Chapter 5 Gases

Download Report

Transcript Chapter 5 Gases

Chemistry: A Molecular Approach

, 1

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

•

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 Tro, Chemistry: A Molecular Approach 2

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 exert  14.7 lbs./in 2  so if our pump could get a perfect vacuum, the maximum height the column could rise is 10.3 m Tro, Chemistry: A Molecular Approach Pressure  Force Area 3

Gases 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 Tro, Chemistry: A Molecular Approach 4

• •

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’s path, the gas will try to push it along as the gas flows Tro, Chemistry: A Molecular Approach 5

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 you  at 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 Tro, Chemistry: A Molecular Approach 6

Pressure Imbalance in Ear

If there is a difference in pressure across the eardrum membrane, the membrane will be pushed out – what we commonly call a “popped eardrum.” Tro, Chemistry: A Molecular Approach 7

•

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 factors  number of gas particles in a given volume  volume of the container  average speed of the gas particles Tro, Chemistry: A Molecular Approach 8

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 Tro, Chemistry: A Molecular Approach 9

Common Units of Pressure

Unit

pascal (Pa), 1 Pa  1 N m 2 kilopascal (kPa) atmosphere (atm) millimeters of mercury (mmHg) inches of mercury (inHg) torr (torr) pounds per square inch (psi, lbs./in 2 ) Tro, Chemistry: A Molecular Approach

Average Air Pressure at Sea Level

101,325 101.325

1 (exactly) 760 (exactly) 29.92

760 (exactly) 14.7

Example 5.1 – A high-performance bicycle tire has a pressure of 132 psi. What is the pressure in mmHg?

Given:

132 psi

Find: Concept Plan:

mmHg

Relationships:

psi 1 atm 14.7

psi atm 760 mmHg 1 atm mmHg 1 atm = 14.7 psi, 1 atm = 760 mmHg

Solution:

132 psi  1 atm 14.7

psi  760 mmHg 1 atm  6 .82

 10 3 mmHg

Check:

since mmHg are smaller than psi, the answer makes sense

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 Tro, Chemistry: A Molecular Approach 12

Manometer

for this sample, the gas has a larger pressure than the atmosphere, so Pressure gas  Pressure atmosphere  Pressure h Pressure gas (mmHg)  Pressure atmosphere (mmHg)  difference in Hg levels (mm) Tro, Chemistry: A Molecular Approach 13

Boyle’s Law

• pressure of a gas is inversely proportional to its volume  constant

and amount of gas  graph

is curve  graph

vs 1/

is straight line • as

increases,

decreases by the same factor • •

= constant

1 x

1 =

2 x

2 Tro, Chemistry: A Molecular Approach 14

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)

48 44 40 36 32 28 24 22

Difference in Hg Levels (in)

0.0

2.8

6.2

10.1

15.1

21.2

29.7

35.0

Tro, Chemistry: A Molecular Approach 15

Boyle's Expt.

140 120 100 80 60 40 20 0 0 10 Tro, Chemistry: A Molecular Approach 20 30

Volume of Air, in 3

40 50 60 16

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

140 120 100 80 60 40 20 0 0 0.01

0.02

0.03

0.04

0.05

Inv. Volume, in -3

0.06

0.07

Tro, Chemistry: A Molecular Approach 0.08

0.09

Boyle’s Experiment,

x

Pressure Volume 29.13

33.50

41.63

50.31

61.31

74.13

87.88

115.56

48 42 34 28 23 19 16 12

1400 1400 1400 1400 1400 1400 1400 1400 Tro, Chemistry: A Molecular Approach 18

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) Tro, Chemistry: A Molecular Approach 19

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 atm  at 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 Tro, Chemistry: A Molecular Approach 20

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?

