is 630 mmHg at this temperature

33

COMBINEDGAS LAW

All pressure-volume-temperature relationships can be combined into a single relationship called the combined gas law

1 1 2 2

1 2

P V P V=

T T

Initial condition

This law is useful for studying the effect of changes in two variables

Final condition

34

COMBINEDGAS LAW

The individual gas laws studied previously are embodied in the combined gas law

1 1 2 2

1 2

P V P V=

T T

Boylersquos LawCharlesrsquos

LawGay-Lussacrsquos

Law

35

Example 1

A 250-mL sample of gas has a pressure of 400 atm at a temperature of 10C What is the volume of the gas at a pressure of 100 atm and a temperature of 18C

1 22 1

2 1

P TV = V x x

P T

V2 = 103 mL

P1= 400 atm V1= 250 mL T1= 283 K P2= 100 atm V2= T2 = 291 K

2

400 atm 291 KV = 250 mL x x

100 atm 283 K

1 1 2 2

1 2

P V P V=

T TBoth pressure

and temp change

Must use combined

gas law

36

AVOGADROrsquoSLAW

The relationship of moles and volume in gases is called Avogadrorsquos Law

At constant temperature and pressure the volume of a fixed amount of gas is directly proportional to the number of moles

As number of moles increases

the volume of the gas

increases

Number of moles and volume of a gas are directly

proportional

Tros Introductory Chemistry Chapter

37

Avogadrorsquos Law

38

AVOGADROrsquoSLAW

As a result of Avogadrorsquos Law equal volumes of different gases at the same temp and pressure contain equal number of moles (molecules)

2 Tanks of gas of equal volume at the same T amp P contain the same number of molecules

39

AVOGADROrsquoSLAW

For example

2 H2 (g) + 1 O2 (g) 2 H2O (g)

2 molecules 1 molecule 2 molecules

2 moles 1 mole 2 moles

2 Liters 1 Liter 2 Liters

Avogadrorsquos Law also allows chemists to relate volumes and moles of a gas in a chemical reaction

This relationship is only valid for

gaseous substances

40

Example 1

A sample of helium gas with a mass of 180 g occupies 16 Liters at a particular temperature and pressure What mass of oxygen would occupy16 L at the same temperature and pressure

1 mol= 180 g x

400 gmol of He = 450 mol

mol of O2 = mol of He = 450 mol

Same Temp amp Pressure

mass of O2

320 g= 450 mol x

1 mol= 144 g

41

Example 2

How many Liters of NH3 can be produced from reaction of 18 L of H2 with excess N2 as shown below

18 L H23

2

L NHx

L H= 12 L NH3

N2 (g) + 3 H2 (g) 2 NH3 (g)

32

42

STP amp MOLAR VOLUME

To better understand the factors that affect gas behavior a set of standard conditions have been chosen for use and are referred to as Standard Temperature and Pressure (STP)

STP =

1 atm (760 mmHg)

0ordmC (273 K)

43

STP amp MOLAR VOLUME

At STP conditions one mole of any gas is observed to occupy a volume of 224 L

V = 224 L

Molar Volume at

STP

44

Example 1

If 200 L of a gas at STP has a mass of 323 g what is the molar mass of the gas

mol200 L x

Lmol of gas =

2241

= 00893 mol

g

molMolar mass =

323 g=

00893 mol= 362 gmol Molar

volume at STP

45

Example 2

A sample of gas has a volume of 250 L at 730 mmHg and 20C What is the volume of this gas at STP

1 22 1

2 1

P TV = V x x

P T

V2 = 224 L

P1= 730 mmHg V1= 250 L T1= 293 K P2= 760 mmHg V2= T2 = 273 K

2

730 mmHg 273 KV = 250 L x x

760 mmHg 293 K

46

IDEAL GAS LAW

Combining all the laws that describe the behavior of gases one can obtain a useful relationship that relates the volume of a gas to the temperature pressure and number of moles

n R TV =

P

Universal gas constant

R = 00821 L atmmol K

47

IDEAL GAS LAW

This relationship is called the Ideal Gas Law and commonly written as

P V = n R T

Pressure in atm Volume

in Liters

Number of moles

Temp in K

48

Example 1

A sample of H2 gas has a volume of 856 L at a temp of 0C and pressure of 15 atm Calculate the moles of gas present

PVn =

RT

= 057 mol

P = 15 atm V = 856 L T = 273 K n = R = 00821

PV = nRT

(15 atm)(856 L)=

L atm(00821 )(273 K)

mol K

49

Example 2

What volume does 400 g of N2 gas occupy at 10C and 750 mmHg

P =750 mmHg V = T = 283 K n = not given m = 400 g R = 00821

mol of N2 =1 mol

400 g x280 g

= 143 mol

P = 750 mmHg1 atm

x760 mmHg

Moles must be calculated

from mass

Pressure must be

converted to atm

P = 0987 atm

50

Example 2

What volume does 400 g of N2 gas occupy at 10C and 750 mmHg

nRTV =

P

V = 337 L

P =0987 atm V = T = 283 K n = 143 R = 00821

L atm(143 mol)(00821 )(283 K)

mol KV = 0987 atm

Tros Introductory Chemistry Chapter

51

Ideal vs Real Gases

bull Real gases often do not behave like ideal gases at high pressure or low temperature

bull Ideal gas laws assume1) no attractions between gas molecules2) gas molecules do not take up space based on the Kinetic-Molecular Theory

bull at low temperatures and high pressures these assumptions are not valid

Tros Introductory Chemistry Chapter

52

Ideal vs Real Gases

53

PARTIAL PRESSURES

Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases

Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture

In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas

54

DALTONrsquoSLAW

In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture

This is called Daltonrsquos law of partial pressures

Total pressure of a gas mixture

Ptotal = P1 + P2 + P3 +

Sum of the partial pressures of the gases in the mixture

=

55

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = PHe + PAr

Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture

= 60 atm

56

PARTIAL PRESSURES

The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture

For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container

57

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = 60 atm

The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture

Twice as many moles of Ar

compared to He

PAr = 2 PHe

58

Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank

Ptotal = Poxygen + Phelium

Poxygen = 1140 mmHg x

1 atm

760 mmHg= 150 atm

Phelium

= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm

59

Example 2

A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture

Ptotal = Poxygen + Pnitrogen

Pnitrogen = 2 Poxygen

Twice as many moles of N2

compared to O2

Pnitrogen = 23 (Ptotal ) = 20 atm

Poxygen = 13 (Ptotal ) = 10 atm

60

THE END

• Slide 1
• CHAPTER OUTLINE
• PROPERTIES OF GASES
• PROPERTIES OF GASES (2)
• KINETIC-MOLECULAR THEORY
• KINETIC-MOLECULAR THEORY (2)
• Kinetic Molecular Theory
• PRESSURE amp ITS MEASUREMENT
• Common Units of Pressure
• PRESSURE amp ITS MEASUREMENT (2)
• Atmospheric Pressure amp Altitude
• Pressure Imbalance in Ear
• Example 1
• Example 2
• PRESSURE amp MOLES OF A GAS
• BOYLErsquoS LAW
• BOYLErsquoS LAW (2)
• Example 1 (2)
• Example 2 (2)
• CHARLESrsquoS LAW
• CHARLESrsquoS LAW (2)
• Example 1 (3)
• Example 2 (3)
• GAY-LUSSACrsquoS LAW
• GAY-LUSSACrsquoS LAW (2)
• Example 1 (4)
• Example 1 (5)
• Example 2 (4)
• VAPOR PRESSURE amp BOILING POINT
• VAPOR PRESSURE amp BOILING POINT (2)
• VAPOR PRESSURE amp BOILING POINT (3)
• VAPOR PRESSURE amp BOILING POINT (4)
• COMBINED GAS LAW
• COMBINED GAS LAW (2)
• Example 1 (6)
• AVOGADROrsquoS LAW
• Avogadrorsquos Law
• AVOGADROrsquoS LAW (2)
• AVOGADROrsquoS LAW (3)
• Example 1 (7)
• Example 2 (5)
• STP amp MOLAR VOLUME
• STP amp MOLAR VOLUME (2)
• Example 1 (8)
• Example 2 (6)
• IDEAL GAS LAW
• IDEAL GAS LAW (2)
• Example 1 (9)
• Example 2 (7)
• Example 2 (8)
• Ideal vs Real Gases
• Ideal vs Real Gases (2)
• PARTIAL PRESSURES
• DALTONrsquoS LAW
• PARTIAL PRESSURES (2)
• PARTIAL PRESSURES (3)
• PARTIAL PRESSURES (4)
• Example 1 (10)
• Example 2 (9)
• Slide 60

26

Example 1

An aerosol spray can has a pressure of 40 atm at 25C What pressure will the can have if it is placed in a fire and reaches temperature of 400C

P1 = 40 atm

P2 =

T1 = 25ordmC

T2 = 400ordmC

T1= 25ordmC + 273 = 298 K

T2= 400ordmC + 273 = 673 K

Temperatures must be in K

27

Example 1

An aerosol spray can has a pressure of 40 atm at 25C What pressure will the can have if it is placed in a fire and reaches temperature of 400C

22 1

1

TP = P x

T

P2 = 90 atm

P1 = 40 atm

P2 =

T1 = 298 K

T2 = 673 K

673 K= 40 L x

298 K

Pressure increases

1 2

1 2

P P=

T T

Temperature increases

28

Example 2

A cylinder of gas with a volume of 150-L and a pressure of 965 mmHg is stored at a temperature of 55C To what temperature must the cylinder be cooled to reach a pressure of 850 mmHg

22 1

1

PT = T x

P

T2 = 289 K

P1 = 965 mmHg

P2 = 850 mmHg

T1 = 55ordmC

T2 =

850 mmHg= 328 K x

965 mmHg

T1= 55ordmC + 273 = 328 K

T2 = 16ordmC

Temperature decreases

Pressure decreases

29

VAPOR PRESSUREamp BOILING POINT

In an open container liquid molecules at the surface that possess sufficient energy can break away from the surface and become gas particles or vapor

In a closed container these gas particles can accumulate and create pressure called vapor pressure

Vapor pressure is defined as the pressure above a liquid at a given temperature Vapor pressure varies with each liquid and increases with temperature

30

VAPOR PRESSUREamp BOILING POINT

Listed below is the vapor pressure of water at various temperatures

31

VAPOR PRESSUREamp BOILING POINT

A liquid reaches its boiling point when its vapor pressure becomes equal to the external pressure (atmospheric pressure)

For example at sea level water reaches its boiling point at 100C since its vapor pressure is 760 mmHg at this temperature

32

VAPOR PRESSUREamp BOILING POINT

At higher altitudes where atmospheric pressure is lower water reaches boiling point at temperatures lower than 100C

For example in Denver where atmospheric pressure is 630 mmHg water boils at 95C since its vapor pressure is 630 mmHg at this temperature

33

COMBINEDGAS LAW

All pressure-volume-temperature relationships can be combined into a single relationship called the combined gas law

1 1 2 2

1 2

P V P V=

T T

Initial condition

This law is useful for studying the effect of changes in two variables

Final condition

34

COMBINEDGAS LAW

The individual gas laws studied previously are embodied in the combined gas law

1 1 2 2

1 2

P V P V=

T T

Boylersquos LawCharlesrsquos

LawGay-Lussacrsquos

Law

35

Example 1

A 250-mL sample of gas has a pressure of 400 atm at a temperature of 10C What is the volume of the gas at a pressure of 100 atm and a temperature of 18C

1 22 1

2 1

P TV = V x x

P T

V2 = 103 mL

P1= 400 atm V1= 250 mL T1= 283 K P2= 100 atm V2= T2 = 291 K

2

400 atm 291 KV = 250 mL x x

100 atm 283 K

1 1 2 2

1 2

P V P V=

T TBoth pressure

and temp change

Must use combined

gas law

36

AVOGADROrsquoSLAW

The relationship of moles and volume in gases is called Avogadrorsquos Law

At constant temperature and pressure the volume of a fixed amount of gas is directly proportional to the number of moles

As number of moles increases

the volume of the gas

increases

Number of moles and volume of a gas are directly

proportional

Tros Introductory Chemistry Chapter

37

Avogadrorsquos Law

38

AVOGADROrsquoSLAW

As a result of Avogadrorsquos Law equal volumes of different gases at the same temp and pressure contain equal number of moles (molecules)

2 Tanks of gas of equal volume at the same T amp P contain the same number of molecules

39

AVOGADROrsquoSLAW

For example

2 H2 (g) + 1 O2 (g) 2 H2O (g)

2 molecules 1 molecule 2 molecules

2 moles 1 mole 2 moles

2 Liters 1 Liter 2 Liters

Avogadrorsquos Law also allows chemists to relate volumes and moles of a gas in a chemical reaction

This relationship is only valid for

gaseous substances

40

Example 1

A sample of helium gas with a mass of 180 g occupies 16 Liters at a particular temperature and pressure What mass of oxygen would occupy16 L at the same temperature and pressure

1 mol= 180 g x

400 gmol of He = 450 mol

mol of O2 = mol of He = 450 mol

Same Temp amp Pressure

mass of O2

320 g= 450 mol x

1 mol= 144 g

41

Example 2

How many Liters of NH3 can be produced from reaction of 18 L of H2 with excess N2 as shown below

18 L H23

2

L NHx

L H= 12 L NH3

N2 (g) + 3 H2 (g) 2 NH3 (g)

32

42

STP amp MOLAR VOLUME

To better understand the factors that affect gas behavior a set of standard conditions have been chosen for use and are referred to as Standard Temperature and Pressure (STP)

STP =

1 atm (760 mmHg)

0ordmC (273 K)

43

STP amp MOLAR VOLUME

At STP conditions one mole of any gas is observed to occupy a volume of 224 L

V = 224 L

Molar Volume at

STP

44

Example 1

If 200 L of a gas at STP has a mass of 323 g what is the molar mass of the gas

mol200 L x

Lmol of gas =

2241

= 00893 mol

g

molMolar mass =

323 g=

00893 mol= 362 gmol Molar

volume at STP

45

Example 2

A sample of gas has a volume of 250 L at 730 mmHg and 20C What is the volume of this gas at STP

1 22 1

2 1

P TV = V x x

P T

V2 = 224 L

P1= 730 mmHg V1= 250 L T1= 293 K P2= 760 mmHg V2= T2 = 273 K

2

730 mmHg 273 KV = 250 L x x

760 mmHg 293 K

46

IDEAL GAS LAW

Combining all the laws that describe the behavior of gases one can obtain a useful relationship that relates the volume of a gas to the temperature pressure and number of moles

n R TV =

P

Universal gas constant

R = 00821 L atmmol K

47

IDEAL GAS LAW

This relationship is called the Ideal Gas Law and commonly written as

P V = n R T

Pressure in atm Volume

in Liters

Number of moles

Temp in K

48

Example 1

A sample of H2 gas has a volume of 856 L at a temp of 0C and pressure of 15 atm Calculate the moles of gas present

PVn =

RT

= 057 mol

P = 15 atm V = 856 L T = 273 K n = R = 00821

PV = nRT

(15 atm)(856 L)=

L atm(00821 )(273 K)

mol K

49

Example 2

What volume does 400 g of N2 gas occupy at 10C and 750 mmHg

P =750 mmHg V = T = 283 K n = not given m = 400 g R = 00821

mol of N2 =1 mol

400 g x280 g

= 143 mol

P = 750 mmHg1 atm

x760 mmHg

Moles must be calculated

from mass

Pressure must be

converted to atm

P = 0987 atm

50

Example 2

What volume does 400 g of N2 gas occupy at 10C and 750 mmHg

nRTV =

P

V = 337 L

P =0987 atm V = T = 283 K n = 143 R = 00821

L atm(143 mol)(00821 )(283 K)

mol KV = 0987 atm

Tros Introductory Chemistry Chapter

51

Ideal vs Real Gases

bull Real gases often do not behave like ideal gases at high pressure or low temperature

bull Ideal gas laws assume1) no attractions between gas molecules2) gas molecules do not take up space based on the Kinetic-Molecular Theory

bull at low temperatures and high pressures these assumptions are not valid

Tros Introductory Chemistry Chapter

52

Ideal vs Real Gases

53

PARTIAL PRESSURES

Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases

Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture

In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas

54

DALTONrsquoSLAW

In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture

This is called Daltonrsquos law of partial pressures

Total pressure of a gas mixture

Ptotal = P1 + P2 + P3 +

Sum of the partial pressures of the gases in the mixture

=

55

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = PHe + PAr

Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture

= 60 atm

56

PARTIAL PRESSURES

The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture

For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container

57

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = 60 atm

The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture

Twice as many moles of Ar

compared to He

PAr = 2 PHe

58

Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank

Ptotal = Poxygen + Phelium

Poxygen = 1140 mmHg x

1 atm

760 mmHg= 150 atm

Phelium

= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm

59

Example 2

A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture

Ptotal = Poxygen + Pnitrogen

Pnitrogen = 2 Poxygen

Twice as many moles of N2

compared to O2

Pnitrogen = 23 (Ptotal ) = 20 atm

Poxygen = 13 (Ptotal ) = 10 atm

60

THE END

• Slide 1
• CHAPTER OUTLINE
• PROPERTIES OF GASES
• PROPERTIES OF GASES (2)
• KINETIC-MOLECULAR THEORY
• KINETIC-MOLECULAR THEORY (2)
• Kinetic Molecular Theory
• PRESSURE amp ITS MEASUREMENT
• Common Units of Pressure
• PRESSURE amp ITS MEASUREMENT (2)
• Atmospheric Pressure amp Altitude
• Pressure Imbalance in Ear
• Example 1
• Example 2
• PRESSURE amp MOLES OF A GAS
• BOYLErsquoS LAW
• BOYLErsquoS LAW (2)
• Example 1 (2)
• Example 2 (2)
• CHARLESrsquoS LAW
• CHARLESrsquoS LAW (2)
• Example 1 (3)
• Example 2 (3)
• GAY-LUSSACrsquoS LAW
• GAY-LUSSACrsquoS LAW (2)
• Example 1 (4)
• Example 1 (5)
• Example 2 (4)
• VAPOR PRESSURE amp BOILING POINT
• VAPOR PRESSURE amp BOILING POINT (2)
• VAPOR PRESSURE amp BOILING POINT (3)
• VAPOR PRESSURE amp BOILING POINT (4)
• COMBINED GAS LAW
• COMBINED GAS LAW (2)
• Example 1 (6)
• AVOGADROrsquoS LAW
• Avogadrorsquos Law
• AVOGADROrsquoS LAW (2)
• AVOGADROrsquoS LAW (3)
• Example 1 (7)
• Example 2 (5)
• STP amp MOLAR VOLUME
• STP amp MOLAR VOLUME (2)
• Example 1 (8)
• Example 2 (6)
• IDEAL GAS LAW
• IDEAL GAS LAW (2)
• Example 1 (9)
• Example 2 (7)
• Example 2 (8)
• Ideal vs Real Gases
• Ideal vs Real Gases (2)
• PARTIAL PRESSURES
• DALTONrsquoS LAW
• PARTIAL PRESSURES (2)
• PARTIAL PRESSURES (3)
• PARTIAL PRESSURES (4)
• Example 1 (10)
• Example 2 (9)
• Slide 60

27

Example 1

An aerosol spray can has a pressure of 40 atm at 25C What pressure will the can have if it is placed in a fire and reaches temperature of 400C

22 1

1

TP = P x

T

P2 = 90 atm

P1 = 40 atm

P2 =

T1 = 298 K

T2 = 673 K

673 K= 40 L x

298 K

Pressure increases

1 2

1 2

P P=

T T

Temperature increases

28

Example 2

A cylinder of gas with a volume of 150-L and a pressure of 965 mmHg is stored at a temperature of 55C To what temperature must the cylinder be cooled to reach a pressure of 850 mmHg

22 1

1

PT = T x

P

T2 = 289 K

P1 = 965 mmHg

P2 = 850 mmHg

T1 = 55ordmC

T2 =

850 mmHg= 328 K x

965 mmHg

T1= 55ordmC + 273 = 328 K

T2 = 16ordmC

Temperature decreases

Pressure decreases

29

VAPOR PRESSUREamp BOILING POINT

In an open container liquid molecules at the surface that possess sufficient energy can break away from the surface and become gas particles or vapor

In a closed container these gas particles can accumulate and create pressure called vapor pressure

Vapor pressure is defined as the pressure above a liquid at a given temperature Vapor pressure varies with each liquid and increases with temperature

30

VAPOR PRESSUREamp BOILING POINT

Listed below is the vapor pressure of water at various temperatures

31

VAPOR PRESSUREamp BOILING POINT

A liquid reaches its boiling point when its vapor pressure becomes equal to the external pressure (atmospheric pressure)

For example at sea level water reaches its boiling point at 100C since its vapor pressure is 760 mmHg at this temperature

32

VAPOR PRESSUREamp BOILING POINT

At higher altitudes where atmospheric pressure is lower water reaches boiling point at temperatures lower than 100C

For example in Denver where atmospheric pressure is 630 mmHg water boils at 95C since its vapor pressure is 630 mmHg at this temperature

33

COMBINEDGAS LAW

All pressure-volume-temperature relationships can be combined into a single relationship called the combined gas law

1 1 2 2

1 2

P V P V=

T T

Initial condition

This law is useful for studying the effect of changes in two variables

Final condition

34

COMBINEDGAS LAW

The individual gas laws studied previously are embodied in the combined gas law

1 1 2 2

1 2

P V P V=

T T

Boylersquos LawCharlesrsquos

LawGay-Lussacrsquos

Law

35

Example 1

A 250-mL sample of gas has a pressure of 400 atm at a temperature of 10C What is the volume of the gas at a pressure of 100 atm and a temperature of 18C

1 22 1

2 1

P TV = V x x

P T

V2 = 103 mL

P1= 400 atm V1= 250 mL T1= 283 K P2= 100 atm V2= T2 = 291 K

2

400 atm 291 KV = 250 mL x x

100 atm 283 K

1 1 2 2

1 2

P V P V=

T TBoth pressure

and temp change

Must use combined

gas law

36

AVOGADROrsquoSLAW

The relationship of moles and volume in gases is called Avogadrorsquos Law

At constant temperature and pressure the volume of a fixed amount of gas is directly proportional to the number of moles

As number of moles increases

the volume of the gas

increases

Number of moles and volume of a gas are directly

proportional

Tros Introductory Chemistry Chapter

37

Avogadrorsquos Law

38

AVOGADROrsquoSLAW

As a result of Avogadrorsquos Law equal volumes of different gases at the same temp and pressure contain equal number of moles (molecules)

2 Tanks of gas of equal volume at the same T amp P contain the same number of molecules

39

AVOGADROrsquoSLAW

For example

2 H2 (g) + 1 O2 (g) 2 H2O (g)

2 molecules 1 molecule 2 molecules

2 moles 1 mole 2 moles

2 Liters 1 Liter 2 Liters

Avogadrorsquos Law also allows chemists to relate volumes and moles of a gas in a chemical reaction

This relationship is only valid for

gaseous substances

40

Example 1

A sample of helium gas with a mass of 180 g occupies 16 Liters at a particular temperature and pressure What mass of oxygen would occupy16 L at the same temperature and pressure

1 mol= 180 g x

400 gmol of He = 450 mol

mol of O2 = mol of He = 450 mol

Same Temp amp Pressure

mass of O2

320 g= 450 mol x

1 mol= 144 g

41

Example 2

How many Liters of NH3 can be produced from reaction of 18 L of H2 with excess N2 as shown below

18 L H23

2

L NHx

L H= 12 L NH3

N2 (g) + 3 H2 (g) 2 NH3 (g)

32

42

STP amp MOLAR VOLUME

To better understand the factors that affect gas behavior a set of standard conditions have been chosen for use and are referred to as Standard Temperature and Pressure (STP)

STP =

1 atm (760 mmHg)

0ordmC (273 K)

43

STP amp MOLAR VOLUME

At STP conditions one mole of any gas is observed to occupy a volume of 224 L

V = 224 L

Molar Volume at

STP

44

Example 1

If 200 L of a gas at STP has a mass of 323 g what is the molar mass of the gas

mol200 L x

Lmol of gas =

2241

= 00893 mol

g

molMolar mass =

323 g=

00893 mol= 362 gmol Molar

volume at STP

45

Example 2

A sample of gas has a volume of 250 L at 730 mmHg and 20C What is the volume of this gas at STP

1 22 1

2 1

P TV = V x x

P T

V2 = 224 L

P1= 730 mmHg V1= 250 L T1= 293 K P2= 760 mmHg V2= T2 = 273 K

2

730 mmHg 273 KV = 250 L x x

760 mmHg 293 K

46

IDEAL GAS LAW

Combining all the laws that describe the behavior of gases one can obtain a useful relationship that relates the volume of a gas to the temperature pressure and number of moles

n R TV =

P

Universal gas constant

R = 00821 L atmmol K

47

IDEAL GAS LAW

This relationship is called the Ideal Gas Law and commonly written as

P V = n R T

Pressure in atm Volume

in Liters

Number of moles

Temp in K

48

Example 1

A sample of H2 gas has a volume of 856 L at a temp of 0C and pressure of 15 atm Calculate the moles of gas present

