Chapter 5 Gases

Report
Chemistry: A Molecular Approach, 1st 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
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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
Tro, Chemistry: A Molecular Approach
Force
Pressure 
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
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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
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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
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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.”
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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
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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
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gravity
9
Common Units of Pressure
Unit
pascal (Pa), 1 Pa  1
Average Air Pressure at
Sea Level
N
m
2
101,325
kilopascal (kPa)
101.325
atmosphere (atm)
1 (exactly)
millimeters of mercury (mmHg)
inches of mercury (inHg)
torr (torr)
pounds per square inch (psi, lbs./in2)
Tro, Chemistry: A Molecular Approach
760 (exactly)
29.92
760 (exactly)
14.7
10
Example 5.1 – A high-performance bicycle tire has a pressure
of 132 psi. What is the pressure in mmHg?
Given:
132 psi
Find:
mmHg
Concept Plan:
psi
atm
1 atm
14.7 psi
Relationships:
mmHg
760 mmHg
1 atm
1 atm = 14.7 psi, 1 atm = 760 mmHg
Solution:
1 atm 760 mmHg
132 psi 

 6.82103 mmHg
14.7psi
1 atm
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
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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)
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Boyle’s 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
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Boyle’s Experiment
• added Hg to a J-tube with
•
air trapped inside
used length of air column
as a measure of volume
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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
15
Boyle's Expt.
140
120
Pressure, inHg
100
80
60
40
20
0
0
10
20
30
40
50
60
3
Volume of Air, in
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Inverse Volume vs Pressure of Air, Boyle's Expt.
140
120
Pressure, inHg
100
80
60
40
20
0
0
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0.09
Inv. Volume, in-3
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Boyle’s Experiment, P x V
Pressure Volume P x V
29.13
48 1400
33.50
42 1400
41.63
34 1400
50.31
28 1400
61.31
23 1400
74.13
19 1400
87.88
16 1400
115.56
12 1400
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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)
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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
if your tank
contained air at 1
atm pressure you
would not be able to
inhale it into your
lungs
at 20 m the total
pressure is 3 atm
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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: V1 =7.25 L, P1 = 4.52 atm, P2 = 1.21 atm
Find: V2, L
Concept Plan:
V1, P1, P2
V2 
Relationships: P1
Solution:
P1  V1
P2
V2
∙ V1 = P2 ∙ V2
P1  V1
V2 
P2

4.52atm  7.25L 

 27.1 L
1.21atm
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?
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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: V2 =2780 mL, P1 = 762 torr, P2 = 0.500 atm
Find: V1, mL
Concept Plan:
V1, P1, P2
V1 
Relationships: P1
P2  V2
P1
∙ V1 = P2 ∙ V2 , 1 atm = 760 torr (exactly)
Solution:
782 torr 
1 atm
 1.03 atm
760 torr
Check:
V2
P2  V2
V1 
P1

0.500atm  2780L 

 1350mL
1.03atm
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
V1 V2

T1 T2
if T measured in Kelvin
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Charles’ Law – A Molecular View
• the pressure of gas inside
and outside the balloon
are the same
• at high
low temperatures,
temperatures,the
the
gas molecules are moving
not
movingsoasthey
faster,
fast,hit
sothe
they
sides
don’t
of
the hit
balloon
the sides
harder
of the
–
balloon as
causing
thehard
volume
– to
thereforelarger
become
the volume is
small
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Charles' Law & Absolute Zero
0.6
Volume (L) of 1 g O2 @ 1500 torr
Volume (L) of 1 g O2 @ 2500 torr
0.5
Volume (L) of 0.5 g O2 @ 1500 torr
Volume (L) of 0.5 g SO2 @ 1500
torr
0.4
Volume, L
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
0.3
0.2
0.1
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: V1 =2.57 L, V2 = 2.80 L, t2 = 0.00°C
Find: t1, K and °C
Concept Plan:
V1, V2, T2
T1  T2 
V1
V2
Relationships: T(K) = t(°C) + 273.15,
T1
V1
V2