Given:

V 1 =7.25 L, P 1 = 4.52 atm, P 2 = 1.21 atm

Find:

V 2 , L

Concept Plan: Relationships: Solution:

V 1 , P 1 , P 2 V 2  P 1  V 1 P 2 P 1 ∙ V 1 = P 2 ∙ V 2 V 2 V 2   P 1  V 1 P 2  4.52

atm  7.25

L 1.21

atm    27 .

1 L

Check:

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

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?

Tro, Chemistry: A Molecular Approach 22

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?

Given:

V 2 =2780 mL, P 1 = 762 torr, P 2 = 0.500 atm

Find:

V 1 , mL

Concept Plan: Relationships: Solution:

P 1 ∙ V 1 , P 1 , P 2 V 1  V 1 = P 2 P 2  V 2 V 2 P 1 ∙ V 2 , 1 atm = 760 torr (exactly) 782 torr  1 atm 760 torr  1 .

03 atm V 1   P 2  V 2 P 1  0.500

atm  1.03

atm 2780 L    1350 mL

Check:

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

Charles’ Law

• • • • volume is directly proportional to temperature  constant P and amount of gas  graph 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 V

1

 V

2

Tro, Chemistry: A Molecular Approach 24

• •

Charles’ Law – A Molecular View

• • moving as fast, so they of the balloon harder – small Tro, Chemistry: A Molecular Approach 25

0.6

0.5

0.4

0.3

0.2

0.1

Charles' Law & Absolute Zero

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

Temperature, °C

0 50 100 150 26

Example 5.3 – A gas has a volume of 2.57 L at 0.00°C. What was the temperature at 2.80 L?

Given:

V 1 =2.57 L, V 2 = 2.80 L, t 2 = 0.00°C

Find:

t 1 , K and °C

Concept Plan: Relationships: Solution:

T 2  0.00

 273.15

T 2  273.15

K V 1 , V 2 , T 2 T 1  T 2  V V 2 1 T(K) = t(°C) + 273.15, T 1 V 1 T 1 T 1   T 2  V 1 V 2  273.15

K  2.80

L  2.57

L   29 7 .

6 K  V 2 T 2 t t 1 1   T 1  273.15

29 7 .

6  273.15

t 1  2 4  C

Check:

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

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? Tro, Chemistry: A Molecular Approach 28

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?

Given:

V 1 =10.0 L, t 1 = 25.0°C L, t 2 = 250.0°C

Find:

V 2 , L

Concept Plan: Relationships:

V 1 , T 1 , T 2 T(K) = t(°C) + 273.15,

Solution:

T 1  2 5 .

0 T 1  298.2

 273.15

K T 2  2 50 .

0  273.15

T 2  523.2

K V 2  V 1  T 2 T 1 V 2 V 1 T 1  V 2   T 2  T 1  523.2

K  V 1 10.0

L 298.2

K   V 2 T 2  17 .

5 L

Check:

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

•

Avogadro’s Law

volume directly proportional to the number of gas molecules 

= constant x

 constant P and T  more gas molecules = larger volume • • count number of gas molecules by

moles

equal volumes of gases contain equal numbers of molecules 

the gas doesn’t matter

V 1  n 1 V 2 n 2 Tro, Chemistry: A Molecular Approach 30

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?

Given:

V 1 =4.65 L, V 2 = 6.48 L, n 1 = 0.225 mol

Find:

n 2 , and added moles

Concept Plan: Relationships:

V 1 , V 2 , n 1 n 1  V 2 V 1 mol added = n 2  n 2 – n 1 , n 2 V 1 n 1  V 2 n 2

Solution:

n 2   n 1  V 1 V 2  0.225

mol    4.65

L  6.48

L  moles added  0 .

314  0 .

225 moles added  0 .

089 mol  0 .

314 mol

Check:

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

Ideal Gas Law

• • By combing the gas laws we can write a general equation

is called the

gas constant

• the value of

depends on the units of P and V • we will use 0.08206 and convert P to atm and V to L mol   L K • 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

       

 R or PV  nRT Tro, Chemistry: A Molecular Approach 32

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?