PVn =

RT

= 057 mol

P = 15 atm V = 856 L T = 273 K n = R = 00821

PV = nRT

(15 atm)(856 L)=

L atm(00821 )(273 K)

mol K

49

Example 2

What volume does 400 g of N2 gas occupy at 10C and 750 mmHg

P =750 mmHg V = T = 283 K n = not given m = 400 g R = 00821

mol of N2 =1 mol

400 g x280 g

= 143 mol

P = 750 mmHg1 atm

x760 mmHg

Moles must be calculated

from mass

Pressure must be

converted to atm

P = 0987 atm

50

Example 2

What volume does 400 g of N2 gas occupy at 10C and 750 mmHg

nRTV =

P

V = 337 L

P =0987 atm V = T = 283 K n = 143 R = 00821

L atm(143 mol)(00821 )(283 K)

mol KV = 0987 atm

Tros Introductory Chemistry Chapter

51

Ideal vs Real Gases

bull Real gases often do not behave like ideal gases at high pressure or low temperature

bull Ideal gas laws assume1) no attractions between gas molecules2) gas molecules do not take up space based on the Kinetic-Molecular Theory

bull at low temperatures and high pressures these assumptions are not valid

Tros Introductory Chemistry Chapter

52

Ideal vs Real Gases

53

PARTIAL PRESSURES

Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases

Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture

In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas

54

DALTONrsquoSLAW

In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture

This is called Daltonrsquos law of partial pressures

Total pressure of a gas mixture

Ptotal = P1 + P2 + P3 +

Sum of the partial pressures of the gases in the mixture

=

55

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = PHe + PAr

Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture

= 60 atm

56

PARTIAL PRESSURES

The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture

For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container

57

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = 60 atm

The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture

Twice as many moles of Ar

compared to He

PAr = 2 PHe

58

Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank

Ptotal = Poxygen + Phelium

Poxygen = 1140 mmHg x

1 atm

760 mmHg= 150 atm

Phelium

= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm

59

Example 2

A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture

Ptotal = Poxygen + Pnitrogen

Pnitrogen = 2 Poxygen

Twice as many moles of N2

compared to O2

Pnitrogen = 23 (Ptotal ) = 20 atm

Poxygen = 13 (Ptotal ) = 10 atm

60

THE END

• Slide 1
• CHAPTER OUTLINE
• PROPERTIES OF GASES
• PROPERTIES OF GASES (2)
• KINETIC-MOLECULAR THEORY
• KINETIC-MOLECULAR THEORY (2)
• Kinetic Molecular Theory
• PRESSURE amp ITS MEASUREMENT
• Common Units of Pressure
• PRESSURE amp ITS MEASUREMENT (2)
• Atmospheric Pressure amp Altitude
• Pressure Imbalance in Ear
• Example 1
• Example 2
• PRESSURE amp MOLES OF A GAS
• BOYLErsquoS LAW
• BOYLErsquoS LAW (2)
• Example 1 (2)
• Example 2 (2)
• CHARLESrsquoS LAW
• CHARLESrsquoS LAW (2)
• Example 1 (3)
• Example 2 (3)
• GAY-LUSSACrsquoS LAW
• GAY-LUSSACrsquoS LAW (2)
• Example 1 (4)
• Example 1 (5)
• Example 2 (4)
• VAPOR PRESSURE amp BOILING POINT
• VAPOR PRESSURE amp BOILING POINT (2)
• VAPOR PRESSURE amp BOILING POINT (3)
• VAPOR PRESSURE amp BOILING POINT (4)
• COMBINED GAS LAW
• COMBINED GAS LAW (2)
• Example 1 (6)
• AVOGADROrsquoS LAW
• Avogadrorsquos Law
• AVOGADROrsquoS LAW (2)
• AVOGADROrsquoS LAW (3)
• Example 1 (7)
• Example 2 (5)
• STP amp MOLAR VOLUME
• STP amp MOLAR VOLUME (2)
• Example 1 (8)
• Example 2 (6)
• IDEAL GAS LAW
• IDEAL GAS LAW (2)
• Example 1 (9)
• Example 2 (7)
• Example 2 (8)
• Ideal vs Real Gases
• Ideal vs Real Gases (2)
• PARTIAL PRESSURES
• DALTONrsquoS LAW
• PARTIAL PRESSURES (2)
• PARTIAL PRESSURES (3)
• PARTIAL PRESSURES (4)
• Example 1 (10)
• Example 2 (9)
• Slide 60

28

Example 2

A cylinder of gas with a volume of 150-L and a pressure of 965 mmHg is stored at a temperature of 55C To what temperature must the cylinder be cooled to reach a pressure of 850 mmHg

22 1

1

PT = T x

P

T2 = 289 K

P1 = 965 mmHg

P2 = 850 mmHg

T1 = 55ordmC

T2 =

850 mmHg= 328 K x

965 mmHg

T1= 55ordmC + 273 = 328 K

T2 = 16ordmC

Temperature decreases

Pressure decreases

29

VAPOR PRESSUREamp BOILING POINT

In an open container liquid molecules at the surface that possess sufficient energy can break away from the surface and become gas particles or vapor

In a closed container these gas particles can accumulate and create pressure called vapor pressure

Vapor pressure is defined as the pressure above a liquid at a given temperature Vapor pressure varies with each liquid and increases with temperature

30

VAPOR PRESSUREamp BOILING POINT

Listed below is the vapor pressure of water at various temperatures

31

VAPOR PRESSUREamp BOILING POINT

A liquid reaches its boiling point when its vapor pressure becomes equal to the external pressure (atmospheric pressure)

For example at sea level water reaches its boiling point at 100C since its vapor pressure is 760 mmHg at this temperature

32

VAPOR PRESSUREamp BOILING POINT

At higher altitudes where atmospheric pressure is lower water reaches boiling point at temperatures lower than 100C

For example in Denver where atmospheric pressure is 630 mmHg water boils at 95C since its vapor pressure is 630 mmHg at this temperature

33

COMBINEDGAS LAW

All pressure-volume-temperature relationships can be combined into a single relationship called the combined gas law

1 1 2 2

1 2

P V P V=

T T

Initial condition

This law is useful for studying the effect of changes in two variables

Final condition

34

COMBINEDGAS LAW

The individual gas laws studied previously are embodied in the combined gas law

1 1 2 2

1 2

P V P V=

T T

Boylersquos LawCharlesrsquos

LawGay-Lussacrsquos

Law

35

Example 1

A 250-mL sample of gas has a pressure of 400 atm at a temperature of 10C What is the volume of the gas at a pressure of 100 atm and a temperature of 18C

1 22 1

2 1

P TV = V x x

P T

V2 = 103 mL

P1= 400 atm V1= 250 mL T1= 283 K P2= 100 atm V2= T2 = 291 K

2

400 atm 291 KV = 250 mL x x

100 atm 283 K

1 1 2 2

1 2

P V P V=

T TBoth pressure

and temp change

Must use combined

gas law

36

AVOGADROrsquoSLAW

The relationship of moles and volume in gases is called Avogadrorsquos Law

At constant temperature and pressure the volume of a fixed amount of gas is directly proportional to the number of moles

As number of moles increases

the volume of the gas

increases

Number of moles and volume of a gas are directly

proportional

Tros Introductory Chemistry Chapter

37

Avogadrorsquos Law

38

AVOGADROrsquoSLAW

As a result of Avogadrorsquos Law equal volumes of different gases at the same temp and pressure contain equal number of moles (molecules)

2 Tanks of gas of equal volume at the same T amp P contain the same number of molecules

39

AVOGADROrsquoSLAW

For example

2 H2 (g) + 1 O2 (g) 2 H2O (g)

2 molecules 1 molecule 2 molecules

2 moles 1 mole 2 moles

2 Liters 1 Liter 2 Liters

Avogadrorsquos Law also allows chemists to relate volumes and moles of a gas in a chemical reaction

This relationship is only valid for

gaseous substances

40

Example 1

A sample of helium gas with a mass of 180 g occupies 16 Liters at a particular temperature and pressure What mass of oxygen would occupy16 L at the same temperature and pressure

1 mol= 180 g x

400 gmol of He = 450 mol

mol of O2 = mol of He = 450 mol

Same Temp amp Pressure

mass of O2

320 g= 450 mol x

1 mol= 144 g

41

Example 2

How many Liters of NH3 can be produced from reaction of 18 L of H2 with excess N2 as shown below

18 L H23

2

L NHx

L H= 12 L NH3

N2 (g) + 3 H2 (g) 2 NH3 (g)

32

42

STP amp MOLAR VOLUME

To better understand the factors that affect gas behavior a set of standard conditions have been chosen for use and are referred to as Standard Temperature and Pressure (STP)

STP =

1 atm (760 mmHg)

0ordmC (273 K)

43

STP amp MOLAR VOLUME

At STP conditions one mole of any gas is observed to occupy a volume of 224 L

V = 224 L

Molar Volume at

STP

44

Example 1

If 200 L of a gas at STP has a mass of 323 g what is the molar mass of the gas

mol200 L x

Lmol of gas =

2241

= 00893 mol

g

molMolar mass =

323 g=

00893 mol= 362 gmol Molar

volume at STP

45

Example 2

A sample of gas has a volume of 250 L at 730 mmHg and 20C What is the volume of this gas at STP

1 22 1

2 1

P TV = V x x

P T

V2 = 224 L

P1= 730 mmHg V1= 250 L T1= 293 K P2= 760 mmHg V2= T2 = 273 K

2

730 mmHg 273 KV = 250 L x x

760 mmHg 293 K

46

IDEAL GAS LAW

Combining all the laws that describe the behavior of gases one can obtain a useful relationship that relates the volume of a gas to the temperature pressure and number of moles

n R TV =

P

Universal gas constant

R = 00821 L atmmol K

47

IDEAL GAS LAW

This relationship is called the Ideal Gas Law and commonly written as

P V = n R T

Pressure in atm Volume

in Liters

Number of moles

Temp in K

48

Example 1

A sample of H2 gas has a volume of 856 L at a temp of 0C and pressure of 15 atm Calculate the moles of gas present

PVn =

RT

= 057 mol

P = 15 atm V = 856 L T = 273 K n = R = 00821

PV = nRT

(15 atm)(856 L)=

L atm(00821 )(273 K)

mol K

49

Example 2

What volume does 400 g of N2 gas occupy at 10C and 750 mmHg

P =750 mmHg V = T = 283 K n = not given m = 400 g R = 00821

mol of N2 =1 mol

400 g x280 g

= 143 mol

P = 750 mmHg1 atm

x760 mmHg

Moles must be calculated

from mass

Pressure must be

converted to atm

P = 0987 atm

50

Example 2

What volume does 400 g of N2 gas occupy at 10C and 750 mmHg

nRTV =

P

V = 337 L

P =0987 atm V = T = 283 K n = 143 R = 00821

L atm(143 mol)(00821 )(283 K)

mol KV = 0987 atm

Tros Introductory Chemistry Chapter

51

Ideal vs Real Gases

bull Real gases often do not behave like ideal gases at high pressure or low temperature

bull Ideal gas laws assume1) no attractions between gas molecules2) gas molecules do not take up space based on the Kinetic-Molecular Theory

bull at low temperatures and high pressures these assumptions are not valid

Tros Introductory Chemistry Chapter

52

Ideal vs Real Gases

53

PARTIAL PRESSURES

Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases

Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture

In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas

54

DALTONrsquoSLAW

In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture

This is called Daltonrsquos law of partial pressures

Total pressure of a gas mixture

Ptotal = P1 + P2 + P3 +

Sum of the partial pressures of the gases in the mixture

=

55

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = PHe + PAr

Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture

= 60 atm

56

PARTIAL PRESSURES

The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture

For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container

57

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = 60 atm

The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture

Twice as many moles of Ar

compared to He

PAr = 2 PHe

58

Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank

Ptotal = Poxygen + Phelium

Poxygen = 1140 mmHg x

1 atm

760 mmHg= 150 atm

Phelium

= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm

59

Example 2

A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture

Ptotal = Poxygen + Pnitrogen

Pnitrogen = 2 Poxygen

Twice as many moles of N2

compared to O2

Pnitrogen = 23 (Ptotal ) = 20 atm

Poxygen = 13 (Ptotal ) = 10 atm

60

THE END

• Slide 1
• CHAPTER OUTLINE
• PROPERTIES OF GASES
• PROPERTIES OF GASES (2)
• KINETIC-MOLECULAR THEORY
• KINETIC-MOLECULAR THEORY (2)
• Kinetic Molecular Theory
• PRESSURE amp ITS MEASUREMENT
• Common Units of Pressure
• PRESSURE amp ITS MEASUREMENT (2)
• Atmospheric Pressure amp Altitude
• Pressure Imbalance in Ear
• Example 1
• Example 2
• PRESSURE amp MOLES OF A GAS
• BOYLErsquoS LAW
• BOYLErsquoS LAW (2)
• Example 1 (2)
• Example 2 (2)
• CHARLESrsquoS LAW
• CHARLESrsquoS LAW (2)
• Example 1 (3)
• Example 2 (3)
• GAY-LUSSACrsquoS LAW
• GAY-LUSSACrsquoS LAW (2)
• Example 1 (4)
• Example 1 (5)
• Example 2 (4)
• VAPOR PRESSURE amp BOILING POINT
• VAPOR PRESSURE amp BOILING POINT (2)
• VAPOR PRESSURE amp BOILING POINT (3)
• VAPOR PRESSURE amp BOILING POINT (4)
• COMBINED GAS LAW
• COMBINED GAS LAW (2)
• Example 1 (6)
• AVOGADROrsquoS LAW
• Avogadrorsquos Law
• AVOGADROrsquoS LAW (2)
• AVOGADROrsquoS LAW (3)
• Example 1 (7)
• Example 2 (5)
• STP amp MOLAR VOLUME
• STP amp MOLAR VOLUME (2)
• Example 1 (8)
• Example 2 (6)
• IDEAL GAS LAW
• IDEAL GAS LAW (2)
• Example 1 (9)
• Example 2 (7)
• Example 2 (8)
• Ideal vs Real Gases
• Ideal vs Real Gases (2)
• PARTIAL PRESSURES
• DALTONrsquoS LAW
• PARTIAL PRESSURES (2)
• PARTIAL PRESSURES (3)
• PARTIAL PRESSURES (4)
• Example 1 (10)
• Example 2 (9)
• Slide 60

29

VAPOR PRESSUREamp BOILING POINT

In an open container liquid molecules at the surface that possess sufficient energy can break away from the surface and become gas particles or vapor

In a closed container these gas particles can accumulate and create pressure called vapor pressure

Vapor pressure is defined as the pressure above a liquid at a given temperature Vapor pressure varies with each liquid and increases with temperature

30

VAPOR PRESSUREamp BOILING POINT

Listed below is the vapor pressure of water at various temperatures

31

VAPOR PRESSUREamp BOILING POINT

A liquid reaches its boiling point when its vapor pressure becomes equal to the external pressure (atmospheric pressure)

For example at sea level water reaches its boiling point at 100C since its vapor pressure is 760 mmHg at this temperature

32

VAPOR PRESSUREamp BOILING POINT

At higher altitudes where atmospheric pressure is lower water reaches boiling point at temperatures lower than 100C

For example in Denver where atmospheric pressure is 630 mmHg water boils at 95C since its vapor pressure is 630 mmHg at this temperature

33

COMBINEDGAS LAW

All pressure-volume-temperature relationships can be combined into a single relationship called the combined gas law

1 1 2 2

1 2

P V P V=

T T

Initial condition

This law is useful for studying the effect of changes in two variables

Final condition

34

COMBINEDGAS LAW

The individual gas laws studied previously are embodied in the combined gas law

1 1 2 2

1 2

P V P V=

T T

Boylersquos LawCharlesrsquos

LawGay-Lussacrsquos

Law

35

Example 1

A 250-mL sample of gas has a pressure of 400 atm at a temperature of 10C What is the volume of the gas at a pressure of 100 atm and a temperature of 18C

1 22 1

2 1

P TV = V x x

P T

V2 = 103 mL

P1= 400 atm V1= 250 mL T1= 283 K P2= 100 atm V2= T2 = 291 K

2

400 atm 291 KV = 250 mL x x

100 atm 283 K

1 1 2 2

1 2

P V P V=

T TBoth pressure

and temp change

Must use combined

gas law

36

AVOGADROrsquoSLAW

The relationship of moles and volume in gases is called Avogadrorsquos Law

At constant temperature and pressure the volume of a fixed amount of gas is directly proportional to the number of moles

As number of moles increases

the volume of the gas

increases

Number of moles and volume of a gas are directly

proportional

Tros Introductory Chemistry Chapter

37

Avogadrorsquos Law

38

AVOGADROrsquoSLAW

As a result of Avogadrorsquos Law equal volumes of different gases at the same temp and pressure contain equal number of moles (molecules)

2 Tanks of gas of equal volume at the same T amp P contain the same number of molecules

39

AVOGADROrsquoSLAW

For example

2 H2 (g) + 1 O2 (g) 2 H2O (g)

2 molecules 1 molecule 2 molecules

2 moles 1 mole 2 moles

2 Liters 1 Liter 2 Liters

Avogadrorsquos Law also allows chemists to relate volumes and moles of a gas in a chemical reaction

This relationship is only valid for

gaseous substances

40

Example 1

A sample of helium gas with a mass of 180 g occupies 16 Liters at a particular temperature and pressure What mass of oxygen would occupy16 L at the same temperature and pressure

1 mol= 180 g x

400 gmol of He = 450 mol

mol of O2 = mol of He = 450 mol

Same Temp amp Pressure

mass of O2

320 g= 450 mol x

1 mol= 144 g

41

Example 2

How many Liters of NH3 can be produced from reaction of 18 L of H2 with excess N2 as shown below

18 L H23

2

L NHx

L H= 12 L NH3

N2 (g) + 3 H2 (g) 2 NH3 (g)

32

42

STP amp MOLAR VOLUME

To better understand the factors that affect gas behavior a set of standard conditions have been chosen for use and are referred to as Standard Temperature and Pressure (STP)

STP =

1 atm (760 mmHg)

0ordmC (273 K)

43

STP amp MOLAR VOLUME

At STP conditions one mole of any gas is observed to occupy a volume of 224 L

V = 224 L

Molar Volume at

STP

44

Example 1

If 200 L of a gas at STP has a mass of 323 g what is the molar mass of the gas

mol200 L x

Lmol of gas =

2241

= 00893 mol

g

molMolar mass =

323 g=

00893 mol= 362 gmol Molar

volume at STP

45

Example 2

A sample of gas has a volume of 250 L at 730 mmHg and 20C What is the volume of this gas at STP

1 22 1

2 1

P TV = V x x

P T

V2 = 224 L

P1= 730 mmHg V1= 250 L T1= 293 K P2= 760 mmHg V2= T2 = 273 K

2

730 mmHg 273 KV = 250 L x x

760 mmHg 293 K

46

IDEAL GAS LAW

Combining all the laws that describe the behavior of gases one can obtain a useful relationship that relates the volume of a gas to the temperature pressure and number of moles

n R TV =

P

Universal gas constant

R = 00821 L atmmol K

47

IDEAL GAS LAW

This relationship is called the Ideal Gas Law and commonly written as

P V = n R T

Pressure in atm Volume

in Liters

Number of moles

Temp in K

48

Example 1

A sample of H2 gas has a volume of 856 L at a temp of 0C and pressure of 15 atm Calculate the moles of gas present

PVn =

RT

= 057 mol

P = 15 atm V = 856 L T = 273 K n = R = 00821

PV = nRT

(15 atm)(856 L)=

L atm(00821 )(273 K)

mol K

49

Example 2

What volume does 400 g of N2 gas occupy at 10C and 750 mmHg

P =750 mmHg V = T = 283 K n = not given m = 400 g R = 00821

mol of N2 =1 mol

400 g x280 g

= 143 mol

P = 750 mmHg1 atm

x760 mmHg

Moles must be calculated

from mass

Pressure must be

converted to atm

P = 0987 atm

50

Example 2

What volume does 400 g of N2 gas occupy at 10C and 750 mmHg

nRTV =

P

V = 337 L

P =0987 atm V = T = 283 K n = 143 R = 00821

L atm(143 mol)(00821 )(283 K)

mol KV = 0987 atm

Tros Introductory Chemistry Chapter

51

Ideal vs Real Gases

bull Real gases often do not behave like ideal gases at high pressure or low temperature

bull Ideal gas laws assume1) no attractions between gas molecules2) gas molecules do not take up space based on the Kinetic-Molecular Theory

bull at low temperatures and high pressures these assumptions are not valid

Tros Introductory Chemistry Chapter

52

Ideal vs Real Gases

53

PARTIAL PRESSURES

Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases

Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture

In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas

54

DALTONrsquoSLAW

In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture

This is called Daltonrsquos law of partial pressures

Total pressure of a gas mixture

Ptotal = P1 + P2 + P3 +

Sum of the partial pressures of the gases in the mixture

=

55

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = PHe + PAr

Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture

= 60 atm

56

PARTIAL PRESSURES

The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture

For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container

57

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = 60 atm

The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture

Twice as many moles of Ar

compared to He

PAr = 2 PHe

58

Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank

Ptotal = Poxygen + Phelium

Poxygen = 1140 mmHg x

1 atm

760 mmHg= 150 atm

Phelium

= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm

59

Example 2

A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture

Ptotal = Poxygen + Pnitrogen

Pnitrogen = 2 Poxygen

Twice as many moles of N2

compared to O2

Pnitrogen = 23 (Ptotal ) = 20 atm

Poxygen = 13 (Ptotal ) = 10 atm

60

THE END

• Slide 1
• CHAPTER OUTLINE
• PROPERTIES OF GASES
• PROPERTIES OF GASES (2)
• KINETIC-MOLECULAR THEORY
• KINETIC-MOLECULAR THEORY (2)
• Kinetic Molecular Theory
• PRESSURE amp ITS MEASUREMENT
• Common Units of Pressure
• PRESSURE amp ITS MEASUREMENT (2)
• Atmospheric Pressure amp Altitude
• Pressure Imbalance in Ear
• Example 1
• Example 2
• PRESSURE amp MOLES OF A GAS
• BOYLErsquoS LAW
• BOYLErsquoS LAW (2)
• Example 1 (2)
• Example 2 (2)
• CHARLESrsquoS LAW
• CHARLESrsquoS LAW (2)
• Example 1 (3)
• Example 2 (3)
• GAY-LUSSACrsquoS LAW
• GAY-LUSSACrsquoS LAW (2)
• Example 1 (4)
• Example 1 (5)
• Example 2 (4)
• VAPOR PRESSURE amp BOILING POINT
• VAPOR PRESSURE amp BOILING POINT (2)
• VAPOR PRESSURE amp BOILING POINT (3)
• VAPOR PRESSURE amp BOILING POINT (4)
• COMBINED GAS LAW
• COMBINED GAS LAW (2)
• Example 1 (6)
• AVOGADROrsquoS LAW
• Avogadrorsquos Law
• AVOGADROrsquoS LAW (2)
• AVOGADROrsquoS LAW (3)
• Example 1 (7)
• Example 2 (5)
• STP amp MOLAR VOLUME
• STP amp MOLAR VOLUME (2)
• Example 1 (8)
• Example 2 (6)
• IDEAL GAS LAW
• IDEAL GAS LAW (2)
• Example 1 (9)
• Example 2 (7)
• Example 2 (8)
• Ideal vs Real Gases
• Ideal vs Real Gases (2)
• PARTIAL PRESSURES
• DALTONrsquoS LAW
• PARTIAL PRESSURES (2)
• PARTIAL PRESSURES (3)
• PARTIAL PRESSURES (4)
• Example 1 (10)
• Example 2 (9)
• Slide 60

30

VAPOR PRESSUREamp BOILING POINT

Listed below is the vapor pressure of water at various temperatures

31

VAPOR PRESSUREamp BOILING POINT

A liquid reaches its boiling point when its vapor pressure becomes equal to the external pressure (atmospheric pressure)

For example at sea level water reaches its boiling point at 100C since its vapor pressure is 760 mmHg at this temperature

32

VAPOR PRESSUREamp BOILING POINT

At higher altitudes where atmospheric pressure is lower water reaches boiling point at temperatures lower than 100C

For example in Denver where atmospheric pressure is 630 mmHg water boils at 95C since its vapor pressure is 630 mmHg at this temperature

33

COMBINEDGAS LAW

All pressure-volume-temperature relationships can be combined into a single relationship called the combined gas law

1 1 2 2

1 2

P V P V=

T T

Initial condition

This law is useful for studying the effect of changes in two variables

Final condition

34

COMBINEDGAS LAW

The individual gas laws studied previously are embodied in the combined gas law

1 1 2 2

1 2

P V P V=

T T

Boylersquos LawCharlesrsquos

LawGay-Lussacrsquos

Law

35

Example 1

A 250-mL sample of gas has a pressure of 400 atm at a temperature of 10C What is the volume of the gas at a pressure of 100 atm and a temperature of 18C

1 22 1

2 1

P TV = V x x

P T

V2 = 103 mL

P1= 400 atm V1= 250 mL T1= 283 K P2= 100 atm V2= T2 = 291 K

2

400 atm 291 KV = 250 mL x x

100 atm 283 K

1 1 2 2

1 2

P V P V=

T TBoth pressure

and temp change

Must use combined

gas law

36

AVOGADROrsquoSLAW

The relationship of moles and volume in gases is called Avogadrorsquos Law

At constant temperature and pressure the volume of a fixed amount of gas is directly proportional to the number of moles

As number of moles increases

the volume of the gas

increases

Number of moles and volume of a gas are directly

proportional

Tros Introductory Chemistry Chapter

37

Avogadrorsquos Law

38

AVOGADROrsquoSLAW

As a result of Avogadrorsquos Law equal volumes of different gases at the same temp and pressure contain equal number of moles (molecules)

2 Tanks of gas of equal volume at the same T amp P contain the same number of molecules

39

AVOGADROrsquoSLAW

For example

2 H2 (g) + 1 O2 (g) 2 H2O (g)

2 molecules 1 molecule 2 molecules

2 moles 1 mole 2 moles

2 Liters 1 Liter 2 Liters

Avogadrorsquos Law also allows chemists to relate volumes and moles of a gas in a chemical reaction

This relationship is only valid for

gaseous substances

40

Example 1

A sample of helium gas with a mass of 180 g occupies 16 Liters at a particular temperature and pressure What mass of oxygen would occupy16 L at the same temperature and pressure

1 mol= 180 g x

400 gmol of He = 450 mol

mol of O2 = mol of He = 450 mol

Same Temp amp Pressure

mass of O2

320 g= 450 mol x

1 mol= 144 g

41

Example 2

How many Liters of NH3 can be produced from reaction of 18 L of H2 with excess N2 as shown below

18 L H23

2

L NHx

L H= 12 L NH3

N2 (g) + 3 H2 (g) 2 NH3 (g)

32

42

STP amp MOLAR VOLUME

To better understand the factors that affect gas behavior a set of standard conditions have been chosen for use and are referred to as Standard Temperature and Pressure (STP)

STP =

1 atm (760 mmHg)

0ordmC (273 K)

43

STP amp MOLAR VOLUME

At STP conditions one mole of any gas is observed to occupy a volume of 224 L

V = 224 L

Molar Volume at

STP

44

Example 1

If 200 L of a gas at STP has a mass of 323 g what is the molar mass of the gas

mol200 L x

Lmol of gas =

2241

= 00893 mol

g

molMolar mass =

323 g=

00893 mol= 362 gmol Molar

volume at STP

45

Example 2

A sample of gas has a volume of 250 L at 730 mmHg and 20C What is the volume of this gas at STP

1 22 1

2 1

P TV = V x x

P T

V2 = 224 L

P1= 730 mmHg V1= 250 L T1= 293 K P2= 760 mmHg V2= T2 = 273 K

2

730 mmHg 273 KV = 250 L x x

760 mmHg 293 K

46

IDEAL GAS LAW

Combining all the laws that describe the behavior of gases one can obtain a useful relationship that relates the volume of a gas to the temperature pressure and number of moles

n R TV =

P

Universal gas constant

R = 00821 L atmmol K

47

IDEAL GAS LAW

This relationship is called the Ideal Gas Law and commonly written as

P V = n R T

Pressure in atm Volume

in Liters

Number of moles

Temp in K

48

Example 1

A sample of H2 gas has a volume of 856 L at a temp of 0C and pressure of 15 atm Calculate the moles of gas present

