Basic Principles of Chemistry Online Southeast Missouri State University Cape Girardeau, MO...

Basic Principles of Chemistry OnlineSoutheast Missouri State University

Cape Girardeau, MO

Introductory Chemistry, 3rd EditionNivaldo Tro

Chapter 11Gases

2009, Prentice Hall

Tro's Introductory Chemistry, Chapter 11

2

Properties of Gases

• Expand to completely fill their container.• Take the shape of their container.• Low density.

Much less than solid or liquid state.• Compressible.• Mixtures of gases are always homogeneous.• Fluid.


3

The Structure of a Gas

• Gases are composed of particles that are flying around very fast in their container(s).

• They move in straight lines until they encounter either the container wall or another particle, then they bounce off.

• If you were able to take a snapshot of the particles in a gas, you would find that there is a lot of empty space in there.


4

Kinetic Molecular Theory

• The particles of the gas (either atoms or molecules) are constantly moving.

• The attraction between particles is negligible.• When the moving particles hit another particle or

the container, they do not stick, but they bounce off and continue moving in another direction.Like billiard balls.


5

Kinetic Molecular Theory of Gases

• There is a lot of empty space between the particles in a gas.Compared to the size of the particles.

• The average kinetic energy of the particles is directly proportional to the Kelvin temperature.As you raise the temperature of the gas, the average

speed of the particles increases.But don’t be fooled into thinking all the particles are

moving at the same speed!!


6

Kinetic Molecular Theory


7

Gas Particles Pushing• Gas molecules are constantly in motion.• As they move and strike a surface, they

push on that surface.Push = force.

• If we could measure the total amount of force exerted by gas molecules hitting the entire surface at any one instant, we would know the pressure the gas is exerting.Pressure = force per unit area.


8

The Effect of Gas Pressure• The pressure exerted by a gas can cause

some amazing and startling effects.• Whenever there is a pressure difference, a

gas will flow from area of high pressure to low pressure.The bigger the difference in pressure, the

stronger the flow of the gas.• If there is something in the gas’ path, the gas

will try to push it along as the gas flows.


9

Which Way Would Air Flow?

Two filled balloons are connected with a long pipe. One of the balloons is plunged down into the water. Which way will the air flow? Will air flow from the lower balloon toward the top balloon; or will it flow from the top balloon to the bottom one?


10

Is This Possible at a Depth of 20 m?


11

Soda Straws and Gas Pressure

The pressure of the air inside the straw is the same as the pressureof the air outsidethe straw—so liquid levels arethe same on bothsides.

The pressure of theair inside the straw

is lower than the pressure

of the air outsidethe straw—so

liquid is pushedup the straw bythe outside air.


12

Gas Properties Explained• Gases have taken the shape and volume of their

container(s) because the particles don’t stick together, allowing them to move and fill the container(s) they’re in. In solids and liquids, the particles are attracted to each

other strongly enough so they stick together. • Gases are compressible and have low density

because of the large amount of unoccupied space between the particles.


13

Properties—Indefinite Shape and Indefinite Volume

Because the gasmolecules have enough kineticenergy to overcomeattractions, theykeep moving aroundand spreading outuntil they fill the container.

As a result, gasestake the shape andthe volume of thecontainer they are in.


14

Properties—Compressibility

Because there is a lot of unoccupied space in the structureof a gas, the gas molecules can be squeezed closer together.


15

Gas Properties Explained—Low Density

Because there is a lot of unoccupied space in the structure of a gas, gases do not have a lot of mass in a given volume, the result is that they have low density.


16

The Pressure of a Gas• Pressure is the result of the

constant movement of the gas molecules and their collisions with the surfaces around them.

• The pressure of a gas depends on several factors:Number of gas particles in a

given volume.Volume of the container.Average speed of the gas

particles.


17

Density and Pressure • Pressure is the result of the

constant movement of the gas molecules and their collisions with the surfaces around them.

• When more molecules are added, more molecules hit the container at any one instant, resulting in higher pressure.Also higher density.


18

Air Pressure• The atmosphere exerts a pressure

on everything it contacts.On average 14.7 psi.The atmosphere goes up about 370

miles, but 80% is in the first 10 miles from Earth’s surface.

• This is the same pressure that a column of water would exert if it were about 10.3 m high.


19

Measuring Air Pressure• Use a barometer.• Column of mercury

supported by air pressure.

