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CHAPTER 12GASES

The Gas Laws Describe HOW gases behave. Can be predicted by the theory. Amount of change can be calculated

with mathematical equations.

The effect of adding gas. When we blow up a balloon we are

adding gas molecules. Doubling the the number of gas

particles doubles the pressure.

(of the same volume at the same temperature).

Pressure and the number of molecules are directly related

More molecules means more collisions.

Fewer molecules means fewer collisions.

Gases naturally move from areas of high pressure to low pressure because there is empty space to move in.

1 atm

If you double the number of molecules

If you double the number of molecules

You double the pressure.

2 atm

As you remove molecules from a container

4 atm

As you remove molecules from a container the pressure decreases

2 atm

As you remove molecules from a container the pressure decreases

Until the pressure inside equals the pressure outside

Molecules naturally move from high to low pressure

1 atm

Changing the size of the container

In a smaller container molecules have less room to move.

Hit the sides of the container more often.

As volume decreases pressure increases.

1 atm

4 Liters

As the pressure on a gas increases

2 atm

2 Liters

As the pressure on a gas increases the volume decreases

Pressure and volume are inversely related

Temperature Raising the temperature of a gas

increases the pressure if the volume is held constant.

The molecules hit the walls harder.

The only way to increase the temperature at constant pressure is to increase the volume.

If you start with 1 liter of gas at 1 atm pressure and 300 K

and heat it to 600 K one of 2 things happens

300 K

Either the volume will increase to 2 liters at 1 atm

300 K600 K

300 K 600 K

•Or the pressure will increase to 2 atm.•Or someplace in between

18

Kinetic Molecular Theory Particles in an ideal gas…

• have no volume.

• have elastic collisions.

• are in constant, random, straight-line motion.

• don’t attract or repel each other.

• have an avg. KE directly related to Kelvin temperature.

19

Real Gases Particles in a REAL gas…

• have their own volume

• attract each other

Gas behavior is most ideal…

• at low pressures

• at high temperatures

• in nonpolar atoms/molecules

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Characteristics of Gases Gases expand to fill any container.

• random motion, no attraction

Gases are fluids (like liquids).• no attraction

Gases have very low densities.• no volume = lots of empty space

21

Characteristics of Gases Gases can be compressed.

• no volume = lots of empty space

Gases undergo diffusion & effusion.• random motion

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Temperature

ºF

ºC

K

-459 32 212

-273 0 100

0 273 373

32FC 95 K = ºC + 273

Always use absolute temperature (Kelvin) when working with gases.

23

Pressure

area

forcepressure

Which shoes create the most pressure?

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Pressure Barometer

• measures atmospheric pressure

Mercury Barometer

Aneroid Barometer

25

Pressure Manometer

• measures contained gas pressure

U-tube Manometer Bourdon-tube gauge

26

Pressure

2m

NkPa

KEY UNITS AT SEA LEVEL

101.325 kPa (kilopascal)

1 atm

760 mm Hg

760 torr

14.7 psi

27

STP

Standard Temperature & PressureStandard Temperature & Pressure

0°C 273 K

1 atm 101.325 kPa-OR-

STP

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Boyle’s Law

P

V

PV = k

29

Boyle’s Law The pressure and volume of

a gas are inversely related

• at constant mass & temp

P

V

PV = k

30

kT

VV

T

Charles’ Law

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kT

VV

T

Charles’ Law The volume and absolute

temperature (K) of a gas are directly related • at constant mass & pressure

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kT

PP

T

Gay-Lussac’s Law

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kT

PP

T

Gay-Lussac’s Law The pressure and absolute

temperature (K) of a gas are directly related • at constant mass & volume

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= kPVPTVT

PVT

Combined Gas Law

P1V1

T1

=P2V2

T2

P1V1T2 = P2V2T1

35

GIVEN:

V1 = 473 cm3

T1 = 36°C = 309K

V2 = ?

T2 = 94°C = 367K

WORK:

P1V1T2 = P2V2T1

Gas Law Problems A gas occupies 473 cm3 at 36°C. Find

its volume at 94°C.