PVn =

RT

= 057 mol

P = 15 atm V = 856 L T = 273 K n = R = 00821

PV = nRT

(15 atm)(856 L)=

L atm(00821 )(273 K)

mol K

49

Example 2

What volume does 400 g of N2 gas occupy at 10C and 750 mmHg

P =750 mmHg V = T = 283 K n = not given m = 400 g R = 00821

mol of N2 =1 mol

400 g x280 g

= 143 mol

P = 750 mmHg1 atm

x760 mmHg

Moles must be calculated

from mass

Pressure must be

converted to atm

P = 0987 atm

50

Example 2

What volume does 400 g of N2 gas occupy at 10C and 750 mmHg

nRTV =

P

V = 337 L

P =0987 atm V = T = 283 K n = 143 R = 00821

L atm(143 mol)(00821 )(283 K)

mol KV = 0987 atm

Tros Introductory Chemistry Chapter

51

Ideal vs Real Gases

bull Real gases often do not behave like ideal gases at high pressure or low temperature

bull Ideal gas laws assume1) no attractions between gas molecules2) gas molecules do not take up space based on the Kinetic-Molecular Theory

bull at low temperatures and high pressures these assumptions are not valid

Tros Introductory Chemistry Chapter

52

Ideal vs Real Gases

53

PARTIAL PRESSURES

Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases

Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture

In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas

54

DALTONrsquoSLAW

In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture

This is called Daltonrsquos law of partial pressures

Total pressure of a gas mixture

Ptotal = P1 + P2 + P3 +

Sum of the partial pressures of the gases in the mixture

=

55

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = PHe + PAr

Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture

= 60 atm

56

PARTIAL PRESSURES

The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture

For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container

57

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = 60 atm

The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture

Twice as many moles of Ar

compared to He

PAr = 2 PHe

58

Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank

Ptotal = Poxygen + Phelium

Poxygen = 1140 mmHg x

1 atm

760 mmHg= 150 atm

Phelium

= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm

59

Example 2

A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture

Ptotal = Poxygen + Pnitrogen

Pnitrogen = 2 Poxygen

Twice as many moles of N2

compared to O2

Pnitrogen = 23 (Ptotal ) = 20 atm

Poxygen = 13 (Ptotal ) = 10 atm

60

THE END

• Slide 1
• CHAPTER OUTLINE
• PROPERTIES OF GASES
• PROPERTIES OF GASES (2)
• KINETIC-MOLECULAR THEORY
• KINETIC-MOLECULAR THEORY (2)
• Kinetic Molecular Theory
• PRESSURE amp ITS MEASUREMENT
• Common Units of Pressure
• PRESSURE amp ITS MEASUREMENT (2)
• Atmospheric Pressure amp Altitude
• Pressure Imbalance in Ear
• Example 1
• Example 2
• PRESSURE amp MOLES OF A GAS
• BOYLErsquoS LAW
• BOYLErsquoS LAW (2)
• Example 1 (2)
• Example 2 (2)
• CHARLESrsquoS LAW
• CHARLESrsquoS LAW (2)
• Example 1 (3)
• Example 2 (3)
• GAY-LUSSACrsquoS LAW
• GAY-LUSSACrsquoS LAW (2)
• Example 1 (4)
• Example 1 (5)
• Example 2 (4)
• VAPOR PRESSURE amp BOILING POINT
• VAPOR PRESSURE amp BOILING POINT (2)
• VAPOR PRESSURE amp BOILING POINT (3)
• VAPOR PRESSURE amp BOILING POINT (4)
• COMBINED GAS LAW
• COMBINED GAS LAW (2)
• Example 1 (6)
• AVOGADROrsquoS LAW
• Avogadrorsquos Law
• AVOGADROrsquoS LAW (2)
• AVOGADROrsquoS LAW (3)
• Example 1 (7)
• Example 2 (5)
• STP amp MOLAR VOLUME
• STP amp MOLAR VOLUME (2)
• Example 1 (8)
• Example 2 (6)
• IDEAL GAS LAW
• IDEAL GAS LAW (2)
• Example 1 (9)
• Example 2 (7)
• Example 2 (8)
• Ideal vs Real Gases
• Ideal vs Real Gases (2)
• PARTIAL PRESSURES
• DALTONrsquoS LAW
• PARTIAL PRESSURES (2)
• PARTIAL PRESSURES (3)
• PARTIAL PRESSURES (4)
• Example 1 (10)
• Example 2 (9)
• Slide 60

31

VAPOR PRESSUREamp BOILING POINT

A liquid reaches its boiling point when its vapor pressure becomes equal to the external pressure (atmospheric pressure)

For example at sea level water reaches its boiling point at 100C since its vapor pressure is 760 mmHg at this temperature

32

VAPOR PRESSUREamp BOILING POINT

At higher altitudes where atmospheric pressure is lower water reaches boiling point at temperatures lower than 100C

For example in Denver where atmospheric pressure is 630 mmHg water boils at 95C since its vapor pressure is 630 mmHg at this temperature

33

COMBINEDGAS LAW

All pressure-volume-temperature relationships can be combined into a single relationship called the combined gas law

1 1 2 2

1 2

P V P V=

T T

Initial condition

This law is useful for studying the effect of changes in two variables

Final condition

34

COMBINEDGAS LAW

The individual gas laws studied previously are embodied in the combined gas law

1 1 2 2

1 2

P V P V=

T T

Boylersquos LawCharlesrsquos

LawGay-Lussacrsquos

Law

35

Example 1

A 250-mL sample of gas has a pressure of 400 atm at a temperature of 10C What is the volume of the gas at a pressure of 100 atm and a temperature of 18C

1 22 1

2 1

P TV = V x x

P T

V2 = 103 mL

P1= 400 atm V1= 250 mL T1= 283 K P2= 100 atm V2= T2 = 291 K

2

400 atm 291 KV = 250 mL x x

100 atm 283 K

1 1 2 2

1 2

P V P V=

T TBoth pressure

and temp change

Must use combined

gas law

36

AVOGADROrsquoSLAW

The relationship of moles and volume in gases is called Avogadrorsquos Law

At constant temperature and pressure the volume of a fixed amount of gas is directly proportional to the number of moles

As number of moles increases

the volume of the gas

increases

Number of moles and volume of a gas are directly

proportional

Tros Introductory Chemistry Chapter

37

Avogadrorsquos Law

38

AVOGADROrsquoSLAW

As a result of Avogadrorsquos Law equal volumes of different gases at the same temp and pressure contain equal number of moles (molecules)

2 Tanks of gas of equal volume at the same T amp P contain the same number of molecules

39

AVOGADROrsquoSLAW

For example

2 H2 (g) + 1 O2 (g) 2 H2O (g)

2 molecules 1 molecule 2 molecules

2 moles 1 mole 2 moles

2 Liters 1 Liter 2 Liters

Avogadrorsquos Law also allows chemists to relate volumes and moles of a gas in a chemical reaction

This relationship is only valid for

gaseous substances

40

Example 1

A sample of helium gas with a mass of 180 g occupies 16 Liters at a particular temperature and pressure What mass of oxygen would occupy16 L at the same temperature and pressure

1 mol= 180 g x

400 gmol of He = 450 mol

mol of O2 = mol of He = 450 mol

Same Temp amp Pressure

mass of O2

320 g= 450 mol x

1 mol= 144 g

41

Example 2

How many Liters of NH3 can be produced from reaction of 18 L of H2 with excess N2 as shown below

18 L H23

2

L NHx

L H= 12 L NH3

N2 (g) + 3 H2 (g) 2 NH3 (g)

32

42

STP amp MOLAR VOLUME

To better understand the factors that affect gas behavior a set of standard conditions have been chosen for use and are referred to as Standard Temperature and Pressure (STP)

STP =

1 atm (760 mmHg)

0ordmC (273 K)

43

STP amp MOLAR VOLUME

At STP conditions one mole of any gas is observed to occupy a volume of 224 L

V = 224 L

Molar Volume at

STP

44

Example 1

If 200 L of a gas at STP has a mass of 323 g what is the molar mass of the gas

mol200 L x

Lmol of gas =

2241

= 00893 mol

g

molMolar mass =

323 g=

00893 mol= 362 gmol Molar

volume at STP

45

Example 2

A sample of gas has a volume of 250 L at 730 mmHg and 20C What is the volume of this gas at STP

1 22 1

2 1

P TV = V x x

P T

V2 = 224 L

P1= 730 mmHg V1= 250 L T1= 293 K P2= 760 mmHg V2= T2 = 273 K

2

730 mmHg 273 KV = 250 L x x

760 mmHg 293 K

46

IDEAL GAS LAW

Combining all the laws that describe the behavior of gases one can obtain a useful relationship that relates the volume of a gas to the temperature pressure and number of moles

n R TV =

P

Universal gas constant

R = 00821 L atmmol K

47

IDEAL GAS LAW

This relationship is called the Ideal Gas Law and commonly written as

P V = n R T

Pressure in atm Volume

in Liters

Number of moles

Temp in K

48

Example 1

A sample of H2 gas has a volume of 856 L at a temp of 0C and pressure of 15 atm Calculate the moles of gas present

PVn =

RT

= 057 mol

P = 15 atm V = 856 L T = 273 K n = R = 00821

PV = nRT

(15 atm)(856 L)=

L atm(00821 )(273 K)

mol K

49

Example 2

What volume does 400 g of N2 gas occupy at 10C and 750 mmHg

P =750 mmHg V = T = 283 K n = not given m = 400 g R = 00821

mol of N2 =1 mol

400 g x280 g

= 143 mol

P = 750 mmHg1 atm

x760 mmHg

Moles must be calculated

from mass

Pressure must be

converted to atm

P = 0987 atm

50

Example 2

What volume does 400 g of N2 gas occupy at 10C and 750 mmHg

nRTV =

P

V = 337 L

P =0987 atm V = T = 283 K n = 143 R = 00821

L atm(143 mol)(00821 )(283 K)

mol KV = 0987 atm

Tros Introductory Chemistry Chapter

51

Ideal vs Real Gases

bull Real gases often do not behave like ideal gases at high pressure or low temperature

bull Ideal gas laws assume1) no attractions between gas molecules2) gas molecules do not take up space based on the Kinetic-Molecular Theory

bull at low temperatures and high pressures these assumptions are not valid

Tros Introductory Chemistry Chapter

52

Ideal vs Real Gases

53

PARTIAL PRESSURES

Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases

Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture

In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas

54

DALTONrsquoSLAW

In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture

This is called Daltonrsquos law of partial pressures

Total pressure of a gas mixture

Ptotal = P1 + P2 + P3 +

Sum of the partial pressures of the gases in the mixture

=

55

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = PHe + PAr

Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture

= 60 atm

56

PARTIAL PRESSURES

The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture

For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container

57

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = 60 atm

The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture

Twice as many moles of Ar

compared to He

PAr = 2 PHe

58

Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank

Ptotal = Poxygen + Phelium

Poxygen = 1140 mmHg x

1 atm

760 mmHg= 150 atm

Phelium

= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm

59

Example 2

A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture

Ptotal = Poxygen + Pnitrogen

Pnitrogen = 2 Poxygen

Twice as many moles of N2

compared to O2

Pnitrogen = 23 (Ptotal ) = 20 atm

Poxygen = 13 (Ptotal ) = 10 atm

60

THE END

• Slide 1
• CHAPTER OUTLINE
• PROPERTIES OF GASES
• PROPERTIES OF GASES (2)
• KINETIC-MOLECULAR THEORY
• KINETIC-MOLECULAR THEORY (2)
• Kinetic Molecular Theory
• PRESSURE amp ITS MEASUREMENT
• Common Units of Pressure
• PRESSURE amp ITS MEASUREMENT (2)
• Atmospheric Pressure amp Altitude
• Pressure Imbalance in Ear
• Example 1
• Example 2
• PRESSURE amp MOLES OF A GAS
• BOYLErsquoS LAW
• BOYLErsquoS LAW (2)
• Example 1 (2)
• Example 2 (2)
• CHARLESrsquoS LAW
• CHARLESrsquoS LAW (2)
• Example 1 (3)
• Example 2 (3)
• GAY-LUSSACrsquoS LAW
• GAY-LUSSACrsquoS LAW (2)
• Example 1 (4)
• Example 1 (5)
• Example 2 (4)
• VAPOR PRESSURE amp BOILING POINT
• VAPOR PRESSURE amp BOILING POINT (2)
• VAPOR PRESSURE amp BOILING POINT (3)
• VAPOR PRESSURE amp BOILING POINT (4)
• COMBINED GAS LAW
• COMBINED GAS LAW (2)
• Example 1 (6)
• AVOGADROrsquoS LAW
• Avogadrorsquos Law
• AVOGADROrsquoS LAW (2)
• AVOGADROrsquoS LAW (3)
• Example 1 (7)
• Example 2 (5)
• STP amp MOLAR VOLUME
• STP amp MOLAR VOLUME (2)
• Example 1 (8)
• Example 2 (6)
• IDEAL GAS LAW
• IDEAL GAS LAW (2)
• Example 1 (9)
• Example 2 (7)
• Example 2 (8)
• Ideal vs Real Gases
• Ideal vs Real Gases (2)
• PARTIAL PRESSURES
• DALTONrsquoS LAW
• PARTIAL PRESSURES (2)
• PARTIAL PRESSURES (3)
• PARTIAL PRESSURES (4)
• Example 1 (10)
• Example 2 (9)
• Slide 60

32

VAPOR PRESSUREamp BOILING POINT

At higher altitudes where atmospheric pressure is lower water reaches boiling point at temperatures lower than 100C

For example in Denver where atmospheric pressure is 630 mmHg water boils at 95C since its vapor pressure is 630 mmHg at this temperature

33

COMBINEDGAS LAW

All pressure-volume-temperature relationships can be combined into a single relationship called the combined gas law

1 1 2 2

1 2

P V P V=

T T

Initial condition

This law is useful for studying the effect of changes in two variables

Final condition

34

COMBINEDGAS LAW

The individual gas laws studied previously are embodied in the combined gas law

1 1 2 2

1 2

P V P V=

T T

Boylersquos LawCharlesrsquos

LawGay-Lussacrsquos

Law

35

Example 1

A 250-mL sample of gas has a pressure of 400 atm at a temperature of 10C What is the volume of the gas at a pressure of 100 atm and a temperature of 18C

1 22 1

2 1

P TV = V x x

P T

V2 = 103 mL

P1= 400 atm V1= 250 mL T1= 283 K P2= 100 atm V2= T2 = 291 K

2

400 atm 291 KV = 250 mL x x

100 atm 283 K

1 1 2 2

1 2

P V P V=

T TBoth pressure

and temp change

Must use combined

gas law

36

AVOGADROrsquoSLAW

The relationship of moles and volume in gases is called Avogadrorsquos Law

At constant temperature and pressure the volume of a fixed amount of gas is directly proportional to the number of moles

As number of moles increases

the volume of the gas

increases

Number of moles and volume of a gas are directly

proportional

Tros Introductory Chemistry Chapter

37

Avogadrorsquos Law

38

AVOGADROrsquoSLAW

As a result of Avogadrorsquos Law equal volumes of different gases at the same temp and pressure contain equal number of moles (molecules)

2 Tanks of gas of equal volume at the same T amp P contain the same number of molecules

39

AVOGADROrsquoSLAW

For example

2 H2 (g) + 1 O2 (g) 2 H2O (g)

2 molecules 1 molecule 2 molecules

2 moles 1 mole 2 moles

2 Liters 1 Liter 2 Liters

Avogadrorsquos Law also allows chemists to relate volumes and moles of a gas in a chemical reaction

This relationship is only valid for

gaseous substances

40

Example 1

A sample of helium gas with a mass of 180 g occupies 16 Liters at a particular temperature and pressure What mass of oxygen would occupy16 L at the same temperature and pressure

1 mol= 180 g x

400 gmol of He = 450 mol

mol of O2 = mol of He = 450 mol

Same Temp amp Pressure

mass of O2

320 g= 450 mol x

1 mol= 144 g

41

Example 2

How many Liters of NH3 can be produced from reaction of 18 L of H2 with excess N2 as shown below

18 L H23

2

L NHx

L H= 12 L NH3

N2 (g) + 3 H2 (g) 2 NH3 (g)

32

42

STP amp MOLAR VOLUME

To better understand the factors that affect gas behavior a set of standard conditions have been chosen for use and are referred to as Standard Temperature and Pressure (STP)

STP =

1 atm (760 mmHg)

0ordmC (273 K)

43

STP amp MOLAR VOLUME

At STP conditions one mole of any gas is observed to occupy a volume of 224 L

V = 224 L

Molar Volume at

STP

44

Example 1

If 200 L of a gas at STP has a mass of 323 g what is the molar mass of the gas

mol200 L x

Lmol of gas =

2241

= 00893 mol

g

molMolar mass =

323 g=

00893 mol= 362 gmol Molar

volume at STP

45

Example 2

A sample of gas has a volume of 250 L at 730 mmHg and 20C What is the volume of this gas at STP

1 22 1

2 1

P TV = V x x

P T

V2 = 224 L

P1= 730 mmHg V1= 250 L T1= 293 K P2= 760 mmHg V2= T2 = 273 K

2

730 mmHg 273 KV = 250 L x x

760 mmHg 293 K

46

IDEAL GAS LAW

Combining all the laws that describe the behavior of gases one can obtain a useful relationship that relates the volume of a gas to the temperature pressure and number of moles

n R TV =

P

Universal gas constant

R = 00821 L atmmol K

47

IDEAL GAS LAW

This relationship is called the Ideal Gas Law and commonly written as

P V = n R T

Pressure in atm Volume

in Liters

Number of moles

Temp in K

48

Example 1

A sample of H2 gas has a volume of 856 L at a temp of 0C and pressure of 15 atm Calculate the moles of gas present

PVn =

RT

= 057 mol

P = 15 atm V = 856 L T = 273 K n = R = 00821

PV = nRT

(15 atm)(856 L)=

L atm(00821 )(273 K)

mol K

49

Example 2

What volume does 400 g of N2 gas occupy at 10C and 750 mmHg

P =750 mmHg V = T = 283 K n = not given m = 400 g R = 00821

mol of N2 =1 mol

400 g x280 g

= 143 mol

P = 750 mmHg1 atm

x760 mmHg

Moles must be calculated

from mass

Pressure must be

converted to atm

P = 0987 atm

50

Example 2

What volume does 400 g of N2 gas occupy at 10C and 750 mmHg

nRTV =

P

V = 337 L

P =0987 atm V = T = 283 K n = 143 R = 00821

L atm(143 mol)(00821 )(283 K)

mol KV = 0987 atm

Tros Introductory Chemistry Chapter

51

Ideal vs Real Gases

bull Real gases often do not behave like ideal gases at high pressure or low temperature

bull Ideal gas laws assume1) no attractions between gas molecules2) gas molecules do not take up space based on the Kinetic-Molecular Theory

bull at low temperatures and high pressures these assumptions are not valid

Tros Introductory Chemistry Chapter

52

Ideal vs Real Gases

53

PARTIAL PRESSURES

Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases

Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture

In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas

54

DALTONrsquoSLAW

In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture

This is called Daltonrsquos law of partial pressures

Total pressure of a gas mixture

Ptotal = P1 + P2 + P3 +

Sum of the partial pressures of the gases in the mixture

=

55

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = PHe + PAr

Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture

= 60 atm

56

PARTIAL PRESSURES

The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture

For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container

57

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = 60 atm

The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture

Twice as many moles of Ar

compared to He

PAr = 2 PHe

58

Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank

Ptotal = Poxygen + Phelium

Poxygen = 1140 mmHg x

1 atm

760 mmHg= 150 atm

Phelium

= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm

59

Example 2

A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture

Ptotal = Poxygen + Pnitrogen

Pnitrogen = 2 Poxygen

Twice as many moles of N2

compared to O2

Pnitrogen = 23 (Ptotal ) = 20 atm

Poxygen = 13 (Ptotal ) = 10 atm

60

THE END

• Slide 1
• CHAPTER OUTLINE
• PROPERTIES OF GASES
• PROPERTIES OF GASES (2)
• KINETIC-MOLECULAR THEORY
• KINETIC-MOLECULAR THEORY (2)
• Kinetic Molecular Theory
• PRESSURE amp ITS MEASUREMENT
• Common Units of Pressure
• PRESSURE amp ITS MEASUREMENT (2)
• Atmospheric Pressure amp Altitude
• Pressure Imbalance in Ear
• Example 1
• Example 2
• PRESSURE amp MOLES OF A GAS
• BOYLErsquoS LAW
• BOYLErsquoS LAW (2)
• Example 1 (2)
• Example 2 (2)
• CHARLESrsquoS LAW
• CHARLESrsquoS LAW (2)
• Example 1 (3)
• Example 2 (3)
• GAY-LUSSACrsquoS LAW
• GAY-LUSSACrsquoS LAW (2)
• Example 1 (4)
• Example 1 (5)
• Example 2 (4)
• VAPOR PRESSURE amp BOILING POINT
• VAPOR PRESSURE amp BOILING POINT (2)
• VAPOR PRESSURE amp BOILING POINT (3)
• VAPOR PRESSURE amp BOILING POINT (4)
• COMBINED GAS LAW
• COMBINED GAS LAW (2)
• Example 1 (6)
• AVOGADROrsquoS LAW
• Avogadrorsquos Law
• AVOGADROrsquoS LAW (2)
• AVOGADROrsquoS LAW (3)
• Example 1 (7)
• Example 2 (5)
• STP amp MOLAR VOLUME
• STP amp MOLAR VOLUME (2)
• Example 1 (8)
• Example 2 (6)
• IDEAL GAS LAW
• IDEAL GAS LAW (2)
• Example 1 (9)
• Example 2 (7)
• Example 2 (8)
• Ideal vs Real Gases
• Ideal vs Real Gases (2)
• PARTIAL PRESSURES
• DALTONrsquoS LAW
• PARTIAL PRESSURES (2)
• PARTIAL PRESSURES (3)
• PARTIAL PRESSURES (4)
• Example 1 (10)
• Example 2 (9)
• Slide 60

33

COMBINEDGAS LAW

All pressure-volume-temperature relationships can be combined into a single relationship called the combined gas law

1 1 2 2

1 2

P V P V=

T T

Initial condition

This law is useful for studying the effect of changes in two variables

Final condition

34

COMBINEDGAS LAW

The individual gas laws studied previously are embodied in the combined gas law

1 1 2 2

1 2

P V P V=

T T

Boylersquos LawCharlesrsquos

LawGay-Lussacrsquos

Law

35

Example 1

A 250-mL sample of gas has a pressure of 400 atm at a temperature of 10C What is the volume of the gas at a pressure of 100 atm and a temperature of 18C

1 22 1

2 1

P TV = V x x

P T

V2 = 103 mL

P1= 400 atm V1= 250 mL T1= 283 K P2= 100 atm V2= T2 = 291 K

2

400 atm 291 KV = 250 mL x x

100 atm 283 K

1 1 2 2

1 2

P V P V=

T TBoth pressure

and temp change

Must use combined

gas law

36

AVOGADROrsquoSLAW

The relationship of moles and volume in gases is called Avogadrorsquos Law

At constant temperature and pressure the volume of a fixed amount of gas is directly proportional to the number of moles

As number of moles increases

the volume of the gas

increases

Number of moles and volume of a gas are directly

proportional

Tros Introductory Chemistry Chapter

37

Avogadrorsquos Law

38

AVOGADROrsquoSLAW

As a result of Avogadrorsquos Law equal volumes of different gases at the same temp and pressure contain equal number of moles (molecules)

2 Tanks of gas of equal volume at the same T amp P contain the same number of molecules

39

AVOGADROrsquoSLAW

For example

2 H2 (g) + 1 O2 (g) 2 H2O (g)

2 molecules 1 molecule 2 molecules

2 moles 1 mole 2 moles

2 Liters 1 Liter 2 Liters

Avogadrorsquos Law also allows chemists to relate volumes and moles of a gas in a chemical reaction

This relationship is only valid for

gaseous substances

40

Example 1

A sample of helium gas with a mass of 180 g occupies 16 Liters at a particular temperature and pressure What mass of oxygen would occupy16 L at the same temperature and pressure

1 mol= 180 g x

400 gmol of He = 450 mol

mol of O2 = mol of He = 450 mol

Same Temp amp Pressure

mass of O2

320 g= 450 mol x

1 mol= 144 g

41

Example 2

How many Liters of NH3 can be produced from reaction of 18 L of H2 with excess N2 as shown below

18 L H23

2

L NHx

L H= 12 L NH3

N2 (g) + 3 H2 (g) 2 NH3 (g)

32

42

STP amp MOLAR VOLUME

To better understand the factors that affect gas behavior a set of standard conditions have been chosen for use and are referred to as Standard Temperature and Pressure (STP)

STP =

1 atm (760 mmHg)

0ordmC (273 K)

43

STP amp MOLAR VOLUME

At STP conditions one mole of any gas is observed to occupy a volume of 224 L

V = 224 L

Molar Volume at

STP

44

Example 1

If 200 L of a gas at STP has a mass of 323 g what is the molar mass of the gas

mol200 L x

Lmol of gas =

2241

= 00893 mol

g

molMolar mass =

323 g=

00893 mol= 362 gmol Molar

volume at STP

45

Example 2

A sample of gas has a volume of 250 L at 730 mmHg and 20C What is the volume of this gas at STP

1 22 1

2 1

P TV = V x x

P T

V2 = 224 L

P1= 730 mmHg V1= 250 L T1= 293 K P2= 760 mmHg V2= T2 = 273 K

2

730 mmHg 273 KV = 250 L x x

760 mmHg 293 K

46

IDEAL GAS LAW

Combining all the laws that describe the behavior of gases one can obtain a useful relationship that relates the volume of a gas to the temperature pressure and number of moles

n R TV =

P

Universal gas constant

R = 00821 L atmmol K

47

IDEAL GAS LAW

This relationship is called the Ideal Gas Law and commonly written as

P V = n R T

Pressure in atm Volume

in Liters

Number of moles

Temp in K

48

Example 1

A sample of H2 gas has a volume of 856 L at a temp of 0C and pressure of 15 atm Calculate the moles of gas present