• Force of the air on the surface of the mercury balanced by the pull of gravity on the column of mercury.

gravity


20

Atmospheric Pressure and Altitude

• The higher up in the atmosphere you go, the lower the atmospheric pressure is around you.At the surface, the atmospheric pressure is 14.7

psi, but at 10,000 ft is is only 10.0 psi.• Rapid changes in atmospheric pressure may

cause your ears to “pop” due to an imbalance in pressure on either side of your ear drum.


21

Pressure Imbalance in Ear

If there is a differencein pressure acrossthe eardrum membrane,the membrane will bepushed out—what we commonly call a “popped eardrum.”


22

Common Units of PressureUnit Average air pressure at

sea levelPascal (Pa) 101,325

Kilopascal (kPa) 101.325

Atmosphere (atm) 1 (exactly)

Millimeters of mercury (mmHg) 760 (exactly)

Inches of mercury (inHg) 29.92

Torr (torr) 760 (exactly)

Pounds per square inch (psi, lbs./in2) 14.7

Example 11.1—A High-Performance Bicycle Tire Has a Pressure of 125 psi. What Is the Pressure in mmHg?

Since mmHg are smaller than psi, the answer makes sense.

1 atm = 14.7 psi, 1 atm = 760 mmHg

125 psimmHg

Check:

Solution:

Solution Map:

Relationships:

Given:

Find:

psi 14.7atm 1

mmHg 10.466 atm 1mmHg 760

psi 14.7atm 1psi 125 3

atm 1mmHg 760

psi atm mmHg


31

Practice—Convert 45.5 psi into kPa.

Unit Average air pressure at sea level

Pascal (Pa) 101,325

Kilopascal (kPa) 101.325

Atmosphere (atm) 1 (exactly)

Millimeters of mercury (mmHg) 760 (exactly)

Inches of mercury (inHg) 29.92

Torr (torr) 760 (exactly)

Pounds per square inch (psi, lbs./in2) 14.7

Practice—Convert 45.5 psi into kPa, Continued

Since kPa are smaller than psi, the answer makes sense.

1 atm = 14.7 psi, 1 atm = 101.325 kPa

45.5 psikPa

Check:

Solution:

Concept Plan:

Relationships:

Given:

Find:

psi 14.7atm 1

kPa 413 atm 1

kPa 325.101psi 14.7

atm 1psi 5.54

atm 1mmHg 760

psi atm kPa


33

Boyle’s Law• Pressure of a gas is inversely

proportional to its volume.Constant T and amount of gas.Graph P vs. V is curved.Graph P vs. 1/V is in a straight

line.• As P increases, V decreases

by the same factor.• P x V = constant.• P1 x V1 = P2 x V2 .


34

Boyle’s Experiment• Added Hg to a J-tube with

air trapped inside.• Used length of air column

as a measure of volume.

Length of air in column

(in)

Difference in Hg levels

(in) 48 0.0 44 2.8 40 6.2 36 10.1 32 15.1 28 21.2 24 29.7 22 35.0


35

Boyle's Experiment

0

20

40

60

80

100

120

140

0 10 20 30 40 50 60

Volume of Air, in3

Pres

sure

, inH

g


36

Inverse Volume vs. Pressure of Air, Boyle's Experiment

0

20

40

60

80

100

120

140

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09

Inv. Volume, in-3

Pres

sure

, inH

g


37

Boyle’s Experiment, P x VPressure Volume P x V

29.13 48 140033.50 42 140041.63 34 140050.31 28 140061.31 23 140074.13 19 140087.88 16 1400

115.56 12 1400


38

When you double the pressure on a gas,the volume is cut in half (as long as the

temperature and amount of gas do not change).


39

Gas Laws Explained— Boyle’s Law

• Boyle’s law says that the volume of a gas is inversely proportional to the pressure.

• Decreasing the volume forces the molecules into a smaller space.

• More molecules will collide with the container at any one instant, increasing the pressure.


40

Boyle’s Law and Diving• Since water is more dense

than air, for each 10 m you dive below the surface, the pressure on your lungs increases 1 atm.At 20 m the total pressure is

3 atm.• If your tank contained air at

1 atm of pressure, you would not be able to inhale it into your lungs.You can only generate

enough force to overcome about 1.06 atm.

Scuba tanks have a regulator so that the air from the tank is delivered at the same pressure as the water surrounding you.This allows you to take in air even when the outside pressure is large.


41

Boyle’s Law and Diving, Continued

• If a diver holds her breath and rises to the surface quickly, the outside pressure drops to 1 atm.

• According to Boyle’s law, what should happen to the volume of air in the lungs?

