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Chapter 5 Gases Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA 2008, Prentice Hall Chemistry: A Molecular Approach , 1 st Ed. Nivaldo Tro

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Chemistry: A Molecular Approach , 1 st Ed. Nivaldo Tro. Chapter 5 Gases. Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA. 2008, Prentice Hall. Air Pressure & Shallow Wells. water for many homes is supplied by a well less than 30 ft. deep with a pump at the surface - PowerPoint PPT Presentation

TRANSCRIPT

Page 1: Chapter 5 Gases

Chapter 5Gases

Roy Kennedy

Massachusetts Bay Community College

Wellesley Hills, MA2008, Prentice Hall

Chemistry: A Molecular Approach, 1st Ed.Nivaldo Tro

Page 2: Chapter 5 Gases

Tro, Chemistry: A Molecular Approach 2

Air Pressure & Shallow Wells• water for many homes is

supplied by a well less than 30 ft. deep with a pump at the surface

• the pump removes air from the pipe, decreasing the air pressure in the pipe

• the outside air pressure then pushes the water up the pipe

• the maximum height the water will rise is related to the amount of pressure the air exerts

Page 3: Chapter 5 Gases

Tro, Chemistry: A Molecular Approach 3

Atmospheric Pressure• pressure is the force

exerted over an area• on average, the air

exerts the same pressure that a column of water 10.3 m high would exert14.7 lbs./in2

so if our pump could get a perfect vacuum, the maximum height the column could rise is 10.3 m Area

Force Pressure

Page 4: Chapter 5 Gases

Tro, Chemistry: A Molecular Approach 4

Gases Pushing• gas molecules are constantly in motion

• as they move and strike a surface, they push on that surfacepush = 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 exertingpressure = force per unit area

Page 5: Chapter 5 Gases

Tro, Chemistry: A Molecular Approach 5

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 pressurethe bigger the difference in pressure, the stronger

the flow of the gas

• if there is something in the gas’s path, the gas will try to push it along as the gas flows

Page 6: Chapter 5 Gases

Tro, Chemistry: A Molecular Approach 6

Atmospheric Pressure Effects• differences in air pressure result in weather

and wind patterns

• the higher up in the atmosphere you climb, the lower the atmospheric pressure is around youat the surface the atmospheric pressure is 14.7 psi,

but at 10,000 ft it 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

Page 7: Chapter 5 Gases

Tro, Chemistry: A Molecular Approach 7

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.”

Page 8: Chapter 5 Gases

Tro, Chemistry: A Molecular Approach 8

The Pressure of a Gas• 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 factorsnumber of gas particles in a

given volumevolume of the containeraverage speed of the gas

particles

Page 9: Chapter 5 Gases

Tro, Chemistry: A Molecular Approach 9

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

Page 10: Chapter 5 Gases

Tro, Chemistry: A Molecular Approach 10

Common Units of PressureUnit 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

2m

N 1 Pa 1

Page 11: Chapter 5 Gases

Example 5.1 – A high-performance bicycle tire has a pressure of 132 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

132 psi

mmHg

Check:

Solution:

Concept Plan:

Relationships:

Given:

Find:

psi 14.7

atm 1

mmHg 10.826 atm 1

mmHg 760

psi 14.7

atm 1psi 132 3

atm 1

mmHg 760

psi atm mmHg

Page 12: Chapter 5 Gases

Tro, Chemistry: A Molecular Approach 12

Manometers• the pressure of a gas trapped in a container can be

measured with an instrument called a manometer

• manometers are U-shaped tubes, partially filled with a liquid, connected to the gas sample on one side and open to the air on the other

• a competition is established between the pressure of the atmosphere and the gas

• the difference in the liquid levels is a measure of the difference in pressure between the gas and the atmosphere

Page 13: Chapter 5 Gases

Tro, Chemistry: A Molecular Approach 13

Manometer

for this sample, the gas has a larger pressure than the atmosphere, so

(mm) levels Hgin difference (mmHg)Pressure (mmHg)Pressure

Pressure Pressure Pressure

atmospheregas

hatmospheregas

Page 14: Chapter 5 Gases

Tro, Chemistry: A Molecular Approach 14

Boyle’s Law• pressure of a gas is inversely proportional

to its volumeconstant T and amount of gasgraph P vs V is curvegraph P vs 1/V is straight line

• as P increases, V decreases by the same factor

• P x V = constant

• P1 x V1 = P2 x V2

Page 15: Chapter 5 Gases

Tro, Chemistry: A Molecular Approach 15

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 Airin Column

(in)

Difference inHg Levels

(in)48 0.044 2.840 6.236 10.132 15.128 21.224 29.722 35.0

Page 16: Chapter 5 Gases

Tro, Chemistry: A Molecular Approach 16

Boyle's Expt.

0

20

40

60

80

100

120

140

0 10 20 30 40 50 60

Volume of Air, in3

Pre

ssu

re, in

Hg

Page 17: Chapter 5 Gases

Tro, Chemistry: A Molecular Approach 17

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

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

Pre

ss

ure

, in

Hg

Page 18: Chapter 5 Gases

Tro, Chemistry: A Molecular Approach 18

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

Page 19: Chapter 5 Gases

Tro, Chemistry: A Molecular Approach 19

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)

Page 20: Chapter 5 Gases

Tro, Chemistry: A Molecular Approach 20

Boyle’s Law and Diving• since water is denser

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

pressure is 3 atm

if your tank contained air at 1 atm pressure you would not be able to inhale it into your lungs

Page 21: Chapter 5 Gases

P1 ∙ V1 = P2 ∙ V2

Example 5.2 – A cylinder with a movable piston has a volume of 7.25 L at 4.52 atm. What is the volume at 1.21 atm?

