Introductory Chemistry, 3rd Edition
Nivaldo Tro
Chapter 11
Gases
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
2009, Prentice Hall
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's Introductory Chemistry,
Chapter 11
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The Structure of a Gas
• Gases are composed of particles that are
flying around very fast in their container(s).
• They move in straight lines until they
encounter either the container wall or
another particle, then they bounce off.
• If you were able to take a snapshot of the
particles in a gas, you would find that there
is a lot of empty space in there.
Tro's Introductory Chemistry,
Chapter 11
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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's Introductory Chemistry,
Chapter 11
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Kinetic Molecular Theory of Gases
• There is a lot of empty space between the
particles in a gas.
Compared to the size of the particles.
• The average kinetic energy of the particles is
directly proportional to the Kelvin temperature.
As you raise the temperature of the gas, the average
speed of the particles increases.
But don’t be fooled into thinking all the particles are
moving at the same speed!!
Tro's Introductory Chemistry,
Chapter 11
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Kinetic Molecular Theory
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Gas Particles Pushing
• Gas molecules are constantly in
motion.
• As they move and strike a surface,
they push on that surface.
 Push = force.
• If we could measure the total amount
of force exerted by gas molecules
hitting the entire surface at any one
instant, we would know the pressure
the gas is exerting.
 Pressure = force per unit area.
Tro's Introductory Chemistry,
Chapter 11
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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’ path, the gas
will try to push it along as the gas flows.
Tro's Introductory Chemistry,
Chapter 11
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Which Way Would Air Flow?
Two filled balloons are
connected with a long pipe.
One of the balloons is
plunged down into the
water. Which way will the
air flow? Will air flow
from the lower balloon
toward the top balloon; or
will it flow from the top
balloon to the bottom one?
Tro's Introductory Chemistry, Chapter
11
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Is This Possible at a Depth of 20 m?
Tro's Introductory Chemistry, Chapter
11
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Soda Straws and Gas Pressure
The pressure of
the air inside the
straw is the same
as the pressure
of the air outside
the straw—so
liquid levels are
the same on both
sides.
The pressure of the
air inside the straw
is lower than the
pressure
of the air outside
the straw—so
liquid is pushed
up the straw by
the outside air.
Tro's Introductory Chemistry, Chapter
11
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Gas Properties Explained
• Gases have taken the shape and volume of their
container(s) because the particles don’t stick
together, allowing them to move and fill the
container(s) they’re in.
 In solids and liquids, the particles are attracted to each
other strongly enough so they stick together.
• Gases are compressible and have low density
because of the large amount of unoccupied space
between the particles.
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Properties—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's Introductory Chemistry, Chapter
11
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Properties—Compressibility
Because there is a lot of unoccupied space in the structure
of a gas, the gas molecules can be squeezed closer together.
Tro's Introductory Chemistry, Chapter
11
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Gas Properties Explained—
Low Density
Because there is a lot of
unoccupied space in the
structure of a gas, gases do
not have a lot of mass in a
given volume, the result is
that they have low density.
Tro's Introductory Chemistry, Chapter
11
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The Pressure of a Gas
• Pressure is the result of the
constant movement of the gas
molecules and their collisions
with the surfaces around them.
• The pressure of a gas depends
on several factors:
Number of gas particles in a
given volume.
Volume of the container.
Average speed of the gas
particles.
Tro's Introductory Chemistry,
Chapter 11
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Density and Pressure
• Pressure is the result of the
constant movement of the gas
molecules and their collisions
with the surfaces around them.
• When more molecules are
added, more molecules hit the
container at any one instant,
resulting in higher pressure.
Also higher density.
Tro's Introductory Chemistry,
Chapter 11
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Air Pressure
• The atmosphere exerts a pressure
on everything it contacts.
 On average 14.7 psi.
 The atmosphere goes up about 370
miles, but 80% is in the first 10 miles
from Earth’s surface.
• This is the same pressure that a
column of water would exert if it
were about 10.3 m high.
Tro's Introductory Chemistry,
Chapter 11
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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.
Tro's Introductory Chemistry, Chapter
11
gravity
19
Atmospheric Pressure and
Altitude
• The higher up in the atmosphere you go, the
lower the atmospheric pressure is around you.
At the surface, the atmospheric pressure is 14.7
psi, but at 10,000 ft is is only 10.0 psi.
• Rapid changes in atmospheric pressure may
cause your ears to “pop” due to an imbalance
in pressure on either side of your ear drum.
Tro's Introductory Chemistry,
Chapter 11
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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.”
Tro's Introductory Chemistry, Chapter
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Common Units of Pressure
Unit
Average air pressure at
sea level
101,325
Pascal (Pa)
Kilopascal (kPa)
101.325
Atmosphere (atm)
1 (exactly)
Millimeters of mercury (mmHg)
Inches of mercury (inHg)
Torr (torr)
760 (exactly)
29.92
760 (exactly)
Pounds per square inch (psi, lbs./in2)
Tro's Introductory Chemistry, Chapter
11
14.7
22
Example 11.1—A High-Performance Bicycle Tire Has a
Pressure of 125 psi. What Is the Pressure in mmHg?
Given:
125 psi
Find:
mmHg
Solution Map:
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
125 psi 

