Chapter 17
Temperature and the Kinetic Theory of Gases
Conceptual Problems
1
(a)
(b)
(c)
•
True or false:
The zeroth law of thermodynamics states that two objects in thermal
equilibrium with each other must be in thermal equilibrium with a third
object.
The Fahrenheit and Celsius temperature scales differ only in the choice of
the ice-point temperature.
The Celsius degree and the kelvin are the same size.
(a) False. If two objects are in thermal equilibrium with a third, then they are in
thermal equilibrium with each other.
(b) False. The Fahrenheit and Celsius temperature scales also differ in the number
of intervals between the ice-point temperature and the steam-point temperature.
(c) True.
2
•
How can you determine if two objects are in thermal equilibrium with
each other when putting them into physical contact with each other would have
undesirable effects? (For example, if you put a piece of sodium in contact with
water there would be a violent chemical reaction.)
Determine the Concept Put each in thermal equilibrium with a third body; that
is, a thermometer. If each body is in thermal equilibrium with the third, then they
are in thermal equilibrium with each other.
3
•
[SSM] ″Yesterday I woke up and it was 20 ºF in my bedroom,″ said
Mert to his old friend Mort. ″That’s nothing,″ replied Mort. ″My room was
–5.0 ºC.″ Who had the colder room, Mert or Mort?
Picture the Problem We can decide which room was colder by converting 20°F
to the equivalent Celsius temperature.
Using the Fahrenheit-Celsius
conversion, convert 20°F to the
equivalent Celsius temperature:
tC =
5
9
(t F − 32°) = 95 (20° − 32°)
= −6.7°C
so Mert' s room was colder.
1713
1714
Chapter 17
4
•
Two identical vessels contain different ideal gases at the same
pressure and temperature. It follows that (a) the number of gas molecules is the
same in both vessels, (b) the total mass of gas is the same in both vessels, (c) the
average speed of the gas molecules is the same in both vessels, (d) None of the
above.
Determine the Concept Because the vessels are identical (have the same volume)
and the two ideal gases are at the same pressure and temperature, the ideal-gas law
( PV = NkT ) tells us that the number of gas molecules must be the same in both
vessels. (a ) is correct.
5
•
[SSM] Figure 17-18 shows a plot of volume versus absolute
temperature for a process that takes a fixed amount of an ideal gas from point A
to point B. What happens to the pressure of the gas during this process?
Determine the Concept From the ideal-gas law, we have P = nR T V . In the
process depicted, both the temperature and the volume increase, but the
temperature increases faster than does the volume. Hence, the pressure increases.
6
•
Figure 17-19 shows a plot of pressure versus absolute temperature for
a process that takes a sample of an ideal gas from point A to point B. What
happens to the volume of the gas during this process?
Determine the Concept From the ideal-gas law, we have V = nR T P. In the
process depicted, both the temperature and the pressure increase, but the pressure
increases faster than does the temperature. Hence, the volume decreases.
7
•
If a vessel contains equal amounts, by mass, of helium and argon,
which of the following are true?
(a)
(b)
(c)
(d)
The partial pressure exerted by each of the two gases on the walls of the
container is the same.
The average speed of a helium atom is the same as that of an argon atom.
The number of helium atoms and argon atoms in the vessel are equal.
None of the above.
Determine the Concept
(a) False. The partial pressure exerted by each gas in the mixture is the pressure it
would exert if it alone occupied the container. Because the densities of helium and
argon are not the same, these gases occupy different volumes and, hence, their
partial pressures are not the same.
(b) False. Assuming the gasses have been in the vessel for some time, their
average kinetic energies would be the same. Because the densities of helium and
Temperature and the Kinetic Theory of Gases 1715
argon are not the same, their average speeds must be different.
(c) False. We know that the volumes of the two gasses are equal (they both
occupy the full volume of the container) and that their temperatures are equal.
Because their pressures are different, the number of atoms of each gas must be
different, too.
(d) Because none of the above is true,
(d ) is true.
8
•
By what factor must the absolute temperature of a gas be increased to
double the rms speed of its molecules?