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

V1 =7.25 L, P1 = 4.52 atm, P2 = 1.21 atm

V2, L

Check:

Solution:

Concept Plan:

Relationships:

Given:

Find:

2

112 P

VPV

V1, P1, P2 V2

L 1.27

atm 1.21

L 7.25atm 4.52

P

VPV

2

112

Page 22: Chapter 5 Gases

Tro, Chemistry: A Molecular Approach 22

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?

Page 23: Chapter 5 Gases

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

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?

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:

Concept Plan:

Relationships:

Given:

Find:

1

221 P

VPV

V1, P1, P2 V2

mL 1350

atm 1.03

L 2780atm 0.500

P

VPV

1

221

atm 03.1 torr760

atm 1 torr782

Page 24: Chapter 5 Gases

Tro, Chemistry: A Molecular Approach 24

Charles’ Law• volume is directly proportional to

temperatureconstant P and amount of gasgraph of V vs T is straight line

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

if T measured in Kelvin

2

2

1

1

T

V

T

V

Page 25: Chapter 5 Gases

Tro, Chemistry: A Molecular Approach 25

Charles’ Law – A Molecular View• the pressure of gas inside

and outside the balloon are the same

• at low temperatures, the gas molecules are not moving as fast, so they don’t hit the sides of the balloon as hard – therefore the volume is small

• the pressure of gas inside and outside the balloon are the same

• at high temperatures, the gas molecules are moving faster, so they hit the sides of the balloon harder – causing the volume to become larger

Page 26: Chapter 5 Gases

26

Charles' 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

Vo

lum

e, 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 @ 1500torr

The data fall on a straight line.If the lines are extrapolated back to a volume of “0,” they all show the same temperature, -273.15°C, called absolute zero

Page 27: Chapter 5 Gases

T(K) = t(°C) + 273.15, 2

2

1

1

T

V

T

V

Example 5.3 – A gas has a volume of 2.57 L at 0.00°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 =2.57 L, V2 = 2.80 L, t2 = 0.00°C

t1, K and °C

Check:

Solution:

Concept Plan:

Relationships:

Given:

Find:

2

121 V

VTT

V1, V2, T2 T1

K 6.729

L 2.80

L 2.57K 273.15

V

VTT

2

121

K 273.15T

273.150.00T

2

2

C 42t

273.156.729t

273.15Tt

1

1

11

Page 28: Chapter 5 Gases

Tro, Chemistry: A Molecular Approach 28

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?

Page 29: Chapter 5 Gases

T(K) = t(°C) + 273.15, 2

2

1

1

T

V

T

V

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?

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:

Concept Plan:

Relationships:

Given:

Find:

1

212 T

TVV

V1, T1, T2 V2

L 5.17

K 298.2

L 10.0K 523.2

T

VTV

1

122

K 523.2T

273.150.502T

K 298.2T

273.150.52T

2

2

1

1

Page 30: Chapter 5 Gases

Tro, Chemistry: A Molecular Approach 30

Avogadro’s Law• volume directly proportional to

the number of gas moleculesV = constant x nconstant P and Tmore gas molecules = larger

volume

• count number of gas molecules by moles

• equal volumes of gases contain equal numbers of moleculesthe gas doesn’t matter

2

2

1

1

n

V

n

V

Page 31: Chapter 5 Gases

mol added = n2 – n1, 2

2

1

1

n

V

n

V

Example 5.4 – A 0.225 mol sample of He has a volume of 4.65 L. How many moles must be added to give 6.48 L?

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

V1 =4.65 L, V2 = 6.48 L, n1 = 0.225 mol

n2, and added moles

Check:

Solution:

Concept Plan:

Relationships:

Given:

Find:

21

21 n

V

Vn

V1, V2, n1 n2

mol 314.0

L 4.65

L 6.48mol 0.225

V

Vnn

1

212

mol 089.0added moles

225.0314.0added moles

Page 32: Chapter 5 Gases

Tro, Chemistry: A Molecular Approach 32

Ideal Gas Law• By combing 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 0.08206 and convert P to atm and V to L• the other gas laws are found in the ideal gas law if two variables are kept constant• allows us to find one of the variables if we know the other 3

Kmol

Latm

nRTPVor R

Tn

VP

Page 33: Chapter 5 Gases

1 atm = 14.7 psi

T(K) = t(°C) + 273.15 Kmol

Latm 0.08206 R nRT,PV

Example 5.6 – How many moles of gas are in a basketball with total pressure 24.3 psi, volume of 3.24 L at 25°C?