Given:

V = 3.24 L, P = 24.3 psi, t = 25 °C,

Find:

n, mol

Concept Plan: Relationships:

P, V, T, R 1 atm = 14.7 psi n  PV RT T(K) = t(°C) + 273.15

Solution:

24.3

psi  1 atm 14.7

psi  1.6

T(K)  25  C  273.15

T  298 K 5 31 atm n PV  nRT, R  0.08206

atm  L mol  K n   P  V R  T  1.6

5 31 atm  0.08206

atm  L mol  K   3 .

24 L  2 98 K    0 .

219 mol

Check:

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

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 conditions

 STP standard pressure = 1 atm standard temperature = 273 K  0°C Tro, Chemistry: A Molecular Approach 34

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

Tro, Chemistry: A Molecular Approach 35

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

Given:

V 1 = 10.0 L, P 1 = 44.1 psi, t 1 = 27 °C, P 2 = 1.00 atm, t 2 = 0°C

Find:

V 2 , L

Concept Plan: Relationships:

P 1 , V 1 , T 1 , R n  PV RT 1 atm = 14.7 psi T(K) = t(°C) + 273.15

44.1



Solution:

n  psi R P   T  1  1 .

2 1 9 mol T(K)  27  C atm 14.7

    psi  0.08206

3 .00

atm  .

00 atm  L m ol  K  atm   2 73 K  T 1  .

3 00 .

K n n PV P 2 , n, T 2 , R V   nRT, nRT V P R  0.08206

atm  L mol  K 2   P  V  R  T 3.00

atm  0.08206

atm  m ol  L K   1 0.0

L   3 00.

K   1 .

2 1 9 mol

Check:

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

Molar Volume

• solving the ideal gas equation for the volume of 1 mol of gas at STP gives 22.4 L  6.022 x 10 23 molecules of gas  notice: the gas is immaterial • we call the volume of 1 mole of gas at STP the

molar volume

 it is important to recognize that one mole of different gases have different masses, even though they have the same volume Tro, Chemistry: A Molecular Approach 37

Molar Volume

Tro, Chemistry: A Molecular Approach 38

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

Density



Molar Mass, 22.4

L g

Tro, Chemistry: A Molecular Approach 39

Gas Density

mass  1 mol  moles  moles molar mass density  mass in grams volume in liters  mass molar mass P  V  n  R  T V P mass   V  molar density mass  P  mass  (molar R R   T T mass) • density is directly proportional to molar mass Tro, Chemistry: A Molecular Approach 40

Example 5.7 – Calculate the density of N 2 at 125°C and 755 mmHg

Given:

P = 755 mmHg, t = 125 °C,

Find:

d N2 , g/L

Concept Plan: Relationships: Solution:

755 mmHg  1 atm 760 mmHg T(K)  125  C  273.15

P, MM, T, R  0 .

99 3 42 atm d  P  R d   MM   T d d  P  MM  T  T  398 K 0 .

852 g/L

Check:

since the density of N 2 is 1.25 g/L at STP, we expect the density to be lower when the temperature is raised, and it is

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 Molar Mass  mass in grams moles Tro, Chemistry: A Molecular Approach 42

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

Given: Find: Concept Plan: Relationships:

P, V, T, R n  P R   V T 1 atm = 760 mmHg, n T(K) = t(°C) + 273.15

n, m PV  nRT, MM  m n MM MM  m R n  0.08206

atm  L mol  K n    

Solution:

P mmHg R 1.1

 6  V T 58  atm 760   328 K atm  mol  L K 1   0 atm mmHg  .225

3 28 L K    1 .

1 6 58 atm  MM  m  0 .

311 g n 9 .