PVn =

RT

= 057 mol

P = 15 atm V = 856 L T = 273 K n = R = 00821

PV = nRT

(15 atm)(856 L)=

L atm(00821 )(273 K)

mol K

49

Example 2

What volume does 400 g of N2 gas occupy at 10C and 750 mmHg

P =750 mmHg V = T = 283 K n = not given m = 400 g R = 00821

mol of N2 =1 mol

400 g x280 g

= 143 mol

P = 750 mmHg1 atm

x760 mmHg

Moles must be calculated

from mass

Pressure must be

converted to atm

P = 0987 atm

50

Example 2

What volume does 400 g of N2 gas occupy at 10C and 750 mmHg

nRTV =

P

V = 337 L

P =0987 atm V = T = 283 K n = 143 R = 00821

L atm(143 mol)(00821 )(283 K)

mol KV = 0987 atm

Tros Introductory Chemistry Chapter

51

Ideal vs Real Gases

bull Real gases often do not behave like ideal gases at high pressure or low temperature

bull Ideal gas laws assume1) no attractions between gas molecules2) gas molecules do not take up space based on the Kinetic-Molecular Theory

bull at low temperatures and high pressures these assumptions are not valid

Tros Introductory Chemistry Chapter

52

Ideal vs Real Gases

53

PARTIAL PRESSURES

Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases

Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture

In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas

54

DALTONrsquoSLAW

In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture

This is called Daltonrsquos law of partial pressures

Total pressure of a gas mixture

Ptotal = P1 + P2 + P3 +

Sum of the partial pressures of the gases in the mixture

=

55

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = PHe + PAr

Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture

= 60 atm

56

PARTIAL PRESSURES

The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture

For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container

57

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = 60 atm

The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture

Twice as many moles of Ar

compared to He

PAr = 2 PHe

58

Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank

Ptotal = Poxygen + Phelium

Poxygen = 1140 mmHg x

1 atm

760 mmHg= 150 atm

Phelium

= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm

59

Example 2

A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture

Ptotal = Poxygen + Pnitrogen

Pnitrogen = 2 Poxygen

Twice as many moles of N2

compared to O2

Pnitrogen = 23 (Ptotal ) = 20 atm

Poxygen = 13 (Ptotal ) = 10 atm

60

THE END

• Slide 1
• CHAPTER OUTLINE
• PROPERTIES OF GASES
• PROPERTIES OF GASES (2)
• KINETIC-MOLECULAR THEORY
• KINETIC-MOLECULAR THEORY (2)
• Kinetic Molecular Theory
• PRESSURE amp ITS MEASUREMENT
• Common Units of Pressure
• PRESSURE amp ITS MEASUREMENT (2)
• Atmospheric Pressure amp Altitude
• Pressure Imbalance in Ear
• Example 1
• Example 2
• PRESSURE amp MOLES OF A GAS
• BOYLErsquoS LAW
• BOYLErsquoS LAW (2)
• Example 1 (2)
• Example 2 (2)
• CHARLESrsquoS LAW
• CHARLESrsquoS LAW (2)
• Example 1 (3)
• Example 2 (3)
• GAY-LUSSACrsquoS LAW
• GAY-LUSSACrsquoS LAW (2)
• Example 1 (4)
• Example 1 (5)
• Example 2 (4)
• VAPOR PRESSURE amp BOILING POINT
• VAPOR PRESSURE amp BOILING POINT (2)
• VAPOR PRESSURE amp BOILING POINT (3)
• VAPOR PRESSURE amp BOILING POINT (4)
• COMBINED GAS LAW
• COMBINED GAS LAW (2)
• Example 1 (6)
• AVOGADROrsquoS LAW
• Avogadrorsquos Law
• AVOGADROrsquoS LAW (2)
• AVOGADROrsquoS LAW (3)
• Example 1 (7)
• Example 2 (5)
• STP amp MOLAR VOLUME
• STP amp MOLAR VOLUME (2)
• Example 1 (8)
• Example 2 (6)
• IDEAL GAS LAW
• IDEAL GAS LAW (2)
• Example 1 (9)
• Example 2 (7)
• Example 2 (8)
• Ideal vs Real Gases
• Ideal vs Real Gases (2)
• PARTIAL PRESSURES
• DALTONrsquoS LAW
• PARTIAL PRESSURES (2)
• PARTIAL PRESSURES (3)
• PARTIAL PRESSURES (4)
• Example 1 (10)
• Example 2 (9)
• Slide 60

34

COMBINEDGAS LAW

The individual gas laws studied previously are embodied in the combined gas law

1 1 2 2

1 2

P V P V=

T T

Boylersquos LawCharlesrsquos

LawGay-Lussacrsquos

Law

35

Example 1

A 250-mL sample of gas has a pressure of 400 atm at a temperature of 10C What is the volume of the gas at a pressure of 100 atm and a temperature of 18C

1 22 1

2 1

P TV = V x x

P T

V2 = 103 mL

P1= 400 atm V1= 250 mL T1= 283 K P2= 100 atm V2= T2 = 291 K

2

400 atm 291 KV = 250 mL x x

100 atm 283 K

1 1 2 2

1 2

P V P V=

T TBoth pressure

and temp change

Must use combined

gas law

36

AVOGADROrsquoSLAW

The relationship of moles and volume in gases is called Avogadrorsquos Law

At constant temperature and pressure the volume of a fixed amount of gas is directly proportional to the number of moles

As number of moles increases

the volume of the gas

increases

Number of moles and volume of a gas are directly

proportional

Tros Introductory Chemistry Chapter

37

Avogadrorsquos Law

38

AVOGADROrsquoSLAW

As a result of Avogadrorsquos Law equal volumes of different gases at the same temp and pressure contain equal number of moles (molecules)

2 Tanks of gas of equal volume at the same T amp P contain the same number of molecules

39

AVOGADROrsquoSLAW

For example

2 H2 (g) + 1 O2 (g) 2 H2O (g)

2 molecules 1 molecule 2 molecules

2 moles 1 mole 2 moles

2 Liters 1 Liter 2 Liters

Avogadrorsquos Law also allows chemists to relate volumes and moles of a gas in a chemical reaction

This relationship is only valid for

gaseous substances

40

Example 1

A sample of helium gas with a mass of 180 g occupies 16 Liters at a particular temperature and pressure What mass of oxygen would occupy16 L at the same temperature and pressure

1 mol= 180 g x

400 gmol of He = 450 mol

mol of O2 = mol of He = 450 mol

Same Temp amp Pressure

mass of O2

320 g= 450 mol x

1 mol= 144 g

41

Example 2

How many Liters of NH3 can be produced from reaction of 18 L of H2 with excess N2 as shown below

18 L H23

2

L NHx

L H= 12 L NH3

N2 (g) + 3 H2 (g) 2 NH3 (g)

32

42

STP amp MOLAR VOLUME

To better understand the factors that affect gas behavior a set of standard conditions have been chosen for use and are referred to as Standard Temperature and Pressure (STP)

STP =

1 atm (760 mmHg)

0ordmC (273 K)

43

STP amp MOLAR VOLUME

At STP conditions one mole of any gas is observed to occupy a volume of 224 L

V = 224 L

Molar Volume at

STP

44

Example 1

If 200 L of a gas at STP has a mass of 323 g what is the molar mass of the gas

mol200 L x

Lmol of gas =

2241

= 00893 mol

g

molMolar mass =

323 g=

00893 mol= 362 gmol Molar

volume at STP

45

Example 2

A sample of gas has a volume of 250 L at 730 mmHg and 20C What is the volume of this gas at STP

1 22 1

2 1

P TV = V x x

P T

V2 = 224 L

P1= 730 mmHg V1= 250 L T1= 293 K P2= 760 mmHg V2= T2 = 273 K

2

730 mmHg 273 KV = 250 L x x

760 mmHg 293 K

46

IDEAL GAS LAW

Combining all the laws that describe the behavior of gases one can obtain a useful relationship that relates the volume of a gas to the temperature pressure and number of moles

n R TV =

P

Universal gas constant

R = 00821 L atmmol K

47

IDEAL GAS LAW

This relationship is called the Ideal Gas Law and commonly written as

P V = n R T

Pressure in atm Volume

in Liters

Number of moles

Temp in K

48

Example 1

A sample of H2 gas has a volume of 856 L at a temp of 0C and pressure of 15 atm Calculate the moles of gas present

PVn =

RT

= 057 mol

P = 15 atm V = 856 L T = 273 K n = R = 00821

PV = nRT

(15 atm)(856 L)=

L atm(00821 )(273 K)

mol K

49

Example 2

What volume does 400 g of N2 gas occupy at 10C and 750 mmHg

P =750 mmHg V = T = 283 K n = not given m = 400 g R = 00821

mol of N2 =1 mol

400 g x280 g

= 143 mol

P = 750 mmHg1 atm

x760 mmHg

Moles must be calculated

from mass

Pressure must be

converted to atm

P = 0987 atm

50

Example 2

What volume does 400 g of N2 gas occupy at 10C and 750 mmHg

nRTV =

P

V = 337 L

P =0987 atm V = T = 283 K n = 143 R = 00821

L atm(143 mol)(00821 )(283 K)

mol KV = 0987 atm

Tros Introductory Chemistry Chapter

51

Ideal vs Real Gases

bull Real gases often do not behave like ideal gases at high pressure or low temperature

bull Ideal gas laws assume1) no attractions between gas molecules2) gas molecules do not take up space based on the Kinetic-Molecular Theory

bull at low temperatures and high pressures these assumptions are not valid

Tros Introductory Chemistry Chapter

52

Ideal vs Real Gases

53

PARTIAL PRESSURES

Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases

Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture

In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas

54

DALTONrsquoSLAW

In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture

This is called Daltonrsquos law of partial pressures

Total pressure of a gas mixture

Ptotal = P1 + P2 + P3 +

Sum of the partial pressures of the gases in the mixture

=

55

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = PHe + PAr

Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture

= 60 atm

56

PARTIAL PRESSURES

The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture

For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container

57

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = 60 atm

The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture

Twice as many moles of Ar

compared to He

PAr = 2 PHe

58

Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank

Ptotal = Poxygen + Phelium

Poxygen = 1140 mmHg x

1 atm

760 mmHg= 150 atm

Phelium

= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm

59

Example 2

A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture

Ptotal = Poxygen + Pnitrogen

Pnitrogen = 2 Poxygen

Twice as many moles of N2

compared to O2

Pnitrogen = 23 (Ptotal ) = 20 atm

Poxygen = 13 (Ptotal ) = 10 atm

60

THE END

• Slide 1
• CHAPTER OUTLINE
• PROPERTIES OF GASES
• PROPERTIES OF GASES (2)
• KINETIC-MOLECULAR THEORY
• KINETIC-MOLECULAR THEORY (2)
• Kinetic Molecular Theory
• PRESSURE amp ITS MEASUREMENT
• Common Units of Pressure
• PRESSURE amp ITS MEASUREMENT (2)
• Atmospheric Pressure amp Altitude
• Pressure Imbalance in Ear
• Example 1
• Example 2
• PRESSURE amp MOLES OF A GAS
• BOYLErsquoS LAW
• BOYLErsquoS LAW (2)
• Example 1 (2)
• Example 2 (2)
• CHARLESrsquoS LAW
• CHARLESrsquoS LAW (2)
• Example 1 (3)
• Example 2 (3)
• GAY-LUSSACrsquoS LAW
• GAY-LUSSACrsquoS LAW (2)
• Example 1 (4)
• Example 1 (5)
• Example 2 (4)
• VAPOR PRESSURE amp BOILING POINT
• VAPOR PRESSURE amp BOILING POINT (2)
• VAPOR PRESSURE amp BOILING POINT (3)
• VAPOR PRESSURE amp BOILING POINT (4)
• COMBINED GAS LAW
• COMBINED GAS LAW (2)
• Example 1 (6)
• AVOGADROrsquoS LAW
• Avogadrorsquos Law
• AVOGADROrsquoS LAW (2)
• AVOGADROrsquoS LAW (3)
• Example 1 (7)
• Example 2 (5)
• STP amp MOLAR VOLUME
• STP amp MOLAR VOLUME (2)
• Example 1 (8)
• Example 2 (6)
• IDEAL GAS LAW
• IDEAL GAS LAW (2)
• Example 1 (9)
• Example 2 (7)
• Example 2 (8)
• Ideal vs Real Gases
• Ideal vs Real Gases (2)
• PARTIAL PRESSURES
• DALTONrsquoS LAW
• PARTIAL PRESSURES (2)
• PARTIAL PRESSURES (3)
• PARTIAL PRESSURES (4)
• Example 1 (10)
• Example 2 (9)
• Slide 60

35

Example 1

A 250-mL sample of gas has a pressure of 400 atm at a temperature of 10C What is the volume of the gas at a pressure of 100 atm and a temperature of 18C

1 22 1

2 1

P TV = V x x

P T

V2 = 103 mL

P1= 400 atm V1= 250 mL T1= 283 K P2= 100 atm V2= T2 = 291 K

2

400 atm 291 KV = 250 mL x x

100 atm 283 K

1 1 2 2

1 2

P V P V=

T TBoth pressure

and temp change

Must use combined

gas law

36

AVOGADROrsquoSLAW

The relationship of moles and volume in gases is called Avogadrorsquos Law

At constant temperature and pressure the volume of a fixed amount of gas is directly proportional to the number of moles

As number of moles increases

the volume of the gas

increases

Number of moles and volume of a gas are directly

proportional

Tros Introductory Chemistry Chapter

37

Avogadrorsquos Law

38

AVOGADROrsquoSLAW

As a result of Avogadrorsquos Law equal volumes of different gases at the same temp and pressure contain equal number of moles (molecules)

2 Tanks of gas of equal volume at the same T amp P contain the same number of molecules

39

AVOGADROrsquoSLAW

For example

2 H2 (g) + 1 O2 (g) 2 H2O (g)

2 molecules 1 molecule 2 molecules

2 moles 1 mole 2 moles

2 Liters 1 Liter 2 Liters

Avogadrorsquos Law also allows chemists to relate volumes and moles of a gas in a chemical reaction

This relationship is only valid for

gaseous substances

40

Example 1

A sample of helium gas with a mass of 180 g occupies 16 Liters at a particular temperature and pressure What mass of oxygen would occupy16 L at the same temperature and pressure

1 mol= 180 g x

400 gmol of He = 450 mol

mol of O2 = mol of He = 450 mol

Same Temp amp Pressure

mass of O2

320 g= 450 mol x

1 mol= 144 g

41

Example 2

How many Liters of NH3 can be produced from reaction of 18 L of H2 with excess N2 as shown below

18 L H23

2

L NHx

L H= 12 L NH3

N2 (g) + 3 H2 (g) 2 NH3 (g)

32

42

STP amp MOLAR VOLUME

To better understand the factors that affect gas behavior a set of standard conditions have been chosen for use and are referred to as Standard Temperature and Pressure (STP)

STP =

1 atm (760 mmHg)

0ordmC (273 K)

43

STP amp MOLAR VOLUME

At STP conditions one mole of any gas is observed to occupy a volume of 224 L

V = 224 L

Molar Volume at

STP

44

Example 1

If 200 L of a gas at STP has a mass of 323 g what is the molar mass of the gas

mol200 L x

Lmol of gas =

2241

= 00893 mol

g

molMolar mass =

323 g=

00893 mol= 362 gmol Molar

volume at STP

45

Example 2

A sample of gas has a volume of 250 L at 730 mmHg and 20C What is the volume of this gas at STP

1 22 1

2 1

P TV = V x x

P T

V2 = 224 L

P1= 730 mmHg V1= 250 L T1= 293 K P2= 760 mmHg V2= T2 = 273 K

2

730 mmHg 273 KV = 250 L x x

760 mmHg 293 K

46

IDEAL GAS LAW

Combining all the laws that describe the behavior of gases one can obtain a useful relationship that relates the volume of a gas to the temperature pressure and number of moles

n R TV =

P

Universal gas constant

R = 00821 L atmmol K

47

IDEAL GAS LAW

This relationship is called the Ideal Gas Law and commonly written as

P V = n R T

Pressure in atm Volume

in Liters

Number of moles

Temp in K

48

Example 1

A sample of H2 gas has a volume of 856 L at a temp of 0C and pressure of 15 atm Calculate the moles of gas present

PVn =

RT

= 057 mol

P = 15 atm V = 856 L T = 273 K n = R = 00821

PV = nRT

(15 atm)(856 L)=

L atm(00821 )(273 K)

mol K

49

Example 2

What volume does 400 g of N2 gas occupy at 10C and 750 mmHg

P =750 mmHg V = T = 283 K n = not given m = 400 g R = 00821

mol of N2 =1 mol

400 g x280 g

= 143 mol

P = 750 mmHg1 atm

x760 mmHg

Moles must be calculated

from mass

Pressure must be

converted to atm

P = 0987 atm

50

Example 2

What volume does 400 g of N2 gas occupy at 10C and 750 mmHg

nRTV =

P

V = 337 L

P =0987 atm V = T = 283 K n = 143 R = 00821

L atm(143 mol)(00821 )(283 K)

mol KV = 0987 atm

Tros Introductory Chemistry Chapter

51

Ideal vs Real Gases

bull Real gases often do not behave like ideal gases at high pressure or low temperature

bull Ideal gas laws assume1) no attractions between gas molecules2) gas molecules do not take up space based on the Kinetic-Molecular Theory

bull at low temperatures and high pressures these assumptions are not valid

Tros Introductory Chemistry Chapter

52

Ideal vs Real Gases

53

PARTIAL PRESSURES

Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases

Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture

In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas

54

DALTONrsquoSLAW

In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture

This is called Daltonrsquos law of partial pressures

Total pressure of a gas mixture

Ptotal = P1 + P2 + P3 +

Sum of the partial pressures of the gases in the mixture

=

55

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = PHe + PAr

Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture

= 60 atm

56

PARTIAL PRESSURES

The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture

For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container

57

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = 60 atm

The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture

Twice as many moles of Ar

compared to He

PAr = 2 PHe

58

Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank

Ptotal = Poxygen + Phelium

Poxygen = 1140 mmHg x

1 atm

760 mmHg= 150 atm

Phelium

= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm

59

Example 2

A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture

Ptotal = Poxygen + Pnitrogen

Pnitrogen = 2 Poxygen

Twice as many moles of N2

compared to O2

Pnitrogen = 23 (Ptotal ) = 20 atm

Poxygen = 13 (Ptotal ) = 10 atm

60

THE END

• Slide 1
• CHAPTER OUTLINE
• PROPERTIES OF GASES
• PROPERTIES OF GASES (2)
• KINETIC-MOLECULAR THEORY
• KINETIC-MOLECULAR THEORY (2)
• Kinetic Molecular Theory
• PRESSURE amp ITS MEASUREMENT
• Common Units of Pressure
• PRESSURE amp ITS MEASUREMENT (2)
• Atmospheric Pressure amp Altitude
• Pressure Imbalance in Ear
• Example 1
• Example 2
• PRESSURE amp MOLES OF A GAS
• BOYLErsquoS LAW
• BOYLErsquoS LAW (2)
• Example 1 (2)
• Example 2 (2)
• CHARLESrsquoS LAW
• CHARLESrsquoS LAW (2)
• Example 1 (3)
• Example 2 (3)
• GAY-LUSSACrsquoS LAW
• GAY-LUSSACrsquoS LAW (2)
• Example 1 (4)
• Example 1 (5)
• Example 2 (4)
• VAPOR PRESSURE amp BOILING POINT
• VAPOR PRESSURE amp BOILING POINT (2)
• VAPOR PRESSURE amp BOILING POINT (3)
• VAPOR PRESSURE amp BOILING POINT (4)
• COMBINED GAS LAW
• COMBINED GAS LAW (2)
• Example 1 (6)
• AVOGADROrsquoS LAW
• Avogadrorsquos Law
• AVOGADROrsquoS LAW (2)
• AVOGADROrsquoS LAW (3)
• Example 1 (7)
• Example 2 (5)
• STP amp MOLAR VOLUME
• STP amp MOLAR VOLUME (2)
• Example 1 (8)
• Example 2 (6)
• IDEAL GAS LAW
• IDEAL GAS LAW (2)
• Example 1 (9)
• Example 2 (7)
• Example 2 (8)
• Ideal vs Real Gases
• Ideal vs Real Gases (2)
• PARTIAL PRESSURES
• DALTONrsquoS LAW
• PARTIAL PRESSURES (2)
• PARTIAL PRESSURES (3)
• PARTIAL PRESSURES (4)
• Example 1 (10)
• Example 2 (9)
• Slide 60

36

AVOGADROrsquoSLAW

The relationship of moles and volume in gases is called Avogadrorsquos Law

At constant temperature and pressure the volume of a fixed amount of gas is directly proportional to the number of moles

As number of moles increases

the volume of the gas

increases

Number of moles and volume of a gas are directly

proportional

Tros Introductory Chemistry Chapter

37

Avogadrorsquos Law

38

AVOGADROrsquoSLAW

As a result of Avogadrorsquos Law equal volumes of different gases at the same temp and pressure contain equal number of moles (molecules)

2 Tanks of gas of equal volume at the same T amp P contain the same number of molecules

39

AVOGADROrsquoSLAW

For example

2 H2 (g) + 1 O2 (g) 2 H2O (g)

2 molecules 1 molecule 2 molecules

2 moles 1 mole 2 moles

2 Liters 1 Liter 2 Liters

Avogadrorsquos Law also allows chemists to relate volumes and moles of a gas in a chemical reaction

This relationship is only valid for

gaseous substances

40

Example 1

A sample of helium gas with a mass of 180 g occupies 16 Liters at a particular temperature and pressure What mass of oxygen would occupy16 L at the same temperature and pressure

1 mol= 180 g x

400 gmol of He = 450 mol

mol of O2 = mol of He = 450 mol

Same Temp amp Pressure

mass of O2

320 g= 450 mol x

1 mol= 144 g

41

Example 2

How many Liters of NH3 can be produced from reaction of 18 L of H2 with excess N2 as shown below

18 L H23

2

L NHx

L H= 12 L NH3

N2 (g) + 3 H2 (g) 2 NH3 (g)

32

42

STP amp MOLAR VOLUME

To better understand the factors that affect gas behavior a set of standard conditions have been chosen for use and are referred to as Standard Temperature and Pressure (STP)

STP =

1 atm (760 mmHg)

0ordmC (273 K)

43

STP amp MOLAR VOLUME

At STP conditions one mole of any gas is observed to occupy a volume of 224 L

V = 224 L

Molar Volume at

STP

44

Example 1

If 200 L of a gas at STP has a mass of 323 g what is the molar mass of the gas

mol200 L x

Lmol of gas =

2241

= 00893 mol

g

molMolar mass =

323 g=

00893 mol= 362 gmol Molar

volume at STP

45

Example 2

A sample of gas has a volume of 250 L at 730 mmHg and 20C What is the volume of this gas at STP

1 22 1

2 1

P TV = V x x

P T

V2 = 224 L

P1= 730 mmHg V1= 250 L T1= 293 K P2= 760 mmHg V2= T2 = 273 K

2

730 mmHg 273 KV = 250 L x x

760 mmHg 293 K

46

IDEAL GAS LAW

Combining all the laws that describe the behavior of gases one can obtain a useful relationship that relates the volume of a gas to the temperature pressure and number of moles

n R TV =

P

Universal gas constant

R = 00821 L atmmol K

47

IDEAL GAS LAW

This relationship is called the Ideal Gas Law and commonly written as

P V = n R T

Pressure in atm Volume

in Liters

Number of moles

Temp in K

48

Example 1

A sample of H2 gas has a volume of 856 L at a temp of 0C and pressure of 15 atm Calculate the moles of gas present

PVn =

RT

= 057 mol

P = 15 atm V = 856 L T = 273 K n = R = 00821

PV = nRT

(15 atm)(856 L)=

L atm(00821 )(273 K)

mol K

49

Example 2

What volume does 400 g of N2 gas occupy at 10C and 750 mmHg

P =750 mmHg V = T = 283 K n = not given m = 400 g R = 00821

mol of N2 =1 mol

400 g x280 g

= 143 mol

P = 750 mmHg1 atm

x760 mmHg

Moles must be calculated

from mass

Pressure must be

converted to atm

P = 0987 atm

50

Example 2

What volume does 400 g of N2 gas occupy at 10C and 750 mmHg

nRTV =

P

V = 337 L

P =0987 atm V = T = 283 K n = 143 R = 00821

L atm(143 mol)(00821 )(283 K)

mol KV = 0987 atm

Tros Introductory Chemistry Chapter

51

Ideal vs Real Gases

bull Real gases often do not behave like ideal gases at high pressure or low temperature

bull Ideal gas laws assume1) no attractions between gas molecules2) gas molecules do not take up space based on the Kinetic-Molecular Theory

bull at low temperatures and high pressures these assumptions are not valid

Tros Introductory Chemistry Chapter

52

Ideal vs Real Gases

53

PARTIAL PRESSURES

Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases

Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture

In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas

54

DALTONrsquoSLAW

In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture

This is called Daltonrsquos law of partial pressures

Total pressure of a gas mixture

Ptotal = P1 + P2 + P3 +

Sum of the partial pressures of the gases in the mixture

=

55

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = PHe + PAr

Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture

= 60 atm

56

PARTIAL PRESSURES

The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture

For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container

57

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = 60 atm

The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture

Twice as many moles of Ar

compared to He

PAr = 2 PHe

58

Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank

Ptotal = Poxygen + Phelium

Poxygen = 1140 mmHg x

1 atm

760 mmHg= 150 atm

Phelium

= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm

59

Example 2

A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture

Ptotal = Poxygen + Pnitrogen

Pnitrogen = 2 Poxygen

Twice as many moles of N2

compared to O2

Pnitrogen = 23 (Ptotal ) = 20 atm

Poxygen = 13 (Ptotal ) = 10 atm

60

THE END

• Slide 1
• CHAPTER OUTLINE
• PROPERTIES OF GASES
• PROPERTIES OF GASES (2)
• KINETIC-MOLECULAR THEORY
• KINETIC-MOLECULAR THEORY (2)
• Kinetic Molecular Theory
• PRESSURE amp ITS MEASUREMENT
• Common Units of Pressure
• PRESSURE amp ITS MEASUREMENT (2)
• Atmospheric Pressure amp Altitude
• Pressure Imbalance in Ear
• Example 1
• Example 2
• PRESSURE amp MOLES OF A GAS
• BOYLErsquoS LAW
• BOYLErsquoS LAW (2)
• Example 1 (2)
• Example 2 (2)
• CHARLESrsquoS LAW
• CHARLESrsquoS LAW (2)
• Example 1 (3)
• Example 2 (3)
• GAY-LUSSACrsquoS LAW
• GAY-LUSSACrsquoS LAW (2)
• Example 1 (4)
• Example 1 (5)
• Example 2 (4)
• VAPOR PRESSURE amp BOILING POINT
• VAPOR PRESSURE amp BOILING POINT (2)
• VAPOR PRESSURE amp BOILING POINT (3)
• VAPOR PRESSURE amp BOILING POINT (4)
• COMBINED GAS LAW
• COMBINED GAS LAW (2)
• Example 1 (6)
• AVOGADROrsquoS LAW
• Avogadrorsquos Law
• AVOGADROrsquoS LAW (2)
• AVOGADROrsquoS LAW (3)
• Example 1 (7)
• Example 2 (5)
• STP amp MOLAR VOLUME
• STP amp MOLAR VOLUME (2)
• Example 1 (8)
• Example 2 (6)
• IDEAL GAS LAW
• IDEAL GAS LAW (2)
• Example 1 (9)
• Example 2 (7)
• Example 2 (8)
• Ideal vs Real Gases
• Ideal vs Real Gases (2)
• PARTIAL PRESSURES
• DALTONrsquoS LAW
• PARTIAL PRESSURES (2)
• PARTIAL PRESSURES (3)
• PARTIAL PRESSURES (4)
• Example 1 (10)
• Example 2 (9)
• Slide 60