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

V = 3.24 L, P = 24.3 psi, t = 25 °C,

n, mol

Check:

Solution:

Concept Plan:

Relationships:

Given:

Find:

RT

PV n

P, V, T, R n

mol 219.0

K 9820.08206

L 24.3atm 3151.6TR

VPn

Kmol

Latm

atm 3151.6psi 14.7

atm 1psi 24.3

K 298T

273.15C25 T(K)

Page 34: Chapter 5 Gases

Tro, Chemistry: A Molecular Approach 34

Standard Conditions• since the volume of a gas varies with pressure

and temperature, chemists have agreed on a set of conditions to report our measurements so that comparison is easy – we call these standard conditionsSTP

• standard pressure = 1 atm• standard temperature = 273 K

0°C

Page 35: Chapter 5 Gases

Tro, Chemistry: A Molecular Approach 35

Practice – A gas occupies 10.0 L at 44.1 psi and 27°C. What volume will it occupy at standard conditions?

Page 36: Chapter 5 Gases

L 3.27

atm .001

K 7320.08206mol 912.1P

TRnV

Kmol

Latm

mol 912.1K 00.30.08206

L 0.01atm 3.00TR

VPn

Kmol

Latm

1 atm = 14.7 psi

T(K) = t(°C) + 273.15 Kmol

Latm 0.08206 R nRT,PV

A gas occupies 10.0 L at 44.1 psi and 27°C. What volume will it occupy at standard conditions?

1 mole at STP occupies 22.4 L, since there is more than 1 mole, we expect more than 22.4 L of gas

V1 = 10.0 L, P1 = 44.1 psi, t1 = 27 °C, P2 = 1.00 atm, t2 = 0°C

V2, L

Check:

Solution:

Concept Plan:

Relationships:

Given:

Find:

RT

PV n

P1, V1, T1, R n

atm 00.3psi 14.7

atm 1psi 44.1

K .003T

273.15C27 T(K)

1

P2, n, T2, R V2

P

nRT V

Page 37: Chapter 5 Gases

Tro, Chemistry: A Molecular Approach 37

Molar Volume

• solving the ideal gas equation for the volume of 1 mol of gas at STP gives 22.4 L6.022 x 1023 molecules of gasnotice: the gas is immaterial

• we call the volume of 1 mole of gas at STP the molar volumeit is important to recognize that one mole of

different gases have different masses, even though they have the same volume

Page 38: Chapter 5 Gases

Tro, Chemistry: A Molecular Approach 38

Molar Volume

Page 39: Chapter 5 Gases

Tro, Chemistry: A Molecular Approach 39

Density at Standard Conditions

• density is the ratio of mass-to-volume

• density of a gas is generally given in g/L

• the mass of 1 mole = molar mass

• the volume of 1 mole at STP = 22.4 L

L 22.4

g Mass,Molar Density

Page 40: Chapter 5 Gases

Tro, Chemistry: A Molecular Approach 40

Gas Density

TRmass)(molar P

density V

mass

TRmassmolar

massVP

TRnVP

litersin volumegramsin mass

density

massmolar mass

moles moles massmolar

mol 1mass

• density is directly proportional to molar mass

Page 41: Chapter 5 Gases

1 atm = 760 mmHg, MM = 28.01 g

T(K) = t(°C) + 273.15Kmol

Latm 0.08206 R

TR

MMPd

Example 5.7 – Calculate the density of N2 at 125°C and 755 mmHg

since the density of N2 is 1.25 g/L at STP, we expect the density to be lower when the temperature is raised, and it is

P = 755 mmHg, t = 125 °C,

dN2, g/L

Check:

Solution:

Concept Plan:

Relationships:

Given:

Find:

TR

MMP d

P, MM, T, R d

g/L 852.0

K 9830.08206

8.012atm 4230.99

TR

MMPd

Kmol

Latmmol

g

atm 42399.0mmHg 760

atm 1mmHg 755

K 398T

273.15C125 T(K)

Page 42: Chapter 5 Gases

Tro, Chemistry: A Molecular Approach 42

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

moles

gramsin massMassMolar

Page 43: Chapter 5 Gases

Example 5.8 – 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:

Concept Plan:

Relationships:

Given:

Find:

TR

VP n

n, m MM

mol 105447.9

K 2830.08206

L .2250atm 5861.1TR

VPn

3

Kmol

Latm

atm 5861.1mmHg 760

atm 1mmHg 886

K 328T

273.15C55 T(K)

P, V, T, R n

n

m MM

g/mol 31.9 mol 109.7454

g 311.0

n

mMM

3-

1 atm = 760 mmHg, T(K) = t(°C) + 273.15

Kmol

Latm 0.08206 R nRT,PV

n

m MM

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

molar mass, g/mol

Page 44: Chapter 5 Gases

Tro, Chemistry: A Molecular Approach 44

Practice - Calculate the density of a gas at 775 torr and 27°C if 0.250 moles weighs 9.988 g

Page 45: Chapter 5 Gases

L 5530.6

atm 1.0197

K .0030.08206mol .2500P

TRnV

Kmol

Latm

Calculate the density of a gas at 775 torr and 27°C if 0.250 moles weighs 9.988 g

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:

Concept Plan:

Relationships:

Given:

Find:

P

TRn V

V, m d

atm 9710.1 torr760

atm 1 torr775

K 300.T

273.15C27 T(K)

