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Chapter 5 Gases

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Title: Chapter 5 Gases


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

2008, Prentice Hall
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

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 exert
  • 14.7 lbs./in2
  • so if our pump could get a perfect vacuum, the
    maximum height the column could rise is 10.3 m

4
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

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
    pressure
  • the bigger the difference in pressure, the
    stronger the flow of the gas
  • if there is something in the gass path, the gas
    will try to push it along as the gas flows

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 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

7
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.
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 factors
  • number of gas particles in a given volume
  • volume of the container
  • average speed of the gas particles

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

10
Gas Pressure
  • Pressure is defined as the force per unit area,
    and is usually measured in Pascals, which are
    N/m2.
  • We measure pressure in mmHg or torr.
  • These units of pressure are equivalent come
    from measurements using a Torricellian barometer.

P pressure, F Force mass (g), A (area)
cm2, d density (g/cm3), h height
11
Common Units of Pressure
12
Example 5.1 A high-performance bicycle tire has
a pressure of 132 psi. What is the pressure in
mmHg?
132 psi mmHg
Given Find
1 atm 14.7 psi, 1 atm 760 mmHg
Concept Plan Relationships
Solution
since mmHg are smaller than psi, the answer makes
sense
Check
13
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

14
Manometer
for this sample, the gas has a larger pressure
than the atmosphere, so
15
Boyles Law
  • pressure of a gas is inversely proportional to
    its volume
  • constant T and amount of gas
  • graph P vs V is curve
  • graph 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

16
Boyles Experiment
  • added Hg to a J-tube with air trapped inside
  • used length of air column as a measure of volume

17
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18
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19
Boyles Experiment, P x V
20
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)
21
Boyles 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
22
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?
V1 7.25 L, P1 4.52 atm, P2 1.21 atm V2, L
Given Find
P1 V1 P2 V2
Concept Plan Relationships
Solution
since P and V are inversely proportional, when
the pressure decreases 4x, the volume should
increase 4x, and it does
Check
23
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?

24
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?
V2 2780 mL, P1 762 torr, P2 0.500 atm V1, mL
Given Find
P1 V1 P2 V2 , 1 atm 760 torr
(exactly)
Concept Plan Relationships
Solution
since P and V are inversely proportional, when
the pressure decreases 2x, the volume should
increase 2x, and it does
Check
25
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

26
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 dont 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

27
The data fall on a straight line. If the lines
are extrapolated back to a volume of 0, they
all show the same temperature, -273.15C, called
absolute zero
28
Example 5.3 A gas has a volume of 2.57 L at
0.00C. What was the temperature at 2.80 L?
V1 2.57 L, V2 2.80 L, t2 0.00C t1, K and C
Given Find
Concept Plan Relationships
Solution
since T and V are directly proportional, when the
volume decreases, the temperature should
decrease, and it does
Check
29
Practice The temperature inside a balloon is
raised from 25.0C to 250.0C. If the volume of
cold air was 10.0 L, what is the volume of hot
air?

30
The temperature inside a balloon is raised from
25.0C to 250.0C. If the volume of cold air was
10.0 L, what is the volume of hot air?
V1 10.0 L, t1 25.0C L, t2 250.0C V2, L
Given Find
Concept Plan Relationships
Solution
since T and V are directly proportional, when the
temperature increases, the volume should
increase, and it does
Check
31
Avogadros Law
  • volume 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
  • equal volumes of gases contain equal numbers of
    molecules
  • the gas doesnt matter

32
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?
V1 4.65 L, V2 6.48 L, n1 0.225 mol n2, and
added moles
Given Find
Concept Plan Relationships
Solution
since n and V are directly proportional, when the
volume increases, the moles should increase, and
it does
Check
33
Ideal Gas Law
34
Example 5.6 How many moles of gas are in a
basketball with total pressure 24.3 psi, volume
of 3.24 L at 25C?
V 3.24 L, P 24.3 psi, t 25 C, n, mol
Given Find
Concept Plan Relationships
Solution
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
Check
35
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
  • 0C

36
Practice A gas occupies 10.0 L at 44.1 psi and
27C. What volume will it occupy at standard
conditions?

37
A gas occupies 10.0 L at 44.1 psi and 27C. What
volume will it occupy at standard conditions?
V1 10.0 L, P1 44.1 psi, t1 27 C, P2 1.00
atm, t2 0C V2, L
Given Find
Concept Plan Relationships
Solution
1 mole at STP occupies 22.4 L, since there is
more than 1 mole, we expect more than 22.4 L of
gas
Check
38
Molar Volume
  • solving the ideal gas equation for the volume of
    1 mol of gas at STP gives 22.4 L
  • 6.022 x 1023 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