 6.46  103 mmHg
14.7 psi
1 atm
Check:
Since mmHg are smaller than psi, the answer
makes sense.
Practice—Convert 45.5 psi into kPa.
Unit
Average air pressure at
sea level
101,325
Pascal (Pa)
Kilopascal (kPa)
101.325
Atmosphere (atm)
1 (exactly)
Millimeters of mercury (mmHg)
Inches of mercury (inHg)
Torr (torr)
760 (exactly)
29.92
760 (exactly)
Pounds per square inch (psi, lbs./in2)
Tro's Introductory Chemistry, Chapter
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14.7
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Practice—Convert 45.5 psi into kPa, Continued
Given:
Find:
Concept Plan:
45.5 psi
kPa
psi
atm
1 atm
14.7 psi
Relationships:
kPa
760 mmHg
1 atm
1 atm = 14.7 psi, 1 atm = 101.325 kPa
Solution:
1 atm 101.325 kPa
45.5 psi 

 314 kPa
14.7 psi
1 atm
Check:
Since kPa are smaller than psi, the answer
makes sense.
Boyle’s Law
• Pressure of a gas is inversely
proportional to its volume.
Constant T and amount of gas.
Graph P vs. V is curved.
Graph P vs. 1/V is in a straight
line.
• As P increases, V decreases
by the same factor.
• P x V = constant.
• P1 x V1 = P2 x V2 .
Tro's Introductory Chemistry,
Chapter 11
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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.
Length of air
in column
(in)
48
44
40
36
32
28
24
22
Tro's Introductory Chemistry, Chapter
11
Difference in
Hg levels
(in)
0.0
2.8
6.2
10.1
15.1
21.2
29.7
35.0
34
Boyle's Experiment
140
120
Pressure, inHg
100
80
60
40
20
0
0
10
20
30
40
50
60
Volume of Air, in3
Tro's Introductory Chemistry, Chapter
11
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Inverse Volume vs. Pressure of Air,
Boyle's Experiment
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
Tro's Introductory Chemistry, Chapter
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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
Tro's Introductory Chemistry, Chapter
11
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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).
Tro's Introductory Chemistry, Chapter
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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's Introductory Chemistry,
Chapter 11
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Boyle’s Law and Diving
• Since water is more dense
than air, for each 10 m you
dive below the surface, the
pressure on your lungs
increases 1 atm.
Scuba tanks have a
regulator so that the
air from the tank is
delivered at the same
 At 20 m the total pressure is
pressure as the water
3 atm.
• If your tank contained air at surrounding you.
1 atm of pressure, you
This allows you to
would not be able to inhale take in air even when
it into your lungs.
the outside pressure is
 You can only generate
large.
enough force to overcome
about 1.06 atm.
Tro's Introductory Chemistry,
Chapter 11
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Boyle’s Law and Diving,
Continued
• If a diver holds her breath
and rises to the surface
quickly, the outside
pressure drops to 1 atm.
• According to Boyle’s law,
what should happen to the
volume of air in the lungs?
• Since the pressure is decreasing
by a factor of 3, the volume will
expand by a factor of 3, causing
damage to internal organs.
Always Exhale When Rising!!
Tro's Introductory Chemistry,
Chapter 11
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Example 11.2—A Cylinder with a Movable Piston Has a Volume
of 6.0 L at 4.0 atm. What Is the Volume at 1.0 atm?
Given: V1 =6.0 L, P1 = 4.0 atm, P2 = 1.0 atm
Find: V2, L
Solution Map:
V1, P1, P2
V2 
Relationships: P1 ∙ V1 = P2 ∙ V2
Solution:
P1  V1
P2
V2
P1  V1
V2 
P2
4.0 atm   6.0 L 