P, n, T, R V

V

m d

g/L 1.65

L 553.06

g .9889

V

md

1 atm = 760 mmHg, T(K) = t(°C) + 273.15

Kmol

Latm 0.08206 R nRT,PV

V

m d

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

density, g/L

Page 46: Chapter 5 Gases

Tro, Chemistry: A Molecular Approach 46

Mixtures of Gases• when gases are mixed together, their molecules

behave independent of each other all the gases in the mixture have the same volume

all completely fill the container each gas’s volume = the volume of the container

all gases in the mixture are at the same temperature therefore they have the same average kinetic energy

• therefore, in certain applications, the mixture can be thought of as one gas even though air is a mixture, we can measure the pressure,

volume, and temperature of air as if it were a pure substance we can calculate the total moles of molecules in an air sample,

knowing P, V, and T, even though they are different molecules

Page 47: Chapter 5 Gases

Tro, Chemistry: A Molecular Approach 47

Partial Pressure• the pressure of a single gas in a mixture of gases is

called its partial pressure

• we can calculate the partial pressure of a gas ifwe know what fraction of the mixture it composes and the

total pressureor, we know the number of moles of the gas in a container of

known volume and temperature

• the sum of the partial pressures of all the gases in the mixture equals the total pressureDalton’s Law of Partial Pressuresbecause the gases behave independently

Page 48: Chapter 5 Gases

Tro, Chemistry: A Molecular Approach 48

Composition of Dry Air

Page 49: Chapter 5 Gases

Tro, Chemistry: A Molecular Approach 49

The partial pressure of each gas in a mixture can be calculated using the ideal gas law

V

T x R x n P P P

n n n

same theare mixture in the

everything of volumeand re temperatutheV

T x R x n P

V

T x R x n P

togethermixed B, andA gases, for two

totalBAtotal

BAtotal

BB

AA

Page 50: Chapter 5 Gases

mol 10125.1

K 2980.08206

L 00.1atm .2750TR

VPn

2

Kmol

Latm

Example 5.9 – Determine the mass of Ar in the mixture

the units are correct, the value is reasonable

PHe=341 mmHg, PNe=112 mmHg, Ptot = 662 mmHg, V = 1.00 L, T=298 KmassAr, g

Check:

Solution:

Concept Plan:

Relationships:

Given:

Find:

TR

VP n

atm 275.0mmHg 760

atm 1mmHg 209

Ptot, PHe, PNe PAr

MMn m

Ar g 0.449 mol 1

g 39.95mol 10125.1 2

Ptot = Pa + Pb + etc., 1 atm = 760 mmHg, MMAr = 39.95 g/mol

Kmol

Latm 0.08206 R nRT,PV

n

m MM

PAr = 0.275 atm, V = 1.00 L, T=298 K

massAr, g

PAr, V, T nAr mAr

PAr = Ptot – (PHe + PNe)

mmHg 209

mmHg 112341662PAr

Page 51: Chapter 5 Gases

Tro, Chemistry: A Molecular Approach 51

Practice – Find the partial pressure of neon in a mixture with total pressure 3.9 atm, volume 8.7 L, temperature

598 K, and 0.17 moles Xe.

Page 52: Chapter 5 Gases

atm 8959.0L 8.7

K 9850.08206mol 0.17V

TRnP

Kmol

Latm

XeXe

XeNetotal ,Kmol

Latm P P P 0.08206 R nRT,PV

Find the partial pressure of neon in a mixture with total pressure 3.9 atm, volume 8.7 L, temperature 598 K, and 0.17 moles Xe

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:

Concept Plan:

Relationships:

Given:

Find:

V

TRn P Xe

Xe

nXe, V, T, R PXe

atm 2.9

atm 8950.9 atm 9.3

PPP XetotalNe

Ptot, PXe PNe

XetotalNe PP P

Page 53: Chapter 5 Gases

Tro, Chemistry: A Molecular Approach 53

Mole Fractionthe fraction of the total pressure that a single gas contributes is equal to the fraction of the total number of moles that a single gas contributes

total

A

total

A

n

n

P

P

the ratio of the moles of a single component to the total number of moles in the mixture is called the mole fraction,

for gases, = volume % / 100%

total

AA n

n

the partial pressure of a gas is equal to the mole fraction of that gas times the total pressure

totalAA PP

Page 54: Chapter 5 Gases

Tro, Chemistry: A Molecular Approach 54

Mountain Climbing & Partial Pressure• our bodies are adapted to breathe O2 at

a partial pressure of 0.21 atmSherpa, people native to the Himalaya

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

• partial pressures of O2 lower than 0.1 atm will lead to hypoxiaunconsciousness or death

• climbers of Mt Everest carry O2 in cylinders to prevent hypoxiaon top of Mt Everest, Pair = 0.311 atm, so

PO2 = 0.065 atm

Page 55: Chapter 5 Gases

Tro, Chemistry: A Molecular Approach 55

Deep Sea Divers & Partial Pressure• its also possible to have too much O2, a condition called oxygen

toxicityPO2 > 1.4 atmoxygen toxicity can lead to muscle spasms, tunnel vision, and

convulsions

• its also possible to have too much N2, a condition called nitrogen narcosisalso known as Rapture of the Deep