7 4 54  10  3 mol  31.9

g/mol 9.7454

 10 3 mol

Check:

the value 31.9 g/mol is reasonable

Practice - Calculate the density of a gas at 775 torr and 27°C if 0.250 moles weighs 9.988 g Tro, Chemistry: A Molecular Approach 44

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

Given: Find: Concept Plan: Relationships:

P, n, T, R V  n  R  T P V 1 atm = 760 mmHg, T(K) = t(°C) + 273.15

V, m d d  m V PV  d  nRT, m V R  0.08206

atm  L mol  K V    0

Solution:

n T  .250

 R P   mol  300.

K  760 torr   1 atm C  0.08206

1.0197

 1 .

0 1 97 atm atm  m ol  atm  L K    3 0 0 .

K   6 .

0 3 55 L d  m  V  1.65

g/L 9 .988

g 6 .0

3 55 L

Check:

the value 1.65 g/L is reasonable

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 Tro, Chemistry: A Molecular Approach 46

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 if  we know what fraction of the mixture it composes and the total pressure  or, 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 pressure  Dalton’s Law of Partial Pressures  because the gases behave independently Tro, Chemistry: A Molecular Approach 47

Composition of Dry Air

Tro, Chemistry: A Molecular Approach 48

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

for two gases, A and B, mixed together P A  n A x R x T V P B  n B x R x T V the temperatu re and volume of everything in the mixture are the same n total  n A  n B P total  P A  P B  n total x R x T V Tro, Chemistry: A Molecular Approach 49

Example 5.9 – Determine the mass of Ar in the mixture

Given: Find:

P He = 0.275 atm, V = 1.00 L, T=298 K V = 1.00 L, T=298 K = 662 mmHg, mass Ar , g

Concept Plan:

P tot , P He , P Ne P Ar = P tot – (P He P Ar + P Ne ) P Ar , V, T n  P R   V T n A r m  n m A r  MM

Relationships:

P tot = P a + P b + etc., PV 1 atm = 760 mmHg, MM Ar  nRT, R  = 39.95 g/mol 0.08206

MM  m n atm  L mol  K P Ar  209 209 n 

Solution:

     mmHg 0 .275

 341 atm 760 1 atm  L     112 atm  1  .

 00 mmHg mmHg L K .

   275 1 .

125 atm 1 .

125  10  2 mol  0.449

g Ar  10  2 mol  39.95

g 1 mol

Check:

the units are correct, the value is reasonable

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. Tro, Chemistry: A Molecular Approach 51

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

Given:

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

Find:

P Ne , atm

Concept Plan:

n Xe , V, T, R P Xe  n Xe  R  T V P Xe P tot , P Xe P Ne  P total P Ne  P Xe

Relationships:

Solution:

Ne  P total   3 .

9 atm 2.9

atm  PV P Xe  nRT,  0.9

5 89 atm R  0.08206

atm  L mol  K , P total  P Ne P Xe   n Xe  R  T  0.17

mol V   0.08206

8.7

L  0 .

9 5 89 atm atm  L mo l  K  P Xe   5 98 K 

Check:

the unit is correct, the value is reasonable

Mole Fraction

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

mole fraction,

c for gases, = volume % / 100% the partial pressure of a gas is equal to the mole fraction of that gas times the total pressure Tro, Chemistry: A Molecular Approach P A P total  n A n total c A  n A n total P A  c A  P total 53

• • • Mountain Climbing & Partial Pressure

our bodies are adapted to breathe O 2 at a partial pressure of 0.21 atm  Sherpa, people native to the Himalaya mountains, are adapted to the much lower partial pressure of oxygen in their air partial pressures of O 2 lower than 0.1 atm will lead to

hypoxia

 unconsciousness or death climbers of Mt Everest carry O 2 cylinders to prevent hypoxia  on top of Mt Everest, P air so P O2 = 0.065 atm in = 0.311 atm, Tro, Chemistry: A Molecular Approach 54