 24 L
1.0 atm 
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?
(1 atm = 760 torr)
Tro's Introductory Chemistry,
Chapter 11
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Practice—A Balloon Is Put in a Bell Jar and the Pressure Is
Reduced from 782 torr to 0.500 atm. If the Volume of the Balloon
Is Now 2780 mL, What Was It Originally?, Continued
Given: V2 =2780 mL, P1 = 762 torr, P2 = 0.500 atm
Find: V1, mL
Solution Map:
V2, P1, P2
P2  V2
V1 
P1
V1
Relationships: P1 ∙ V1 = P2 ∙ V2 , 1 atm = 760 torr (exactly)
Solution:
P2  V2
V1 
P1
1 atm
782 torr 
 1.03 atm
760 torr
0.500 atm   2780 L 

 1350 mL
1.03 atm 
Check:
Since P and V are inversely proportional, when the pressure
decreases ~2x, the volume should increase ~2x, and it does.
Temperature Scales
100°C
373 K
0°C
212°F
671 R
BP Water
273 K
32°F
459 R
MP Ice
-38.9°C
234.1 K
-38°F
421 R
BP Mercury
-183°C
90 K
-297°F
162 R
BP Oxygen
-269°C
-273°C 4 K
0 K -452°F
-459 °F 7 R
Celsius
Kelvin
Fahrenheit
BP Helium
0 R Absolute
Rankine Zero
Gas Laws and Temperature
• Gases expand when heated and contract when cooled,
so there is a relationship between volume and
temperature.
• Gas molecules move faster when heated, causing them
to strike surfaces with more force, so there is a
relationship between pressure and temperature.
• In order for the relationships to be proportional, the
temperature must be measured on an absolute scale.
• When doing gas problems, always convert your
temperatures to kelvins.
K = °C + 273
°F = 1.8 °C + 32
&
&
°C = K - 273
°C = 0.556(°F-32)
Tro's Introductory Chemistry,
Chapter 11
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Standard Conditions
• Common reference points for comparing.
• Standard pressure = 1.00 atm.
• Standard temperature = 0 °C.
273 K.
• STP.
Tro's Introductory Chemistry,
Chapter 11
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Volume and Temperature
• In a rigid container, raising the temperature
increases the pressure.
• For a cylinder with a piston, the pressure
outside and inside stay the same.
• To keep the pressure from rising, the piston
moves out increasing the volume of the
cylinder.
As volume increases, pressure decreases.
Tro's Introductory Chemistry,
Chapter 11
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Volume and Temperature, Continued
As a gas is heated, it expands.
This causes the density of the
gas to decrease.
Because the hot air in the
balloon is less dense than the
surrounding air, it rises.
Tro's Introductory Chemistry, Chapter
11
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Charles’s Law
• Volume is directly proportional to
temperature.
Constant P and amount of gas.
Graph of V vs. T is a straight line.
• As T increases, V also increases.
• Kelvin T = Celsius T + 273.
• V = constant x T.
V1 V2

T1 T2
If T is measured in kelvin.
Tro's Introductory Chemistry, Chapter
11
57
Charle's Law & Absolute Zero
0.6
0.5
Volume (L) of 1 g O2 @ 1500 torr
Volume, L
0.4
Volume (L) of 1 g O2 @ 2500 torr
Volume (L) of 0.5 g O2 @ 1500 torr
0.3
Volume (L) of 0.5 g SO2 @ 1500 torr
0.2
0.1
0
-300
-250
-200
-150
-100
-50
Temperature, °C
0
50
100
150
58
We’re losing altitude.
Quick, Professor, give your
lecture on Charles’s law!
Absolute Zero
• Theoretical temperature at which a gas
would have zero volume and no pressure.
Kelvin calculated by extrapolation.
• 0 K = -273.15 °C = -459 °F = 0 R.
• Never attainable.
Though we’ve gotten real close!
• All gas law problems use the Kelvin
temperature scale.
Tro's Introductory Chemistry,
Chapter 11
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Determining Absolute Zero
William Thomson,
the Lord of Kelvin,
extrapolated the
line graphs of
volume vs. temperature to determine
the theoretical
temperature that
a gas would have
given a volume of 0.
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Example 11.3—A Gas Has a Volume of 2.57 L at 0 °C. What
Was the Temperature at 2.80 L?
Given: V1 =2.80 L, V2 = 2.57 L, t2 = 0°C
Find: t1, K and °C
Solution Map:
V1, V2, T2
V1
T1  T2 
V2
Relationships: T(K) = t(°C) + 273,
T1
V1
V2