T1
T2
T2  V1
T1 
V2
T2  0.00 273.15
T2  273.15K

273.15K   2.57 L 

 297.6 K
2.80L
Solution:
Check:
t1  T1  273.15
t1  297.6  273.15
t1  24 C
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?
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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: V1 =10.0 L, t1 = 25.0°C L, t2 = 250.0°C
Find: V2, L
Concept Plan:
V1, T1, T2
V2  V1 
V2
T2
T1
Relationships: T(K) = t(°C) + 273.15,
Solution:
T1  25.0  273.15
T1  298.2K
T2  250.0  273.15
V2 
V1
V2

T1
T2
T2  V1
T1

523.2K   10.0L 

 17.5 L
298.2K 
T2  523.2K
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
V = constant x n
constant P and T
more gas molecules = larger
volume
V 1 V2

n1 n 2
• count number of gas molecules
by moles
• equal volumes of gases contain
equal numbers of molecules
the gas doesn’t matter
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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: V1 =4.65 L, V2 = 6.48 L, n1 = 0.225 mol
Find: n2, and added moles
Concept Plan:
V1, V2, n1
V2
n1 
 n2
V1
n2
V1
V2

n1
n2
Relationships: mol added = n2 – n1,
Solution:
n V
n2  1 2
V1
molesadded  0.314 0.225
molesadded  0.089 mol

0.225mol  6.48L 

 0.314 mol
4.65L
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
• R is called the gas constant
• the value of R depends on the units of P and V
atm  L
• we will use 0.08206 mol  Kand 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
P   V   R
n   T 
Tro, Chemistry: A Molecular Approach
or PV  nRT
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:
P, V, T, R
Relationships: 1 atm = 14.7 psi
PV
n
RT
n
PV  nRT, R  0.08206
T(K) = t(°C) + 273.15
atmL
molK
PV
n
1 atm
24.3 psi 
 1.6531 atm
R T
14.7 psi

1.6531atm  3.24 L 

 0.219 mol
T(K)  25C  273.15
atm L
0.08206molK  298 K 
Solution:
T  298 K
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
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Practice – A gas occupies 10.0 L at 44.1 psi and 27°C.
What volume will it occupy at standard conditions?
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A gas occupies 10.0 L at 44.1 psi and 27°C. What
volume will it occupy at standard conditions?
Given: V1 = 10.0 L, P1 = 44.1 psi, t1 = 27 °C, P2 = 1.00 atm, t2 = 0°C
Find: V2, L
Concept Plan:
P1, V1, T1, R
Relationships: 1 atm = 14.7 psi
PV
n
RT
n
n  R  T1 atm
V  psi 
44.1
 3.00 atm
P 14.7 psi
atm  L

 273 K 
1.219 mol  0.08206mol
K

T(K)  27C 273.15
1.00 at m
127
L .K
T
 .300
nRT
V
P
PV  nRT, R  0.08206
T(K) = t(°C) + 273.15
Solution:
P2, n, T2, R
V2
atmL
molK
PV
R T
3.00atm  10.0 L   1.219 mol

atm  L
0.08206mol
 300.K 
K
n
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 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
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Molar Volume
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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
Molar Mass, g
Density 
22.4 L
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Gas Density
1 mol
mass
mass 
 moles  moles 
molar mass
molar mass
mass in grams
density 
volume in liters
PV  nR T
mass
PV 
RT
molar mass
mass
P  (molar mass)
 density 
V
RT
• density is directly proportional to molar mass
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Example 5.7 – Calculate the density of N2
at 125°C and 755 mmHg
Given:
Find:
Concept Plan:
P = 755 mmHg, t = 125 °C,
dN2, g/L
P, MM, T, R
d
P  MM 
R T
1 atm = 760 mmHg, MM = 28.01 g
d
Relationships: T(K) = t(°C) + 273.15
Solution:
755 mmHg 
1 atm
 0.99342 atm
760 mmHg
T(K)  125C  273.15
T  398 K
Check:
P  MM
R T
atm L
R  0.08206mol
K
d
P  MM
R T
0.99342 atm  28.01molg
d