39
Molar Volume
40
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

41
Gas Density
  • density is directly proportional to molar mass

42
Example 5.7 Calculate the density of N2 at
125C and 755 mmHg
P 755 mmHg, t 125 C, dN2, g/L
Given Find
Concept Plan Relationships
Solution
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
Check
43
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

44
Example 5.8 Calculate the molar mass of a gas
with mass 0.311 g that has a volume of 0.225 L at
55C and 886 mmHg
m0.311g, V0.225 L, P886 mmHg, t55C, molar
mass, g/mol
Given Find
m0.311g, V0.225 L, P1.1658 atm, T328 K,
molar mass, g/mol
Concept Plan Relationships
Solution
Check
the value 31.9 g/mol is reasonable
45
Practice - Calculate the density of a gas at 775
torr and 27C if 0.250 moles weighs 9.988 g

46
Calculate the density of a gas at 775 torr and
27C if 0.250 moles weighs 9.988 g
m9.988g, n0.250 mol, P775 mmHg, t27C,
density, g/L
Given Find
m9.988g, n0.250 mol, P1.0197 atm, T300. K
density, g/L
Concept Plan Relationships
Solution
Check
the value 1.65 g/L is reasonable
47
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 gass
    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

48
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
  • Daltons Law of Partial Pressures
  • because the gases behave independently

49
Composition of Dry Air

50
The partial pressure of each gas in a mixture can
be calculated using the ideal gas law
51
Example 5.9 Determine the mass of Ar in the
mixture
PHe341 mmHg, PNe112 mmHg, Ptot 662 mmHg, V
1.00 L, T298 K massAr, g
PAr 0.275 atm, V 1.00 L, T298 K massAr, g
Given Find
Concept Plan Relationships
PAr Ptot (PHe PNe)
Solution
Check
the units are correct, the value is reasonable
52
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.

53
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
Ptot 3.9 atm, V 8.7 L, T 598 K, Xe 0.17
mol PNe, atm
Given Find
Concept Plan Relationships
Solution
the unit is correct, the value is reasonable
Check
54
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
55
Sample Problems
  • A mixture of 2.50 moles neon, 1.45 moles helium,
    and 2.80 moles argon has a pressure of 1.45 atm.
    What are the partial pressure of all the gases in
    this system?

PA XAPTotal
56
Sample Problems
A mixture of 2.50 moles neon, 1.45 moles helium,
and 2.80 moles argon has a pressure of 1.45 atm.
What are the partial pressure of all the gases in
this system?
2.50 mol Ne 1.45 mol He 2.8 mol Ar 6.75
mole total
57
Sample Problems
A mixture of 2.50 moles neon, 1.45 moles helium,
and 2.80 moles argon has a pressure of 1.45 atm.
What are the partial pressure of all the gases in
this system?
58
Sample Problems
A mixture of 2.50 moles neon, 1.45 moles helium,
and 2.80 moles argon has a pressure of 1.45 atm.
What are the partial pressure of all the gases in
this system?
Check Your Answer PNe PHe PAr .537 atm .311
atm .601 atm 1.45 atm total
59
Mountain Climbing Partial Pressure
  • our bodies are adapted to breathe O2 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 O2 lower than 0.1 atm will
    lead to hypoxia
  • unconsciousness or death
  • climbers of Mt Everest carry O2 in cylinders to
    prevent hypoxia
  • on top of Mt Everest, Pair 0.311 atm, so PO2
    0.065 atm

60
Deep Sea Divers Partial Pressure
  • its also possible to have too much O2, a
    condition called oxygen toxicity
  • PO2 gt 1.4 atm
  • oxygen toxicity can lead to muscle spasms, tunnel
    vision, and convulsions
  • its 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 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

61
Partial Pressure Diving
62
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 cHe, cO2, PHe, atm, PO2, atm, Ptotal, atm
Given Find
nHe 6.05 mol, nO2 0.135 mol V 12.5 L, T
298 K cHe0.97817, cO20.021827, PHe, atm, PO2,
atm, Ptotal, atm
Concept Plan Relationships
Solution
63
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 25C, 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

64
Vapor Pressure of Water

65
Collecting Gas by Water Displacement
66
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.
V1.02 L, P755.2 mmHg, T293 K mass O2, g
Given Find
V1.02 L, PO2737.65 mmHg, T293 K mass O2, g
Concept Plan Relationships
Solution
67
Practice 0.12 moles of H2 is collected over
water in a 10.0 L container at 323 K. Find the
total pressure.