T1
T2
T2  V1
t1  T1  273
T1 
V2
T2  0  273
t1  297.6  273
T2  273 K  273 K   2.57 L   297.6 K
t

2
4

C
1
2.80 L 
Solution:
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's Introductory Chemistry,
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Practice—The Temperature Inside a Balloon Is Raised from
25.0 °C to 250.0 °C. If the Volume of Cold Air Was 10.0 L, What Is the
Volume of Hot Air?, Continued
Given: V1 =10.0 L, t1 = 25.0 °C L, t2 = 250.0 °C
Find: V2, L
Solution Map:
V1, T1, T2
V2
T2
V2  V1 
T1
Relationships: T(K) = t(°C) + 273.15,
Solution:
T1  25.0  273.15
T1  298.2 K
T2  250.0  273.15
T2  523.2 K
Check:
T2  V1
V2 
T1
V1
V2

T1
T2
523.2 K   10.0 L 

 17.5 L
298.2 K 
Since T and V are directly proportional, when the temperature
increases, the volume should increase, and it does.
The Combined Gas Law
• Boyle’s law shows the relationship between pressure and
volume.
 At constant temperature.
• Charles’s law shows the relationship between volume and
absolute temperature.
 At constant pressure.
• The two laws can be combined together to give a law that
predicts what happens to the volume of a sample of gas when
both the pressure and temperature change.
 As long as the amount of gas stays constant.
 P1   V1   P2   V2 

T1 
T2 
73
Example 11.4—A Sample of Gas Has an Initial Volume of 158 mL at a
Pressure of 735 mmHg and a Temperature of 34 °C. If the Gas Is
Compressed to 108 mL and Heated to 85 °C, What Is the Final Pressure?
Given: V1 = 158 mL, t1 = 34 °C L, P1 = 735 mmHg
V2 = 108 mL, t2 = 85 °C
Find: P2, mmHg
Solution Map:
P1,V1, V2, T1, T2
P2 
Relationships: T(K) = t(°C) + 273,
Solution:
T1  34  273
T1  307 K
T2  85  273
T2  358 K
Check:
P1  V1  T2
V2  T1
P2
P1  V1
P  V2
 2
T1
T2
P1  V1  T2
P2 
V2  T1
735 mmHg   158 mL   358 K 
P2 
108 mL   307 K 
P2  1.25  103 mmHg
Since T increases and V decreases we expect the
pressure should increase, and it does.
Practice—A Gas Occupies 10.0 L When Its Pressure
Is 3.00 atm and Temperature Is 27 °C. What
Volume Will the Gas Occupy Under Standard
Conditions?
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Practice—A Gas Occupies 10.0 L When Its Pressure Is 3.00
Atm and Temperature Is 27 °C. What Volume Will the Gas
Occupy Under Standard Conditions?, Continued
Given: V1 = 10.0 L, t1 = 27 °C L, P1 = 3.00 atm
t2 = 0 °C, P2 = 1.00 atm
Find: V2, L
Solution Map:
V1, P1,P2, T1, T2
V2 
Relationships: T(K) = t(°C) + 273,
Solution:
T1  27  273
T1  300 K
T2  0  273
T2  273 K
Check:
P1  V1  T2
P2  T1
V2 
V2 
V2
P1  V1
P  V2
 2
T1
T2
P1  V1  T2
P2  T1
3.00 atm   10.0 L   273 K 
1.00 atm   300 K 
V2  27.3 L
When T decreases, V should decrease; when P decreases, V should
increase—opposite trends make it hard to evaluate our answer.
Avogadro’s Law
• Volume is directly proportional to
the number of gas molecules.
V = constant x n.
Constant P and T.
More gas molecules = larger
volume.
• Count number of gas molecules
by moles, n.
• Equal volumes of gases contain
equal numbers of molecules.
V1 V2

n1 n2
The gas doesn’t matter.
Tro's Introductory Chemistry,
Chapter 11
84
Avogadro’s Law, Continued
Tro's Introductory Chemistry, Chapter
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Example 11.5—A 0.22 Mol Sample of He Has a Volume of 4.8 L.
How Many Moles Must Be Added to Give 6.4 L?
Given: V1 =4.8 L, V2 = 6.4 L, n1 = 0.22 mol
Find: n2, and added moles
Solution Map:
V1, V2, n1
V2
n1   n2
V1
Relationships: mol added = n2 – n1,
Solution: n  V
n2  1 2
V1
n2
V1
V2

n1
n2
moles added  0.29  0.22
moles added  0.07 mol
0.22 mol  6.4 L 

 0.29 mol
4.8 L 
Check:
Since n and V are directly proportional, when the volume
increases, the moles should increase, and it does.
Practice—If 1.00 Mole of a Gas Occupies 22.4 L at
STP, What Volume Would 0.750 Moles Occupy?
Tro's Introductory Chemistry, Chapter
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Practice—If 1.00 Mole of a Gas Occupies 22.4 L at STP,
What Volume Would 0.750 Moles Occupy?, Continued
Given: V1 =22.4 L, n1 = 1.00 mol, n2 = 0.750 mol
Find: V2
Solution Map:
V1, n1, n2
n2
V1   V2
n1
Relationships:
Solution:
V1  n2
V2 
n1
V2
V1
V2