Tros Introductory Chemistry Chapter

37

Avogadrorsquos Law

38

AVOGADROrsquoSLAW

As a result of Avogadrorsquos Law equal volumes of different gases at the same temp and pressure contain equal number of moles (molecules)

2 Tanks of gas of equal volume at the same T amp P contain the same number of molecules

39

AVOGADROrsquoSLAW

For example

2 H2 (g) + 1 O2 (g) 2 H2O (g)

2 molecules 1 molecule 2 molecules

2 moles 1 mole 2 moles

2 Liters 1 Liter 2 Liters

Avogadrorsquos Law also allows chemists to relate volumes and moles of a gas in a chemical reaction

This relationship is only valid for

gaseous substances

40

Example 1

A sample of helium gas with a mass of 180 g occupies 16 Liters at a particular temperature and pressure What mass of oxygen would occupy16 L at the same temperature and pressure

1 mol= 180 g x

400 gmol of He = 450 mol

mol of O2 = mol of He = 450 mol

Same Temp amp Pressure

mass of O2

320 g= 450 mol x

1 mol= 144 g

41

Example 2

How many Liters of NH3 can be produced from reaction of 18 L of H2 with excess N2 as shown below

18 L H23

2

L NHx

L H= 12 L NH3

N2 (g) + 3 H2 (g) 2 NH3 (g)

32

42

STP amp MOLAR VOLUME

To better understand the factors that affect gas behavior a set of standard conditions have been chosen for use and are referred to as Standard Temperature and Pressure (STP)

STP =

1 atm (760 mmHg)

0ordmC (273 K)

43

STP amp MOLAR VOLUME

At STP conditions one mole of any gas is observed to occupy a volume of 224 L

V = 224 L

Molar Volume at

STP

44

Example 1

If 200 L of a gas at STP has a mass of 323 g what is the molar mass of the gas

mol200 L x

Lmol of gas =

2241

= 00893 mol

g

molMolar mass =

323 g=

00893 mol= 362 gmol Molar

volume at STP

45

Example 2

A sample of gas has a volume of 250 L at 730 mmHg and 20C What is the volume of this gas at STP

1 22 1

2 1

P TV = V x x

P T

V2 = 224 L

P1= 730 mmHg V1= 250 L T1= 293 K P2= 760 mmHg V2= T2 = 273 K

2

730 mmHg 273 KV = 250 L x x

760 mmHg 293 K

46

IDEAL GAS LAW

Combining all the laws that describe the behavior of gases one can obtain a useful relationship that relates the volume of a gas to the temperature pressure and number of moles

n R TV =

P

Universal gas constant

R = 00821 L atmmol K

47

IDEAL GAS LAW

This relationship is called the Ideal Gas Law and commonly written as

P V = n R T

Pressure in atm Volume

in Liters

Number of moles

Temp in K

48

Example 1

A sample of H2 gas has a volume of 856 L at a temp of 0C and pressure of 15 atm Calculate the moles of gas present

PVn =

RT

= 057 mol

P = 15 atm V = 856 L T = 273 K n = R = 00821

PV = nRT

(15 atm)(856 L)=

L atm(00821 )(273 K)

mol K

49

Example 2

What volume does 400 g of N2 gas occupy at 10C and 750 mmHg

P =750 mmHg V = T = 283 K n = not given m = 400 g R = 00821

mol of N2 =1 mol

400 g x280 g

= 143 mol

P = 750 mmHg1 atm

x760 mmHg

Moles must be calculated

from mass

Pressure must be

converted to atm

P = 0987 atm

50

Example 2

What volume does 400 g of N2 gas occupy at 10C and 750 mmHg

nRTV =

P

V = 337 L

P =0987 atm V = T = 283 K n = 143 R = 00821

L atm(143 mol)(00821 )(283 K)

mol KV = 0987 atm

Tros Introductory Chemistry Chapter

51

Ideal vs Real Gases

bull Real gases often do not behave like ideal gases at high pressure or low temperature

bull Ideal gas laws assume1) no attractions between gas molecules2) gas molecules do not take up space based on the Kinetic-Molecular Theory

bull at low temperatures and high pressures these assumptions are not valid

Tros Introductory Chemistry Chapter

52

Ideal vs Real Gases

53

PARTIAL PRESSURES

Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases

Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture

In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas

54

DALTONrsquoSLAW

In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture

This is called Daltonrsquos law of partial pressures

Total pressure of a gas mixture

Ptotal = P1 + P2 + P3 +

Sum of the partial pressures of the gases in the mixture

=

55

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = PHe + PAr

Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture

= 60 atm

56

PARTIAL PRESSURES

The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture

For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container

57

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = 60 atm

The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture

Twice as many moles of Ar

compared to He

PAr = 2 PHe

58

Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank

Ptotal = Poxygen + Phelium

Poxygen = 1140 mmHg x

1 atm

760 mmHg= 150 atm

Phelium

= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm

59

Example 2

A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture

Ptotal = Poxygen + Pnitrogen

Pnitrogen = 2 Poxygen

Twice as many moles of N2

compared to O2

Pnitrogen = 23 (Ptotal ) = 20 atm

Poxygen = 13 (Ptotal ) = 10 atm

60

THE END

• Slide 1
• CHAPTER OUTLINE
• PROPERTIES OF GASES
• PROPERTIES OF GASES (2)
• KINETIC-MOLECULAR THEORY
• KINETIC-MOLECULAR THEORY (2)
• Kinetic Molecular Theory
• PRESSURE amp ITS MEASUREMENT
• Common Units of Pressure
• PRESSURE amp ITS MEASUREMENT (2)
• Atmospheric Pressure amp Altitude
• Pressure Imbalance in Ear
• Example 1
• Example 2
• PRESSURE amp MOLES OF A GAS
• BOYLErsquoS LAW
• BOYLErsquoS LAW (2)
• Example 1 (2)
• Example 2 (2)
• CHARLESrsquoS LAW
• CHARLESrsquoS LAW (2)
• Example 1 (3)
• Example 2 (3)
• GAY-LUSSACrsquoS LAW
• GAY-LUSSACrsquoS LAW (2)
• Example 1 (4)
• Example 1 (5)
• Example 2 (4)
• VAPOR PRESSURE amp BOILING POINT
• VAPOR PRESSURE amp BOILING POINT (2)
• VAPOR PRESSURE amp BOILING POINT (3)
• VAPOR PRESSURE amp BOILING POINT (4)
• COMBINED GAS LAW
• COMBINED GAS LAW (2)
• Example 1 (6)
• AVOGADROrsquoS LAW
• Avogadrorsquos Law
• AVOGADROrsquoS LAW (2)
• AVOGADROrsquoS LAW (3)
• Example 1 (7)
• Example 2 (5)
• STP amp MOLAR VOLUME
• STP amp MOLAR VOLUME (2)
• Example 1 (8)
• Example 2 (6)
• IDEAL GAS LAW
• IDEAL GAS LAW (2)
• Example 1 (9)
• Example 2 (7)
• Example 2 (8)
• Ideal vs Real Gases
• Ideal vs Real Gases (2)
• PARTIAL PRESSURES
• DALTONrsquoS LAW
• PARTIAL PRESSURES (2)
• PARTIAL PRESSURES (3)
• PARTIAL PRESSURES (4)
• Example 1 (10)
• Example 2 (9)
• Slide 60

38

AVOGADROrsquoSLAW

As a result of Avogadrorsquos Law equal volumes of different gases at the same temp and pressure contain equal number of moles (molecules)

2 Tanks of gas of equal volume at the same T amp P contain the same number of molecules

39

AVOGADROrsquoSLAW

For example

2 H2 (g) + 1 O2 (g) 2 H2O (g)

2 molecules 1 molecule 2 molecules

2 moles 1 mole 2 moles

2 Liters 1 Liter 2 Liters

Avogadrorsquos Law also allows chemists to relate volumes and moles of a gas in a chemical reaction

This relationship is only valid for

gaseous substances

40

Example 1

A sample of helium gas with a mass of 180 g occupies 16 Liters at a particular temperature and pressure What mass of oxygen would occupy16 L at the same temperature and pressure

1 mol= 180 g x

400 gmol of He = 450 mol

mol of O2 = mol of He = 450 mol

Same Temp amp Pressure

mass of O2

320 g= 450 mol x

1 mol= 144 g

41

Example 2

How many Liters of NH3 can be produced from reaction of 18 L of H2 with excess N2 as shown below

18 L H23

2

L NHx

L H= 12 L NH3

N2 (g) + 3 H2 (g) 2 NH3 (g)

32

42

STP amp MOLAR VOLUME

To better understand the factors that affect gas behavior a set of standard conditions have been chosen for use and are referred to as Standard Temperature and Pressure (STP)

STP =

1 atm (760 mmHg)

0ordmC (273 K)

43

STP amp MOLAR VOLUME

At STP conditions one mole of any gas is observed to occupy a volume of 224 L

V = 224 L

Molar Volume at

STP

44

Example 1

If 200 L of a gas at STP has a mass of 323 g what is the molar mass of the gas

mol200 L x

Lmol of gas =

2241

= 00893 mol

g

molMolar mass =

323 g=

00893 mol= 362 gmol Molar

volume at STP

45

Example 2

A sample of gas has a volume of 250 L at 730 mmHg and 20C What is the volume of this gas at STP

1 22 1

2 1

P TV = V x x

P T

V2 = 224 L

P1= 730 mmHg V1= 250 L T1= 293 K P2= 760 mmHg V2= T2 = 273 K

2

730 mmHg 273 KV = 250 L x x

760 mmHg 293 K

46

IDEAL GAS LAW

Combining all the laws that describe the behavior of gases one can obtain a useful relationship that relates the volume of a gas to the temperature pressure and number of moles

n R TV =

P

Universal gas constant

R = 00821 L atmmol K

47

IDEAL GAS LAW

This relationship is called the Ideal Gas Law and commonly written as

P V = n R T

Pressure in atm Volume

in Liters

Number of moles

Temp in K

48

Example 1

A sample of H2 gas has a volume of 856 L at a temp of 0C and pressure of 15 atm Calculate the moles of gas present

PVn =

RT

= 057 mol

P = 15 atm V = 856 L T = 273 K n = R = 00821

PV = nRT

(15 atm)(856 L)=

L atm(00821 )(273 K)

mol K

49

Example 2

What volume does 400 g of N2 gas occupy at 10C and 750 mmHg

P =750 mmHg V = T = 283 K n = not given m = 400 g R = 00821

mol of N2 =1 mol

400 g x280 g

= 143 mol

P = 750 mmHg1 atm

x760 mmHg

Moles must be calculated

from mass

Pressure must be

converted to atm

P = 0987 atm

50

Example 2

What volume does 400 g of N2 gas occupy at 10C and 750 mmHg

nRTV =

P

V = 337 L

P =0987 atm V = T = 283 K n = 143 R = 00821

L atm(143 mol)(00821 )(283 K)

mol KV = 0987 atm

Tros Introductory Chemistry Chapter

51

Ideal vs Real Gases

bull Real gases often do not behave like ideal gases at high pressure or low temperature

bull Ideal gas laws assume1) no attractions between gas molecules2) gas molecules do not take up space based on the Kinetic-Molecular Theory

bull at low temperatures and high pressures these assumptions are not valid

Tros Introductory Chemistry Chapter

52

Ideal vs Real Gases

53

PARTIAL PRESSURES

Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases

Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture

In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas

54

DALTONrsquoSLAW

In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture

This is called Daltonrsquos law of partial pressures

Total pressure of a gas mixture

Ptotal = P1 + P2 + P3 +

Sum of the partial pressures of the gases in the mixture

=

55

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = PHe + PAr

Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture

= 60 atm

56

PARTIAL PRESSURES

The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture

For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container

57

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = 60 atm

The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture

Twice as many moles of Ar

compared to He

PAr = 2 PHe

58

Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank

Ptotal = Poxygen + Phelium

Poxygen = 1140 mmHg x

1 atm

760 mmHg= 150 atm

Phelium

= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm

59

Example 2

A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture

Ptotal = Poxygen + Pnitrogen

Pnitrogen = 2 Poxygen

Twice as many moles of N2

compared to O2

Pnitrogen = 23 (Ptotal ) = 20 atm

Poxygen = 13 (Ptotal ) = 10 atm

60

THE END

• Slide 1
• CHAPTER OUTLINE
• PROPERTIES OF GASES
• PROPERTIES OF GASES (2)
• KINETIC-MOLECULAR THEORY
• KINETIC-MOLECULAR THEORY (2)
• Kinetic Molecular Theory
• PRESSURE amp ITS MEASUREMENT
• Common Units of Pressure
• PRESSURE amp ITS MEASUREMENT (2)
• Atmospheric Pressure amp Altitude
• Pressure Imbalance in Ear
• Example 1
• Example 2
• PRESSURE amp MOLES OF A GAS
• BOYLErsquoS LAW
• BOYLErsquoS LAW (2)
• Example 1 (2)
• Example 2 (2)
• CHARLESrsquoS LAW
• CHARLESrsquoS LAW (2)
• Example 1 (3)
• Example 2 (3)
• GAY-LUSSACrsquoS LAW
• GAY-LUSSACrsquoS LAW (2)
• Example 1 (4)
• Example 1 (5)
• Example 2 (4)
• VAPOR PRESSURE amp BOILING POINT
• VAPOR PRESSURE amp BOILING POINT (2)
• VAPOR PRESSURE amp BOILING POINT (3)
• VAPOR PRESSURE amp BOILING POINT (4)
• COMBINED GAS LAW
• COMBINED GAS LAW (2)
• Example 1 (6)
• AVOGADROrsquoS LAW
• Avogadrorsquos Law
• AVOGADROrsquoS LAW (2)
• AVOGADROrsquoS LAW (3)
• Example 1 (7)
• Example 2 (5)
• STP amp MOLAR VOLUME
• STP amp MOLAR VOLUME (2)
• Example 1 (8)
• Example 2 (6)
• IDEAL GAS LAW
• IDEAL GAS LAW (2)
• Example 1 (9)
• Example 2 (7)
• Example 2 (8)
• Ideal vs Real Gases
• Ideal vs Real Gases (2)
• PARTIAL PRESSURES
• DALTONrsquoS LAW
• PARTIAL PRESSURES (2)
• PARTIAL PRESSURES (3)
• PARTIAL PRESSURES (4)
• Example 1 (10)
• Example 2 (9)
• Slide 60

39

AVOGADROrsquoSLAW

For example

2 H2 (g) + 1 O2 (g) 2 H2O (g)

2 molecules 1 molecule 2 molecules

2 moles 1 mole 2 moles

2 Liters 1 Liter 2 Liters

Avogadrorsquos Law also allows chemists to relate volumes and moles of a gas in a chemical reaction

This relationship is only valid for

gaseous substances

40

Example 1

A sample of helium gas with a mass of 180 g occupies 16 Liters at a particular temperature and pressure What mass of oxygen would occupy16 L at the same temperature and pressure

1 mol= 180 g x

400 gmol of He = 450 mol

mol of O2 = mol of He = 450 mol

Same Temp amp Pressure

mass of O2

320 g= 450 mol x

1 mol= 144 g

41

Example 2

How many Liters of NH3 can be produced from reaction of 18 L of H2 with excess N2 as shown below

18 L H23

2

L NHx

L H= 12 L NH3

N2 (g) + 3 H2 (g) 2 NH3 (g)

32

42

STP amp MOLAR VOLUME

To better understand the factors that affect gas behavior a set of standard conditions have been chosen for use and are referred to as Standard Temperature and Pressure (STP)

STP =

1 atm (760 mmHg)

0ordmC (273 K)

43

STP amp MOLAR VOLUME

At STP conditions one mole of any gas is observed to occupy a volume of 224 L

V = 224 L

Molar Volume at

STP

44

Example 1

If 200 L of a gas at STP has a mass of 323 g what is the molar mass of the gas

mol200 L x

Lmol of gas =

2241

= 00893 mol

g

molMolar mass =

323 g=

00893 mol= 362 gmol Molar

volume at STP

45

Example 2

A sample of gas has a volume of 250 L at 730 mmHg and 20C What is the volume of this gas at STP

1 22 1

2 1

P TV = V x x

P T

V2 = 224 L

P1= 730 mmHg V1= 250 L T1= 293 K P2= 760 mmHg V2= T2 = 273 K

2

730 mmHg 273 KV = 250 L x x

760 mmHg 293 K

46

IDEAL GAS LAW

Combining all the laws that describe the behavior of gases one can obtain a useful relationship that relates the volume of a gas to the temperature pressure and number of moles

n R TV =

P

Universal gas constant

R = 00821 L atmmol K

47

IDEAL GAS LAW

This relationship is called the Ideal Gas Law and commonly written as

P V = n R T

Pressure in atm Volume

in Liters

Number of moles

Temp in K

48

Example 1

A sample of H2 gas has a volume of 856 L at a temp of 0C and pressure of 15 atm Calculate the moles of gas present

PVn =

RT

= 057 mol

P = 15 atm V = 856 L T = 273 K n = R = 00821

PV = nRT

(15 atm)(856 L)=

L atm(00821 )(273 K)

mol K

49

Example 2

What volume does 400 g of N2 gas occupy at 10C and 750 mmHg

P =750 mmHg V = T = 283 K n = not given m = 400 g R = 00821

mol of N2 =1 mol

400 g x280 g

= 143 mol

P = 750 mmHg1 atm

x760 mmHg

Moles must be calculated

from mass

Pressure must be

converted to atm

P = 0987 atm

50

Example 2

What volume does 400 g of N2 gas occupy at 10C and 750 mmHg

nRTV =

P

V = 337 L

P =0987 atm V = T = 283 K n = 143 R = 00821

L atm(143 mol)(00821 )(283 K)

mol KV = 0987 atm

Tros Introductory Chemistry Chapter

51

Ideal vs Real Gases

bull Real gases often do not behave like ideal gases at high pressure or low temperature

bull Ideal gas laws assume1) no attractions between gas molecules2) gas molecules do not take up space based on the Kinetic-Molecular Theory

bull at low temperatures and high pressures these assumptions are not valid

Tros Introductory Chemistry Chapter

52

Ideal vs Real Gases

53

PARTIAL PRESSURES

Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases

Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture

In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas

54

DALTONrsquoSLAW

In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture

This is called Daltonrsquos law of partial pressures

Total pressure of a gas mixture

Ptotal = P1 + P2 + P3 +

Sum of the partial pressures of the gases in the mixture

=

55

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = PHe + PAr

Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture

= 60 atm

56

PARTIAL PRESSURES

The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture

For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container

57

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = 60 atm

The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture

Twice as many moles of Ar

compared to He

PAr = 2 PHe

58

Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank

Ptotal = Poxygen + Phelium

Poxygen = 1140 mmHg x

1 atm

760 mmHg= 150 atm

Phelium

= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm

59

Example 2

A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture

Ptotal = Poxygen + Pnitrogen

Pnitrogen = 2 Poxygen

Twice as many moles of N2

compared to O2

Pnitrogen = 23 (Ptotal ) = 20 atm

Poxygen = 13 (Ptotal ) = 10 atm

60

THE END

• Slide 1
• CHAPTER OUTLINE
• PROPERTIES OF GASES
• PROPERTIES OF GASES (2)
• KINETIC-MOLECULAR THEORY
• KINETIC-MOLECULAR THEORY (2)
• Kinetic Molecular Theory
• PRESSURE amp ITS MEASUREMENT
• Common Units of Pressure
• PRESSURE amp ITS MEASUREMENT (2)
• Atmospheric Pressure amp Altitude
• Pressure Imbalance in Ear
• Example 1
• Example 2
• PRESSURE amp MOLES OF A GAS
• BOYLErsquoS LAW
• BOYLErsquoS LAW (2)
• Example 1 (2)
• Example 2 (2)
• CHARLESrsquoS LAW
• CHARLESrsquoS LAW (2)
• Example 1 (3)
• Example 2 (3)
• GAY-LUSSACrsquoS LAW
• GAY-LUSSACrsquoS LAW (2)
• Example 1 (4)
• Example 1 (5)
• Example 2 (4)
• VAPOR PRESSURE amp BOILING POINT
• VAPOR PRESSURE amp BOILING POINT (2)
• VAPOR PRESSURE amp BOILING POINT (3)
• VAPOR PRESSURE amp BOILING POINT (4)
• COMBINED GAS LAW
• COMBINED GAS LAW (2)
• Example 1 (6)
• AVOGADROrsquoS LAW
• Avogadrorsquos Law
• AVOGADROrsquoS LAW (2)
• AVOGADROrsquoS LAW (3)
• Example 1 (7)
• Example 2 (5)
• STP amp MOLAR VOLUME
• STP amp MOLAR VOLUME (2)
• Example 1 (8)
• Example 2 (6)
• IDEAL GAS LAW
• IDEAL GAS LAW (2)
• Example 1 (9)
• Example 2 (7)
• Example 2 (8)
• Ideal vs Real Gases
• Ideal vs Real Gases (2)
• PARTIAL PRESSURES
• DALTONrsquoS LAW
• PARTIAL PRESSURES (2)
• PARTIAL PRESSURES (3)
• PARTIAL PRESSURES (4)
• Example 1 (10)
• Example 2 (9)
• Slide 60

40

Example 1

A sample of helium gas with a mass of 180 g occupies 16 Liters at a particular temperature and pressure What mass of oxygen would occupy16 L at the same temperature and pressure

1 mol= 180 g x

400 gmol of He = 450 mol

mol of O2 = mol of He = 450 mol

Same Temp amp Pressure

mass of O2

320 g= 450 mol x

1 mol= 144 g

41

Example 2

How many Liters of NH3 can be produced from reaction of 18 L of H2 with excess N2 as shown below

18 L H23

2

L NHx

L H= 12 L NH3

N2 (g) + 3 H2 (g) 2 NH3 (g)

32

42

STP amp MOLAR VOLUME

To better understand the factors that affect gas behavior a set of standard conditions have been chosen for use and are referred to as Standard Temperature and Pressure (STP)

STP =

1 atm (760 mmHg)

0ordmC (273 K)

43

STP amp MOLAR VOLUME

At STP conditions one mole of any gas is observed to occupy a volume of 224 L

V = 224 L

Molar Volume at

STP

44

Example 1

If 200 L of a gas at STP has a mass of 323 g what is the molar mass of the gas

mol200 L x

Lmol of gas =

2241

= 00893 mol

g

molMolar mass =

323 g=

00893 mol= 362 gmol Molar

volume at STP

45

Example 2

A sample of gas has a volume of 250 L at 730 mmHg and 20C What is the volume of this gas at STP

1 22 1

2 1

P TV = V x x

P T

V2 = 224 L

P1= 730 mmHg V1= 250 L T1= 293 K P2= 760 mmHg V2= T2 = 273 K

2

730 mmHg 273 KV = 250 L x x

760 mmHg 293 K

46

IDEAL GAS LAW

Combining all the laws that describe the behavior of gases one can obtain a useful relationship that relates the volume of a gas to the temperature pressure and number of moles

n R TV =

P

Universal gas constant

R = 00821 L atmmol K

47

IDEAL GAS LAW

This relationship is called the Ideal Gas Law and commonly written as

P V = n R T

Pressure in atm Volume

in Liters

Number of moles

Temp in K

48

Example 1

A sample of H2 gas has a volume of 856 L at a temp of 0C and pressure of 15 atm Calculate the moles of gas present

PVn =

RT

= 057 mol

P = 15 atm V = 856 L T = 273 K n = R = 00821

PV = nRT

(15 atm)(856 L)=

L atm(00821 )(273 K)

mol K

49

Example 2

What volume does 400 g of N2 gas occupy at 10C and 750 mmHg

P =750 mmHg V = T = 283 K n = not given m = 400 g R = 00821

mol of N2 =1 mol

400 g x280 g

= 143 mol

P = 750 mmHg1 atm

x760 mmHg

Moles must be calculated

from mass

Pressure must be

converted to atm

P = 0987 atm

50

Example 2

What volume does 400 g of N2 gas occupy at 10C and 750 mmHg

nRTV =

P

V = 337 L

P =0987 atm V = T = 283 K n = 143 R = 00821

L atm(143 mol)(00821 )(283 K)

mol KV = 0987 atm

Tros Introductory Chemistry Chapter

51

Ideal vs Real Gases

bull Real gases often do not behave like ideal gases at high pressure or low temperature

bull Ideal gas laws assume1) no attractions between gas molecules2) gas molecules do not take up space based on the Kinetic-Molecular Theory

bull at low temperatures and high pressures these assumptions are not valid

Tros Introductory Chemistry Chapter

52

Ideal vs Real Gases

53

PARTIAL PRESSURES

Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases

Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture

In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas

54

DALTONrsquoSLAW

In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture

This is called Daltonrsquos law of partial pressures

Total pressure of a gas mixture

Ptotal = P1 + P2 + P3 +

Sum of the partial pressures of the gases in the mixture

=

55

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = PHe + PAr

Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture

= 60 atm

56

PARTIAL PRESSURES

The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture

For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container

57

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = 60 atm

The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture

Twice as many moles of Ar

compared to He

PAr = 2 PHe

58

Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank

Ptotal = Poxygen + Phelium

Poxygen = 1140 mmHg x

1 atm

760 mmHg= 150 atm

Phelium

= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm

59

Example 2

A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture

Ptotal = Poxygen + Pnitrogen

Pnitrogen = 2 Poxygen

Twice as many moles of N2

compared to O2

Pnitrogen = 23 (Ptotal ) = 20 atm

Poxygen = 13 (Ptotal ) = 10 atm

60

THE END

• Slide 1
• CHAPTER OUTLINE
• PROPERTIES OF GASES
• PROPERTIES OF GASES (2)
• KINETIC-MOLECULAR THEORY
• KINETIC-MOLECULAR THEORY (2)
• Kinetic Molecular Theory
• PRESSURE amp ITS MEASUREMENT
• Common Units of Pressure
• PRESSURE amp ITS MEASUREMENT (2)
• Atmospheric Pressure amp Altitude
• Pressure Imbalance in Ear
• Example 1
• Example 2
• PRESSURE amp MOLES OF A GAS
• BOYLErsquoS LAW
• BOYLErsquoS LAW (2)
• Example 1 (2)
• Example 2 (2)
• CHARLESrsquoS LAW
• CHARLESrsquoS LAW (2)
• Example 1 (3)
• Example 2 (3)
• GAY-LUSSACrsquoS LAW
• GAY-LUSSACrsquoS LAW (2)
• Example 1 (4)
• Example 1 (5)
• Example 2 (4)
• VAPOR PRESSURE amp BOILING POINT
• VAPOR PRESSURE amp BOILING POINT (2)
• VAPOR PRESSURE amp BOILING POINT (3)
• VAPOR PRESSURE amp BOILING POINT (4)
• COMBINED GAS LAW
• COMBINED GAS LAW (2)
• Example 1 (6)
• AVOGADROrsquoS LAW
• Avogadrorsquos Law
• AVOGADROrsquoS LAW (2)
• AVOGADROrsquoS LAW (3)
• Example 1 (7)
• Example 2 (5)
• STP amp MOLAR VOLUME
• STP amp MOLAR VOLUME (2)
• Example 1 (8)
• Example 2 (6)
• IDEAL GAS LAW
• IDEAL GAS LAW (2)
• Example 1 (9)
• Example 2 (7)
• Example 2 (8)
• Ideal vs Real Gases
• Ideal vs Real Gases (2)
• PARTIAL PRESSURES
• DALTONrsquoS LAW
• PARTIAL PRESSURES (2)
• PARTIAL PRESSURES (3)
• PARTIAL PRESSURES (4)
• Example 1 (10)
• Example 2 (9)
• Slide 60