• Since the pressure is decreasing by a factor of 3, the volume will expand by a factor of 3, causing damage to internal organs. Always Exhale When Rising!!

P1 ∙ V1 = P2 ∙ V2

Example 11.2—A Cylinder with a Movable Piston Has a Volume of 6.0 L at 4.0 atm. What Is the Volume at 1.0 atm?

Since P and V are inversely proportional, when the pressure decreases ~4x, the volume should increase ~4x, and it does.

V1 =6.0 L, P1 = 4.0 atm, P2 = 1.0 atm

V2, L

Check:

Solution:

Solution Map:

Relationships:

Given:

Find:

2

112 P

VPV

V1, P1, P2 V2

L 24atm 1.0

L 6.0atm 4.02

112

PVPV


50

Practice—A Balloon Is Put in a Bell Jar and the Pressure Is Reduced from 782 torr to 0.500 atm. If the Volume of the Balloon Is Now 2780 mL, What Was It Originally?

(1 atm = 760 torr)

P1 ∙ V1 = P2 ∙ V2 , 1 atm = 760 torr (exactly)

Practice—A Balloon Is Put in a Bell Jar and the Pressure Is Reduced from 782 torr to 0.500 atm. If the Volume of the Balloon

Is Now 2780 mL, What Was It Originally?, Continued

Since P and V are inversely proportional, when the pressure decreases ~2x, the volume should increase ~2x, and it does.

V2 =2780 mL, P1 = 762 torr, P2 = 0.500 atm

V1, mL

Check:

Solution:

Solution Map:

Relationships:

Given:

Find:

1

221 P

VPV

V2, P1, P2 V1

mL 1350atm 1.03

L 2780atm 0.5001

221

PVPV

atm 03.1 torr760

atm 1 torr782

Temperature Scales

Celsius Kelvin Fahrenheit-273°C-269°C

-183°C

-38.9°C

0°C

100°C

0 K4 K

90 K

234.1 K

273 K

373 K

-459 °F-452°F

-297°F

-38°F

32°F

212°F

Absolute Zero

BP Helium

BP Oxygen

BP Mercury

MP Ice

BP Water

0 R7 R

162 R

421 R

459 R

671 R

Rankine


53

Gas Laws and Temperature• Gases expand when heated and contract when cooled,

so there is a relationship between volume and temperature.

• Gas molecules move faster when heated, causing them to strike surfaces with more force, so there is a relationship between pressure and temperature.

• In order for the relationships to be proportional, the temperature must be measured on an absolute scale.

• When doing gas problems, always convert your temperatures to kelvins.

K = °C + 273 & °C = K - 273°F = 1.8 °C + 32 & °C = 0.556(°F-32)


54

Standard Conditions

• Common reference points for comparing.• Standard pressure = 1.00 atm.• Standard temperature = 0 °C.

273 K.• STP.


55

Volume and Temperature• In a rigid container, raising the temperature

increases the pressure.• For a cylinder with a piston, the pressure

outside and inside stay the same.• To keep the pressure from rising, the piston

moves out increasing the volume of the cylinder.As volume increases, pressure decreases.


56

Volume and Temperature, Continued

Because the hot air in theballoon is less dense than thesurrounding air, it rises.

As a gas is heated, it expands.This causes the density of thegas to decrease.


57

Charles’s Law• Volume is directly proportional to

temperature.Constant P and amount of gas.Graph of V vs. T is a straight line.

• As T increases, V also increases.• Kelvin T = Celsius T + 273.• V = constant x T.

If T is measured in kelvin.

2

2

1

1

TV

TV

58

Charle's Law & Absolute Zero

0

0.1

0.2

0.3

0.4

0.5

0.6

-300 -250 -200 -150 -100 -50 0 50 100 150

Temperature, °C

Volu

me,

L

Volume (L) of 1 g O2 @ 1500 torr

Volume (L) of 1 g O2 @ 2500 torr

Volume (L) of 0.5 g O2 @ 1500 torr

Volume (L) of 0.5 g SO2 @ 1500 torr

We’re losing altitude.Quick, Professor, give yourlecture on Charles’s law!


60

Absolute Zero

• Theoretical temperature at which a gas would have zero volume and no pressure.Kelvin calculated by extrapolation.

• 0 K = -273.15 °C = -459 °F = 0 R.• Never attainable.

Though we’ve gotten real close!• All gas law problems use the Kelvin

temperature scale.


61

Determining Absolute Zero

William Thomson,the Lord of Kelvin,extrapolated theline graphs ofvolume vs. temp-erature to determine the theoretical temperature thata gas would havegiven a volume of 0.