n1
n2
22.4 L   0.750 mol

 16.8 L
1.00 mol
Check:
Since n and V are directly proportional, when the moles
decreases, the volume should decrease, and it does.
Ideal Gas Law
• By combining the gas laws, we can write a general
equation.
• R is called the Gas Constant.
• The value of R depends on the units of P and V.
 We will use 0.0821
atm  L
and
mol  K
convert P to atm and V to L.
• Use the ideal gas law when you have a gas at one
condition, use the combined gas law when you have a
gas whose condition is changing.
 P   V 
 R or PV  nRT
n   T 
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Chapter 11
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Example 11.7—How Many Moles of Gas Are in a Basketball
with Total Pressure 24.2 Psi, Volume of 3.2 L at 25 °C?
Given: V = 3.2 L, P = 24.2 psi, t = 25 °C,
Find: n, mol
Solution Map:
P, V, T, R
PV
n 
RT
Relationships: 1 atm = 14.7 psi
T(K) = t(°C) + 273
n
atmL
PV  nRT, R  0.0821 mol K
P V
n
1 atm
24.2 psi 
 1.6462 atm
R T
14.7 psi
1.6462 atm   3.2 L 
T (K)  25C  273

 0.22 mol
atmL
0.0821 molK   298 K 
T  298 K
Solution:
Check:
1 mole at STP occupies 22.4 L at STP; since there is a much
smaller volume than 22.4 L, we expect less than 1 mole of gas.
Practice—Calculate the Volume Occupied by 637 g
of SO2 (MM 64.07) at 6.08 x 103 mmHg and –23 °C.
Tro's Introductory Chemistry, Chapter
11
107
Practice—Calculate the Volume Occupied by 637 g of SO2
(MM 64.07) at 6.08 x 103 mmHg and –23 °C, Continued.
Given: mSO2 = 637 g, P = 6.08 x 103 mmHg, t = −23 °C,
Find: V, L
Solution Map:
g
Relationships:
n
1 mol
64.07 g
1 atm = 760 mmHg
P, n, T, R
nRT
V 
P
V
atmL
PV  nRT, R  0.0821 mol K
T(K) = t(°C) + 273, 1 mol SO2 = 64.07 g
Solution:
1 atm
6.08 10 mmHg 
 80.0 atm
760 mmHg
3
T (K)  - 23 C  273
T  250 K
637 g SO 2 
1 mol SO 2
 9.942 mol SO 2
64.07 g
n  R T
P
atmL
9.942 mol  0.0821 mol

 250 K 
K

 2.55 L
80.0 atm 
V
Practice—Calculate the Density of a Gas at 775 torr
and 27 °C if 0.250 moles Weighs 9.988 g.
Tro's Introductory Chemistry, Chapter
11
109
Practice—Calculate the Density of a Gas at 775 torr and
27 °C if 0.250 moles Weighs 9.988 g, Continued
Given: m
= 9.988g, n=0.250
n = 0.250mol,
mol, P=775
P = 1.0197
atm, T
= 300. K
m=9.988g,
mmHg,
t=27°C,
Find: density, g/L
Solution Map:
P, n, T, R
V 
V
V, m
n  R T
P
m
d 
V
d
atmL
1 atm = 760 mmHg, PV  nRT, R  0.0821
m
mol K
Relationships: T(K) = t(°C) + 273
d 
V
Solution:
m 9.988 g
1 atm
n

R

T
d 
 1.0197 atm
V 775 torr 
V 6.0355 L
P 760 torr
T (K)  27 C  273atmL
0.250 mol  0.082 molK   300 K 
 1.65 g/L
 T  300 K
 6.0355 L
1.0197 atm 
Check:
The value 1.65 g/L is reasonable.
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
Tro's Introductory Chemistry,
Chapter 11
111
Example 11.8—Calculate the Molar Mass of a Gas with Mass
0.311 g that Has a Volume of 0.225 L at 55 °C and 886 mmHg.
Given: m=0.311g,
m = 0.311g,V=0.225
V = 0.225
L, L,
P=886
P = 1.1658
mmHg,atm,
t=55°C,
T = 328 K
Find: Molar Mass, g/mol
Solution Map:
P, V, T, R
n 
P V
R T
n
n, m
MM 
MM
m
n
atmL
1 atm = 760 mmHg, PV  nRT, R  0.0821
m
mol K
Relationships: T(K) = t(°C) + 273
MM 
n
Solution:
m
0.311 g
MM  
P V
n 886 mmHg  1 atm  1.1658 atm
n 9.7454  10-3 mol
R T
760 mmHg
 31.9 g/mol
 55 C0.225
273L 
1.1T6(K)
58 atm
3 mol