• when diving deep, the pressure of the air divers breathe increases – so the partial pressure of the oxygen increasesat a depth of 55 m the partial pressure of O2 is 1.4 atmdivers that go below 50 m use a mixture of He and O2 called heliox that

contains a lower percentage of O2 than air

Page 56: Chapter 5 Gases

Tro, Chemistry: A Molecular Approach 56

Partial Pressure & Diving

Page 57: Chapter 5 Gases

22

2 O mol 135.0g 32.00

O mol 1O g 4.32

atm 990.12L 12.5

K 9820.08206mol 586.1V

TRnP

Kmol

Latm

totaltotal

Ex 5.10 – Find the mole fractions and partial pressures in a 12.5 L tank with 24.2 g He and 4.32 g O2 at 298 K

mHe = 24.2 g, mO2 = 43.2 g V = 12.5 L, T = 298 K

He, O2, PHe, atm, PO2, atm, Ptotal, atm

Solution:

Concept Plan:

Relationships:

Given:

Find:

V

TRn P total

total

He mol 05.6g 4.00

He mol 1He g 24.2

ntot, V, T, R Ptotmgas ngas gas

total

A gasA gas n

n

gas, Ptotal Pgas

totalAA ,Kmol

Latm P P 0.08206 R nRT,PV

MMHe = 4.00 g/mol

MMO2 = 32.00 g/mol

nHe = 6.05 mol, nO2 = 0.135 mol V = 12.5 L, T = 298 K

He=0.97817, O2=0.021827, PHe, atm, PO2, atm, Ptotal, atm

17897.0O mol 0.135He mol 6.05

He mol 6.05

2He

278021.0O mol 0.135He mol 6.05

O mol 0.135

2

2O2

totalAA P P

atm 11.8

atm 990.1217897.0

P P totalHeHe

atm 264.0

atm 990.12278021.0

P P totalOO 22

Page 58: Chapter 5 Gases

Tro, Chemistry: A Molecular Approach 58

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

758.2 mmHg* at 25°C, the partial pressure of the water vapor will be 23.78 mmHg – so the partial pressure of the dry gas will be 734.4 mmHg Table 5.4*

Page 59: Chapter 5 Gases

Tro, Chemistry: A Molecular Approach 59

Vapor Pressure of Water

Page 60: Chapter 5 Gases

Tro, Chemistry: A Molecular Approach 60

Collecting Gas by Water Displacement

Page 61: Chapter 5 Gases

mol 107511.4

K .2930.08206

L 02.1atm 950.970TR

VPn

2

Kmol

Latm

atm 95970.0mmHg 760

atm 1mmHg 56737.

Ex 5.11 – 1.02 L of O2 collected over water at 293 K with a total pressure of 755.2 mmHg. Find mass O2.

V=1.02 L, P=755.2 mmHg, T=293 K

mass O2, g

Solution:

Concept Plan:

Relationships:

Given:

Find:

TR

VP n

Ptot, PH2O PO2

mmHg 56.377P

5.4) (Table 17.5555.27 P

2

2

O

O

C20 @ OHtotalO 22PPP

g 1.32 mol 1

g 2.003mol 107511.4 2

1 atm = 760 mmHg, Ptotal = PA + PB, O2 = 32.00 g/mol

Kmol

Latm 0.08206 R nRT,PV

V=1.02 L, PO2=737.65 mmHg, T=293 K

mass O2, g

PO2,V,T nO2 gO2

Page 62: Chapter 5 Gases

Tro, Chemistry: A Molecular Approach 62

Practice – 0.12 moles of H2 is collected over water in a 10.0 L container at 323 K. Find the total pressure.

Page 63: Chapter 5 Gases

atm 8113.0

L 10.0

K 3230.08206mol 0.12V

TRnP

Kmol

Latm

2H

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

Ptotal, atm

0.12 moles of H2 is collected over water in a 10.0 L container at 323 K. Find the total pressure.

Solution:

Concept Plan:

Relationships:

Given:

Find:

V

TRn P

PH2, PH2O Ptotal

mmHg 330P

5.4) (Table 6.291.842 P

total

total

C50 @ OHHtotal 22PPP

1 atm = 760 mmHgPtotal = PA + PB,

Kmol

Latm 0.08206 R nRT,PV

nH2,V,T PH2

mmHg 8.142atm 1

mmHg 760atm 8110.3

Page 64: Chapter 5 Gases

Tro, Chemistry: A Molecular Approach 64

Reactions Involving Gases• the principles of reaction stoichiometry from Chapter 4

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 molesas we’ve seen, 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

• when gases are at STP, use 1 mol = 22.4 L

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

Page 65: Chapter 5 Gases

L 9.66

atm 0510.97

K 3550.08206mol 842.22P

TRnV

Kmol

Latm

Ex 5.12 – What volume of H2 is needed to make 35.7 g of CH3OH at 738 mmHg and 355 K?