T(K) = t(°C) + 273, 2

2

1

1

TV

TV

Example 11.3—A Gas Has a Volume of 2.57 L at 0 °C. What Was the Temperature at 2.80 L?

Since T and V are directly proportional, when the volume decreases, the temperature should decrease, and it does.

V1 =2.80 L, V2 = 2.57 L, t2 = 0°C

t1, K and °C

Check:

Solution:

Solution Map:

Relationships:

Given:

Find:

2

121 V

VTT

V1, V2, T2 T1

K 6.729L 2.80

L 2.57K 2732

121

VVTT

K 2732730

2

2

TT

C 422736729

273

1

1

11

t.t

Tt


71

Practice—The Temperature Inside a Balloon Is Raised from 25.0 °C to 250.0 °C. If the Volume of Cold Air Was

10.0 L, What Is the Volume of Hot Air?

T(K) = t(°C) + 273.15, 2

2

1

1

TV

TV

Practice—The Temperature Inside a Balloon Is Raised from 25.0 °C to 250.0 °C. If the Volume of Cold Air Was 10.0 L, What Is the

Volume of Hot Air?, Continued

Since T and V are directly proportional, when the temperature increases, the volume should increase, and it does.

V1 =10.0 L, t1 = 25.0 °C L, t2 = 250.0 °C

V2, L

Check:

Solution:

Solution Map:

Relationships:

Given:

Find:

1

212 T

TVV

V1, T1, T2 V2

L 5.17K 298.2

L 10.0K 523.21

122

TVTV

K 523.2273.150.502

K 298.2273.150.52

2

2

1

1

TTTT

73

The Combined Gas Law• Boyle’s law shows the relationship between pressure and volume.

At constant temperature.• Charles’s law shows the relationship between volume and

absolute temperature.At constant pressure.

• The two laws can be combined together to give a law that predicts what happens to the volume of a sample of gas when both the pressure and temperature change.As long as the amount of gas stays constant.

2

22

1

11

TVP

TVP

mmHg 1025.1K 307mL 108

K 358mL 158mmHg 357

32

2

12

2112

P

P

TV

TVPP

T(K) = t(°C) + 273, 2

22

1

11

TVP

TVP

Example 11.4—A Sample of Gas Has an Initial Volume of 158 mL at a Pressure of 735 mmHg and a Temperature of 34 °C. If the Gas Is

Compressed to 108 mL and Heated to 85 °C, What Is the Final Pressure?

Since T increases and V decreases we expect the pressure should increase, and it does.

V1 = 158 mL, t1 = 34 °C L, P1 = 735 mmHg V2 = 108 mL, t2 = 85 °CP2, mmHg

Check:

Solution:

Solution Map:

Relationships:

Given:

Find:

12

2112 TV

TVPP

P1,V1, V2, T1, T2 P2

K 58327358

K 07327334

2

2

1

1

TTTT


82

Practice—A Gas Occupies 10.0 L When Its Pressure Is 3.00 atm and Temperature Is 27 °C. What Volume Will the Gas Occupy Under Standard

Conditions?

L 3.27K 300atm 1.00

K 273L 0.01atm .003

2

2

12

2112

V

V

TP

TVPV

T(K) = t(°C) + 273, 2

22

1

11

TVP

TVP

Practice—A Gas Occupies 10.0 L When Its Pressure Is 3.00 Atm and Temperature Is 27 °C. What Volume Will the Gas

Occupy Under Standard Conditions?, Continued

When T decreases, V should decrease; when P decreases, V should increase—opposite trends make it hard to evaluate our answer.

V1 = 10.0 L, t1 = 27 °C L, P1 = 3.00 atm t2 = 0 °C, P2 = 1.00 atmV2, L

Check:

Solution:

Solution Map:

Relationships:

Given:

Find:

12

2112 TP

TVPV

V1, P1,P2, T1, T2 V2

K 2732730K 00327327

2

2

1

1

TTTT


84

Avogadro’s Law• Volume is directly proportional to

the number of gas molecules.V = constant x n.Constant P and T.More gas molecules = larger

volume.• Count number of gas molecules

by moles, n.• Equal volumes of gases contain

equal numbers of molecules.The gas doesn’t matter.

2

2

1

1

nV

nV


85

Avogadro’s Law, Continued

mol added = n2 – n1, 2

2

1

1

nV

nV

Example 11.5—A 0.22 Mol Sample of He Has a Volume of 4.8 L. How Many Moles Must Be Added to Give 6.4 L?