CHARLES’ LAW

T V

(473 cm3)(367 K)=V2(309 K)

V2 = 562 cm3

36

GIVEN:

V1 = 100. mL

P1 = 150. kPa

V2 = ?

P2 = 200. kPa

WORK:

P1V1T2 = P2V2T1

Gas Law Problems A gas occupies 100. mL at 150. kPa.

Find its volume at 200. kPa.

BOYLE’S LAW

P V

(150.kPa)(100.mL)=(200.kPa)V2

V2 = 75.0 mL

37

GIVEN:

V1 = 7.84 cm3

P1 = 71.8 kPa

T1 = 25°C = 298 K

V2 = ?

P2 = 101.325 kPa

T2 = 273 K

WORK:

P1V1T2 = P2V2T1

(71.8 kPa)(7.84 cm3)(273 K)

=(101.325 kPa) V2 (298 K)

V2 = 5.09 cm3

Gas Law Problems A gas occupies 7.84 cm3 at 71.8 kPa & 25°C. Find its volume at STP.

P T VCOMBINED GAS LAW

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GIVEN:

P1 = 765 torr

T1 = 23°C = 296K

P2 = 560. torr

T2 = ?

WORK:

P1V1T2 = P2V2T1

Gas Law Problems A gas’ pressure is 765 torr at 23°C.

At what temperature will the pressure be 560. torr?

GAY-LUSSAC’S LAW

P T

(765 torr)T2 = (560. torr)(309K)

T2 = 226 K = -47°C

A balloon is filled with 25 L of air at 1.0 atm pressure. If the pressure is change to 1.5 atm what is the new volume?

A balloon is filled with 73 L of air at 1.3 atm pressure. What pressure is needed to change to volume to 43 L?

Your Turn

Your Turn What is the temperature of a gas that

is expanded from 2.5 L at 25ºC to 4.1L at constant pressure.

What is the final volume of a gas that starts at 8.3 L and 17ºC and is heated to 96ºC?

Your Turn What is the pressure inside a 0.250 L

can of deodorant that starts at 25ºC and 1.2 atm if the temperature is raised to 100ºC?

At what temperature will the can above have a pressure of 2.2 atm?

Your Turn A 15 L cylinder of gas at 4.8 atm

pressure at 25ºC is heated to 75ºC and compressed to 17 atm. What is the new volume?

If 6.2 L of gas at 723 mm Hg at 21ºC is compressed to 2.2 L at 4117 mm Hg, what is the temperature of the gas?

43

kn

VV

n

Avogadro’s Principle Equal volumes of gases contain

equal numbers of moles• at constant temp & pressure• true for any gas

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PV

TVn

PVnT

Ideal Gas Law

= k

UNIVERSAL GAS CONSTANT

R=0.0821 Latm/molKR=8.315

dm3kPa/molK

= R

You don’t need to memorize these values!

Merge the Combined Gas Law with Avogadro’s Principle:

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Ideal Gas Law

UNIVERSAL GAS CONSTANT

R=0.0821 Latm/molKR=8.315

dm3kPa/molK

PV=nRT

You don’t need to memorize these values!

The Ideal Gas Law Pressure times Volume equals the

number of moles times the Ideal Gas Constant (R) times the temperature in Kelvin.

R does not depend on anything, it is really constant

R = 0.0821 (L atm)/(mol K) R = 62.4 (L mm Hg)/(K mol)

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GIVEN:

P = ? atm

n = 0.412 mol

T = 16°C = 289 K

V = 3.25 LR = 0.0821Latm/molK

WORK:

PV = nRT

P(3.25)=(0.412)(0.0821)(289) L mol Latm/molK K

P = 3.01 atm

Ideal Gas Law Problems Calculate the pressure in atmospheres of

0.412 mol of He at 16°C & occupying 3.25 L.

48

GIVEN:

V = ?

n = 85 g

T = 25°C = 298 K

P = 104.5 kPaR = 8.315 dm3kPa/molK

Ideal Gas Law Problems Find the volume of 85 g of O2 at 25°C

and 104.5 kPa.