Deep Sea Divers & Partial Pressure •

its also possible to have too much O 2 , a condition called

oxygen toxicity

 P O2  > 1.4 atm oxygen toxicity can lead to muscle spasms, tunnel vision, and convulsions

•

its also possible to have too much N 2 , a condition called

nitrogen narcosis

 also known as Rapture of the Deep

•

when diving deep, the pressure of the air divers breathe increases – so the partial pressure of the oxygen increases  at a depth of 55 m the partial pressure of O 2  heliox that contains a lower percentage of O 2 is 1.4 atm divers that go below 50 m use a mixture of He and O than air 2 called Tro, Chemistry: A Molecular Approach 55

Partial Pressure & Diving

Tro, Chemistry: A Molecular Approach 56

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 O 2 at 298 K

Given:

m He = 24.2 g, m O2

Find:

c He , c O2 , P He , atm, P O2 , atm, P total , atm O2 , atm, P total , atm

Concept Plan:

m gas n gas c gas A  c gas n gas A n total n tot , V, T, R P total  n total  R V  T P tot c gas , P total P A  c A P gas  P total

Relationships:

PV  nRT, R  0.08206

atm  L mol  K , P A  c A  P total MM He MM O2 = 4.00 g/mol = 32.00 g/mol P   g

Solution:

 He 8 2 n 5  mol  V R He   32.00

g 2   T 6 .

05 mol 12.5

L He atm  L 0 .

135 mol  O 2   2 98 K c  c  12 .

0 99 atm He  6.05

P 6.05

mol 0 .

He c mol  17  He t ot al 12 .

mol O  atm O 2   P O 2 11.8

c atm mol 2 6.05

mol .

He  8 27 O mol  12 .

0 O atm  0 .

264 atm 0 .

97 8 17 0 .

021 8 27

• • • • 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* Tro, Chemistry: A Molecular Approach 58

Vapor Pressure of Water

Tro, Chemistry: A Molecular Approach 59

Collecting Gas by Water Displacement

Tro, Chemistry: A Molecular Approach 60

Ex 5.11 – 1.02 L of O 2 collected over water at 293 K with a total pressure of 755.2 mmHg. Find mass O 2 .

Given: Find:

mass O 2 , g

Concept Plan: Relationships:

P tot , P H2O P O 2  P total P O2  P H 2 O @ 20  C P O2 ,V,T n  P R   V T n O2 g O2 1 atm = 760 mmHg, P total = P A + P B , O 2 PV  nRT, R  0.08206

atm  L mol  K = 32.00 g/mol P O n P O  2

Solution:

2     6 5 P R   V 0.970

T 6 5 mmHg 5 9 mmHg 0.08206

17.55

(Table atm atm  L mo l  K     1 .

02 293 .

5.4) L K    0 .

970 5 9 atm  4 .

1 1 75  10  2 mol 4 .

1 1 75  10  2 mol  3 2.00

g 1 mol  1.32

Practice – 0.12 moles of H 2 is collected over water in a 10.0 L container at 323 K. Find the total pressure.

Tro, Chemistry: A Molecular Approach 62

0.12 moles of H 2 is collected over water in a 10.0 L container at 323 K. Find the total pressure.

Given:

V=10.0 L, n H2 =0.12 mol, T=323 K

Find:

P total , atm

Concept Plan: Relationships:

n H2 ,V,T P H2 P  n  R  T V 1 atm = 760 mmHg P total = P A + P B , P H2 , P H2O P total  P H 2 P PV  nRT, R  0.08206

atm  L mol  K total  P H 2 O @ 50  C P H  2

Solution:

 n  R  T V  0.12

mol   0.08206

 10.0

L  atm  L mo l  K  0 .

3 1 81 atm 0.3

1 81 atm    323 K  760 mmHg 1 atm  2 4 1 .

8 mmHg P total  2 4 1.8

P total  330  9 2 mmHg .