9
.
7
4
54

10
atmL
T  328
K  328 K 
0.0821
molK
Check:
The value 31.9 g/mol is reasonable.
Practice—What Is the Molar Mass of a Gas if
12.0 g Occupies 197 L at 380 torr and 127 °C?
Tro's Introductory Chemistry, Chapter
11
121
Practice—What Is the Molar Mass of a Gas if 12.0 g
Occupies 197 L at 380 torr and 127 °C?, Continued
Given: m=12.0
m = 12.0g,
g, V=
V =197
197L,L,P=380
P = 0.50
torr,atm,
t=127°C,
T =400 K,
Find: molar mass, g/mol
Solution Map:
P, V, T, R
n
P V
R T
1 atm = 760 torr,
Relationships: T(K) = t(°C) + 273
Solution:
n
n, m
MM 
atmL
PV  nRT,
R  0.0821 mol K
m
MM 
P  V1 atm
38n0 
torr 
 0.50 atm
R 760
T torr
T (K)  127 C  273
0.50 atm   197 L 
 3.0 mol
T  400 K atmL
0.0821 molK   400 K 
Check:
m
n
MM
n
m 12.0 g

n 3.0 mol
 4.0 g/mol
MM 
The value 31.9 g/mol is reasonable.
Mixtures of Gases
• According to the kinetic molecular theory, the particles
in a gas behave independently.
• Air is a mixture, yet we can treat it as a single gas.
• Also, we can think of each gas in the mixture as
independent of the other gases.
 All gases in the mixture have the same volume and
temperature.
 All gases completely occupy the container, so all gases in the mixture
have the volume of the container.
Gas
Nitrogen, N2
Oxygen, O2
% in Air,
Gas
by volume
78
Argon, Ar
21
Carbon dioxide, CO2
% in Air,
by volume
0.9
0.03
Partial Pressure
• Each gas in the mixture exerts a pressure
independent of the other gases in the mixture.
• The pressure of a component gas in a mixture
is called a partial pressure.
• The sum of the partial pressures of all the
gases in a mixture equals the total pressure.
Dalton’s law of partial pressures.
Ptotal = Pgas A + Pgas B + Pgas C +...
Pair  PN2  PO2  PAr  0.78 atm  0.21 atm  0.01 atm  1.00 atm
Tro's Introductory Chemistry,
Chapter 11
124
Example 11.9—A Mixture of He, Ne, and Ar Has a Total Pressure
of 558 MmHg. If the Partial Pressure of He Is 341 MmHg and Ne
Is 112 MmHg, Determine the Partial Pressure of Ar in the Mixture.
Given: PHe= 341 mmHg, PNe= 112 mmHg, Ptot = 558 mmHg
Find: PAr, mmHg
Solution Map:
Ptot, PHe, PNe
PAr
PAr = Ptot – (PHe + PNe)
Relationships: Ptot = Pa + Pb + etc.
Solution:
PAr  558  341  112  mmHg
 105 mmHg
Check:
The units are correct, the value is reasonable.
Finding Partial Pressure
• To find the partial pressure of a
gas, multiply the total pressure of
the mixture by the fractional
composition of the gas.
• For example, in a gas mixture that
is 80.0% He and 20.0% Ne that
has a total pressure of 1.0 atm, the
partial pressure of He would be:
PHe = (0.800)(1.0 atm) = 0.80 atm
 Fractional composition = percentage
divided by 100.
Tro's Introductory Chemistry,
Chapter 11
126
The Partial Pressure of Each Gas in a Mixture,
or the Total Pressure of a Mixture, Can Be
Calculated Using the Ideal Gas Law
for gases A and B in a mixture
nA x R x T
nB x R x T
PA 
PB 
V
V
the temperature and volume of everything
in the mixture are the same
ntotal  nA  nB
Ptotal
ntotal x R x T

 PA  PB
V
127
Practice—Find the Partial Pressure of Neon in a Mixture
of Ne and Xe with Total Pressure 3.9 atm, Volume 8.7 L,
Temperature 598 K, and 0.17 moles Xe.
Tro's Introductory Chemistry, Chapter
11
128
Practice—Find the Partial Pressure of Neon in a Mixture of Ne and Xe
with Total Pressure 3.9 Atm, Volume 8.7 L, Temperature 598 K, and
0.17 Moles Xe, Continued.
Given:
Find:
Solution Map:
Ptot = 3.9 atm, V = 8.7 L, T = 598 K, Xe = 0.17 mol
PNe, atm
nXe, V, T, R
PXe 
Relationships:
Solution:
PXe
nXe  R  T
V
Ptot, PXe
PNe
PNe  Ptotal  PXe
atmL
PV  nRT, R  0.0821 mol K , Ptotal  PNe  PXe
PNe  Ptotal  PXe
nXe  R  T
 3.9 atm  0.9589 atm
V
atmL
0.17 mol  0.0821 mol