Since n and V are directly proportional, when the volume increases, the moles should increase, and it does.

V1 =4.8 L, V2 = 6.4 L, n1 = 0.22 mol

n2, and added moles

Check:

Solution:

Solution Map:

Relationships:

Given:

Find:

21

21 n

VVn

V1, V2, n1 n2

mol 29.0L 4.8

L 6.4mol 0.221

212

VVnn

mol 07.0added moles22.029.0added moles


95

Practice—If 1.00 Mole of a Gas Occupies 22.4 L at STP, What Volume Would 0.750 Moles Occupy?

2

2

1

1

nV

nV

Practice—If 1.00 Mole of a Gas Occupies 22.4 L at STP, What Volume Would 0.750 Moles Occupy?, Continued

Since n and V are directly proportional, when the moles decreases, the volume should decrease, and it does.

V1 =22.4 L, n1 = 1.00 mol, n2 = 0.750 mol

V2

Check:

Solution:

Solution Map:

Relationships:

Given:

Find:

21

21 V

nnV

V1, n1, n2 V2

L 8.16mol 1.00

mol 0.750L 22.41

212

nnVV


97

Ideal Gas Law• By combining the gas laws, we can write a general

equation.• R is called the Gas Constant.• The value of R depends on the units of P and V.

We will use 0.0821 and convert P to atm and V to L.• Use the ideal gas law when you have a gas at one

condition, use the combined gas law when you have a gas whose condition is changing.

KmolLatm

nRT PVR or TnVP

1 atm = 14.7 psiT(K) = t(°C) + 273 Kmol

Latm 0.0821 nRT, RPV

Example 11.7—How Many Moles of Gas Are in a Basketball with Total Pressure 24.2 Psi, Volume of 3.2 L at 25 °C?

1 mole at STP occupies 22.4 L at STP; since there is a much smaller volume than 22.4 L, we expect less than 1 mole of gas.

V = 3.2 L, P = 24.2 psi, t = 25 °C,

n, mol

Check:

Solution:

Solution Map:

Relationships:

Given:

Find:

RTPV n

P, V, T, R n

mol 22.0K 9820.0821L 2.3atm 6241.6

KmolLatm

TRVPn

atm 6241.6psi 14.7

atm 1psi 24.2

K 298273C25 (K)

TT

Practice—Calculate the Volume Occupied by 637 g of SO2 (MM 64.07) at 6.08 x 103 mmHg and –23 °C.


107

1 atm = 760 mmHgT(K) = t(°C) + 273, 1 mol SO2 = 64.07 g Kmol


Practice—Calculate the Volume Occupied by 637 g of SO2 (MM 64.07) at 6.08 x 103 mmHg and –23 °C, Continued.

mSO2 = 637 g, P = 6.08 x 103 mmHg, t = −23 °C,

V, L

Solution:

Solution Map:

Relationships:

Given:

Find:

PnRT V

P, n, T, R V

L 55.2atm 0.80

K 0520.0821mol 249.9 KmolLatm

PTRnV

atm 0.08mmHg 760atm 1mmHg 106.08 3

K 025273C 23- (K)

TT

g n

g 64.07mol 1

22

2 SO mol 249.9 g 64.07

SO mol 1SO g 637

Practice—Calculate the Density of a Gas at 775 torr and 27 °C if 0.250 moles Weighs 9.988 g.


109

m = 9.988g, n = 0.250 mol, P = 1.0197 atm, T = 300. K

density, g/L

L 5530.6atm 9711.0

K 0030.082mol .2500 KmolLatm

PTRnV

Solution:

Practice—Calculate the Density of a Gas at 775 torr and 27 °C if 0.250 moles Weighs 9.988 g, Continued

The value 1.65 g/L is reasonable.

m=9.988g, n=0.250 mol, P=775 mmHg, t=27°C,

density, g/L

Check:

Solution Map:

Relationships:

Given:

Find:

PTRn V

V, m d

atm 9710.1 torr760

atm 1 torr775

K 030273C 27 (K)

TT

P, n, T, R V

Vm d

g/L 1.65 L 553.06g .9889

Vmd

1 atm = 760 mmHg, T(K) = t(°C) + 273 Kmol


Vm d


111

Molar Mass of a Gas

• One of the methods chemists use to determine the molar mass of an unknown substance is to heat a weighed sample until it becomes a gas, measure the temperature, pressure, and volume, and use the ideal gas law.

molesgramsin massMassMolar

m = 0.311g, V = 0.225 L, P = 1.1658 atm, T = 328 K

Molar Mass, g/mol

mol 105447.9K 2830.0821L .2250atm 5861.1 3

KmolLatm

TRVPn atm 5861.1

mmHg 760atm 1mmHg 886

Example 11.8—Calculate the Molar Mass of a Gas with Mass 0.311 g that Has a Volume of 0.225 L at 55 °C and 886 mmHg.