CO(g) + 2 H2(g) → CH3OH(g)mCH3OH = 37.5g, P=738 mmHg, T=355 K

VH2, L

Solution:

Concept Plan:

Relationships:

Given:

Find:

P

TRn V

atm 05197.0mmHg 760

atm 1mmHg 738

2

3

233

H mol 8422.2

OHCH mol 1

H mol 2

g 32.04

OHCH mol 1OHCH g 5.37

P, n, T, R V

1 atm = 760 mmHg, CH3OH = 32.04 g/mol1 mol CH3OH : 2 mol H2 Kmol

Latm 0.08206 R nRT,PV

g CH3OH mol CH3OH mol H2

OHCH mol 1

H mol 2

3

2

nH2 = 2.2284 mol, P=0.97105 atm, T=355 K

VH2, L

g 32.04

OHCH mol 1 3

Page 66: Chapter 5 Gases

Ex 5.13 – 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:

Concept Plan:

Relationships:

Given:

Find:

OH mol 1

g 02.18

2

OH g 998.0

OH mol 1

OH g 8.021

H mol 2

OH mol 2

H L 22.4

H 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 2

H mol 2

2

2

L 22.4

H mol 1 2

g H2OL H2 mol H2 mol H2O

Page 67: Chapter 5 Gases

Tro, Chemistry: A Molecular Approach 67

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)

Page 68: Chapter 5 Gases

L 791.0

atm 0.750

K3130.08206mol 850023.0P

TRnV

Kmol

Latm

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

VO2, L

Solution:

Concept Plan:

Relationships:

Given:

Find:

P

TRn V

2

2

O mol 850023.0

HgO mol 2

O mol 1

g 216.59

HgO 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

Latm 0.08206 R nRT,PV

g HgO mol HgO mol O2

HgO mol 2

O mol 1 2

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

VO2, L

g 216.59

HgO mol 1

Page 69: Chapter 5 Gases

Tro, Chemistry: A Molecular Approach 69

Properties of Gases

• expand to completely fill their container

• take the shape of their container

• low densitymuch less than solid or liquid state

• compressible

• mixtures of gases are always homogeneous

• fluid

Page 70: Chapter 5 Gases

Tro, Chemistry: A Molecular Approach 70

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 directionlike billiard balls

Page 71: Chapter 5 Gases

Tro, Chemistry: A Molecular Approach 71

Kinetic Molecular Theory• there is a lot of empty space

between the particlescompared to the size of the particles

• the average kinetic energy of the particles is directly proportional to the Kelvin temperatureas you raise the temperature of the

gas, the average speed of the particles increasesbut don’t be fooled into thinking all the

particles are moving at the same speed!!

Page 72: Chapter 5 Gases

Tro, Chemistry: A Molecular Approach 72

Gas Properties Explained – 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.

Page 73: Chapter 5 Gases

Tro, Chemistry: A Molecular Approach 73

Gas Properties Explained - Compressibility

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

Page 74: Chapter 5 Gases

Tro, Chemistry: A Molecular Approach 74

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 they have low density

Page 75: Chapter 5 Gases

Tro, Chemistry: A Molecular Approach 75

Density & Pressure • 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 pressurealso higher density

Page 76: Chapter 5 Gases

Tro, Chemistry: A Molecular Approach 76

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

Page 77: Chapter 5 Gases

Tro, Chemistry: A Molecular Approach 77

Gas Laws Explained - Charles’s Law

• Charles’s Law says that the volume of a gas is directly proportional to the absolute temperature

• increasing the temperature increases their average speed, causing them to hit the wall harder and more frequentlyon average

• in order to keep the pressure constant, the volume must then increase

Page 78: Chapter 5 Gases

Tro, Chemistry: A Molecular Approach 78

Gas Laws ExplainedAvogadro’s Law

• Avogadro’s Law says that the volume of a gas is directly proportional to the number of gas molecules

• increasing the number of gas molecules causes more of them to hit the wall at the same time

• in order to keep the pressure constant, the volume must then increase

Page 79: Chapter 5 Gases

Tro, Chemistry: A Molecular Approach 79

Gas Laws Explained – Dalton’s Law of Partial Pressures

• Dalton’s Law says that the total pressure of a mixture of gases is the sum of the partial pressures

• kinetic-molecular theory says that the gas molecules are negligibly small and don’t interact

• therefore the molecules behave independent of each other, each gas contributing its own collisions to the container with the same average kinetic energy

• since the average kinetic energy is the same, the total pressure of the collisions is the same

Page 80: Chapter 5 Gases

Tro, Chemistry: A Molecular Approach 80

Dalton’s Law & Pressure

• since the gas molecules are not sticking together, each gas molecule contributes its own force to the total force on the side

Page 81: Chapter 5 Gases

Tro, Chemistry: A Molecular Approach 81

Deriving the Ideal Gas Law from Kinetic-Molecular Theory

• pressure = Forcetotal/Area

• Ftotal = F1 collision x number of collisions in a particular time intervalF1 collision = mass x 2(velocity)/time intervalno. of collisions is proportional to the number of particles

within the distance (velocity x time interval) from the wallFtotal α mass∙velocity2

x Area x no. molecules/Volume

• Pressure α mv2 x n/V• Temperature α mv2

• P α T∙n/V, PV=nRT

Page 82: Chapter 5 Gases

Tro, Chemistry: A Molecular Approach 82

Calculating Gas Pressure

Page 83: Chapter 5 Gases

Tro, Chemistry: A Molecular Approach 83

Molecular Velocities• all the gas molecules in a sample can travel at different

speeds• however, the distribution of speeds follows a pattern

called a Boltzman distribution• we talk about the “average velocity” of the molecules,

but there are different ways to take this kind of average• the method of choice for our average velocity is called

the root-mean-square method, where the rms average velocity, urms, is the square root of the average of the sum of the squares of all the molecule velocities