 598 K 
 2.9 atm
K

 0.9589 atm
8.7 L
PXe 
Check:
The unit is correct, the value is reasonable.
Mountain Climbing and Partial Pressure
• Our bodies are adapted to breathe O2
at a partial pressure of 0.21 atm.
 Sherpa, people native to the Himalaya
mountains, have adapted to the much
lower partial pressure of oxygen in
their air.
• Partial pressures of O2—lower than
0.1 atm—leads to hypoxia.
 Unconsciousness or death.
• Climbers of Mt. Everest must carry
O2 in cylinders to prevent hypoxia.
 On top of Mt. Everest:
Pair = 0.311 atm, so PO2 = 0.065 atm.
Tro's Introductory Chemistry,
Chapter 11
130
Deep Sea Divers and Partial Pressure
• It is also possible to have too much O2, a condition called
oxygen toxicity.
 PO2 > 1.4 atm.
 Oxygen toxicity can lead to muscle spasms, tunnel vision, and
convulsions.
• It is also possible to have too much N2, a condition called
nitrogen narcosis.
 Also known as rapture of the deep.
• When diving deep, the pressure of the air that divers
breathe increases, so the partial pressure of the oxygen
increases.
 At a depth of 55 m, the partial pressure of O2 is 1.4 atm.
 Divers that go below 50 m use a mixture of He and O2 called
heliox that contains a lower percentage of O2 than air.
Partial Pressure vs. Total Pressure
At a depth of 30 m, the total pressure of air in the divers
lungs, and the partial pressure of all the gases in the air,
are quadrupled!
Tro's Introductory Chemistry, Chapter
11
132
Collecting Gases
• Gases are often collected by having them displace
water from a container.
• The problem is that since water evaporates, there
is also water vapor in the collected gas.
• The partial pressure of the water vapor, called the
vapor pressure, depends only on the temperature.
 So you can use a table to find out the partial pressure of
the water vapor in the gas you collect.
• If you collect a gas sample with a total pressure of
758 mmHg at 25 °C, the partial pressure of the
water vapor will be 23.8 mmHg, so the partial
pressure of the dry gas will be 734 mmHg.
Tro's Introductory Chemistry,
Chapter 11
133
Vapor Pressure of Water
Temp., °C
Pressure,
mmHg
10
20
25
9.2
17.5
23.8
30
40
50
60
31.8
55.3
92.5
149.4
70
80
233.7
355.1
Tro's Introductory Chemistry, Chapter
11
134
If the temperature of the water
is 30, the vapor pressure of the
water is 31.8 mmHg.
If the total pressure is 760 mmHg,
the partial pressure of the H2 is
760 − 31.8 mmHg = 728 mmHg.
Tro's Introductory Chemistry, Chapter
11
Zn metal reacts
with HCl(aq) to
produce H2(g).
The gas flows
through the tube
and bubbles into
the jar, where it
displaces the
water in the jar.
Because water
evaporates, some
water vapor gets
mixed in with
the H2.
135
Practice—0.12 moles of H2 Is Collected Over Water in
a 10.0 L Container at 323 K. Find the Total Pressure
(Vapor Pressure of Water at 50 C = 92.6 mmHg).
Tro's Introductory Chemistry, Chapter
11
136
Practice—0.12 moles of H2 Is Collected Over Water in a 10.0 L
Container at 323 K. Find the Total Pressure
(Vapor Pressure of Water at 50 C = 92.6 mmHg), Continued.
Given: V = 10.0 L, nH2 = 0.12 mol, T = 323 K
Find: Ptotal, atm
Solution Map:
nH2,V,T
P
Relationships:
Solution:
n R T
V
PH2
1 atm = 760 mmHg
Ptotal = PA + PB,
Ptotal
Ptotal  PH 2  PH 2O @ 50C
atmL
PV  nRT , R  0.0821 mol K
760 mmHg
0.3181 atm 
 241.8 mmHg
1 atm
n  R T
V
atmL
0.12 mol  0.0821 mol