Determine the Concept We can use vrms = 3RT M to relate the temperature of
a gas to the rms speed of its molecules.
Express the dependence of the rms
speed of the molecules of a gas on
their absolute temperature:
3RT
M
where R is the gas constant, M is the
molar mass, and T is the absolute
temperature.
vrms =
Because v rms ∝ T , the temperature must be quadrupled in order to double the
rms speed of the molecules.
9
•
Two different gases are at the same temperature. What can be said
about the average translational kinetic energies of the molecules? What can be
said about the rms speeds of the gas molecules?
Determine the Concept The average kinetic energies are equal. The ratio of their
rms speeds is equal to the square root of the reciprocal of the ratio of their
molecular masses.
10 •
A vessel holds a mixture of helium (He) and methane (CH4). The ratio
of the rms speed of the He atoms to that of the CH4 molecules is (a) 1, (b) 2, (c) 4,
(d) 16
Picture the Problem We can express the rms speeds of the helium atoms and the
methane molecules using vrms = 3RT M .
Express the rms speed of the helium
atoms:
vrms, He =
3RT
M He
1716
Chapter 17
Express the rms speed of the
methane molecules:
Divide the first of these equations by
the second to obtain:
Use Appendix C to find the molar
masses of helium and methane:
vrms. CH 4 =
vrms, He
vrms. CH 4
vrms, He
vrms. CH 4
3RT
M CH 4
=
M CH 4
=
16 g/mol
=2
4 g/mol
M He
and (b) is correct.
11 •
True or false: If the pressure of a fixed amount of gas increases, the
temperature of the gas must increase.
False. Whether the pressure changes also depends on whether and how the
volume changes. In an isothermal process, the pressure can increase while the
volume decreases and the temperature is constant.
12 •
Why might the Celsius and Fahrenheit scales be more convenient than
the absolute scale for ordinary, nonscientific purposes?
Determine the Concept For the Celsius scale, the ice point (0°C) and the boiling
point of water at 1 atm (100°C) are more convenient than 273 K and 373 K;
temperatures in roughly this range are normally encountered. On the Fahrenheit
scale, the temperature of warm-blooded animals is roughly 100°F; this may be a
more convenient reference than approximately 300 K. Throughout most of the
world, the Celsius scale is the standard for nonscientific purposes.
13 •
An astronomer claims that the temperature at the center of the Sun is
7
about 10 degrees. Do you think that this temperature is in kelvins, degrees
Celsius, or doesn’t it matter?
Determine the Concept Because T = tC + 273.15 K and 107 >> 273, it doesn’t
matter.
14 •
Imagine that you have a fixed amount of ideal gas in a container that
expands to maintain constant pressure. If you double the absolute temperature of
the gas, the average speed of the molecules (a) remains constant, (b) doubles, (c)
quadruples, (d) increases by a factor of 2 .
Determine the Concept The average speed of the molecules in an ideal gas
depends on the square root of the kelvin temperature. Because vav ∝ T , doubling
Temperature and the Kinetic Theory of Gases 1717
the temperature while maintaining constant pressure increases the average speed
by a factor of 2 . (d ) is correct.
15 •
[SSM] Suppose that you compress an ideal gas to half its original
volume, while also halving its absolute temperature. During this process, the
pressure of the gas (a) halves, (b) remains constant, (c) doubles, (d) quadruples.
Determine the Concept From the ideal-gas law, PV = nRT , halving both the
temperature and volume of the gas leaves the pressure unchanged. (b) is
correct.
16 •
The average translational kinetic energy of the molecules of a gas
depends on (a) the number of moles and the temperature, (b) the pressure and the
temperature, (c) the pressure only, (d) the temperature only.
Determine the Concept The average translational kinetic energy of the molecules
of an ideal gas Kav depends on its temperature T according to K av = 32 kT . (d ) is
correct.
17 ••
[SSM] Which speed is greater, the speed of sound in a gas or the
rms speed of the molecules of the gas? Justify your answer, using the appropriate
formulas, and explain why your answer is intuitively plausible.