22

n

vuurms

Page 84: Chapter 5 Gases

Tro, Chemistry: A Molecular Approach 84

Boltzman DistributionDistribution Function

Molecular Speed

Fra

ctio

n of

Mol

ecul

es

O2 @ 300 K

Page 85: Chapter 5 Gases

Tro, Chemistry: A Molecular Approach 85

Kinetic Energy and Molecular Velocities

• average kinetic energy of the gas molecules depends on the average mass and velocityKE = ½mv2

• gases in the same container have the same temperature, the same average kinetic energy

• if they have different masses, the only way for them to have the same kinetic energy is to have different average velocities lighter particles will have a faster average velocity than more

massive particles

Page 86: Chapter 5 Gases

Tro, Chemistry: A Molecular Approach 86

Molecular Speed vs. Molar Mass• in order to have the same average kinetic

energy, heavier molecules must have a slower average speed

Page 87: Chapter 5 Gases

Tro, Chemistry: A Molecular Approach 87

Temperature and Molecular Velocities _• KEavg = ½NAmu2

NA is Avogadro’s number

• KEavg = 1.5RTR is the gas constant in energy units, 8.314 J/mol∙K

1 J = 1 kg∙m2/s2

• equating and solving we get:NA∙mass = molar mass in kg/mol

MM

RT

mN

RTu

A

33rms

• as temperature increases, the average velocity increases

Page 88: Chapter 5 Gases

Tro, Chemistry: A Molecular Approach 88

Temperature vs. Molecular Speed

• as the absolute temperature increases, the average velocity increasesthe distribution

function “spreads out,” resulting in more molecules with faster speeds

Page 89: Chapter 5 Gases

m/s 4821032.00

K 298314.83

MM

3RT

mol

kg3-

Kmols

mkg

rms

2

2

u

T(K) = t(°C) + 273.15, O2 = 32.00 g/mol MM

3RTrms u

Ex 5.14 – Calculate the rms velocity of O2 at 25°C

O2, t = 25°C

urms

Solution:

Concept Plan:

Relationships:

Given:

Find:

MM, T urms

MM

3RTrms u

K 298T

273.1525T

273.15C) t( T(K)

Page 90: Chapter 5 Gases

Tro, Chemistry: A Molecular Approach 90

Mean Free Path• molecules in a gas travel in

straight lines until they collide with another molecule or the container

• the average distance a molecule travels between collisions is called the mean free path

• mean free path decreases as the pressure increases

Page 91: Chapter 5 Gases

Tro, Chemistry: A Molecular Approach 91

Diffusion and Effusion• the process of a collection of molecules spreading out

from high concentration to low concentration is called diffusion

• the process by which a collection of molecules escapes through a small hole into a vacuum is called effusion

• both the rates of diffusion and effusion of a gas are related to its rms average velocity

• for gases at the same temperature, this means that the rate of gas movement is inversely proportional to the square root of the molar mass

MM

1 rate

Page 92: Chapter 5 Gases

Tro, Chemistry: A Molecular Approach 92

Effusion

Page 93: Chapter 5 Gases

Tro, Chemistry: A Molecular Approach 93

Graham’s Law of Effusion• for two different gases at the same temperature,

the ratio of their rates of effusion is given by the following equation:

A gas

B gas

B gas

A gas

MassMolar

MassMolar

rate

rate

Page 94: Chapter 5 Gases

Ex 5.15 – Calculate the molar mass of a gas that effuses at a rate 0.462 times N2

MM, g/mol

Solution:

Concept Plan:

Relationships:

Given:

Find:

rateA/rateB, MMN2 MMunknown

462.0rate

rate

2N

gasunknown

N2 = 28.01 g/mol A gas

B gas

B gas

A gas

MassMolar

MassMolar

rate

rate

2

N

unknown

Nunknown

2

2

rate

rate

MassMolar MassMolar

mol

g2

molg

2

N

unknown

Nunknown 131

0.462

01.28

rate

rate

MassMolar MassMolar

2

2

Page 95: Chapter 5 Gases

95

Ideal vs. Real Gases• Real gases often do not behave like ideal gases

at high pressure or low temperature• Ideal gas laws assume

1) no attractions between gas molecules2) gas molecules do not take up space based on the kinetic-molecular theory

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

Page 96: Chapter 5 Gases

Tro, Chemistry: A Molecular Approach 96

The Effect of Molecular Volume• at high pressure, the amount of space occupied

by the molecules is a significant amount of the total volume

• the molecular volume makes the real volume larger than the ideal gas law would predict

• van der Waals modified the ideal gas equation to account for the molecular volumeb is called a van der Waals constant and is

different for every gas because their molecules are different sizes

bnP

nRTV

Page 97: Chapter 5 Gases

Tro, Chemistry: A Molecular Approach 97

Real Gas Behavior

• because real molecules take up space, the molar volume of a real gas is larger than predicted by the ideal gas law at high pressures

Page 98: Chapter 5 Gases

Tro, Chemistry: A Molecular Approach 98

The Effect of Intermolecular Attractions

• at low temperature, the attractions between the molecules is significant

• the intermolecular attractions makes the real pressure less than the ideal gas law would predict