 323 K 
K

10.0 L 
 0.3181 atm
PH 2 
PH2, PH2O
Ptotal  241.8  92.6
Ptotal  330 mmHg
Reactions Involving Gases
• The principles of reaction involving stoichiometry
from Chapter 8 can be combined with the gas laws
for reactions involving gases.
• In reactions of gases, the amount of a gas is often
given as a volume.
 Instead of moles.
 As we’ve seen, you must state pressure and
temperature.
• The ideal gas law allows us to convert from the
volume of the gas to moles; then, we can use the
coefficients in the equation as a mole ratio.
P, V, T of Gas A
mole A
mole B
Tro's Introductory Chemistry,
Chapter 11
P, V, T of Gas B
138
Example 11.11—How Many Liters of O2 Are Made from 294 g of
KClO3 at 755 mmHg and 305 K?
2 KClO3(s) → 2 KCl(s) + 3 O2(g)
Given: m
nO2
= 3.60
= 294
mol,
g, P
P=755
= 0.99342
mmHg,
atm,
T=305
T = 305
K K
KClO3
Find: VO2, L
Solution Map:
g KClO3
mol KClO3
1 mol KClO 3
122.5 g
mol O2
3 mol O2
2 mol KClO 3
P, n, T, R
V
n  R T
P
V
Relationships: 1 atm = 760 mmHg, KClO3 = 122.5 g/mol
atmL
2 mol KClO3 : 3 mol O2
PV  nRT, R  0.0821
mol K
 RO
T
1 mol KClO 3
3n mol
2
V 
294 g KClO 3 
122.5 g
2 mol KClO
P 3
Solution:
 3.60 mol O 2
1 atm
755 mmHg 
760 mmHg
atmL
3.60 mol  0.0821 mol

 305 K 
K

 0.99342 atm 0.99342 atm 
 90.7 L
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's Introductory Chemistry, Chapter
11
149
Practice—What Volume of O2 at 0.750 atm and 313 K is Generated by
the Thermolysis of 10.0 g of HgO?
2 HgO(s)  2 Hg(l) + O2(g), Continued
Given: n
mO2
= =0.023085
10.0g, P=0.750
mol, P =atm,
0.750
T=313
atm, K
T = 313 K
HgO
Find: VO2, L
Solution Map:
g HgO
mol HgO
1 mol HgO
216.59 g
mol O2
1 mol O 2
2 mol HgO
P, n, T, R
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  0.0821 mol K
n  R T

1 mol V
HgO
1 mol O 2
10.0 g HgO 
 P
216.59 g 2 mol HgO
atmL
0.023085 mol  0.08206 molK   313K 
 0.023085 mol O 2 
0.750 atm 
 0.791 L
Solution:
Calculate the Volume Occupied by 1.00 Mole
of an Ideal Gas at STP.
PxV=nxRxT
L∙atm
moles)(0.0821 mol∙K)(273
(1.00 atm) x V = (1.00
V = 22.4 L
K)
• 1 mole of any gas at STP will occupy 22.4 L.
• This volume is called the molar volume and can
be used as a conversion factor.
 As long as you work at STP.
1 mol  22.4 L
Tro's Introductory Chemistry,
Chapter 11
151
Molar Volume
There is so much
empty space
between molecules
in the gas state that
the volume of the
gas is not effected
by the size of the
molecules (under
ideal conditions).
Tro's Introductory Chemistry, Chapter
11
152
Example 11.12—How Many Grams of H2O Form When 1.24 L H2
Reacts Completely with O2 at STP?
O2(g) + 2 H2(g) → 2 H2O(g)
Given: VH2 = 1.24 L, P = 1.00 atm, T = 273 K
Find: massH2O, g
Solution Map:
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
Tro's Introductory Chemistry, Chapter
11
153
Practice—What Volume of O2 at STP is Generated by the
Thermolysis of 10.0 g of HgO?
2 HgO(s)  2 Hg(l) + O2(g)
(MMHgO = 216.59 g/mol)
Tro's Introductory Chemistry, Chapter
11
161
Practice—What Volume of O2 at STP is Generated by the
Thermolysis of 10.0 g of HgO?
2 HgO(s)  2 Hg(l) + O2(g), Continued
Given: mHgO = 10.0 g, P = 1.00 atm, T = 273 K
Find: VO2, L
Solution Map:
g HgO
mol HgO
1 mol HgO
216 .59 g
mol O2
1 mol O2
2 mol HgO
L O2
22.4 L
1 mol O 2
Relationships: HgO = 216.59 g/mol, 1 mol = 22.4 L at STP
2 mol HgO : 1 mol O2
Solution:
1 mol HgO 1 mol O 2
22.4 L O 2
10.0 g HgO 


216.59 g 2 mol HgO 1 mol O 2
 0.517 L O 2
Tro's Introductory Chemistry, Chapter
11
162