41

Example 2

How many Liters of NH3 can be produced from reaction of 18 L of H2 with excess N2 as shown below

18 L H23

2

L NHx

L H= 12 L NH3

N2 (g) + 3 H2 (g) 2 NH3 (g)

32

42

STP amp MOLAR VOLUME

To better understand the factors that affect gas behavior a set of standard conditions have been chosen for use and are referred to as Standard Temperature and Pressure (STP)

STP =

1 atm (760 mmHg)

0ordmC (273 K)

43

STP amp MOLAR VOLUME

At STP conditions one mole of any gas is observed to occupy a volume of 224 L

V = 224 L

Molar Volume at

STP

44

Example 1

If 200 L of a gas at STP has a mass of 323 g what is the molar mass of the gas

mol200 L x

Lmol of gas =

2241

= 00893 mol

g

molMolar mass =

323 g=

00893 mol= 362 gmol Molar

volume at STP

45

Example 2

A sample of gas has a volume of 250 L at 730 mmHg and 20C What is the volume of this gas at STP

1 22 1

2 1

P TV = V x x

P T

V2 = 224 L

P1= 730 mmHg V1= 250 L T1= 293 K P2= 760 mmHg V2= T2 = 273 K

2

730 mmHg 273 KV = 250 L x x

760 mmHg 293 K

46

IDEAL GAS LAW

Combining all the laws that describe the behavior of gases one can obtain a useful relationship that relates the volume of a gas to the temperature pressure and number of moles

n R TV =

P

Universal gas constant

R = 00821 L atmmol K

47

IDEAL GAS LAW

This relationship is called the Ideal Gas Law and commonly written as

P V = n R T

Pressure in atm Volume

in Liters

Number of moles

Temp in K

48

Example 1

A sample of H2 gas has a volume of 856 L at a temp of 0C and pressure of 15 atm Calculate the moles of gas present

PVn =

RT

= 057 mol

P = 15 atm V = 856 L T = 273 K n = R = 00821

PV = nRT

(15 atm)(856 L)=

L atm(00821 )(273 K)

mol K

49

Example 2

What volume does 400 g of N2 gas occupy at 10C and 750 mmHg

P =750 mmHg V = T = 283 K n = not given m = 400 g R = 00821

mol of N2 =1 mol

400 g x280 g

= 143 mol

P = 750 mmHg1 atm

x760 mmHg

Moles must be calculated

from mass

Pressure must be

converted to atm

P = 0987 atm

50

Example 2

What volume does 400 g of N2 gas occupy at 10C and 750 mmHg

nRTV =

P

V = 337 L

P =0987 atm V = T = 283 K n = 143 R = 00821

L atm(143 mol)(00821 )(283 K)

mol KV = 0987 atm

Tros Introductory Chemistry Chapter

51

Ideal vs Real Gases

bull Real gases often do not behave like ideal gases at high pressure or low temperature

bull Ideal gas laws assume1) no attractions between gas molecules2) gas molecules do not take up space based on the Kinetic-Molecular Theory

bull at low temperatures and high pressures these assumptions are not valid

Tros Introductory Chemistry Chapter

52

Ideal vs Real Gases

53

PARTIAL PRESSURES

Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases

Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture

In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas

54

DALTONrsquoSLAW

In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture

This is called Daltonrsquos law of partial pressures

Total pressure of a gas mixture

Ptotal = P1 + P2 + P3 +

Sum of the partial pressures of the gases in the mixture

=

55

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = PHe + PAr

Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture

= 60 atm

56

PARTIAL PRESSURES

The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture

For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container

57

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = 60 atm

The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture

Twice as many moles of Ar

compared to He

PAr = 2 PHe

58

Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank

Ptotal = Poxygen + Phelium

Poxygen = 1140 mmHg x

1 atm

760 mmHg= 150 atm

Phelium

= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm

59

Example 2

A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture

Ptotal = Poxygen + Pnitrogen

Pnitrogen = 2 Poxygen

Twice as many moles of N2

compared to O2

Pnitrogen = 23 (Ptotal ) = 20 atm

Poxygen = 13 (Ptotal ) = 10 atm

60

THE END

• Slide 1
• CHAPTER OUTLINE
• PROPERTIES OF GASES
• PROPERTIES OF GASES (2)
• KINETIC-MOLECULAR THEORY
• KINETIC-MOLECULAR THEORY (2)
• Kinetic Molecular Theory
• PRESSURE amp ITS MEASUREMENT
• Common Units of Pressure
• PRESSURE amp ITS MEASUREMENT (2)
• Atmospheric Pressure amp Altitude
• Pressure Imbalance in Ear
• Example 1
• Example 2
• PRESSURE amp MOLES OF A GAS
• BOYLErsquoS LAW
• BOYLErsquoS LAW (2)
• Example 1 (2)
• Example 2 (2)
• CHARLESrsquoS LAW
• CHARLESrsquoS LAW (2)
• Example 1 (3)
• Example 2 (3)
• GAY-LUSSACrsquoS LAW
• GAY-LUSSACrsquoS LAW (2)
• Example 1 (4)
• Example 1 (5)
• Example 2 (4)
• VAPOR PRESSURE amp BOILING POINT
• VAPOR PRESSURE amp BOILING POINT (2)
• VAPOR PRESSURE amp BOILING POINT (3)
• VAPOR PRESSURE amp BOILING POINT (4)
• COMBINED GAS LAW
• COMBINED GAS LAW (2)
• Example 1 (6)
• AVOGADROrsquoS LAW
• Avogadrorsquos Law
• AVOGADROrsquoS LAW (2)
• AVOGADROrsquoS LAW (3)
• Example 1 (7)
• Example 2 (5)
• STP amp MOLAR VOLUME
• STP amp MOLAR VOLUME (2)
• Example 1 (8)
• Example 2 (6)
• IDEAL GAS LAW
• IDEAL GAS LAW (2)
• Example 1 (9)
• Example 2 (7)
• Example 2 (8)
• Ideal vs Real Gases
• Ideal vs Real Gases (2)
• PARTIAL PRESSURES
• DALTONrsquoS LAW
• PARTIAL PRESSURES (2)
• PARTIAL PRESSURES (3)
• PARTIAL PRESSURES (4)
• Example 1 (10)
• Example 2 (9)
• Slide 60

42

STP amp MOLAR VOLUME

To better understand the factors that affect gas behavior a set of standard conditions have been chosen for use and are referred to as Standard Temperature and Pressure (STP)

STP =

1 atm (760 mmHg)

0ordmC (273 K)

43

STP amp MOLAR VOLUME

At STP conditions one mole of any gas is observed to occupy a volume of 224 L

V = 224 L

Molar Volume at

STP

44

Example 1

If 200 L of a gas at STP has a mass of 323 g what is the molar mass of the gas

mol200 L x

Lmol of gas =

2241

= 00893 mol

g

molMolar mass =

323 g=

00893 mol= 362 gmol Molar

volume at STP

45

Example 2

A sample of gas has a volume of 250 L at 730 mmHg and 20C What is the volume of this gas at STP

1 22 1

2 1

P TV = V x x

P T

V2 = 224 L

P1= 730 mmHg V1= 250 L T1= 293 K P2= 760 mmHg V2= T2 = 273 K

2

730 mmHg 273 KV = 250 L x x

760 mmHg 293 K

46

IDEAL GAS LAW

Combining all the laws that describe the behavior of gases one can obtain a useful relationship that relates the volume of a gas to the temperature pressure and number of moles

n R TV =

P

Universal gas constant

R = 00821 L atmmol K

47

IDEAL GAS LAW

This relationship is called the Ideal Gas Law and commonly written as

P V = n R T

Pressure in atm Volume

in Liters

Number of moles

Temp in K

48

Example 1

A sample of H2 gas has a volume of 856 L at a temp of 0C and pressure of 15 atm Calculate the moles of gas present

PVn =

RT

= 057 mol

P = 15 atm V = 856 L T = 273 K n = R = 00821

PV = nRT

(15 atm)(856 L)=

L atm(00821 )(273 K)

mol K

49

Example 2

What volume does 400 g of N2 gas occupy at 10C and 750 mmHg

P =750 mmHg V = T = 283 K n = not given m = 400 g R = 00821

mol of N2 =1 mol

400 g x280 g

= 143 mol

P = 750 mmHg1 atm

x760 mmHg

Moles must be calculated

from mass

Pressure must be

converted to atm

P = 0987 atm

50

Example 2

What volume does 400 g of N2 gas occupy at 10C and 750 mmHg

nRTV =

P

V = 337 L

P =0987 atm V = T = 283 K n = 143 R = 00821

L atm(143 mol)(00821 )(283 K)

mol KV = 0987 atm

Tros Introductory Chemistry Chapter

51

Ideal vs Real Gases

bull Real gases often do not behave like ideal gases at high pressure or low temperature

bull Ideal gas laws assume1) no attractions between gas molecules2) gas molecules do not take up space based on the Kinetic-Molecular Theory

bull at low temperatures and high pressures these assumptions are not valid

Tros Introductory Chemistry Chapter

52

Ideal vs Real Gases

53

PARTIAL PRESSURES

Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases

Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture

In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas

54

DALTONrsquoSLAW

In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture

This is called Daltonrsquos law of partial pressures

Total pressure of a gas mixture

Ptotal = P1 + P2 + P3 +

Sum of the partial pressures of the gases in the mixture

=

55

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = PHe + PAr

Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture

= 60 atm

56

PARTIAL PRESSURES

The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture

For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container

57

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = 60 atm

The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture

Twice as many moles of Ar

compared to He

PAr = 2 PHe

58

Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank

Ptotal = Poxygen + Phelium

Poxygen = 1140 mmHg x

1 atm

760 mmHg= 150 atm

Phelium

= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm

59

Example 2

A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture

Ptotal = Poxygen + Pnitrogen

Pnitrogen = 2 Poxygen

Twice as many moles of N2

compared to O2

Pnitrogen = 23 (Ptotal ) = 20 atm

Poxygen = 13 (Ptotal ) = 10 atm

60

THE END

• Slide 1
• CHAPTER OUTLINE
• PROPERTIES OF GASES
• PROPERTIES OF GASES (2)
• KINETIC-MOLECULAR THEORY
• KINETIC-MOLECULAR THEORY (2)
• Kinetic Molecular Theory
• PRESSURE amp ITS MEASUREMENT
• Common Units of Pressure
• PRESSURE amp ITS MEASUREMENT (2)
• Atmospheric Pressure amp Altitude
• Pressure Imbalance in Ear
• Example 1
• Example 2
• PRESSURE amp MOLES OF A GAS
• BOYLErsquoS LAW
• BOYLErsquoS LAW (2)
• Example 1 (2)
• Example 2 (2)
• CHARLESrsquoS LAW
• CHARLESrsquoS LAW (2)
• Example 1 (3)
• Example 2 (3)
• GAY-LUSSACrsquoS LAW
• GAY-LUSSACrsquoS LAW (2)
• Example 1 (4)
• Example 1 (5)
• Example 2 (4)
• VAPOR PRESSURE amp BOILING POINT
• VAPOR PRESSURE amp BOILING POINT (2)
• VAPOR PRESSURE amp BOILING POINT (3)
• VAPOR PRESSURE amp BOILING POINT (4)
• COMBINED GAS LAW
• COMBINED GAS LAW (2)
• Example 1 (6)
• AVOGADROrsquoS LAW
• Avogadrorsquos Law
• AVOGADROrsquoS LAW (2)
• AVOGADROrsquoS LAW (3)
• Example 1 (7)
• Example 2 (5)
• STP amp MOLAR VOLUME
• STP amp MOLAR VOLUME (2)
• Example 1 (8)
• Example 2 (6)
• IDEAL GAS LAW
• IDEAL GAS LAW (2)
• Example 1 (9)
• Example 2 (7)
• Example 2 (8)
• Ideal vs Real Gases
• Ideal vs Real Gases (2)
• PARTIAL PRESSURES
• DALTONrsquoS LAW
• PARTIAL PRESSURES (2)
• PARTIAL PRESSURES (3)
• PARTIAL PRESSURES (4)
• Example 1 (10)
• Example 2 (9)
• Slide 60

43

STP amp MOLAR VOLUME

At STP conditions one mole of any gas is observed to occupy a volume of 224 L

V = 224 L

Molar Volume at

STP

44

Example 1

If 200 L of a gas at STP has a mass of 323 g what is the molar mass of the gas

mol200 L x

Lmol of gas =

2241

= 00893 mol

g

molMolar mass =

323 g=

00893 mol= 362 gmol Molar

volume at STP

45

Example 2

A sample of gas has a volume of 250 L at 730 mmHg and 20C What is the volume of this gas at STP

1 22 1

2 1

P TV = V x x

P T

V2 = 224 L

P1= 730 mmHg V1= 250 L T1= 293 K P2= 760 mmHg V2= T2 = 273 K

2

730 mmHg 273 KV = 250 L x x

760 mmHg 293 K

46

IDEAL GAS LAW

Combining all the laws that describe the behavior of gases one can obtain a useful relationship that relates the volume of a gas to the temperature pressure and number of moles

n R TV =

P

Universal gas constant

R = 00821 L atmmol K

47

IDEAL GAS LAW

This relationship is called the Ideal Gas Law and commonly written as

P V = n R T

Pressure in atm Volume

in Liters

Number of moles

Temp in K

48

Example 1

A sample of H2 gas has a volume of 856 L at a temp of 0C and pressure of 15 atm Calculate the moles of gas present

PVn =

RT

= 057 mol

P = 15 atm V = 856 L T = 273 K n = R = 00821

PV = nRT

(15 atm)(856 L)=

L atm(00821 )(273 K)

mol K

49

Example 2

What volume does 400 g of N2 gas occupy at 10C and 750 mmHg

P =750 mmHg V = T = 283 K n = not given m = 400 g R = 00821

mol of N2 =1 mol

400 g x280 g

= 143 mol

P = 750 mmHg1 atm

x760 mmHg

Moles must be calculated

from mass

Pressure must be

converted to atm

P = 0987 atm

50

Example 2

What volume does 400 g of N2 gas occupy at 10C and 750 mmHg

nRTV =

P

V = 337 L

P =0987 atm V = T = 283 K n = 143 R = 00821

L atm(143 mol)(00821 )(283 K)

mol KV = 0987 atm

Tros Introductory Chemistry Chapter

51

Ideal vs Real Gases

bull Real gases often do not behave like ideal gases at high pressure or low temperature

bull Ideal gas laws assume1) no attractions between gas molecules2) gas molecules do not take up space based on the Kinetic-Molecular Theory

bull at low temperatures and high pressures these assumptions are not valid

Tros Introductory Chemistry Chapter

52

Ideal vs Real Gases

53

PARTIAL PRESSURES

Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases

Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture

In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas

54

DALTONrsquoSLAW

In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture

This is called Daltonrsquos law of partial pressures

Total pressure of a gas mixture

Ptotal = P1 + P2 + P3 +

Sum of the partial pressures of the gases in the mixture

=

55

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = PHe + PAr

Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture

= 60 atm

56

PARTIAL PRESSURES

The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture

For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container

57

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = 60 atm

The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture

Twice as many moles of Ar

compared to He

PAr = 2 PHe

58

Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank

Ptotal = Poxygen + Phelium

Poxygen = 1140 mmHg x

1 atm

760 mmHg= 150 atm

Phelium

= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm

59

Example 2

A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture

Ptotal = Poxygen + Pnitrogen

Pnitrogen = 2 Poxygen

Twice as many moles of N2

compared to O2

Pnitrogen = 23 (Ptotal ) = 20 atm

Poxygen = 13 (Ptotal ) = 10 atm

60

THE END

• Slide 1
• CHAPTER OUTLINE
• PROPERTIES OF GASES
• PROPERTIES OF GASES (2)
• KINETIC-MOLECULAR THEORY
• KINETIC-MOLECULAR THEORY (2)
• Kinetic Molecular Theory
• PRESSURE amp ITS MEASUREMENT
• Common Units of Pressure
• PRESSURE amp ITS MEASUREMENT (2)
• Atmospheric Pressure amp Altitude
• Pressure Imbalance in Ear
• Example 1
• Example 2
• PRESSURE amp MOLES OF A GAS
• BOYLErsquoS LAW
• BOYLErsquoS LAW (2)
• Example 1 (2)
• Example 2 (2)
• CHARLESrsquoS LAW
• CHARLESrsquoS LAW (2)
• Example 1 (3)
• Example 2 (3)
• GAY-LUSSACrsquoS LAW
• GAY-LUSSACrsquoS LAW (2)
• Example 1 (4)
• Example 1 (5)
• Example 2 (4)
• VAPOR PRESSURE amp BOILING POINT
• VAPOR PRESSURE amp BOILING POINT (2)
• VAPOR PRESSURE amp BOILING POINT (3)
• VAPOR PRESSURE amp BOILING POINT (4)
• COMBINED GAS LAW
• COMBINED GAS LAW (2)
• Example 1 (6)
• AVOGADROrsquoS LAW
• Avogadrorsquos Law
• AVOGADROrsquoS LAW (2)
• AVOGADROrsquoS LAW (3)
• Example 1 (7)
• Example 2 (5)
• STP amp MOLAR VOLUME
• STP amp MOLAR VOLUME (2)
• Example 1 (8)
• Example 2 (6)
• IDEAL GAS LAW
• IDEAL GAS LAW (2)
• Example 1 (9)
• Example 2 (7)
• Example 2 (8)
• Ideal vs Real Gases
• Ideal vs Real Gases (2)
• PARTIAL PRESSURES
• DALTONrsquoS LAW
• PARTIAL PRESSURES (2)
• PARTIAL PRESSURES (3)
• PARTIAL PRESSURES (4)
• Example 1 (10)
• Example 2 (9)
• Slide 60

44

Example 1

If 200 L of a gas at STP has a mass of 323 g what is the molar mass of the gas

mol200 L x

Lmol of gas =

2241

= 00893 mol

g

molMolar mass =

323 g=

00893 mol= 362 gmol Molar

volume at STP

45

Example 2

A sample of gas has a volume of 250 L at 730 mmHg and 20C What is the volume of this gas at STP

1 22 1

2 1

P TV = V x x

P T

V2 = 224 L

P1= 730 mmHg V1= 250 L T1= 293 K P2= 760 mmHg V2= T2 = 273 K

2

730 mmHg 273 KV = 250 L x x

760 mmHg 293 K

46

IDEAL GAS LAW

Combining all the laws that describe the behavior of gases one can obtain a useful relationship that relates the volume of a gas to the temperature pressure and number of moles

n R TV =

P

Universal gas constant

R = 00821 L atmmol K

47

IDEAL GAS LAW

This relationship is called the Ideal Gas Law and commonly written as

P V = n R T

Pressure in atm Volume

in Liters

Number of moles

Temp in K

48

Example 1

A sample of H2 gas has a volume of 856 L at a temp of 0C and pressure of 15 atm Calculate the moles of gas present

PVn =

RT

= 057 mol

P = 15 atm V = 856 L T = 273 K n = R = 00821

PV = nRT

(15 atm)(856 L)=

L atm(00821 )(273 K)

mol K

49

Example 2

What volume does 400 g of N2 gas occupy at 10C and 750 mmHg

P =750 mmHg V = T = 283 K n = not given m = 400 g R = 00821

mol of N2 =1 mol

400 g x280 g

= 143 mol

P = 750 mmHg1 atm

x760 mmHg

Moles must be calculated

from mass

Pressure must be

converted to atm

P = 0987 atm

50

Example 2

What volume does 400 g of N2 gas occupy at 10C and 750 mmHg

nRTV =

P

V = 337 L

P =0987 atm V = T = 283 K n = 143 R = 00821

L atm(143 mol)(00821 )(283 K)

mol KV = 0987 atm

Tros Introductory Chemistry Chapter

51

Ideal vs Real Gases

bull Real gases often do not behave like ideal gases at high pressure or low temperature

bull Ideal gas laws assume1) no attractions between gas molecules2) gas molecules do not take up space based on the Kinetic-Molecular Theory

bull at low temperatures and high pressures these assumptions are not valid

Tros Introductory Chemistry Chapter

52

Ideal vs Real Gases

53

PARTIAL PRESSURES

Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases

Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture

In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas

54

DALTONrsquoSLAW

In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture

This is called Daltonrsquos law of partial pressures

Total pressure of a gas mixture

Ptotal = P1 + P2 + P3 +

Sum of the partial pressures of the gases in the mixture

=

55

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = PHe + PAr

Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture

= 60 atm

56

PARTIAL PRESSURES

The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture

For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container

57

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = 60 atm

The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture

Twice as many moles of Ar

compared to He

PAr = 2 PHe

58

Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank

Ptotal = Poxygen + Phelium

Poxygen = 1140 mmHg x

1 atm

760 mmHg= 150 atm

Phelium

= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm

59

Example 2

A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture

Ptotal = Poxygen + Pnitrogen

Pnitrogen = 2 Poxygen

Twice as many moles of N2

compared to O2

Pnitrogen = 23 (Ptotal ) = 20 atm

Poxygen = 13 (Ptotal ) = 10 atm

60

THE END

• Slide 1
• CHAPTER OUTLINE
• PROPERTIES OF GASES
• PROPERTIES OF GASES (2)
• KINETIC-MOLECULAR THEORY
• KINETIC-MOLECULAR THEORY (2)
• Kinetic Molecular Theory
• PRESSURE amp ITS MEASUREMENT
• Common Units of Pressure
• PRESSURE amp ITS MEASUREMENT (2)
• Atmospheric Pressure amp Altitude
• Pressure Imbalance in Ear
• Example 1
• Example 2
• PRESSURE amp MOLES OF A GAS
• BOYLErsquoS LAW
• BOYLErsquoS LAW (2)
• Example 1 (2)
• Example 2 (2)
• CHARLESrsquoS LAW
• CHARLESrsquoS LAW (2)
• Example 1 (3)
• Example 2 (3)
• GAY-LUSSACrsquoS LAW
• GAY-LUSSACrsquoS LAW (2)
• Example 1 (4)
• Example 1 (5)
• Example 2 (4)
• VAPOR PRESSURE amp BOILING POINT
• VAPOR PRESSURE amp BOILING POINT (2)
• VAPOR PRESSURE amp BOILING POINT (3)
• VAPOR PRESSURE amp BOILING POINT (4)
• COMBINED GAS LAW
• COMBINED GAS LAW (2)
• Example 1 (6)
• AVOGADROrsquoS LAW
• Avogadrorsquos Law
• AVOGADROrsquoS LAW (2)
• AVOGADROrsquoS LAW (3)
• Example 1 (7)
• Example 2 (5)
• STP amp MOLAR VOLUME
• STP amp MOLAR VOLUME (2)
• Example 1 (8)
• Example 2 (6)
• IDEAL GAS LAW
• IDEAL GAS LAW (2)
• Example 1 (9)
• Example 2 (7)
• Example 2 (8)
• Ideal vs Real Gases
• Ideal vs Real Gases (2)
• PARTIAL PRESSURES
• DALTONrsquoS LAW
• PARTIAL PRESSURES (2)
• PARTIAL PRESSURES (3)
• PARTIAL PRESSURES (4)
• Example 1 (10)
• Example 2 (9)
• Slide 60

45

Example 2

A sample of gas has a volume of 250 L at 730 mmHg and 20C What is the volume of this gas at STP

1 22 1

2 1

P TV = V x x

P T

V2 = 224 L

P1= 730 mmHg V1= 250 L T1= 293 K P2= 760 mmHg V2= T2 = 273 K

2

730 mmHg 273 KV = 250 L x x

760 mmHg 293 K

46

IDEAL GAS LAW

Combining all the laws that describe the behavior of gases one can obtain a useful relationship that relates the volume of a gas to the temperature pressure and number of moles

n R TV =

P

Universal gas constant

R = 00821 L atmmol K

47

IDEAL GAS LAW

This relationship is called the Ideal Gas Law and commonly written as

P V = n R T

Pressure in atm Volume

in Liters

Number of moles

Temp in K

48

Example 1

A sample of H2 gas has a volume of 856 L at a temp of 0C and pressure of 15 atm Calculate the moles of gas present