= 2.7 mol

WORK:

85 g 1 mol = 2.7 mol

32.00 g

PV = nRT(104.5)V=(2.7) (8.315) (298) kPa mol dm3kPa/molK K

V = 64 dm3

Examples How many moles of air are there in a

2.0 L bottle at 19ºC and 747 mm Hg? What is the pressure exerted by 1.8 g

of H2 gas exert in a 4.3 L balloon at 27ºC?

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Dalton’s Law The total pressure of a mixture

of gases equals the sum of the partial pressures of the individual gases.

Ptotal = P1 + P2 + ...When a H2 gas is collected by water displacement, the gas in the collection bottle is actually a mixture of H2 and water vapor.

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GIVEN:

PH2 = ?

Ptotal = 94.4 kPa

PH2O = 2.72 kPa

WORK:

Ptotal = PH2 + PH2O

94.4 kPa = PH2 + 2.72 kPa

PH2 = 91.7 kPa

Dalton’s Law Hydrogen gas is collected over water at

22.5°C. Find the pressure of the dry gas if the atmospheric pressure is 94.4 kPa.

Look up water-vapor pressure for 22.5°C.

Sig Figs: Round to least number of decimal places.

The total pressure in the collection bottle is equal to atmospheric pressure and is a mixture of H2 and water vapor.

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GIVEN:

Pgas = ?

Ptotal = 742.0 torr

PH2O = 42.2 torr

WORK:

Ptotal = Pgas + PH2O

742.0 torr = PH2 + 42.2 torr

Pgas = 699.8 torr

A gas is collected over water at a temp of 35.0°C when the barometric pressure is 742.0 torr. What is the partial pressure of the dry gas?

Look up water-vapor pressure for 35.0°C.

Sig Figs: Round to least number of decimal places.

Dalton’s Law

The total pressure in the collection bottle is equal to barometric pressure and is a mixture of the “gas” and water vapor.

Density The Molar mass of a gas can be

determined by the density of the gas. D= mass = m

Volume V Molar mass = mass = m

Moles n n = PV

RT

Molar Mass = m (PV/RT)

Molar mass = m RT V P

Molar mass = DRT P

You might need to review Chapter 8

At STP At STP determining the amount of

gas required or produced is easy. 22.4 L = 1 mole For example

How many liters of O2 at STP are required to produce 20.3 g of H2O?

Not At STP Chemical reactions happen in

MOLES. If you know how much gas - change

it to moles Use the Ideal Gas Law n = PV/RT If you want to find how much gas -

use moles to figure out volume V = nRT/P

Example #1 HCl(g) can be formed by the

following reaction 2NaCl(aq) + H2SO4 (aq)

2HCl(g) + Na2SO4(aq) What mass of NaCl is needed to

produce 340 mL of HCl at 1.51 atm at 20ºC?

Example #2 2NaCl(aq) + H2SO4 (aq)

2HCl(g) + Na2SO4 (aq) What volume of HCl gas at 25ºC and

715 mm Hg will be generated if 10.2 g of NaCl react?

Ideal Gases don’t exist Molecules do take up space There are attractive forces otherwise there would be no liquids

Real Gases behave like Ideal Gases

When the molecules are far apart

The molecules do not take up as big a percentage of the space

We can ignore their volume.

This is at low pressure

Real Gases behave like Ideal gases when

When molecules are moving fast. Collisions are harder and faster. Molecules are not next to each other

very long. Attractive forces can’t play a role.

Diffusion

Effusion Gas escaping through a tiny hole in a container.

Depends on the speed of the molecule.

Molecules moving from areas of high concentration to low concentration.

Perfume molecules spreading across the room.

Graham’s Law The rate of effusion and diffusion is

inversely proportional to the square root of the molar mass of the molecules.

Kinetic energy = 1/2 mv2

m is the mass v is the velocity.

Chem Express

bigger molecules move slower at the same temp. (by Square root)

Bigger molecules effuse and diffuse slower

Helium effuses and diffuses faster than air - escapes from balloon.

Graham’s Law

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Graham’s Law DiffusionDiffusion

• Spreading of gas molecules throughout a container until evenly distributed.