Determine the Concept The rms speed of molecules of an ideal gas is given by
3RT
γRT
and the speed of sound in a gas is given by vsound =
.
vrms =
M
M
The rms speed of the molecules of an
ideal-gas is given by:
vrms =
The speed of sound in a gas is given
by:
vsound =
Divide the first of these equations by
the second and simplify to obtain:
For a monatomic gas, γ = 1.67 and:
vrms
=
vsound
3RT
M
γRT
M
3RT
M = 3
γ
γRT
M
vrms, monatomic
vsound
=
3
= 1.34
1.67
1718
Chapter 17
For a diatomic gas, γ = 1.40 and:
vrms, diatomic
vsound
=
3
= 1.46
1.40
In general, the rms speed is always somewhat greater than the speed of sound.
However, it is only the component of the molecular velocities in the direction of
propagation that is relevant to this issue. In addition, In a gas the mean free path is
greater than the average intermolecular distance.
18 •• Imagine that you increase the temperature of a gas while holding its
volume fixed. Explain in terms of molecular motion why the pressure of the gas
on the walls of its container increases.
Determine the Concept The pressure is a measure of the change in momentum
per second of a gas molecule on collision with the wall of the container. When
the gas is heated, the average velocity and the average momentum of the
molecules increase and, as a consequence, the pressure exerted by the molecules
increases.
19 ••
Imagine that you compress a gas while holding it at fixed temperature
(perhaps by immersing the container in cool water). Explain in terms of
molecular motion why the pressure of the gas on the walls of its container
increases.
Determine the Concept If the volume decreases the pressure increases because
more molecules hit a unit of area of the walls in a given time. The fact that the
temperature does not change tells us the molecular speed does not change with
volume.
20 ••
Oxygen has a molar mass of 32 g/mol and nitrogen has a molar mass
of 28 g/mol. The oxygen and nitrogen molecules in a room have
(a)
(b)
(c)
(d)
(e)
(f)
equal average translational kinetic energies, but the oxygen molecules have
a larger average speed than the nitrogen molecules have.
equal average translational kinetic energies, but the oxygen molecules have
a smaller average speed than the nitrogen molecules have.
equal average translational kinetic energies and equal average speeds.
equal average speeds, but the oxygen molecules have a larger average
translational kinetic energy than the nitrogen molecules have.
equal average speeds, but the oxygen molecules have a smaller average
translational kinetic energy than the nitrogen molecules have.
None of the above.
Picture the Problem The average kinetic energies of the molecules are given by
K av = (12 mv 2 )av = 32 kT . Assuming that the room’s temperature distribution is
Temperature and the Kinetic Theory of Gases 1719
uniform, we can conclude that the oxygen and nitrogen molecules have equal
average kinetic energies. Because the oxygen molecules are more massive, they
must be moving slower than the nitrogen molecules. (b) is correct.
21 ••
[SSM] Liquid nitrogen is relatively cheap, while liquid helium is
relatively expensive. One reason for the difference in price is that while nitrogen
is the most common constituent of the atmosphere, only small traces of helium
can be found in the atmosphere. Use ideas from this chapter to explain why it is
that only small traces of helium can be found in the atmosphere.
Determine the Concept The average molecular speed of He gas at 300 K is about
1.4 km/s, so a significant fraction of He molecules have speeds in excess of
earth’s escape velocity (11.2 km/s). Thus, they "leak" away into space. Over
time, the He content of the atmosphere decreases to almost nothing.
Estimation and Approximation
22
•
Estimate the total number of air molecules in your classroom.
Picture the Problem The number of air molecules in your classroom is given by
the ideal-gas law.
From the ideal-gas law, the number
of molecules N in a given volume V
at pressure P and temperature T is
given by:
PV
kT
where k is Boltzmann’s constant.
Assuming that a typical classroom is
a rectangular parallelepiped, its
volume is given by:
V = Awh
N=
Substituting for V yields:
N=
PAwh
kT
Assume atmospheric pressure, room temperature and that the room is 15 m square
and 5 m high to obtain:
101.325 kPa ⎞
⎛
⎜1 atm ×
⎟ (15 m )(15 m )(5 m )
atm
⎝
⎠
N=
≈ 3 × 10 28
1.381× 10 −23 J/K (293 K )
(
)
Chapter 17
1720
23
••
Estimate the density of dry air at sea level on a warm summer day.