PVn =

RT

= 057 mol

P = 15 atm V = 856 L T = 273 K n = R = 00821

PV = nRT

(15 atm)(856 L)=

L atm(00821 )(273 K)

mol K

49

Example 2

What volume does 400 g of N2 gas occupy at 10C and 750 mmHg

P =750 mmHg V = T = 283 K n = not given m = 400 g R = 00821

mol of N2 =1 mol

400 g x280 g

= 143 mol

P = 750 mmHg1 atm

x760 mmHg

Moles must be calculated

from mass

Pressure must be

converted to atm

P = 0987 atm

50

Example 2

What volume does 400 g of N2 gas occupy at 10C and 750 mmHg

nRTV =

P

V = 337 L

P =0987 atm V = T = 283 K n = 143 R = 00821

L atm(143 mol)(00821 )(283 K)

mol KV = 0987 atm

Tros Introductory Chemistry Chapter

51

Ideal vs Real Gases

bull Real gases often do not behave like ideal gases at high pressure or low temperature

bull Ideal gas laws assume1) no attractions between gas molecules2) gas molecules do not take up space based on the Kinetic-Molecular Theory

bull at low temperatures and high pressures these assumptions are not valid

Tros Introductory Chemistry Chapter

52

Ideal vs Real Gases

53

PARTIAL PRESSURES

Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases

Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture

In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas

54

DALTONrsquoSLAW

In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture

This is called Daltonrsquos law of partial pressures

Total pressure of a gas mixture

Ptotal = P1 + P2 + P3 +

Sum of the partial pressures of the gases in the mixture

=

55

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = PHe + PAr

Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture

= 60 atm

56

PARTIAL PRESSURES

The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture

For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container

57

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = 60 atm

The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture

Twice as many moles of Ar

compared to He

PAr = 2 PHe

58

Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank

Ptotal = Poxygen + Phelium

Poxygen = 1140 mmHg x

1 atm

760 mmHg= 150 atm

Phelium

= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm

59

Example 2

A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture

Ptotal = Poxygen + Pnitrogen

Pnitrogen = 2 Poxygen

Twice as many moles of N2

compared to O2

Pnitrogen = 23 (Ptotal ) = 20 atm

Poxygen = 13 (Ptotal ) = 10 atm

60

THE END

• Slide 1
• CHAPTER OUTLINE
• PROPERTIES OF GASES
• PROPERTIES OF GASES (2)
• KINETIC-MOLECULAR THEORY
• KINETIC-MOLECULAR THEORY (2)
• Kinetic Molecular Theory
• PRESSURE amp ITS MEASUREMENT
• Common Units of Pressure
• PRESSURE amp ITS MEASUREMENT (2)
• Atmospheric Pressure amp Altitude
• Pressure Imbalance in Ear
• Example 1
• Example 2
• PRESSURE amp MOLES OF A GAS
• BOYLErsquoS LAW
• BOYLErsquoS LAW (2)
• Example 1 (2)
• Example 2 (2)
• CHARLESrsquoS LAW
• CHARLESrsquoS LAW (2)
• Example 1 (3)
• Example 2 (3)
• GAY-LUSSACrsquoS LAW
• GAY-LUSSACrsquoS LAW (2)
• Example 1 (4)
• Example 1 (5)
• Example 2 (4)
• VAPOR PRESSURE amp BOILING POINT
• VAPOR PRESSURE amp BOILING POINT (2)
• VAPOR PRESSURE amp BOILING POINT (3)
• VAPOR PRESSURE amp BOILING POINT (4)
• COMBINED GAS LAW
• COMBINED GAS LAW (2)
• Example 1 (6)
• AVOGADROrsquoS LAW
• Avogadrorsquos Law
• AVOGADROrsquoS LAW (2)
• AVOGADROrsquoS LAW (3)
• Example 1 (7)
• Example 2 (5)
• STP amp MOLAR VOLUME
• STP amp MOLAR VOLUME (2)
• Example 1 (8)
• Example 2 (6)
• IDEAL GAS LAW
• IDEAL GAS LAW (2)
• Example 1 (9)
• Example 2 (7)
• Example 2 (8)
• Ideal vs Real Gases
• Ideal vs Real Gases (2)
• PARTIAL PRESSURES
• DALTONrsquoS LAW
• PARTIAL PRESSURES (2)
• PARTIAL PRESSURES (3)
• PARTIAL PRESSURES (4)
• Example 1 (10)
• Example 2 (9)
• Slide 60

46

IDEAL GAS LAW

Combining all the laws that describe the behavior of gases one can obtain a useful relationship that relates the volume of a gas to the temperature pressure and number of moles

n R TV =

P

Universal gas constant

R = 00821 L atmmol K

47

IDEAL GAS LAW

This relationship is called the Ideal Gas Law and commonly written as

P V = n R T

Pressure in atm Volume

in Liters

Number of moles

Temp in K

48

Example 1

A sample of H2 gas has a volume of 856 L at a temp of 0C and pressure of 15 atm Calculate the moles of gas present

PVn =

RT

= 057 mol

P = 15 atm V = 856 L T = 273 K n = R = 00821

PV = nRT

(15 atm)(856 L)=

L atm(00821 )(273 K)

mol K

49

Example 2

What volume does 400 g of N2 gas occupy at 10C and 750 mmHg

P =750 mmHg V = T = 283 K n = not given m = 400 g R = 00821

mol of N2 =1 mol

400 g x280 g

= 143 mol

P = 750 mmHg1 atm

x760 mmHg

Moles must be calculated

from mass

Pressure must be

converted to atm

P = 0987 atm

50

Example 2

What volume does 400 g of N2 gas occupy at 10C and 750 mmHg

nRTV =

P

V = 337 L

P =0987 atm V = T = 283 K n = 143 R = 00821

L atm(143 mol)(00821 )(283 K)

mol KV = 0987 atm

Tros Introductory Chemistry Chapter

51

Ideal vs Real Gases

bull Real gases often do not behave like ideal gases at high pressure or low temperature

bull Ideal gas laws assume1) no attractions between gas molecules2) gas molecules do not take up space based on the Kinetic-Molecular Theory

bull at low temperatures and high pressures these assumptions are not valid

Tros Introductory Chemistry Chapter

52

Ideal vs Real Gases

53

PARTIAL PRESSURES

Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases

Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture

In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas

54

DALTONrsquoSLAW

In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture

This is called Daltonrsquos law of partial pressures

Total pressure of a gas mixture

Ptotal = P1 + P2 + P3 +

Sum of the partial pressures of the gases in the mixture

=

55

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = PHe + PAr

Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture

= 60 atm

56

PARTIAL PRESSURES

The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture

For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container

57

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = 60 atm

The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture

Twice as many moles of Ar

compared to He

PAr = 2 PHe

58

Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank

Ptotal = Poxygen + Phelium

Poxygen = 1140 mmHg x

1 atm

760 mmHg= 150 atm

Phelium

= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm

59

Example 2

A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture

Ptotal = Poxygen + Pnitrogen

Pnitrogen = 2 Poxygen

Twice as many moles of N2

compared to O2

Pnitrogen = 23 (Ptotal ) = 20 atm

Poxygen = 13 (Ptotal ) = 10 atm

60

THE END

• Slide 1
• CHAPTER OUTLINE
• PROPERTIES OF GASES
• PROPERTIES OF GASES (2)
• KINETIC-MOLECULAR THEORY
• KINETIC-MOLECULAR THEORY (2)
• Kinetic Molecular Theory
• PRESSURE amp ITS MEASUREMENT
• Common Units of Pressure
• PRESSURE amp ITS MEASUREMENT (2)
• Atmospheric Pressure amp Altitude
• Pressure Imbalance in Ear
• Example 1
• Example 2
• PRESSURE amp MOLES OF A GAS
• BOYLErsquoS LAW
• BOYLErsquoS LAW (2)
• Example 1 (2)
• Example 2 (2)
• CHARLESrsquoS LAW
• CHARLESrsquoS LAW (2)
• Example 1 (3)
• Example 2 (3)
• GAY-LUSSACrsquoS LAW
• GAY-LUSSACrsquoS LAW (2)
• Example 1 (4)
• Example 1 (5)
• Example 2 (4)
• VAPOR PRESSURE amp BOILING POINT
• VAPOR PRESSURE amp BOILING POINT (2)
• VAPOR PRESSURE amp BOILING POINT (3)
• VAPOR PRESSURE amp BOILING POINT (4)
• COMBINED GAS LAW
• COMBINED GAS LAW (2)
• Example 1 (6)
• AVOGADROrsquoS LAW
• Avogadrorsquos Law
• AVOGADROrsquoS LAW (2)
• AVOGADROrsquoS LAW (3)
• Example 1 (7)
• Example 2 (5)
• STP amp MOLAR VOLUME
• STP amp MOLAR VOLUME (2)
• Example 1 (8)
• Example 2 (6)
• IDEAL GAS LAW
• IDEAL GAS LAW (2)
• Example 1 (9)
• Example 2 (7)
• Example 2 (8)
• Ideal vs Real Gases
• Ideal vs Real Gases (2)
• PARTIAL PRESSURES
• DALTONrsquoS LAW
• PARTIAL PRESSURES (2)
• PARTIAL PRESSURES (3)
• PARTIAL PRESSURES (4)
• Example 1 (10)
• Example 2 (9)
• Slide 60

47

IDEAL GAS LAW

This relationship is called the Ideal Gas Law and commonly written as

P V = n R T

Pressure in atm Volume

in Liters

Number of moles

Temp in K

48

Example 1

A sample of H2 gas has a volume of 856 L at a temp of 0C and pressure of 15 atm Calculate the moles of gas present

PVn =

RT

= 057 mol

P = 15 atm V = 856 L T = 273 K n = R = 00821

PV = nRT

(15 atm)(856 L)=

L atm(00821 )(273 K)

mol K

49

Example 2

What volume does 400 g of N2 gas occupy at 10C and 750 mmHg

P =750 mmHg V = T = 283 K n = not given m = 400 g R = 00821

mol of N2 =1 mol

400 g x280 g

= 143 mol

P = 750 mmHg1 atm

x760 mmHg

Moles must be calculated

from mass

Pressure must be

converted to atm

P = 0987 atm

50

Example 2

What volume does 400 g of N2 gas occupy at 10C and 750 mmHg

nRTV =

P

V = 337 L

P =0987 atm V = T = 283 K n = 143 R = 00821

L atm(143 mol)(00821 )(283 K)

mol KV = 0987 atm

Tros Introductory Chemistry Chapter

51

Ideal vs Real Gases

bull Real gases often do not behave like ideal gases at high pressure or low temperature

bull Ideal gas laws assume1) no attractions between gas molecules2) gas molecules do not take up space based on the Kinetic-Molecular Theory

bull at low temperatures and high pressures these assumptions are not valid

Tros Introductory Chemistry Chapter

52

Ideal vs Real Gases

53

PARTIAL PRESSURES

Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases

Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture

In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas

54

DALTONrsquoSLAW

In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture

This is called Daltonrsquos law of partial pressures

Total pressure of a gas mixture

Ptotal = P1 + P2 + P3 +

Sum of the partial pressures of the gases in the mixture

=

55

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = PHe + PAr

Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture

= 60 atm

56

PARTIAL PRESSURES

The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture

For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container

57

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = 60 atm

The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture

Twice as many moles of Ar

compared to He

PAr = 2 PHe

58

Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank

Ptotal = Poxygen + Phelium

Poxygen = 1140 mmHg x

1 atm

760 mmHg= 150 atm

Phelium

= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm

59

Example 2

A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture

Ptotal = Poxygen + Pnitrogen

Pnitrogen = 2 Poxygen

Twice as many moles of N2

compared to O2

Pnitrogen = 23 (Ptotal ) = 20 atm

Poxygen = 13 (Ptotal ) = 10 atm

60

THE END

• Slide 1
• CHAPTER OUTLINE
• PROPERTIES OF GASES
• PROPERTIES OF GASES (2)
• KINETIC-MOLECULAR THEORY
• KINETIC-MOLECULAR THEORY (2)
• Kinetic Molecular Theory
• PRESSURE amp ITS MEASUREMENT
• Common Units of Pressure
• PRESSURE amp ITS MEASUREMENT (2)
• Atmospheric Pressure amp Altitude
• Pressure Imbalance in Ear
• Example 1
• Example 2
• PRESSURE amp MOLES OF A GAS
• BOYLErsquoS LAW
• BOYLErsquoS LAW (2)
• Example 1 (2)
• Example 2 (2)
• CHARLESrsquoS LAW
• CHARLESrsquoS LAW (2)
• Example 1 (3)
• Example 2 (3)
• GAY-LUSSACrsquoS LAW
• GAY-LUSSACrsquoS LAW (2)
• Example 1 (4)
• Example 1 (5)
• Example 2 (4)
• VAPOR PRESSURE amp BOILING POINT
• VAPOR PRESSURE amp BOILING POINT (2)
• VAPOR PRESSURE amp BOILING POINT (3)
• VAPOR PRESSURE amp BOILING POINT (4)
• COMBINED GAS LAW
• COMBINED GAS LAW (2)
• Example 1 (6)
• AVOGADROrsquoS LAW
• Avogadrorsquos Law
• AVOGADROrsquoS LAW (2)
• AVOGADROrsquoS LAW (3)
• Example 1 (7)
• Example 2 (5)
• STP amp MOLAR VOLUME
• STP amp MOLAR VOLUME (2)
• Example 1 (8)
• Example 2 (6)
• IDEAL GAS LAW
• IDEAL GAS LAW (2)
• Example 1 (9)
• Example 2 (7)
• Example 2 (8)
• Ideal vs Real Gases
• Ideal vs Real Gases (2)
• PARTIAL PRESSURES
• DALTONrsquoS LAW
• PARTIAL PRESSURES (2)
• PARTIAL PRESSURES (3)
• PARTIAL PRESSURES (4)
• Example 1 (10)
• Example 2 (9)
• Slide 60

48

Example 1

A sample of H2 gas has a volume of 856 L at a temp of 0C and pressure of 15 atm Calculate the moles of gas present

PVn =

RT

= 057 mol

P = 15 atm V = 856 L T = 273 K n = R = 00821

PV = nRT

(15 atm)(856 L)=

L atm(00821 )(273 K)

mol K

49

Example 2

What volume does 400 g of N2 gas occupy at 10C and 750 mmHg

P =750 mmHg V = T = 283 K n = not given m = 400 g R = 00821

mol of N2 =1 mol

400 g x280 g

= 143 mol

P = 750 mmHg1 atm

x760 mmHg

Moles must be calculated

from mass

Pressure must be

converted to atm

P = 0987 atm

50

Example 2

What volume does 400 g of N2 gas occupy at 10C and 750 mmHg

nRTV =

P

V = 337 L

P =0987 atm V = T = 283 K n = 143 R = 00821

L atm(143 mol)(00821 )(283 K)

mol KV = 0987 atm

Tros Introductory Chemistry Chapter

51

Ideal vs Real Gases

bull Real gases often do not behave like ideal gases at high pressure or low temperature

bull Ideal gas laws assume1) no attractions between gas molecules2) gas molecules do not take up space based on the Kinetic-Molecular Theory

bull at low temperatures and high pressures these assumptions are not valid

Tros Introductory Chemistry Chapter

52

Ideal vs Real Gases

53

PARTIAL PRESSURES

Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases

Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture

In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas

54

DALTONrsquoSLAW

In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture

This is called Daltonrsquos law of partial pressures

Total pressure of a gas mixture

Ptotal = P1 + P2 + P3 +

Sum of the partial pressures of the gases in the mixture

=

55

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = PHe + PAr

Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture

= 60 atm

56

PARTIAL PRESSURES

The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture

For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container

57

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = 60 atm

The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture

Twice as many moles of Ar

compared to He

PAr = 2 PHe

58

Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank

Ptotal = Poxygen + Phelium

Poxygen = 1140 mmHg x

1 atm

760 mmHg= 150 atm

Phelium

= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm

59

Example 2

A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture

Ptotal = Poxygen + Pnitrogen

Pnitrogen = 2 Poxygen

Twice as many moles of N2

compared to O2

Pnitrogen = 23 (Ptotal ) = 20 atm

Poxygen = 13 (Ptotal ) = 10 atm

60

THE END

• Slide 1
• CHAPTER OUTLINE
• PROPERTIES OF GASES
• PROPERTIES OF GASES (2)
• KINETIC-MOLECULAR THEORY
• KINETIC-MOLECULAR THEORY (2)
• Kinetic Molecular Theory
• PRESSURE amp ITS MEASUREMENT
• Common Units of Pressure
• PRESSURE amp ITS MEASUREMENT (2)
• Atmospheric Pressure amp Altitude
• Pressure Imbalance in Ear
• Example 1
• Example 2
• PRESSURE amp MOLES OF A GAS
• BOYLErsquoS LAW
• BOYLErsquoS LAW (2)
• Example 1 (2)
• Example 2 (2)
• CHARLESrsquoS LAW
• CHARLESrsquoS LAW (2)
• Example 1 (3)
• Example 2 (3)
• GAY-LUSSACrsquoS LAW
• GAY-LUSSACrsquoS LAW (2)
• Example 1 (4)
• Example 1 (5)
• Example 2 (4)
• VAPOR PRESSURE amp BOILING POINT
• VAPOR PRESSURE amp BOILING POINT (2)
• VAPOR PRESSURE amp BOILING POINT (3)
• VAPOR PRESSURE amp BOILING POINT (4)
• COMBINED GAS LAW
• COMBINED GAS LAW (2)
• Example 1 (6)
• AVOGADROrsquoS LAW
• Avogadrorsquos Law
• AVOGADROrsquoS LAW (2)
• AVOGADROrsquoS LAW (3)
• Example 1 (7)
• Example 2 (5)
• STP amp MOLAR VOLUME
• STP amp MOLAR VOLUME (2)
• Example 1 (8)
• Example 2 (6)
• IDEAL GAS LAW
• IDEAL GAS LAW (2)
• Example 1 (9)
• Example 2 (7)
• Example 2 (8)
• Ideal vs Real Gases
• Ideal vs Real Gases (2)
• PARTIAL PRESSURES
• DALTONrsquoS LAW
• PARTIAL PRESSURES (2)
• PARTIAL PRESSURES (3)
• PARTIAL PRESSURES (4)
• Example 1 (10)
• Example 2 (9)
• Slide 60

49

Example 2

What volume does 400 g of N2 gas occupy at 10C and 750 mmHg

P =750 mmHg V = T = 283 K n = not given m = 400 g R = 00821

mol of N2 =1 mol

400 g x280 g

= 143 mol

P = 750 mmHg1 atm

x760 mmHg

Moles must be calculated

from mass

Pressure must be

converted to atm

P = 0987 atm

50

Example 2

What volume does 400 g of N2 gas occupy at 10C and 750 mmHg

nRTV =

P

V = 337 L

P =0987 atm V = T = 283 K n = 143 R = 00821

L atm(143 mol)(00821 )(283 K)

mol KV = 0987 atm

Tros Introductory Chemistry Chapter

51

Ideal vs Real Gases

bull Real gases often do not behave like ideal gases at high pressure or low temperature

bull Ideal gas laws assume1) no attractions between gas molecules2) gas molecules do not take up space based on the Kinetic-Molecular Theory

bull at low temperatures and high pressures these assumptions are not valid

Tros Introductory Chemistry Chapter

52

Ideal vs Real Gases

53

PARTIAL PRESSURES

Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases

Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture

In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas

54

DALTONrsquoSLAW

In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture

This is called Daltonrsquos law of partial pressures

Total pressure of a gas mixture

Ptotal = P1 + P2 + P3 +

Sum of the partial pressures of the gases in the mixture

=

55

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = PHe + PAr

Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture

= 60 atm

56

PARTIAL PRESSURES

The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture

For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container

57

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = 60 atm

The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture

Twice as many moles of Ar

compared to He

PAr = 2 PHe

58

Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank

Ptotal = Poxygen + Phelium

Poxygen = 1140 mmHg x

1 atm

760 mmHg= 150 atm

Phelium

= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm

59

Example 2

A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture

Ptotal = Poxygen + Pnitrogen

Pnitrogen = 2 Poxygen

Twice as many moles of N2

compared to O2

Pnitrogen = 23 (Ptotal ) = 20 atm

Poxygen = 13 (Ptotal ) = 10 atm

60

THE END

• Slide 1
• CHAPTER OUTLINE
• PROPERTIES OF GASES
• PROPERTIES OF GASES (2)
• KINETIC-MOLECULAR THEORY
• KINETIC-MOLECULAR THEORY (2)
• Kinetic Molecular Theory
• PRESSURE amp ITS MEASUREMENT
• Common Units of Pressure
• PRESSURE amp ITS MEASUREMENT (2)
• Atmospheric Pressure amp Altitude
• Pressure Imbalance in Ear
• Example 1
• Example 2
• PRESSURE amp MOLES OF A GAS
• BOYLErsquoS LAW
• BOYLErsquoS LAW (2)
• Example 1 (2)
• Example 2 (2)
• CHARLESrsquoS LAW
• CHARLESrsquoS LAW (2)
• Example 1 (3)
• Example 2 (3)
• GAY-LUSSACrsquoS LAW
• GAY-LUSSACrsquoS LAW (2)
• Example 1 (4)
• Example 1 (5)
• Example 2 (4)
• VAPOR PRESSURE amp BOILING POINT
• VAPOR PRESSURE amp BOILING POINT (2)
• VAPOR PRESSURE amp BOILING POINT (3)
• VAPOR PRESSURE amp BOILING POINT (4)
• COMBINED GAS LAW
• COMBINED GAS LAW (2)
• Example 1 (6)
• AVOGADROrsquoS LAW
• Avogadrorsquos Law
• AVOGADROrsquoS LAW (2)
• AVOGADROrsquoS LAW (3)
• Example 1 (7)
• Example 2 (5)
• STP amp MOLAR VOLUME
• STP amp MOLAR VOLUME (2)
• Example 1 (8)
• Example 2 (6)
• IDEAL GAS LAW
• IDEAL GAS LAW (2)
• Example 1 (9)
• Example 2 (7)
• Example 2 (8)
• Ideal vs Real Gases
• Ideal vs Real Gases (2)
• PARTIAL PRESSURES
• DALTONrsquoS LAW
• PARTIAL PRESSURES (2)
• PARTIAL PRESSURES (3)
• PARTIAL PRESSURES (4)
• Example 1 (10)
• Example 2 (9)
• Slide 60

50

Example 2

What volume does 400 g of N2 gas occupy at 10C and 750 mmHg

nRTV =

P

V = 337 L

P =0987 atm V = T = 283 K n = 143 R = 00821

L atm(143 mol)(00821 )(283 K)

mol KV = 0987 atm

Tros Introductory Chemistry Chapter

51

Ideal vs Real Gases

bull Real gases often do not behave like ideal gases at high pressure or low temperature

bull Ideal gas laws assume1) no attractions between gas molecules2) gas molecules do not take up space based on the Kinetic-Molecular Theory

bull at low temperatures and high pressures these assumptions are not valid

Tros Introductory Chemistry Chapter

52

Ideal vs Real Gases

53

PARTIAL PRESSURES

Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases

Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture

In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas

54

DALTONrsquoSLAW

In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture

This is called Daltonrsquos law of partial pressures

Total pressure of a gas mixture

Ptotal = P1 + P2 + P3 +

Sum of the partial pressures of the gases in the mixture

=

55

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = PHe + PAr

Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture

= 60 atm

56

PARTIAL PRESSURES

The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture

For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container

57

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = 60 atm

The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture

Twice as many moles of Ar

compared to He

PAr = 2 PHe

58

Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank

Ptotal = Poxygen + Phelium

Poxygen = 1140 mmHg x

1 atm

760 mmHg= 150 atm

Phelium

= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm

59

Example 2

A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture

Ptotal = Poxygen + Pnitrogen

Pnitrogen = 2 Poxygen

Twice as many moles of N2

compared to O2

Pnitrogen = 23 (Ptotal ) = 20 atm

Poxygen = 13 (Ptotal ) = 10 atm

60

THE END

• Slide 1
• CHAPTER OUTLINE
• PROPERTIES OF GASES
• PROPERTIES OF GASES (2)
• KINETIC-MOLECULAR THEORY
• KINETIC-MOLECULAR THEORY (2)
• Kinetic Molecular Theory
• PRESSURE amp ITS MEASUREMENT
• Common Units of Pressure
• PRESSURE amp ITS MEASUREMENT (2)
• Atmospheric Pressure amp Altitude
• Pressure Imbalance in Ear
• Example 1
• Example 2
• PRESSURE amp MOLES OF A GAS
• BOYLErsquoS LAW
• BOYLErsquoS LAW (2)
• Example 1 (2)
• Example 2 (2)
• CHARLESrsquoS LAW
• CHARLESrsquoS LAW (2)
• Example 1 (3)
• Example 2 (3)
• GAY-LUSSACrsquoS LAW
• GAY-LUSSACrsquoS LAW (2)
• Example 1 (4)
• Example 1 (5)
• Example 2 (4)
• VAPOR PRESSURE amp BOILING POINT
• VAPOR PRESSURE amp BOILING POINT (2)
• VAPOR PRESSURE amp BOILING POINT (3)
• VAPOR PRESSURE amp BOILING POINT (4)
• COMBINED GAS LAW
• COMBINED GAS LAW (2)
• Example 1 (6)
• AVOGADROrsquoS LAW
• Avogadrorsquos Law
• AVOGADROrsquoS LAW (2)
• AVOGADROrsquoS LAW (3)
• Example 1 (7)
• Example 2 (5)
• STP amp MOLAR VOLUME
• STP amp MOLAR VOLUME (2)
• Example 1 (8)
• Example 2 (6)
• IDEAL GAS LAW
• IDEAL GAS LAW (2)
• Example 1 (9)
• Example 2 (7)
• Example 2 (8)
• Ideal vs Real Gases
• Ideal vs Real Gases (2)
• PARTIAL PRESSURES
• DALTONrsquoS LAW
• PARTIAL PRESSURES (2)
• PARTIAL PRESSURES (3)
• PARTIAL PRESSURES (4)
• Example 1 (10)
• Example 2 (9)
• Slide 60