68
0.12 moles of H2 is collected over water in a
10.0 L container at 323 K. Find the total
pressure.
V10.0 L, nH20.12 mol, T323 K Ptotal, atm
Given Find
Concept Plan Relationships
Solution
69
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 weve 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
70
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, P738 mmHg, T355 K VH2, L
Given Find
nH2 2.2284 mol, P0.97105 atm, T355 K VH2, L
Concept Plan Relationships
Solution
71
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, P1.00 atm, T273 K massH2O, g
Given Find
Concept Plan Relationships
H2O 18.02 g/mol, 1 mol 22.4 L _at_ STP 2 mol H2O
2 mol H2
Solution
72
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)

73
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, P0.750 atm, T313 K VO2, L
Given Find
nO2 0.023085 mol, P0.750 atm, T313 K VO2, L
Concept Plan Relationships
Solution
74
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

75
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

76
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 dont be fooled into thinking all the
    particles are moving at the same speed!!

77
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.
78
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
79
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
80
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

81
Gas Laws Explained - Boyles Law
  • Boyles 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

82
Gas Laws Explained - Charless Law
  • Charless 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

83
Gas Laws ExplainedAvogadros Law
  • Avogadros 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

84
Gas Laws Explained Daltons Law of Partial
Pressures
  • Daltons 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 dont 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

85
Daltons Law Pressure
  • since the gas molecules are not sticking
    together, each gas molecule contributes its own
    force to the total force on the side

86
Deriving the Ideal Gas Law from Kinetic-Molecular
Theory
  • pressure Forcetotal/Area
  • Ftotal F1 collision x number of collisions
  • in a particular time interval
  • F1 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
  • Ftotal a massvelocity2 x Area x no.
    molecules/Volume
  • Pressure a mv2 x n/V
  • Temperature a mv2
  • P a Tn/V, ? PVnRT

87
Calculating Gas Pressure
88
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

89
Boltzman Distribution
90
Boltzmann animation
1000 meters/sec 3281 feet/sec speed of sound
at sea level 1116 feet per second or 768 mph or
about one mile in five seconds
91
Kinetic Energy and Molecular Velocities
  • average kinetic energy of the gas molecules
    depends on the average mass and velocity
  • KE 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

92
Molecular Speed vs. Molar Mass
  • in order to have the same average kinetic energy,
    heavier molecules must have a slower average speed

93
Temperature and Molecular Velocities
  • _
  • KEavg ½NAmu2
  • NA is Avogadros number
  • KEavg 1.5RT
  • R is the gas constant in energy units, 8.314
    J/molK
  • 1 J 1 kgm2/s2
  • equating and solving we get
  • NAmass molar mass in kg/mol
  • as temperature increases, the average velocity
    increases

94
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

95
Ex 5.14 Calculate the rms velocity of O2 at 25C
O2, t 25C urms
Given Find
Concept Plan Relationships
Solution
96
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

97
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

98
Effusion
99
Grahams Law of Effusion
  • for two different gases at the same temperature,
    the ratio of their rates of effusion is given by
    the following equation

100
Ex 5.15 Calculate the molar mass of a gas that
effuses at a rate 0.462 times N2
Given Find
MM, g/mol
Concept Plan Relationships
Solution
101
Ideal vs. Real Gases
  • Real gases often do not behave like ideal gases
    at high pressure or low temperature
  • 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

102
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
  • b is called a van der Waals constant and is
    different for every gas because their molecules
    are different sizes

103
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

104
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
  • a is called a van der Waals constant and is
    different for every gas because their molecules
    are different sizes

105
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

106
Van der Waals Equation
  • combining the equations to account for molecular
    volume and intermolecular attractions we get the
    following equation
  • used for real gases
  • a and b are called van der Waal constants and are
    different for each gas

107
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

108
PV/RT Plots
109
Structure of the Atmosphere
  • the atmosphere shows several layers, each with
    its own characteristics
  • the troposphere is the layer closest to the
    earths 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

110
Air Pollution
  • air pollution is materials added to the
    atmosphere that would not be present in the air
    without, or are increased by, mans 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

111
Pollutant Gases, SOx
  • SO2 and SO3, 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 SO2 O2 2 H2O ? 2 H2SO4
  • SO3 H2O ? H2SO4

112
Pollutant Gases, NOx
  • NO and NO2, oxides of nitrogen, come from burning
    of fossil fuels in cars, trucks, and power plants
  • as 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

113
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

114
Pollutant Gases, O3
  • 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
  • O3 is present in the brown haze seen in some
    cities
  • lung and eye irritants
  • strong oxidizer

115
Major Pollutant Levels
  • government regulation has resulted in a decrease
    in the emission levels for most major pollutants

116
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)

117
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 regenerated
  • NO2 also catalyzes O3 destruction
  • Cl O3 ? ClO O2
  • O3 UV light ? O2 O
  • ClO O ? O2 Cl

118
Ozone Holes
  • satellite data over the past 3 decades reveals a
    marked drop in ozone concentration over certain
    regions
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