Picture the Problem We can use the definition of mass density, the definition of
the molar mass of a gas, and the ideal-gas law to estimate the density of air at sea
level on a warm day.
The density of air at sea level is
given by:
ρ=
m
V
The mass of the air molecules
occupying a given volume is the
product of the number of moles n
and the molar mass M of the
molecules. Substitute for m to obtain:
ρ=
nM
V
n=
PV
RT
ρ=
M ⎛ PV ⎞ PM
⎜
⎟=
V ⎝ RT ⎠ RT
From the ideal-gas law, the number
of moles in a given volume depends
on the pressure and temperature
according to:
Substitute in the expression for ρ and
simplify to obtain:
Assuming atmospheric pressure, a
temperature of 27°C (93°F), and
29 g/mol for the molar mass of air,
substitute numerical values and
evaluate ρ:
101.325 kPa ⎞
⎛
⎜1 atm ×
⎟(29 g/mol)
atm
⎝
⎠
ρ=
(8.314 J/mol ⋅ K )(300 K )
= 1.2 kg/m 3
24 ••
A stoppered test tube that has a volume of 10.0 mL has 1.00 mL of
water at its bottom. The water has a temperature of 100ºC and is initially at a
pressure of 1.00 atm. The test tube is held over a flame until the water has
completely boiled away. Estimate the final pressure inside the test tube.
Picture the Problem Assuming the steam to be an ideal gas at a temperature of
373 K, we can use the ideal-gas law to estimate the pressure inside the test tube
when the water is completely boiled away.
Using the ideal-gas law, relate the
pressure inside the test tube to its
volume and the temperature:
P=
NkT
V
Temperature and the Kinetic Theory of Gases 1721
Relate the number of particles N to
the mass of water, its molar mass M,
and Avogadro’s number NA:
N
m M
=
⇒N = m A
N NA
M
Relate the mass of 1.00 mL of water
to its density:
m = ρV = 1.00 × 103 kg/m3
(
(
)
× 1.00 × 10− 6 m3
)
= 1.00 g
Substitute for m, NA, and M
(18 g/mol) and evaluate N:
N = (1.00 g )
6.022 × 1023 particles/mol
18 g/mol
= 3.346 × 1022 particles
Substitute numerical values and evaluate P:
(3.346 ×10
)(
)
particles 1.381× 10 −23 J/K (373 K )
10 − 6 m 3
10.0 mL ×
mL
1
atm
= 1.723 ×10 7 N/m 2 ×
101.325 ×10 3 N/m 2
P=
22
= 170 atm
25 ••
[SSM] In Chapter 11, we found that the escape speed at the surface
of a planet of radius R is ve = 2gR , where g is the acceleration due to gravity at
the surface of the planet. If the rms speed of a gas is greater than about 15 to 20
percent of the escape speed of a planet, virtually all of the molecules of that gas
will escape the atmosphere of the planet.
(a)
(b)
(c)
(d)
At what temperature is vrms for O2 equal to 15 percent of the escape speed
for Earth?
At what temperature is vrms for H2 equal to 15 percent of the escape speed
for Earth?
Temperatures in the upper atmosphere reach 1000 K. How does this help
account for the low abundance of hydrogen in Earth’s atmosphere?
Compute the temperatures for which the rms speeds of O2 and H2 are equal
to 15 percent of the escape speed at the surface of the moon, where g is
about one-sixth of its value on Earth and R = 1738 km. How does this
account for the absence of an atmosphere on the moon?
Picture the Problem We can find the escape temperatures for the earth and the
moon by equating, in turn, 0.15ve and vrms of O2 and H2. We can compare these
1722
Chapter 17
temperatures to explain the absence from Earth’s upper atmosphere and from the
surface of the moon. See Appendix C for the molar masses of O2 and H2.