The value 31.9 g/mol is reasonable.

m=0.311g, V=0.225 L, P=886 mmHg, t=55°C,

Molar Mass, g/mol

Check:

Solution:

Solution Map:

Relationships:

Given:

Find:

TRVP n

n, m MM

K 328273C55 (K)

TT

P, V, T, R n

nm MM

g/mol 31.9 mol 109.7454

g 311.03-

nmMM

1 atm = 760 mmHg, T(K) = t(°C) + 273 Kmol


nm MM

Practice—What Is the Molar Mass of a Gas if 12.0 g Occupies 197 L at 380 torr and 127 °C?


121

m = 12.0g, V = 197 L, P = 0.50 atm, T =400 K,

molar mass, g/mol

Practice—What Is the Molar Mass of a Gas if 12.0 g Occupies 197 L at 380 torr and 127 °C?, Continued

The value 31.9 g/mol is reasonable.

m=12.0 g, V= 197 L, P=380 torr, t=127°C,

molar mass, g/mol

Check:

Solution:

Solution Map:

Relationships:

Given:

Find:

TRVPn

n, m MM

mol 0.3K 0040.0821

L 971atm 0.50

KmolLatm

TRVPn atm 50.0

torr760atm 1 torr083

K 004273C127 (K)

TT

P, V, T, R n

nmMM

g/mol 4.0 mol .03

g 2.01

nmMM

1 atm = 760 torr, T(K) = t(°C) + 273 Kmol


nmMM

Mixtures of Gases• According to the kinetic molecular theory, the particles in

a gas behave independently.• Air is a mixture, yet we can treat it as a single gas.• Also, we can think of each gas in the mixture as

independent of the other gases.All gases in the mixture have the same volume and temperature.

All gases completely occupy the container, so all gases in the mixture have the volume of the container.

Gas % in Air, by volume Gas % in Air,

by volumeNitrogen, N2 78 Argon, Ar 0.9Oxygen, O2 21 Carbon dioxide, CO2 0.03


124

Partial Pressure• Each gas in the mixture exerts a pressure

independent of the other gases in the mixture.• The pressure of a component gas in a mixture

is called a partial pressure.• The sum of the partial pressures of all the

gases in a mixture equals the total pressure.Dalton’s law of partial pressures.Ptotal = Pgas A + Pgas B + Pgas C +...

atm 1.00 atm 0.01 atm 0.21 atm 0.78 Ar2O2Nair PPPP

Example 11.9—A Mixture of He, Ne, and Ar Has a Total Pressure of 558 MmHg. If the Partial Pressure of He Is 341 MmHg and Ne Is 112 MmHg, Determine the Partial Pressure of Ar in the Mixture.

The units are correct, the value is reasonable.

PHe= 341 mmHg, PNe= 112 mmHg, Ptot = 558 mmHg

PAr, mmHg

Check:

Solution:

Solution Map:

Relationships:

Given:

Find:

Ptot, PHe, PNe PAr

Ptot = Pa + Pb + etc.

PAr = Ptot – (PHe + PNe)

mmHg 051

mmHg 112341558Ar

P


126

Finding Partial Pressure• To find the partial pressure of a

gas, multiply the total pressure of the mixture by the fractional composition of the gas.

• For example, in a gas mixture that is 80.0% He and 20.0% Ne that has a total pressure of 1.0 atm, the partial pressure of He would be:

PHe = (0.800)(1.0 atm) = 0.80 atmFractional composition = percentage

divided by 100.

The Partial Pressure of Each Gas in a Mixture, or the Total Pressure of a Mixture, Can Be

Calculated Using the Ideal Gas Law

BAtotal

total

BAtotal

BB

AA

P PV x R x Tn P

n n n

V x R x Tn P

V x R x Tn P

same theare mixture in theeverything of volumeand re temperatuthe

mixture ain B andA gasesfor

127

Practice—Find the Partial Pressure of Neon in a Mixture of Ne and Xe with Total Pressure 3.9 atm, Volume 8.7 L,

Temperature 598 K, and 0.17 moles Xe.