• van der Waals modified the ideal gas equation to account for the intermolecular attractionsa is called a van der Waals constant and is different for

every gas because their molecules are different sizes

2

V

n

V

nRTP

a

Page 99: Chapter 5 Gases

Tro, Chemistry: A Molecular Approach 99

Real Gas Behavior

• because real molecules attract each other, the molar volume of a real gas is smaller than predicted by the ideal gas law at low temperatures

Page 100: Chapter 5 Gases

Tro, Chemistry: A Molecular Approach 100

Van der Waals’ Equation

• combining the equations to account for molecular volume and intermolecular attractions we get the following equationused for real gasesa and b are called van der Waal

constants and are different for each gas

nRTn-VV

nP

2

ba

Page 101: Chapter 5 Gases

Tro, Chemistry: A Molecular Approach 101

Real Gases• a plot of PV/RT vs. P for 1 mole of a gas shows

the difference between real and ideal gases• it reveals a curve that shows the PV/RT ratio for

a real gas is generally lower than ideality for “low” pressures – meaning the most important factor is the intermolecular attractions

• it reveals a curve that shows the PV/RT ratio for a real gas is generally higher than ideality for “high” pressures – meaning the most important factor is the molecular volume

Page 102: Chapter 5 Gases

Tro, Chemistry: A Molecular Approach 102

PV/RT Plots

Page 103: Chapter 5 Gases

Tro, Chemistry: A Molecular Approach 103

Structure of the Atmosphere• the atmosphere shows several

layers, each with its own characteristics

• the troposphere is the layer closest to the earth’s surface circular mixing due to thermal currents

– weather

• the stratosphere is the next layer up less air mixing

• the boundary between the troposphere and stratosphere is called the tropopause

• the ozone layer is located in the stratosphere

Page 104: Chapter 5 Gases

Tro, Chemistry: A Molecular Approach 104

Air Pollution• air pollution is materials added to the atmosphere that

would not be present in the air without, or are increased by, man’s activities though many of the “pollutant” gases have natural sources as

well• pollution added to the troposphere has a direct effect on

human health and the materials we use because we come in contact with itand the air mixing in the troposphere means that we all get a

smell of it!• pollution added to the stratosphere may have indirect

effects on human health caused by depletion of ozoneand the lack of mixing and weather in the stratosphere means

that pollutants last longer before “washing” out

Page 105: Chapter 5 Gases

Tro, Chemistry: A Molecular Approach 105

Pollutant Gases, SOx

• SO2 and SO3, oxides of sulfur, come from coal combustion in power plants and metal refiningas well as volcanoes

• lung and eye irritants

• major contributor to acid rain

2 SO2 + O2 + 2 H2O 2 H2SO4

SO3 + H2O H2SO4

Page 106: Chapter 5 Gases

Tro, Chemistry: A Molecular Approach 106

Pollutant Gases, NOx

• NO and NO2, oxides of nitrogen, come from burning of fossil fuels in cars, trucks, and power plantsas well as lightning storms

• NO2 causes the brown haze seen in some cities• lung and eye irritants• strong oxidizers• major contributor to acid rain

4 NO + 3 O2 + 2 H2O 4 HNO3

4 NO2 + O2 + 2 H2O 4 HNO3

Page 107: Chapter 5 Gases

Tro, Chemistry: A Molecular Approach 107

Pollutant Gases, CO

• CO comes from incomplete burning of fossil fuels in cars, trucks, and power plants

• adheres to hemoglobin in your red blood cells, depleting your ability to acquire O2

• at high levels can cause sensory impairment, stupor, unconsciousness, or death

Page 108: Chapter 5 Gases

Tro, Chemistry: A Molecular Approach 108

Pollutant Gases, O3

• ozone pollution comes from other pollutant gases reacting in the presence of sunlightas well as lightning stormsknown as photochemical smog and ground-level

ozone

• O3 is present in the brown haze seen in some cities

• lung and eye irritants• strong oxidizer

Page 109: Chapter 5 Gases

Tro, Chemistry: A Molecular Approach 109

Major Pollutant Levels

• government regulation has resulted in a decrease in the emission levels for most major pollutants

Page 110: Chapter 5 Gases

Tro, Chemistry: A Molecular Approach 110

Stratospheric Ozone

• ozone occurs naturally in the stratosphere

• stratospheric ozone protects the surface of the earth from over-exposure to UV light from the sun

O3(g) + UV light O2(g) + O(g)

• normally the reverse reaction occurs quickly, but the energy is not UV light

O2(g) + O(g) O3(g)

Page 111: Chapter 5 Gases

Tro, Chemistry: A Molecular Approach 111

Ozone Depletion• chlorofluorocarbons became popular as aerosol

propellants and refrigerants in the 1960s• CFCs pass through the tropopause into the stratosphere• there CFCs can be decomposed by UV light, releasing Cl

atoms

CF2Cl2 + UV light CF2Cl + Cl

• Cl atoms catalyze O3 decomposition and removes O atoms so that O3 cannot be regeneratedNO2 also catalyzes O3 destruction

Cl + O3 ClO + O2

O3 + UV light O2 + O

ClO + O O2 + Cl

Page 112: Chapter 5 Gases

Tro, Chemistry: A Molecular Approach 112

Ozone Holes

• satellite data over the past 3 decades reveals a marked drop in ozone concentration over certain regions