0.08206  398 K 
atm L
mol K
  0.852g/L
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
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
mass in grams
Molar Mass 
moles
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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: m=0.311g, V=0.225 L, P=1.1658
P=886 mmHg,
atm, T=328
t=55°C,K,
Find: molar mass, g/mol
P, V, T, R
Concept Plan:
n
PV
R T
n
n, m
MM 
m
n
MM
atmL
PV  nRT, R  0.08206mol
1 atm = 760 mmHg,
K
m
MM 
Relationships: T(K) = t(°C) + 273.15
n
Solution:
m
0.311g
1 atm
MM  
PV
 1.1658 atm
n886
 mmHg 
n 9.745410-3 mol
760 mmHg
R T
 31.9g/mol
T(K)
 55
C273.15


1.1658
atm
0.225L 
 T  328 KatmL
 9.7454103 mol
0.08206molK  328 K 


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
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44
Calculate the density of a gas at 775 torr and 27°C if
0.250 moles weighs 9.988 g
Given: m=9.988g, n=0.250 mol, P=1.0197
P=775 mmHg,
atm, T=300.
t=27°C,K
Find: density, g/L
Concept Plan:
P, n, T, R
V
nR T
P
V
V, m
m
d
V
d
atmL
PV  nRT, R  0.08206mol
1 atm = 760 mmHg,
K
m
d
Relationships: T(K) = t(°C) + 273.15
V
Solution:
nR
 T 1 atm  1.0197 atm
775
torr
V
760 torr
P
T(K)mol
 27C0.08206
 273.15 atm  L   300. K 

0.250
mol K
 T  300. K
 6.0355 L
1.0197atm
Check:
m 9.988g

V 6.0355 L
 1.65g/L
d
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
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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
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Composition of Dry Air
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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
nA x R x T
nB x R x T
PA 
PB 
V
V
the temperatu re and volume of everything
in the mixture are the same
n total  n A  n B
n total x R x T
Ptotal  PA  PB 
V
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49
Example 5.9 – Determine the mass of Ar in the mixture
= 0.275
mmHg,
atm, V
PNe
= =112
1.00 L,
mmHg,
T=298Ptot
K = 662 mmHg,
Given: PAr
He=341
V = 1.00 L, T=298 K
massAr
,g
Find: mass
Ar, g
PAr
Concept Plan: Ptot, PHe, PNe
PAr = Ptot – (PHe + PNe)
PAr, V, T
PV
n
R T
n Ar
mAr
m  n  MM
atmL
Ptot = Pa + Pb + etc.,
PV  nRT, R  0.08206mol
m K
1
atm
=
760
mmHg,
MM
=
39.95
g/mol
MM 
Relationships:
Ar
n
Solution:
PV
PAr  662
 341  112 mmHg
n
R T
 209 mmHg
1.125 10
2
39.95 g
mol 
1 mol
 0.449 g Ar

0.275atm  1.00 L 
1 atm

 1.125102 mol
209 mmHg

K
0.275
atm
0.08206


 298
760 mmHg
atm  L
mol K
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:
Find:
Concept Plan:
Ptot = 3.9 atm, V = 8.7 L, T = 598 K, Xe = 0.17 mol
PNe, atm
nXe, V, T, R
PXe 
Relationships:
n Xe  R  T
V
n Xe  R  T
V
atm L

0.17 mol  0.08206mol
K   598 K 

8.7 L
 0.9589 at m
PXe 
PNe  Ptotal  PXe
 3.9 atm  0.9589 atm
Check:
Ptot, PXe
PNe
PNe  Ptotal  PXe
atmL
PV  nRT, R  0.08206mol
, Ptotal  PNe  PXe
K
Solution:
 2.9 atm
PXe
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
PA
nA