6 (Table 5.4)

• • • • 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 moles  as 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 Tro, Chemistry: A Molecular Approach 64

Ex 5.12 – What volume of H 2 is needed to make 35.7 g of CH 3 OH at 738 mmHg and 355 K?

Given:

CO(

) + 2 H 2 (

) → CH 3 OH(

) m CH3OH

Find: Concept Plan:

V H2 , L g CH 3 OH mol CH 3 OH 1 mol CH 3 OH mol H 2 2 mol H 2 32.04

g 1 mol CH 3 OH P, n, T, R V  n  R  T V P

Relationships:

1 atm = 760 mmHg, CH 3 OH = 32.04 g/mol 1 mol CH 3 OH : 2 mol H 2 PV  nRT, R  0.08206

atm  L mol  K 

Solution :

37 .

5 g CH 2.2

3 OH 2 84 mol 738 mmHg  H  2 1 mol CH 3 32.04

g OH V    n 2  R  T 2 1 mol P CH 3 OH  2 .2

2 84 mol 1 atm 760 mmHg   .

97 1 05 atm 66 .

9 L    0.08206

0.97

1 05 atm atm  mo l   K L 355 K 

Ex 5.13 – How many grams of H 2 O form when 1.24 L H 2 completely with O 2 at STP?

O 2 (

) + 2 H 2 (

) → 2 H 2 O(

) reacts

Given:

V H2 = 1.24 L, P=1.00 atm, T=273 K

Find: Concept Plan:

mass H2O , g L H 2 1 mol H 2 22.4

L mol H 2 2 mol H 2 mol H 2 O 2 mol H 2 O 18 .

02 g 1 mol H 2 O g H 2 O

Relationships:

H 2 O = 18.02 g/mol, 1 mol = 22.4 L @ STP 2 mol H 2 O : 2 mol H 2

Solution :

1 .24

L H 2  1 mol H 2 22.4

L H 2  0 .

998 g H 2 O  2 mol H 2 O 2 mol H 2  1 8.02

g H 2 O 1 mol H 2 O

Practice – What volume of O 2 (MM HgO at 0.750 atm and 313 K is generated by the thermolysis of 10.0 g of HgO?

2 HgO(

)  2 Hg(

) + O 2 (

) = 216.59 g/mol) Tro, Chemistry: A Molecular Approach 67

What volume of O 2

Given:

the thermolysis of 10.0 g of HgO?

2 HgO(

)  2 Hg(

) + O 2 (

) m HgO at 0.750 atm and 313 K is generated by

Find: Concept Plan:

V O2 , L g HgO 1 mol HgO mol HgO 1 mol O 2 mol O 2 216.59

g 2 mol HgO P, n, T, R V  n  R  T V P

Relationships:

1 atm = 760 mmHg, HgO = 216.59 g/mol 2 mol HgO : 1 mol O 2 PV  nRT, R  0.08206

atm  L mol  K

Solution :

1 0.0

g HgO  1 mol HgO 216.59

g  0 .

023 0 85 mol O 2  1 V  2 2 mol  HgO  n  R P  T 0 .

023 0 85 mol     0.08206

0.750

atm   0 .

791 L atm  L m ol  K    313 K 

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 Tro, Chemistry: A Molecular Approach 69

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 Tro, Chemistry: A Molecular Approach 70

•

Kinetic Molecular Theory

there is a lot of empty space between the particles  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!!

Tro, Chemistry: A Molecular Approach 71

Gas Properties Explained – Indefinite Shape and Indefinite Volume

Because the gas molecules have enough kinetic energy to overcome attractions, they keep moving around and spreading out until they fill the container.

As a result, gases take the shape and the volume of the container they are in.

