1

Gases

Chapter 5

Copyright © The McGraw-Hill Companies, Inc.  Permission required for reproduction or display.

2

Relating temperature, pressure, volume and amount of gases

Heat an aerosol can

Blow up a balloon and place it in a freezer

Adding 5 mols of nitrogen to a tank that already contains 10 mols of nitrogen

Exploding tires on a hot summer day

3

Elements that exist as gases at 250C and 1 atmosphere

4

NO2 gas

5

• Gases assume the volume and shape of their containers.

• Gases are the most compressible state of matter.

• Gases will mix evenly and completely when confined to the same container.

• Gases have much lower densities than liquids and solids.

Physical Characteristics of Gases

solid liquid gas

6

Atmospheric Pressure

Sea level 1 atm

4 miles 0.5 atm

10 miles 0.2 atm

7

Units of Pressure

1 pascal (Pa) = 1 N/m2

1 atm = 760 mmHg = 760 torr

1 atm = 101,325 Pa

Pressure = ForceArea

(force = mass x acceleration)

Barometer

8

The pressure outside of a jet plane flying at high altitude falls considerably below standard atmospheric pressure. Therefore the air inside of the cabin must be pressurized to protect the passengers. What is the pressure in the cabin if the barometer reading is 749 mmHg?

P =1 atm

760 mmHg= 0.986 atm

1 atm = 760 mm Hg

749 mmHg x

9

Problem 5.13

Convert 562 mmHg to atm.

10

The atmospheric pressure in New York City on a given day is 744 mmHg. What was the pressure in kPa?

P =101.325 kPa

760 mmHg= 99.2 kPa

1 atm = 760 mm Hg = 101.325 kPa

744 mmHg x

11

Problem 5.14The atmospheric pressure at the summit of Mt. McKinley is 606 mmHg on a certain day.

What is the pressure in atm? In kPa?

12

Manometers Used to Measure Gas Pressures

closed-tube open-tube

13

Forces Affecting Gases

• .• .• .• .

The Gas Laws

• Boyle’s Law– Robert Boyle (1627-1691)

– Relates pressure and volume at constant temperature

– For a given mass of a gas at constant temperature, the volume of the gas varied inversely with pressure.

15As P (h) increases V decreases

Apparatus for Studying the Relationship BetweenPressure and Volume of a Gas

16

P a 1/V

P x V = k

P1 x V1 = P2 x V2

Boyle’s Law

Constant temperatureConstant amount of gas

17

A sample of chlorine gas occupies a volume of 946 mL at a pressure of 726 mmHg. What is the pressure of the gas (in mmHg) if the volume is reduced at constant temperature to 154 mL?

P1 x V1 = P2 x V2

P1 = 726 mmHg

V1 = 946 mL

P2 = ?

V2 = 154 mL

P2 = P1 x V1

V2

726 mmHg x 946 mL154 mL

= = 4460 mmHg

P x V = constant

The Gas Laws

• Charles's Law– Jacques Charles (1746-1823)

– Relates temperature and volume at constant pressure

– For a given mass of a gas at constant pressure, the volume of the gas varied directly with temperature.

V = b x T

V1 = V2

T1 T2

19As T increases V increases

Variation in Gas Volume with Temperature at Constant Pressure

20

Variation of Gas Volume with Temperatureat Constant Pressure

V a T

V = constant x T

V1/T1 = V2 /T2T (K) = t (0C) + 273.15

Charles’ Law

Temperature must bein Kelvin

21

A sample of carbon monoxide gas occupies 3.20 L at 125 0C. At what temperature will the gas occupy a volume of 1.54 L if the pressure remains constant?

V1 = 3.20 L

T1 = 398.15 K

V2 = 1.54 L

T2 = ?

T2 = V2 x T1

V1

1.54 L x 398.15 K3.20 L

= = 192 K

V1 /T1 = V2 /T2

T1 = 125 (0C) + 273.15 (K) = 398.15 K

22

Problem 5.24Under constant pressure conditions a sample of hydrogen gas initially at 88ºC and 9.6 L is cooled until its final volume is 3.4 L. What is the final temperature (in K)?