EffusionEffusion

• Passing of gas molecules through a tiny opening in a container

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Graham’s Law

KE = ½mv2

Speed of diffusion/effusionSpeed of diffusion/effusion

• Kinetic energy is determined by the temperature of the gas.

• At the same temp & KE, heavier molecules move more slowly.

– Larger m smaller v

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Graham’s Law Graham’s LawGraham’s Law

• Rate of diffusion of a gas is inversely related to the square root of its molar mass.

• The equation shows the ratio of Gas A’s speed to Gas B’s speed.

A

B

B

A

m

m

v

v

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Determine the relative rate of diffusion for krypton and bromine.

1.381

Kr diffuses 1.381 times faster than Br2.

Kr

Br

Br

Kr

m

m

v

v2

2

A

B

B

A

m

m

v

v

g/mol83.80

g/mol159.80

Graham’s Law

The first gas is “Gas A” and the second gas is “Gas B”. Relative rate mean find the ratio “vA/vB”.

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A molecule of oxygen gas has an average speed of 12.3 m/s at a given temp and pressure. What is the average speed of hydrogen molecules at the same conditions?

A

B

B

A

m

m

v

v

2

2

2

2

H

O

O

H

m

m

v

v

g/mol 2.02

g/mol32.00

m/s 12.3

vH 2

Graham’s Law

3.980m/s 12.3

vH 2

m/s49.0 vH 2

Put the gas with the unknown

speed as “Gas A”.

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An unknown gas diffuses 4.0 times faster than O2. Find its molar mass.

Am

g/mol32.00 16

A

B

B

A

m

m

v

v

A

O

O

A

m

m

v

v2

2

Am

g/mol32.00 4.0

16

g/mol32.00 mA

2

Am

g/mol32.00 4.0

g/mol2.0

Graham’s Law

The first gas is “Gas A” and the second gas is “Gas B”. The ratio “vA/vB” is 4.0.

Square both sides to get rid of the square

root sign.

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Gas Stoichiometry Moles Moles Liters of a Gas: Liters of a Gas:

• STP - use 22.4 L/mol • Non-STP - use ideal gas law

Non-Non-STPSTP• Given liters of gas?

– start with ideal gas law

• Looking for liters of gas?– start with stoichiometry conv.

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1 molCaCO3

100.09g CaCO3

Gas Stoichiometry Problem What volume of CO2 forms from 5.25 g of

CaCO3 at 103 kPa & 25ºC?

5.25 gCaCO3

= 1.26 mol CO2

CaCO3 CaO + CO2

1 molCO2

1 molCaCO3

5.25 g ? Lnon-STP

Looking for liters: Start with stoich and calculate moles of CO2.

Plug this into the Ideal Gas Law to find liters.

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WORK:

PV = nRT

(103 kPa)V=(1mol)(8.315dm3kPa/molK)(298K)

V = 1.26 dm3 CO2

Gas Stoichiometry Problem What volume of CO2 forms from 5.25 g

of CaCO3 at 103 kPa & 25ºC?

GIVEN:

P = 103 kPaV = ?

n = 1.26 molT = 25°C = 298 KR = 8.315 dm3kPa/molK

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WORK:

PV = nRT

(97.3 kPa) (15.0 L)= n (8.315dm3kPa/molK) (294K)

n = 0.597 mol O2

Gas Stoichiometry Problem How many grams of Al2O3 are formed from 15.0

L of O2 at 97.3 kPa & 21°C?

GIVEN:

P = 97.3 kPaV = 15.0 L

n = ?T = 21°C = 294 KR = 8.315 dm3kPa/molK

4 Al + 3 O2 2 Al2O3 15.0 L

non-STP ? gGiven liters: Start with

Ideal Gas Law and calculate moles of O2.

NEXT

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2 mol Al2O3

3 mol O2

Gas Stoichiometry Problem How many grams of Al2O3 are formed

from 15.0 L of O2 at 97.3 kPa & 21°C?

0.597mol O2

= 40.6 g Al2O3

4 Al + 3 O2 2 Al2O3

101.96 g Al2O3

1 molAl2O3

15.0Lnon-STP

? gUse stoich to convert moles of O2 to grams Al2O3.