Tro, Chemistry: A Molecular Approach 72

Gas Properties Explained Compressibility

Because there is a lot of unoccupied space in the structure of a gas, the gas molecules can be squeezed closer together Tro, Chemistry: A Molecular Approach 73

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 Tro, Chemistry: A Molecular Approach 74

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 pressure  also higher density Tro, Chemistry: A Molecular Approach 75

• • • 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 Tro, Chemistry: A Molecular Approach 76

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 frequently  on average

•

in order to keep the pressure constant, the volume must then increase Tro, Chemistry: A Molecular Approach 77

Gas Laws Explained Avogadro’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 Tro, Chemistry: A Molecular Approach 78

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 Tro, Chemistry: A Molecular Approach 79

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 Tro, Chemistry: A Molecular Approach 80

• • • • • Deriving the Ideal Gas Law from Kinetic-Molecular Theory

pressure = Force total /Area F total  = F 1 collision x number of collisions in a particular time interval  F 1 collision  = mass x 2(velocity)/time interval no. of collisions is proportional to the number of particles within the distance (velocity x time interval) from the wall  F total α m ass∙ v elocity 2 x A rea x n o. molecules/ V olume Pressure α mv 2 x n/V Temperature α mv 2 P α T∙n/V,  PV=nRT Tro, Chemistry: A Molecular Approach 81

Calculating Gas Pressure

Tro, Chemistry: A Molecular Approach 82

• • • • 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,

u rms ,

is the square root of the average of the sum of the squares of all the molecule velocities

u rms

  v 2  n

2 Tro, Chemistry: A Molecular Approach 83

Boltzman Distribution

Distribution Function

O2 @ 300 K Tro, Chemistry: A Molecular Approach

Molecular Speed

Kinetic Energy and Molecular Velocities • •

average kinetic energy of the gas molecules depends on the average mass and velocity  KE = ½mv 2

•

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 Tro, Chemistry: A Molecular Approach 85

Molecular Speed vs. Molar Mass

• in order to have the same average kinetic energy, heavier molecules must have a slower average speed Tro, Chemistry: A Molecular Approach 86

• • KE avg  N A = ½N A m

2 is Avogadro’s number KE avg = 1.5RT

 R is the gas constant in energy units, 8.314 J/mol∙K  1 J = 1 kg∙m 2 /s 2 • equating and solving we get:  N A ∙mass = molar mass in kg/mol

rms  3

RT N A



 3

RT MM

• as temperature increases, the average velocity increases Tro, Chemistry: A Molecular Approach 87

Temperature vs. Molecular Speed

• as the absolute temperature increases, the average velocity increases  the distribution function “spreads out,” resulting in more molecules with faster speeds Tro, Chemistry: A Molecular Approach 88

Ex 5.14 – Calculate the rms velocity of O 2 at 25°C

Given:

O 2 , t = 25°C

Find:

rms

Concept Plan:

MM, T

rms

rms  3RT MM

Relationships:

T(K) = t(°C) + 273.15, O 2 = 32.00 g/mol

Solution:

T(K)  t(  C)  273.15

T  25  273.15

T  298 K  3RT

rms  MM 3      8 .

314  32.00

kg  m 2 mol  K s 2  10 3      kg mol  298 K    482 m/s

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 Tro, Chemistry: A Molecular Approach 90

• 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 1 rate  MM Tro, Chemistry: A Molecular Approach 91

Tro, Chemistry: A Molecular Approach

Effusion

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:

rate

gas A

rate

gas B 

Molar Mass

gas B

Molar Mass

gas A Tro, Chemistry: A Molecular Approach 93

Ex 5.15 – Calculate the molar mass of a gas that

Given: Find:

effuses at a rate 0.462 times N

2 rate unknown gas rate N 2 MM, g/mol  0 .

462

Concept Plan: Relationships:

N 2 rate A /rate B , MM N2 Molar Mass unknown  = 28.01 g/mol Molar Mass N 2 2 rate unknown MM unknown rate N 2 rate gas A rate gas B  Molar Mass gas B Molar Mass gas A

Solution:

Molar Mass unknown  Molar Mass N 2 2 rate unknown rate N 2   28 .