V = b x T

V1 = V2

T1 T2

The Gas Laws

• Gay-Lussac's Law– Joseph Gay-Lussac (1778-1850)

• Gay-Lussac studied the effect of temperature on the pressure of a gas at constant volume

– For a given mass of a gas at constant volume, the pressure of the gas varied directly with temperature.

P1 = P2

T1 T2

24

A sample of nitrogen gas has a pressure of 2.50 kPa at 25 0C. At what temperature will the gas have a pressure of 1.54 kPa if the volume remains constant?

P1 = 2.50 kPa

T1 = 298.15 K

P2 = 1.54 kPa

T2 = ?

T2 = P2 x T1

P1

1.54 kPa x 298.15 K2.50 kPa

= = 184 K

P1 /T1 = P2 /T2

T1 = 25 (0C) + 273.15 (K) = 298.15 K

25

Problem 5.20At 46ºC a sample of ammonia gas exerts a pressure of 5.3 atm. What is the temperature when the pressure of the gas is reduced to one-tenth of the original pressure at the same volume?

26

Problem 5.22A sample of air occupies 3.8 L when the pressure is 1.2 atm. (a) What volume (in L) does it occupy at 5010mmHg if the temperature remains constant?

(b) At the original pressure and volume the temperature was 101ºC. What temperature is necessary to increase the pressure to 5010mmHg, but keep the volume constant?

27

Explain the following. State the gas law.

• Helium balloon pops at a party on a summer day

• Predict how the size of exhaled bubbles change as a scuba diver approaches the surface of the water

• Aerosol cans have a warning not to put near a flame.

28

Describe the relationships of…

1. Boyle’s Law

2. Charles’s Law

3. Gay-Lussac’s Law

29

Avogadro’s Law

V a number of moles (n)

V = a x n

V1 / n1 = V2 / n2

Constant temperatureConstant pressure

30

Ammonia burns in oxygen to form nitric oxide (NO) and water vapor. How many volumes of NO are obtained from one volume of ammonia at the same temperature and pressure?

4NH3 + 5O2 4NO + 6H2O

1 mole NH3 1 mole NO

At constant T and P

1 volume NH3 1 volume NO

31

Summary of Gas Laws

Boyle’s Law

32

Charles Law

Gay-Lussac’s Law

33

Avogadro’s Law

34

Ideal Gas Equation

Charles’ law: V a T (at constant n and P)

Avogadro’s law: V a n (at constant P and T)

Boyle’s law: P a (at constant n and T)1V

V a nT

P

V = constant x = RnT

P

nT

PR is the gas constant

PV = nRT

R = 0.082057 L • atm / (mol • K)

35

The conditions 0 0C and 1 atm are called standard temperature and pressure (STP).

PV = nRT

R = 0.082057 L • atm / (mol • K)

Calculate the volume of 1 mole of an ideal gas at STP

When the moles are constant,

PV = nR T

Therefore P1V1 = P2V2 T1 T2

36

What is the volume (in liters) occupied by 49.8 g of HCl at STP?

PV = nRT

V = nRTP

T = 0 0C = 273.15 K

P = 1 atm

n = 49.8 g x 1 mol HCl36.45 g HCl

= 1.37 mol

V =1 atm

1.37 mol x 0.0821 x 273.15 KL•atmmol•K

V = 30.7 L

37

Problem 5.40Calculate the volume (in L) of 88.4 g of CO2 at STP.

38

Argon is an inert gas used in lightbulbs to retard the vaporization of the filament. A certain lightbulb containing argon at 1.20 atm and 18 0C is heated to 85 0C at constant volume. What is the final pressure of argon in the lightbulb (in atm)?

PV = nRT n, V and R are constant

nRV

= PT

= constant

P1

T1

P2

T2

=

P1 = 1.20 atm

T1 = 291 K

P2 = ?

T2 = 358 K

P2 = P1 x T2

T1

= 1.20 atm x 358 K291 K

= 1.48 atm

39

Extra Practice: Problem 5.36The temperature of 2.5 L of a gas initially at STP is raised to 250ºC at constant volume. Calculate the final pressure of the gas in atm.