128

atm 8959.0

L 8.7K 9850.0821mol 0.17 Kmol

Latm

XeXe

VTRnP

XeNetotal ,KmolLatm 0.0821 PPPnRT, RPV

Practice—Find the Partial Pressure of Neon in a Mixture of Ne and Xe with Total Pressure 3.9 Atm, Volume 8.7 L, Temperature 598 K, and 0.17

Moles Xe, Continued.

The unit is correct, the value is reasonable.

Ptot = 3.9 atm, V = 8.7 L, T = 598 K, Xe = 0.17 mol

PNe, atm

Check:

Solution:

Solution Map:

Relationships:

Given:

Find:

VTRnP

XeXe

nXe, V, T, R PXe

atm 2.9atm 8950.9 atm 9.3

XetotalNe

PPP

Ptot, PXe PNe

XetotalNe PPP


130

Mountain Climbing and Partial Pressure• Our bodies are adapted to breathe O2 at

a partial pressure of 0.21 atm.Sherpa, people native to the Himalaya

mountains, have adapted to the much lower partial pressure of oxygen in their air.

• Partial pressures of O2—lower than 0.1 atm—leads to hypoxia.Unconsciousness or death.

• Climbers of Mt. Everest must carry O2 in cylinders to prevent hypoxia.On top of Mt. Everest:

Pair = 0.311 atm, so PO2 = 0.065 atm.

Deep Sea Divers and Partial Pressure• It is also possible to have too much O2, a condition called oxygen

toxicity.PO2 > 1.4 atm.Oxygen toxicity can lead to muscle spasms, tunnel vision, and convulsions.

• It is also possible to have too much N2, a condition called nitrogen narcosis.Also known as rapture of the deep.

• When diving deep, the pressure of the air that divers breathe increases, so the partial pressure of the oxygen increases.At a depth of 55 m, the partial pressure of O2 is 1.4 atm.Divers that go below 50 m use a mixture of He and O2 called heliox that

contains a lower percentage of O2 than air.


132

Partial Pressure vs. Total Pressure

At a depth of 30 m, the total pressure of air in the diverslungs, and the partial pressure of all the gases in the air,

are quadrupled!


133

Collecting Gases• Gases are often collected by having them displace

water from a container.• The problem is that since water evaporates, there

is also water vapor in the collected gas.• The partial pressure of the water vapor, called the

vapor pressure, depends only on the temperature. So you can use a table to find out the partial pressure of

the water vapor in the gas you collect.• If you collect a gas sample with a total pressure of

758 mmHg at 25 °C, the partial pressure of the water vapor will be 23.8 mmHg, so the partial pressure of the dry gas will be 734 mmHg.


134

Vapor Pressure of WaterTemp., °C Pressure,

mmHg10 9.220 17.525 23.830 31.840 55.350 92.560 149.470 233.780 355.1


135

Zn metal reactswith HCl(aq) toproduce H2(g).

Because waterevaporates, somewater vapor getsmixed in withthe H2.

The gas flowsthrough the tubeand bubbles intothe jar, where itdisplaces the water in the jar.

If the temperature of the water is 30, the vapor pressure of the water is 31.8 mmHg.

If the total pressure is 760 mmHg, the partial pressure of the H2 is 760 − 31.8 mmHg = 728 mmHg.

Practice—0.12 moles of H2 Is Collected Over Water in a 10.0 L Container at 323 K. Find the Total Pressure

(Vapor Pressure of Water at 50 C = 92.6 mmHg).


136

atm 8113.0L 10.0

K 3230.0821mol 0.12 KmolLatm

2H

VTRnP

V = 10.0 L, nH2 = 0.12 mol, T = 323 K

Ptotal, atm

Practice—0.12 moles of H2 Is Collected Over Water in a 10.0 L Container at 323 K. Find the Total Pressure

(Vapor Pressure of Water at 50 C = 92.6 mmHg), Continued.

Solution:

Solution Map:

Relationships:

Given:

Find:

VTRn P

PH2, PH2O Ptotal

mmHg 330P 6.291.842 P

total

total

C50 @ OHHtotal 22PPP

1 atm = 760 mmHgPtotal = PA + PB, Kmol

Latm 0.0821 , RnRTPV

nH2,V,T PH2

mmHg 8.142atm 1mmHg 760atm 8110.3


138

Reactions Involving Gases• The principles of reaction involving stoichiometry

from Chapter 8 can be combined with the gas laws for reactions involving gases.

• In reactions of gases, the amount of a gas is often given as a volume. Instead of moles.As we’ve seen, you must state pressure and temperature.

• The ideal gas law allows us to convert from the volume of the gas to moles; then, we can use the coefficients in the equation as a mole ratio.