01 g  0.462

mol  2   131 g mol

• • • Ideal vs. Real Gases

Real gases often do not behave like ideal gases at high pressure or low temperature 1) 2)  Ideal gas laws assume no attractions between gas molecules gas molecules do not take up space based on the kinetic-molecular theory at low temperatures and high pressures these assumptions are not valid 95

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 volume 

is called a

van der Waals constant

and is different for every gas because their molecules are different sizes Tro, Chemistry: A Molecular Approach V  nRT P  n

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 Tro, Chemistry: A Molecular Approach 97

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 attractions 

is called a

van der Waals constant

and is different for every gas because their molecules are different sizes 2

P



nRT V

Tro, Chemistry: A Molecular Approach 



n V

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 Tro, Chemistry: A Molecular Approach 99

Van der Waals’ • Equation

combining the equations to account for molecular volume and intermolecular attractions we get the following equation  used for real gases 

and

are called van der Waal constants and are different for each   P  gas

n V 2     V n

  nRT Tro, Chemistry: A Molecular Approach 100

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 Tro, Chemistry: A Molecular Approach 101

PV/RT Plots

Tro, Chemistry: A Molecular Approach 102

• • • • •

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

 less air mixing is the next layer up the boundary between the troposphere and stratosphere is called the

tropopause

the

ozone layer

stratosphere is located in the Tro, Chemistry: A Molecular Approach 103

• • • 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 it  and 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 ozone  and the lack of mixing and weather in the stratosphere means that pollutants last longer before “washing” out Tro, Chemistry: A Molecular Approach 104

Pollutant Gases, SO

x • SO 2 and SO 3 , oxides of sulfur, come from coal combustion in power plants and metal refining  as well as volcanoes • • lung and eye irritants major contributor to acid rain 2 SO 2 + O 2 SO 3 + 2 H 2 O  + H 2 O  H 2 2 H SO 4 2 SO 4 Tro, Chemistry: A Molecular Approach 105

Pollutant Gases, NO

• • • • •

NO and NO 2 , oxides of nitrogen, come from burning of fossil fuels in cars, trucks, and power plants  as well as lightning storms NO 2 causes the brown haze seen in some cities lung and eye irritants strong oxidizers major contributor to acid rain 4 NO + 3 O 2 4 NO 2 + O 2 + 2 H 2 O  + 2 H 2 O  4 HNO 4 HNO 3 3 Tro, Chemistry: A Molecular Approach 106

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 O 2 at high levels can cause sensory impairment, stupor, unconsciousness, or death Tro, Chemistry: A Molecular Approach 107

Pollutant Gases, O

3 • • • • ozone pollution comes from other pollutant gases reacting in the presence of sunlight  as well as lightning storms  known as photochemical smog and ground-level ozone O 3 is present in the brown haze seen in some cities lung and eye irritants strong oxidizer Tro, Chemistry: A Molecular Approach 108

Major Pollutant Levels

• government regulation has resulted in a decrease in the emission levels for most major pollutants Tro, Chemistry: A Molecular Approach 109

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 O 3 (

) + UV light  O 2 (

) + O(

) normally the reverse reaction occurs quickly, but the energy is not UV light O 2 (

) + O(

)  O 3 (

) Tro, Chemistry: A Molecular Approach 110

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 CF 2 Cl 2 + UV light  CF 2 Cl + Cl

•

Cl atoms catalyze O 3 atoms so that O 3 decomposition and removes O cannot be regenerated  NO 2 also catalyzes O 3 O 3 destruction Cl + O 3  ClO + O 2 + UV light  ClO + O  O 2 O 2 + O + Cl Tro, Chemistry: A Molecular Approach 111

Ozone Holes

• satellite data over the past 3 decades reveals a marked drop in ozone concentration over certain regions Tro, Chemistry: A Molecular Approach 112