Ptotal n total
nA
cA 
n total
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
PA  cA  Ptotal
53
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
Tro, Chemistry: A Molecular Approach
54
Deep Sea Divers & Partial Pressure
• its also possible to have too much O2, a condition called
oxygen toxicity
 PO2 > 1.4 atm
 oxygen toxicity can lead to muscle spasms, tunnel vision, and
convulsions
• 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
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 O2 at 298 K
Given:
Find:
Concept Plan:
nmHe
24.2mol,
g, mO2
nO2==43.2
0.135
g Vmol
= 12.5
V = L,
12.5
T =L,298
T =K298 K
He==6.05
cHe=0.97817,
, cO2, PHe, atm,
cO2=0.021827,
PO2, atm, PPtotal
atm, PO2, atm, Ptotal, atm
He, atm
mgas
cgas
ngas
c gas A 
cgas, Ptotal
Relationships:
n gas A
ntot, V, T, R
Ptotal 
n total
n total  R  T
V
Pgas
PA  c A  Ptotal
atmL
PV  nRT, R  0.08206mol
, PA  c A  Ptotal
K
Solution:
Ptot
c He 
MMHe = 4.00 g/mol
MMO2 = 32.00 g/mol
PHe
 cmol
Ptotal
6.05
He He
 0.97817
6.05mol
0.135
 0He
.97
817
 12mol
.099Oatm
2
n total  R  T
1 mol He
P

total
24.2
g He 
V g  6.05 mol He
4.00
 11.8atm
atmL
1
mol
O
6.185 mol   20.08206 molOK 2 298 K 
4.32
PO0.135
 c Omol
 POtotal
2
 g O2  32.00 g  0.135 mol
c

 0.021827
O
12.5 L
6.05mol
He 827
0.135
 0.021
12mol
.099Oatm
2
 12.099 atm
 0.264atm
2
2
2
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*
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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 O2 collected over water at 293 K
with a total pressure of 755.2 mmHg. Find mass O2.
Given: V=1.02 L, P
P=755.2
mmHg,
mmHg,
T=293
T=293
K K
O2=737.65
Find: mass O2, g
Concept Plan:
Ptot, PH2O
PO2
PO2  Ptotal  PH2O @ 20C
PO2,V,T
PV
n
R T
nO2
gO2
atmL
PV  nRT, R  0.08206mol
1 atm = 760 mmHg,
K
Relationships: Ptotal = PA + PB, O2 = 32.00 g/mol
Solution:
PO2 P755.2
 V  17.55 (Table 5.4)
n
PO2 R
737
.65 mmHg
T
0.97059 atm 1atm
1.02 L 

737.65 mmHg atm
L
 0.97059 atm
0.08206mol K760
  mmHg
293. K 
 4.117510 2 mol
4.117510 2 mol
 1.32g
32.00g
1 mol
Practice – 0.12 moles of H2 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 H2 is collected over water in a 10.0 L
container at 323 K. Find the total pressure.
Given: V=10.0 L, nH2=0.12 mol, T=323 K
Find: Ptotal, atm
Concept Plan:
nH2,V,T
P
nR T
V
PH2
PH2, PH2O
Ptotal
Ptotal  PH2  PH2O @ 50C
1 atm = 760 mmHg
Relationships: Ptotal = PA + PB, PV  nRT, R  0.08206atmL
molK
Solution:
760 mmHg
0.3181 atm 
 241.8 mmHg
1 atm
nR T
V
atm L

0.12mol  0.08206mol
K   323 K 

10.0L 
 0.3181at m
PH 2 
Ptotal  241.8  92.6 (Table 5.4)
Ptotal  330 mmHg
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
Tro, Chemistry: A Molecular Approach
mole A
mole B
P, V, T of Gas B
64
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)
Given: n
mH2
= 2.2284
= 37.5g,
mol,P=738
P=0.97105
mmHg,
atm,
T=355
T=355
KK
CH3OH
Find: VH2, L
Concept Plan:
g CH3OH
mol CH3OH
1 mol CH 3OH
32.04g
mol H2
P, n, T, R
2 mol H 2
1 mol CH 3OH
V
V
nR T
P
Relationships: 1 atm = 760 mmHg, CH3OH = 32.04 g/mol
atmL
1 mol CH3OH : 2 mol H2 PV  nRT, R  0.08206mol
K
Solution:
R HT2
1 mol CH3OH n 2 mol
37.5 g CH3OH 
V 
32.04 g
1 molPCH3OH
 2.2284 mol H 2
738 mmHg 