Tros Introductory Chemistry Chapter

51

Ideal vs Real Gases

bull Real gases often do not behave like ideal gases at high pressure or low temperature

bull Ideal gas laws assume1) no attractions between gas molecules2) gas molecules do not take up space based on the Kinetic-Molecular Theory

bull at low temperatures and high pressures these assumptions are not valid

Tros Introductory Chemistry Chapter

52

Ideal vs Real Gases

53

PARTIAL PRESSURES

Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases

Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture

In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas

54

DALTONrsquoSLAW

In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture

This is called Daltonrsquos law of partial pressures

Total pressure of a gas mixture

Ptotal = P1 + P2 + P3 +

Sum of the partial pressures of the gases in the mixture

=

55

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = PHe + PAr

Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture

= 60 atm

56

PARTIAL PRESSURES

The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture

For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container

57

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = 60 atm

The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture

Twice as many moles of Ar

compared to He

PAr = 2 PHe

58

Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank

Ptotal = Poxygen + Phelium

Poxygen = 1140 mmHg x

1 atm

760 mmHg= 150 atm

Phelium

= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm

59

Example 2

A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture

Ptotal = Poxygen + Pnitrogen

Pnitrogen = 2 Poxygen

Twice as many moles of N2

compared to O2

Pnitrogen = 23 (Ptotal ) = 20 atm

Poxygen = 13 (Ptotal ) = 10 atm

60

THE END

• Slide 1
• CHAPTER OUTLINE
• PROPERTIES OF GASES
• PROPERTIES OF GASES (2)
• KINETIC-MOLECULAR THEORY
• KINETIC-MOLECULAR THEORY (2)
• Kinetic Molecular Theory
• PRESSURE amp ITS MEASUREMENT
• Common Units of Pressure
• PRESSURE amp ITS MEASUREMENT (2)
• Atmospheric Pressure amp Altitude
• Pressure Imbalance in Ear
• Example 1
• Example 2
• PRESSURE amp MOLES OF A GAS
• BOYLErsquoS LAW
• BOYLErsquoS LAW (2)
• Example 1 (2)
• Example 2 (2)
• CHARLESrsquoS LAW
• CHARLESrsquoS LAW (2)
• Example 1 (3)
• Example 2 (3)
• GAY-LUSSACrsquoS LAW
• GAY-LUSSACrsquoS LAW (2)
• Example 1 (4)
• Example 1 (5)
• Example 2 (4)
• VAPOR PRESSURE amp BOILING POINT
• VAPOR PRESSURE amp BOILING POINT (2)
• VAPOR PRESSURE amp BOILING POINT (3)
• VAPOR PRESSURE amp BOILING POINT (4)
• COMBINED GAS LAW
• COMBINED GAS LAW (2)
• Example 1 (6)
• AVOGADROrsquoS LAW
• Avogadrorsquos Law
• AVOGADROrsquoS LAW (2)
• AVOGADROrsquoS LAW (3)
• Example 1 (7)
• Example 2 (5)
• STP amp MOLAR VOLUME
• STP amp MOLAR VOLUME (2)
• Example 1 (8)
• Example 2 (6)
• IDEAL GAS LAW
• IDEAL GAS LAW (2)
• Example 1 (9)
• Example 2 (7)
• Example 2 (8)
• Ideal vs Real Gases
• Ideal vs Real Gases (2)
• PARTIAL PRESSURES
• DALTONrsquoS LAW
• PARTIAL PRESSURES (2)
• PARTIAL PRESSURES (3)
• PARTIAL PRESSURES (4)
• Example 1 (10)
• Example 2 (9)
• Slide 60

Tros Introductory Chemistry Chapter

52

Ideal vs Real Gases

53

PARTIAL PRESSURES

Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases

Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture

In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas

54

DALTONrsquoSLAW

In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture

This is called Daltonrsquos law of partial pressures

Total pressure of a gas mixture

Ptotal = P1 + P2 + P3 +

Sum of the partial pressures of the gases in the mixture

=

55

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = PHe + PAr

Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture

= 60 atm

56

PARTIAL PRESSURES

The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture

For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container

57

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = 60 atm

The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture

Twice as many moles of Ar

compared to He

PAr = 2 PHe

58

Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank

Ptotal = Poxygen + Phelium

Poxygen = 1140 mmHg x

1 atm

760 mmHg= 150 atm

Phelium

= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm

59

Example 2

A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture

Ptotal = Poxygen + Pnitrogen

Pnitrogen = 2 Poxygen

Twice as many moles of N2

compared to O2

Pnitrogen = 23 (Ptotal ) = 20 atm

Poxygen = 13 (Ptotal ) = 10 atm

60

THE END

• Slide 1
• CHAPTER OUTLINE
• PROPERTIES OF GASES
• PROPERTIES OF GASES (2)
• KINETIC-MOLECULAR THEORY
• KINETIC-MOLECULAR THEORY (2)
• Kinetic Molecular Theory
• PRESSURE amp ITS MEASUREMENT
• Common Units of Pressure
• PRESSURE amp ITS MEASUREMENT (2)
• Atmospheric Pressure amp Altitude
• Pressure Imbalance in Ear
• Example 1
• Example 2
• PRESSURE amp MOLES OF A GAS
• BOYLErsquoS LAW
• BOYLErsquoS LAW (2)
• Example 1 (2)
• Example 2 (2)
• CHARLESrsquoS LAW
• CHARLESrsquoS LAW (2)
• Example 1 (3)
• Example 2 (3)
• GAY-LUSSACrsquoS LAW
• GAY-LUSSACrsquoS LAW (2)
• Example 1 (4)
• Example 1 (5)
• Example 2 (4)
• VAPOR PRESSURE amp BOILING POINT
• VAPOR PRESSURE amp BOILING POINT (2)
• VAPOR PRESSURE amp BOILING POINT (3)
• VAPOR PRESSURE amp BOILING POINT (4)
• COMBINED GAS LAW
• COMBINED GAS LAW (2)
• Example 1 (6)
• AVOGADROrsquoS LAW
• Avogadrorsquos Law
• AVOGADROrsquoS LAW (2)
• AVOGADROrsquoS LAW (3)
• Example 1 (7)
• Example 2 (5)
• STP amp MOLAR VOLUME
• STP amp MOLAR VOLUME (2)
• Example 1 (8)
• Example 2 (6)
• IDEAL GAS LAW
• IDEAL GAS LAW (2)
• Example 1 (9)
• Example 2 (7)
• Example 2 (8)
• Ideal vs Real Gases
• Ideal vs Real Gases (2)
• PARTIAL PRESSURES
• DALTONrsquoS LAW
• PARTIAL PRESSURES (2)
• PARTIAL PRESSURES (3)
• PARTIAL PRESSURES (4)
• Example 1 (10)
• Example 2 (9)
• Slide 60

53

PARTIAL PRESSURES

Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases

Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture

In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas

54

DALTONrsquoSLAW

In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture

This is called Daltonrsquos law of partial pressures

Total pressure of a gas mixture

Ptotal = P1 + P2 + P3 +

Sum of the partial pressures of the gases in the mixture

=

55

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = PHe + PAr

Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture

= 60 atm

56

PARTIAL PRESSURES

The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture

For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container

57

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = 60 atm

The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture

Twice as many moles of Ar

compared to He

PAr = 2 PHe

58

Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank

Ptotal = Poxygen + Phelium

Poxygen = 1140 mmHg x

1 atm

760 mmHg= 150 atm

Phelium

= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm

59

Example 2

A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture

Ptotal = Poxygen + Pnitrogen

Pnitrogen = 2 Poxygen

Twice as many moles of N2

compared to O2

Pnitrogen = 23 (Ptotal ) = 20 atm

Poxygen = 13 (Ptotal ) = 10 atm

60

THE END

• Slide 1
• CHAPTER OUTLINE
• PROPERTIES OF GASES
• PROPERTIES OF GASES (2)
• KINETIC-MOLECULAR THEORY
• KINETIC-MOLECULAR THEORY (2)
• Kinetic Molecular Theory
• PRESSURE amp ITS MEASUREMENT
• Common Units of Pressure
• PRESSURE amp ITS MEASUREMENT (2)
• Atmospheric Pressure amp Altitude
• Pressure Imbalance in Ear
• Example 1
• Example 2
• PRESSURE amp MOLES OF A GAS
• BOYLErsquoS LAW
• BOYLErsquoS LAW (2)
• Example 1 (2)
• Example 2 (2)
• CHARLESrsquoS LAW
• CHARLESrsquoS LAW (2)
• Example 1 (3)
• Example 2 (3)
• GAY-LUSSACrsquoS LAW
• GAY-LUSSACrsquoS LAW (2)
• Example 1 (4)
• Example 1 (5)
• Example 2 (4)
• VAPOR PRESSURE amp BOILING POINT
• VAPOR PRESSURE amp BOILING POINT (2)
• VAPOR PRESSURE amp BOILING POINT (3)
• VAPOR PRESSURE amp BOILING POINT (4)
• COMBINED GAS LAW
• COMBINED GAS LAW (2)
• Example 1 (6)
• AVOGADROrsquoS LAW
• Avogadrorsquos Law
• AVOGADROrsquoS LAW (2)
• AVOGADROrsquoS LAW (3)
• Example 1 (7)
• Example 2 (5)
• STP amp MOLAR VOLUME
• STP amp MOLAR VOLUME (2)
• Example 1 (8)
• Example 2 (6)
• IDEAL GAS LAW
• IDEAL GAS LAW (2)
• Example 1 (9)
• Example 2 (7)
• Example 2 (8)
• Ideal vs Real Gases
• Ideal vs Real Gases (2)
• PARTIAL PRESSURES
• DALTONrsquoS LAW
• PARTIAL PRESSURES (2)
• PARTIAL PRESSURES (3)
• PARTIAL PRESSURES (4)
• Example 1 (10)
• Example 2 (9)
• Slide 60

54

DALTONrsquoSLAW

In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture

This is called Daltonrsquos law of partial pressures

Total pressure of a gas mixture

Ptotal = P1 + P2 + P3 +

Sum of the partial pressures of the gases in the mixture

=

55

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = PHe + PAr

Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture

= 60 atm

56

PARTIAL PRESSURES

The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture

For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container

57

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = 60 atm

The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture

Twice as many moles of Ar

compared to He

PAr = 2 PHe

58

Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank

Ptotal = Poxygen + Phelium

Poxygen = 1140 mmHg x

1 atm

760 mmHg= 150 atm

Phelium

= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm

59

Example 2

A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture

Ptotal = Poxygen + Pnitrogen

Pnitrogen = 2 Poxygen

Twice as many moles of N2

compared to O2

Pnitrogen = 23 (Ptotal ) = 20 atm

Poxygen = 13 (Ptotal ) = 10 atm

60

THE END

• Slide 1
• CHAPTER OUTLINE
• PROPERTIES OF GASES
• PROPERTIES OF GASES (2)
• KINETIC-MOLECULAR THEORY
• KINETIC-MOLECULAR THEORY (2)
• Kinetic Molecular Theory
• PRESSURE amp ITS MEASUREMENT
• Common Units of Pressure
• PRESSURE amp ITS MEASUREMENT (2)
• Atmospheric Pressure amp Altitude
• Pressure Imbalance in Ear
• Example 1
• Example 2
• PRESSURE amp MOLES OF A GAS
• BOYLErsquoS LAW
• BOYLErsquoS LAW (2)
• Example 1 (2)
• Example 2 (2)
• CHARLESrsquoS LAW
• CHARLESrsquoS LAW (2)
• Example 1 (3)
• Example 2 (3)
• GAY-LUSSACrsquoS LAW
• GAY-LUSSACrsquoS LAW (2)
• Example 1 (4)
• Example 1 (5)
• Example 2 (4)
• VAPOR PRESSURE amp BOILING POINT
• VAPOR PRESSURE amp BOILING POINT (2)
• VAPOR PRESSURE amp BOILING POINT (3)
• VAPOR PRESSURE amp BOILING POINT (4)
• COMBINED GAS LAW
• COMBINED GAS LAW (2)
• Example 1 (6)
• AVOGADROrsquoS LAW
• Avogadrorsquos Law
• AVOGADROrsquoS LAW (2)
• AVOGADROrsquoS LAW (3)
• Example 1 (7)
• Example 2 (5)
• STP amp MOLAR VOLUME
• STP amp MOLAR VOLUME (2)
• Example 1 (8)
• Example 2 (6)
• IDEAL GAS LAW
• IDEAL GAS LAW (2)
• Example 1 (9)
• Example 2 (7)
• Example 2 (8)
• Ideal vs Real Gases
• Ideal vs Real Gases (2)
• PARTIAL PRESSURES
• DALTONrsquoS LAW
• PARTIAL PRESSURES (2)
• PARTIAL PRESSURES (3)
• PARTIAL PRESSURES (4)
• Example 1 (10)
• Example 2 (9)
• Slide 60

55

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = PHe + PAr

Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture

= 60 atm

56

PARTIAL PRESSURES

The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture

For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container

57

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = 60 atm

The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture

Twice as many moles of Ar

compared to He

PAr = 2 PHe

58

Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank

Ptotal = Poxygen + Phelium

Poxygen = 1140 mmHg x

1 atm

760 mmHg= 150 atm

Phelium

= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm

59

Example 2

A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture

Ptotal = Poxygen + Pnitrogen

Pnitrogen = 2 Poxygen

Twice as many moles of N2

compared to O2

Pnitrogen = 23 (Ptotal ) = 20 atm

Poxygen = 13 (Ptotal ) = 10 atm

60

THE END

• Slide 1
• CHAPTER OUTLINE
• PROPERTIES OF GASES
• PROPERTIES OF GASES (2)
• KINETIC-MOLECULAR THEORY
• KINETIC-MOLECULAR THEORY (2)
• Kinetic Molecular Theory
• PRESSURE amp ITS MEASUREMENT
• Common Units of Pressure
• PRESSURE amp ITS MEASUREMENT (2)
• Atmospheric Pressure amp Altitude
• Pressure Imbalance in Ear
• Example 1
• Example 2
• PRESSURE amp MOLES OF A GAS
• BOYLErsquoS LAW
• BOYLErsquoS LAW (2)
• Example 1 (2)
• Example 2 (2)
• CHARLESrsquoS LAW
• CHARLESrsquoS LAW (2)
• Example 1 (3)
• Example 2 (3)
• GAY-LUSSACrsquoS LAW
• GAY-LUSSACrsquoS LAW (2)
• Example 1 (4)
• Example 1 (5)
• Example 2 (4)
• VAPOR PRESSURE amp BOILING POINT
• VAPOR PRESSURE amp BOILING POINT (2)
• VAPOR PRESSURE amp BOILING POINT (3)
• VAPOR PRESSURE amp BOILING POINT (4)
• COMBINED GAS LAW
• COMBINED GAS LAW (2)
• Example 1 (6)
• AVOGADROrsquoS LAW
• Avogadrorsquos Law
• AVOGADROrsquoS LAW (2)
• AVOGADROrsquoS LAW (3)
• Example 1 (7)
• Example 2 (5)
• STP amp MOLAR VOLUME
• STP amp MOLAR VOLUME (2)
• Example 1 (8)
• Example 2 (6)
• IDEAL GAS LAW
• IDEAL GAS LAW (2)
• Example 1 (9)
• Example 2 (7)
• Example 2 (8)
• Ideal vs Real Gases
• Ideal vs Real Gases (2)
• PARTIAL PRESSURES
• DALTONrsquoS LAW
• PARTIAL PRESSURES (2)
• PARTIAL PRESSURES (3)
• PARTIAL PRESSURES (4)
• Example 1 (10)
• Example 2 (9)
• Slide 60

56

PARTIAL PRESSURES

The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture

For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container

57

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = 60 atm

The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture

Twice as many moles of Ar

compared to He

PAr = 2 PHe

58

Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank

Ptotal = Poxygen + Phelium

Poxygen = 1140 mmHg x

1 atm

760 mmHg= 150 atm

Phelium

= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm

59

Example 2

A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture

Ptotal = Poxygen + Pnitrogen

Pnitrogen = 2 Poxygen

Twice as many moles of N2

compared to O2

Pnitrogen = 23 (Ptotal ) = 20 atm

Poxygen = 13 (Ptotal ) = 10 atm

60

THE END

• Slide 1
• CHAPTER OUTLINE
• PROPERTIES OF GASES
• PROPERTIES OF GASES (2)
• KINETIC-MOLECULAR THEORY
• KINETIC-MOLECULAR THEORY (2)
• Kinetic Molecular Theory
• PRESSURE amp ITS MEASUREMENT
• Common Units of Pressure
• PRESSURE amp ITS MEASUREMENT (2)
• Atmospheric Pressure amp Altitude
• Pressure Imbalance in Ear
• Example 1
• Example 2
• PRESSURE amp MOLES OF A GAS
• BOYLErsquoS LAW
• BOYLErsquoS LAW (2)
• Example 1 (2)
• Example 2 (2)
• CHARLESrsquoS LAW
• CHARLESrsquoS LAW (2)
• Example 1 (3)
• Example 2 (3)
• GAY-LUSSACrsquoS LAW
• GAY-LUSSACrsquoS LAW (2)
• Example 1 (4)
• Example 1 (5)
• Example 2 (4)
• VAPOR PRESSURE amp BOILING POINT
• VAPOR PRESSURE amp BOILING POINT (2)
• VAPOR PRESSURE amp BOILING POINT (3)
• VAPOR PRESSURE amp BOILING POINT (4)
• COMBINED GAS LAW
• COMBINED GAS LAW (2)
• Example 1 (6)
• AVOGADROrsquoS LAW
• Avogadrorsquos Law
• AVOGADROrsquoS LAW (2)
• AVOGADROrsquoS LAW (3)
• Example 1 (7)
• Example 2 (5)
• STP amp MOLAR VOLUME
• STP amp MOLAR VOLUME (2)
• Example 1 (8)
• Example 2 (6)
• IDEAL GAS LAW
• IDEAL GAS LAW (2)
• Example 1 (9)
• Example 2 (7)
• Example 2 (8)
• Ideal vs Real Gases
• Ideal vs Real Gases (2)
• PARTIAL PRESSURES
• DALTONrsquoS LAW
• PARTIAL PRESSURES (2)
• PARTIAL PRESSURES (3)
• PARTIAL PRESSURES (4)
• Example 1 (10)
• Example 2 (9)
• Slide 60

57

PARTIAL PRESSURES

PHe = 20 atm

+

PAr = 40 atm PTotal = 60 atm

The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture

Twice as many moles of Ar

compared to He

PAr = 2 PHe

58

Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank

Ptotal = Poxygen + Phelium

Poxygen = 1140 mmHg x

1 atm

760 mmHg= 150 atm

Phelium

= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm

59

Example 2

A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture

Ptotal = Poxygen + Pnitrogen

Pnitrogen = 2 Poxygen

Twice as many moles of N2

compared to O2

Pnitrogen = 23 (Ptotal ) = 20 atm

Poxygen = 13 (Ptotal ) = 10 atm

60

THE END

• Slide 1
• CHAPTER OUTLINE
• PROPERTIES OF GASES
• PROPERTIES OF GASES (2)
• KINETIC-MOLECULAR THEORY
• KINETIC-MOLECULAR THEORY (2)
• Kinetic Molecular Theory
• PRESSURE amp ITS MEASUREMENT
• Common Units of Pressure
• PRESSURE amp ITS MEASUREMENT (2)
• Atmospheric Pressure amp Altitude
• Pressure Imbalance in Ear
• Example 1
• Example 2
• PRESSURE amp MOLES OF A GAS
• BOYLErsquoS LAW
• BOYLErsquoS LAW (2)
• Example 1 (2)
• Example 2 (2)
• CHARLESrsquoS LAW
• CHARLESrsquoS LAW (2)
• Example 1 (3)
• Example 2 (3)
• GAY-LUSSACrsquoS LAW
• GAY-LUSSACrsquoS LAW (2)
• Example 1 (4)
• Example 1 (5)
• Example 2 (4)
• VAPOR PRESSURE amp BOILING POINT
• VAPOR PRESSURE amp BOILING POINT (2)
• VAPOR PRESSURE amp BOILING POINT (3)
• VAPOR PRESSURE amp BOILING POINT (4)
• COMBINED GAS LAW
• COMBINED GAS LAW (2)
• Example 1 (6)
• AVOGADROrsquoS LAW
• Avogadrorsquos Law
• AVOGADROrsquoS LAW (2)
• AVOGADROrsquoS LAW (3)
• Example 1 (7)
• Example 2 (5)
• STP amp MOLAR VOLUME
• STP amp MOLAR VOLUME (2)
• Example 1 (8)
• Example 2 (6)
• IDEAL GAS LAW
• IDEAL GAS LAW (2)
• Example 1 (9)
• Example 2 (7)
• Example 2 (8)
• Ideal vs Real Gases
• Ideal vs Real Gases (2)
• PARTIAL PRESSURES
• DALTONrsquoS LAW
• PARTIAL PRESSURES (2)
• PARTIAL PRESSURES (3)
• PARTIAL PRESSURES (4)
• Example 1 (10)
• Example 2 (9)
• Slide 60

58

Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank

Ptotal = Poxygen + Phelium

Poxygen = 1140 mmHg x

1 atm

760 mmHg= 150 atm

Phelium

= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm

59

Example 2

A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture

Ptotal = Poxygen + Pnitrogen

Pnitrogen = 2 Poxygen

Twice as many moles of N2

compared to O2

Pnitrogen = 23 (Ptotal ) = 20 atm

Poxygen = 13 (Ptotal ) = 10 atm

60

THE END

• Slide 1
• CHAPTER OUTLINE
• PROPERTIES OF GASES
• PROPERTIES OF GASES (2)
• KINETIC-MOLECULAR THEORY
• KINETIC-MOLECULAR THEORY (2)
• Kinetic Molecular Theory
• PRESSURE amp ITS MEASUREMENT
• Common Units of Pressure
• PRESSURE amp ITS MEASUREMENT (2)
• Atmospheric Pressure amp Altitude
• Pressure Imbalance in Ear
• Example 1
• Example 2
• PRESSURE amp MOLES OF A GAS
• BOYLErsquoS LAW
• BOYLErsquoS LAW (2)
• Example 1 (2)
• Example 2 (2)
• CHARLESrsquoS LAW
• CHARLESrsquoS LAW (2)
• Example 1 (3)
• Example 2 (3)
• GAY-LUSSACrsquoS LAW
• GAY-LUSSACrsquoS LAW (2)
• Example 1 (4)
• Example 1 (5)
• Example 2 (4)
• VAPOR PRESSURE amp BOILING POINT
• VAPOR PRESSURE amp BOILING POINT (2)
• VAPOR PRESSURE amp BOILING POINT (3)
• VAPOR PRESSURE amp BOILING POINT (4)
• COMBINED GAS LAW
• COMBINED GAS LAW (2)
• Example 1 (6)
• AVOGADROrsquoS LAW
• Avogadrorsquos Law
• AVOGADROrsquoS LAW (2)
• AVOGADROrsquoS LAW (3)
• Example 1 (7)
• Example 2 (5)
• STP amp MOLAR VOLUME
• STP amp MOLAR VOLUME (2)
• Example 1 (8)
• Example 2 (6)
• IDEAL GAS LAW
• IDEAL GAS LAW (2)
• Example 1 (9)
• Example 2 (7)
• Example 2 (8)
• Ideal vs Real Gases
• Ideal vs Real Gases (2)
• PARTIAL PRESSURES
• DALTONrsquoS LAW
• PARTIAL PRESSURES (2)
• PARTIAL PRESSURES (3)
• PARTIAL PRESSURES (4)
• Example 1 (10)
• Example 2 (9)
• Slide 60

59

Example 2

A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture

Ptotal = Poxygen + Pnitrogen

Pnitrogen = 2 Poxygen

Twice as many moles of N2

compared to O2

Pnitrogen = 23 (Ptotal ) = 20 atm

Poxygen = 13 (Ptotal ) = 10 atm

60

THE END

• Slide 1
• CHAPTER OUTLINE
• PROPERTIES OF GASES
• PROPERTIES OF GASES (2)
• KINETIC-MOLECULAR THEORY
• KINETIC-MOLECULAR THEORY (2)
• Kinetic Molecular Theory
• PRESSURE amp ITS MEASUREMENT
• Common Units of Pressure
• PRESSURE amp ITS MEASUREMENT (2)
• Atmospheric Pressure amp Altitude
• Pressure Imbalance in Ear
• Example 1
• Example 2
• PRESSURE amp MOLES OF A GAS
• BOYLErsquoS LAW
• BOYLErsquoS LAW (2)
• Example 1 (2)
• Example 2 (2)
• CHARLESrsquoS LAW
• CHARLESrsquoS LAW (2)
• Example 1 (3)
• Example 2 (3)
• GAY-LUSSACrsquoS LAW
• GAY-LUSSACrsquoS LAW (2)
• Example 1 (4)
• Example 1 (5)
• Example 2 (4)
• VAPOR PRESSURE amp BOILING POINT
• VAPOR PRESSURE amp BOILING POINT (2)
• VAPOR PRESSURE amp BOILING POINT (3)
• VAPOR PRESSURE amp BOILING POINT (4)
• COMBINED GAS LAW
• COMBINED GAS LAW (2)
• Example 1 (6)
• AVOGADROrsquoS LAW
• Avogadrorsquos Law
• AVOGADROrsquoS LAW (2)
• AVOGADROrsquoS LAW (3)
• Example 1 (7)
• Example 2 (5)
• STP amp MOLAR VOLUME
• STP amp MOLAR VOLUME (2)
• Example 1 (8)
• Example 2 (6)
• IDEAL GAS LAW
• IDEAL GAS LAW (2)
• Example 1 (9)
• Example 2 (7)
• Example 2 (8)
• Ideal vs Real Gases
• Ideal vs Real Gases (2)
• PARTIAL PRESSURES
• DALTONrsquoS LAW
• PARTIAL PRESSURES (2)
• PARTIAL PRESSURES (3)
• PARTIAL PRESSURES (4)
• Example 1 (10)
• Example 2 (9)
• Slide 60

60

THE END

• Slide 1
• CHAPTER OUTLINE
• PROPERTIES OF GASES
• PROPERTIES OF GASES (2)
• KINETIC-MOLECULAR THEORY
• KINETIC-MOLECULAR THEORY (2)
• Kinetic Molecular Theory
• PRESSURE amp ITS MEASUREMENT
• Common Units of Pressure
• PRESSURE amp ITS MEASUREMENT (2)
• Atmospheric Pressure amp Altitude
• Pressure Imbalance in Ear
• Example 1
• Example 2
• PRESSURE amp MOLES OF A GAS
• BOYLErsquoS LAW
• BOYLErsquoS LAW (2)
• Example 1 (2)
• Example 2 (2)
• CHARLESrsquoS LAW
• CHARLESrsquoS LAW (2)
• Example 1 (3)
• Example 2 (3)
• GAY-LUSSACrsquoS LAW
• GAY-LUSSACrsquoS LAW (2)
• Example 1 (4)
• Example 1 (5)
• Example 2 (4)
• VAPOR PRESSURE amp BOILING POINT
• VAPOR PRESSURE amp BOILING POINT (2)
• VAPOR PRESSURE amp BOILING POINT (3)
• VAPOR PRESSURE amp BOILING POINT (4)
• COMBINED GAS LAW
• COMBINED GAS LAW (2)
• Example 1 (6)
• AVOGADROrsquoS LAW
• Avogadrorsquos Law
• AVOGADROrsquoS LAW (2)
• AVOGADROrsquoS LAW (3)
• Example 1 (7)
• Example 2 (5)
• STP amp MOLAR VOLUME
• STP amp MOLAR VOLUME (2)
• Example 1 (8)
• Example 2 (6)
• IDEAL GAS LAW
• IDEAL GAS LAW (2)
• Example 1 (9)
• Example 2 (7)
• Example 2 (8)
• Ideal vs Real Gases
• Ideal vs Real Gases (2)
• PARTIAL PRESSURES
• DALTONrsquoS LAW
• PARTIAL PRESSURES (2)
• PARTIAL PRESSURES (3)
• PARTIAL PRESSURES (4)
• Example 1 (10)
• Example 2 (9)
• Slide 60