40

Extra Practice: Problem 5.32Given that 6.9 moles of carbon monoxide are present in a container of volume 30.4 L, what is the pressure of the gas (in atm) if the temperature is 62ºC?

41

Density (d) Calculations

d = mV =

PMRT

m is the mass of the gas in g

M is the molar mass of the gas

Molar Mass (M ) of a Gaseous Substance

dRTP

M = d is the density of the gas in g/L

n = mM

42

A 2.10-L vessel contains 4.65 g of a gas at 1.00 atm and 27.0 0C. What is the molar mass of the gas?

dRTP

M = d = mV

4.65 g2.10 L

= = 2.21 g

L

M =2.21

g

L

1 atm

x 0.0821 x 300.15 KL•atmmol•K

M = 54.5 g/mol

43

Problem 5.44

At 741 torr and 44ºC 7.10 g of a gas occupy a volume of 5.40 L. What is the molar mass of the gas (in g/mol)?

44

Gas Stoichiometry

What is the volume of CO2 produced at 37 0C and 1.00 atm when 5.60 g of glucose are used up in the reaction:

C6H12O6 (s) + 6O2 (g) 6CO2 (g) + 6H2O (l)

g C6H12O6 mol C6H12O6 mol CO2 V CO2

5.60 g C6H12O6

1 mol C6H12O6

180.12 g C6H12O6

x6 mol CO2

1 mol C6H12O6

x = 0.187 mol CO2

V = nRT

P

0.187 mol x 0.0821 x 310.15 KL•atmmol•K

1.00 atm= = 4.76 L

45

Problem 5.60Ethanol (C2H5OH) burns in air:

C2H5OH(l) + O2(g) CO2(g) + H2O(l)

Determine the volume of air in liters at 35.0ºC and 790 mmHg required to burn 227 g of ethanol. Assume that air is 21.0 percent O2 by volume.

46

Dalton’s Law of Partial PressuresV and T are constant

P1 P2Ptotal = P1 + P2

Studying gaseous mixtures shows that each gas behaves independantly of the others

47

Consider a case in which two gases, A and B, are in a container of volume V.

PA = nART

V

PB = nBRT

V

nA is the number of moles of A

nB is the number of moles of B

PT = PA + PB XA = nA

nA + nB

XB = nB

nA + nB

PA = XA PT PB = XB PT

Pi = Xi PT mole fraction (Xi ) = ni

nT

48

A sample of natural gas contains 8.24 moles of CH4, 0.421 moles of C2H6, and 0.116 moles of C3H8. If the total pressure of the gases is 1.37 atm, what is the partial pressure of propane (C3H8)?

Pi = Xi PT

Xpropane = 0.116

8.24 + 0.421 + 0.116

PT = 1.37 atm

= 0.0132

Ppropane = 0.0132 x 1.37 atm = 0.0181 atm

49

Problem 5.63A mixture of gases contains 0.31 mol CH4, 0.25 mol C2H6, and 0.29 mol C3H8. The total pressure is 1.50 atm. Calculate the partial pressure of each gas.

= _____ = ______ = ______4CHP 62HCP

83HCP

50

2KClO3 (s) 2KCl (s) + 3O2 (g)

PT = PO + PH O2 2

Collecting a Gas over Water

51

Vapor of Water and Temperature

52

Problem 5.66A mixture of helium and neon gasses is collected over water at 28.0ºC and 745 mmHg. If the partial pressure of helium is 368 mmHg, what is the partial pressure of neon (in mmHg)?

(vpH20 = 28.3 mmHg at 28.0ºC)

53

Problem 5.68A sample of zinc metal reacts completely with an excess of hydrochloric acid:

Zn(s) + 2HCl(aq)

The hydrogen gas produced is collected over water at 25.0ºC. The volume of the as is 7.80L, and the pressure is 0.980 atm. Calculate

the amount of zinc metal (in grams) consumed in the reaction.(vpH20 = 28.3 mmHg at 28.0ºC)