2.2284 mol  0.08206   355 K 

0.97105 atm
1 atm
 0.97105 atm
760 mmHg  66.9 L
atm L
mol K
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)
Given: VH2 = 1.24 L, P=1.00 atm, T=273 K
Find: massH2O, g
Concept Plan:
L H2
1 mol H 2
22.4 L
mol H2
mol H2O
2 mol H 2
2 mol H 2O
g H2 O
18.02 g
1 mol H 2O
Relationships: H2O = 18.02 g/mol, 1 mol = 22.4 L @ STP
2 mol H2O : 2 mol H2
Solution:
1 mol H 2 2 mol H 2O 18.02 g H 2O
1.24 L H 2 


22.4 L H 2
2 mol H 2
1 mol H 2O
 0.998 g H 2O
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)
Tro, Chemistry: A Molecular Approach
67
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)
Given: n
mO2
==
0.023085
10.0g, P=0.750
mol, P=0.750
atm, T=313
atm, T=313
K
K
HgO
Find: VO2, L
Concept Plan:
g HgO
mol HgO
1 molHgO
216.59g
mol O2
P, n, T, R
1 molO 2
2 mol HgO
V
V
nR T
P
Relationships: 1 atm = 760 mmHg, HgO = 216.59 g/mol
atmL
2 mol HgO : 1 mol O2
PV  nRT, R  0.08206mol
K
Solution:
nR T
1 mol HgO 1 mol
V O2
10.0 g HgO 

216.59 g 2 mol HgO P
0.023085 mol
 0.023085 mol O 2


 0.791L
  0.08206
0.750at m
atm  L
mol K
 313K 
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
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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
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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.
Tro, Chemistry: A Molecular Approach
As a result, gases
take the shape and
the volume of the
container they
are in.
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 = 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 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
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, urms, is the square root of the average of the
sum of the squares of all the molecule velocities
 v2
urms 
 u2
n
Tro, Chemistry: A Molecular Approach
83
Boltzman Distribution
Distribution Function
Fraction of Molecules
O2 @ 300 K
Molecular Speed
Tro, Chemistry: A Molecular Approach
84
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
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
Temperature _and Molecular Velocities
• KEavg = ½NAmu2
NA is Avogadro’s number
• KEavg = 1.5RT
R 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
urms 
3RT
3RT

NA m
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 O2 at 25°C
Given: O2, t = 25°C
Find: urms
Concept Plan:
MM, T
urms
urm s 
3RT
MM
urm s 
3RT
MM
Relationships: T(K) = t(°C) + 273.15, O2 = 32.00 g/mol
Solution:
T (K) t (C)  273.15
T  25  273.15
T  298 K
3RT
urms 
MM
kg m 2 

2 

s
3   8.314
 298K 

molK



 482 m/s
-3 kg
32.0010

mol

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
rate 
Tro, Chemistry: A Molecular Approach
1
MM
91
Effusion
Tro, Chemistry: A Molecular Approach
92
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:
rategas A
rategas B
Tro, Chemistry: A Molecular Approach

Molar Massgas B
Molar Massgas A
93
Ex 5.15 – Calculate the molar mass of a gas that
effuses at a rate 0.462 times N2
Given: rateunknown gas
 0.462
rateN 2
Find: MM, g/mol
Concept Plan:
rateA/rateB, MMN2
MMunknown
Molar Massunknown 
Relationships: N2 = 28.01 g/mol
Molar MassN2
 rateunknown

 rateN
2





2
rat egas A
rat egas B

Molar Massgas B
Molar Massgas A
Solution:
Molar Massunknown 
Molar MassN 2
 rateunknown

 rateN
2





2

28.01 g m ol

 131g
0.462
2
m ol
•
•
•
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 molecules
2) 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
b is called a van der Waals constant and is
different for every gas because their molecules are
different sizes
nRT
V
Tro, Chemistry: A Molecular Approach
P
 nb
96
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
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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
 a is called a van der Waals constant and is different for
every gas because their molecules are different sizes
nRT  n 
P
 a 
V
V
Tro, Chemistry: A Molecular Approach
2
98
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
 a and b are called van der Waal
constants and are different for each
gas
2

 P  a n    V - nb   nRT


V




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 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
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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, 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
Tro, Chemistry: A Molecular Approach
105
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
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 O2
• at high levels can cause sensory impairment,
stupor, unconsciousness, or death
Tro, Chemistry: A Molecular Approach
107
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
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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
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)
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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
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
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

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