P, V, T of Gas A mole A mole B P, V, T of Gas B

nO2 = 3.60 mol, P = 0.99342 atm, T = 305 K

VO2, L

L 7.90atm 4230.99

K 3050.0821mol .603 KmolLatm

PTRnV

Example 11.11—How Many Liters of O2 Are Made from 294 g of KClO3 at 755 mmHg and 305 K?2 KClO3(s) → 2 KCl(s) + 3 O2(g)

mKClO3 = 294 g, P=755 mmHg, T=305 K

VO2, L

Solution:

Solution Map:

Relationships:

Given:

Find:

PTRnV

atm 24399.0mmHg 760atm 1mmHg 755

2

3

233

O mol 60.3KClO mol 2

O mol 3g 122.5

KClO mol 1KClO g 942

P, n, T, R V

1 atm = 760 mmHg, KClO3 = 122.5 g/mol2 mol KClO3 : 3 mol O2 Kmol


g KClO3 mol KClO3 mol O2

3

2

KClO mol 2O mol 3

g 122.5KClO mol 1 3

Practice—What Volume of O2 at 0.750 atm and 313 K is Generated by the Thermolysis of 10.0 g of HgO?

2 HgO(s) 2 Hg(l) + O2(g)(MMHgO = 216.59 g/mol)


149

mHgO = 10.0g, P=0.750 atm, T=313 K

VO2, L

nO2 = 0.023085 mol, P = 0.750 atm, T = 313 K

VO2, L

L 791.0atm 0.750

K3130.08206mol 850023.0 KmolLatm

PTRnV

Practice—What Volume of O2 at 0.750 atm and 313 K is Generated by the Thermolysis of 10.0 g of HgO?

2 HgO(s) 2 Hg(l) + O2(g), Continued

Solution:

Solution Map:

Relationships:

Given:

Find:

PTRnV

2

2

O mol 850023.0HgO mol 2O mol 1

g 216.59HgO mol 1HgO g 0.01

P, n, T, R V

1 atm = 760 mmHg, HgO = 216.59 g/mol2 mol HgO : 1 mol O2 Kmol


g HgO mol HgO mol O2

HgO mol 2O mol 1 2

g 216.59HgO mol 1


151

Calculate the Volume Occupied by 1.00 Mole of an Ideal Gas at STP.

• 1 mole of any gas at STP will occupy 22.4 L.• This volume is called the molar volume and can

be used as a conversion factor.As long as you work at STP.

1 mol 22.4 L

(1.00 atm) x V = (1.00 moles)(0.0821 )(273 K)L∙atmmol∙K

V = 22.4 L

P x V = n x R x T


152

Molar VolumeThere is so muchempty spacebetween moleculesin the gas state thatthe volume of thegas is not effectedby the size of themolecules (underideal conditions).

Example 11.12—How Many Grams of H2O Form When 1.24 L H2 Reacts Completely with O2 at STP?

O2(g) + 2 H2(g) → 2 H2O(g)VH2 = 1.24 L, P = 1.00 atm, T = 273 K

massH2O, g

Solution:

Solution Map:

Relationships:

Given:

Find:

OH mol 1g 02.18 2

OH g 998.0OH mol 1OH g 8.021

H mol 2OH mol 2

H L 22.4H mol 1H L .241

2

2

2

2

2

2

22

H2O = 18.02 g/mol, 1 mol = 22.4 L @ STP2 mol H2O : 2 mol H2

OH mol 2H mol 2

2

2L 22.4H mol 1 2

g H2OL H2 mol H2 mol H2O


153

Practice—What Volume of O2 at STP is Generated by the Thermolysis of 10.0 g of HgO?

2 HgO(s) 2 Hg(l) + O2(g)(MMHgO = 216.59 g/mol)


161

Practice—What Volume of O2 at STP is Generated by the Thermolysis of 10.0 g of HgO?

2 HgO(s) 2 Hg(l) + O2(g), ContinuedmHgO = 10.0 g, P = 1.00 atm, T = 273 K

VO2, L

Solution:

Solution Map:

Relationships:

Given:

Find:

g 59.216HgO mol 1

2

2

22

O L 517.0O mol 1O L 2.42

HgO mol 2O mol 1

g 216.59HgO mol 1HgO g 0.01

HgO = 216.59 g/mol, 1 mol = 22.4 L at STP2 mol HgO : 1 mol O2

HgO mol 2O mol 1 2

2O mol 1L 22.4

L O2g HgO mol HgO mol O2


162

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