(a) Express vrms for O2:
vrms, O 2 =
3RT
M O2
where R is the gas constant, T is the
absolute temperature, and M O 2 is the
molar mass of oxygen.
Equate 0.15ve and v rms, O 2 :
3RT
M
0.15 2 gRearth =
Solve for T to obtain:
T=
0.045 gRearth M
3R
(1)
Substitute numerical values and evaluate T for O2:
T=
(
)(
)(
)
0.045 9.81m/s 2 6.37 × 106 m 32.0 × 10−3 kg/mol
= 3.60 × 103 K
3(8.314 J/mol ⋅ K )
(b) Substitute numerical values and evaluate T for H2:
T=
(
)(
)(
)
0.045 9.81 m/s 2 6.37 ×106 m 2.02 ×10 −3 kg/mol
= 230 K
3(8.314 J/mol ⋅ K )
(c) Because hydrogen is lighter than air it rises to the top of the atmosphere.
Because the temperature is high there, a greater fraction of the molecules reach
escape speed.
(d) Express equation (1) at the
surface of the moon:
0.045 g moon Rmoon M
3R
1
0.045( 6 g earth )Rmoon M
=
3R
0.0025 g earth Rmoon M
=
R
T=
Substitute numerical values and evaluate T for O2:
(
)(
)(
)
0.0025 9.81 m/s 2 1.738 × 106 m 32.0 × 10−3 kg/mol
T=
= 160 K
8.314 J/mol ⋅ K
Temperature and the Kinetic Theory of Gases 1723
Substitute numerical values and evaluate T for H2:
T=
(
)(
)(
)
0.0025 9.81 m/s 2 1.738 × 106 m 2.02 × 10−3 kg/mol
= 10 K
8.314 J/mol ⋅ K
Because g is less on the moon, the escape speed is lower. Thus, a larger
percentage of the molecules are moving at escape speed.
26 ••
The escape speed for gas molecules in the atmosphere of Mars is
5.0 km/s and the surface temperature of Mars is typically 0ºC. Calculate the rms
speeds for (a) H2, (b) O2, and (c) CO2 at this temperature. (d) Are H2, O2, and CO2
likely to be found in the atmosphere of Mars?
Picture the Problem We can use vrms = 3RT M to calculate the rms speeds of
H2, O2, and CO2 at 273 K and then compare these speeds to 20% of the escape
velocity on Mars to decide the likelihood of finding these gases in the atmosphere
of Mars. See Appendix C for molar masses.
Express the rms speed of an atom as
a function of the temperature:
vrms =
(a) Substitute numerical values
and evaluate vrms for H2:
v rms,H2 =
3RT
M
3(8.314 J/mol ⋅ K )(273 K )
2.02 × 10 −3 kg/mol
= 1.84 km/s
(b) Evaluate vrms for O2:
v rms,O2 =
3(8.314 J/mol ⋅ K )(273 K )
32.0 × 10 −3 kg/mol
= 461 m/s
(c) Evaluate vrms for CO2:
v rms,CO2 =
3(8.314 J/mol ⋅ K )(273 K )
44.0 × 10 −3 kg/mol
= 393 m/s
(d) Calculate 20% of vesc for Mars:
v = 15 vesc =
1
5
(5.0 km/s) = 1.0 km/s
Because v is greater than v rms CO2 and O2 but less than v rms for H2, O2 and CO2,
but not H2 should be present.
1724
Chapter 17
27 ••[SSM] The escape speed for gas molecules in the atmosphere of Jupiter is
60 km/s and the surface temperature of Jupiter is typically –150ºC. Calculate the
rms speeds for (a) H2, (b) O2, and (c) CO2 at this temperature. (d) Are H2, O2, and
CO2 likely to be found in the atmosphere of Jupiter?
Picture the Problem We can use vrms = 3RT M to calculate the rms speeds of
H2, O2, and CO2 at 123 K and then compare these speeds to 20% of the escape
velocity on Jupiter to decide the likelihood of finding these gases in the
atmosphere of Jupiter. See Appendix C for the molar masses of H2, O2, and CO2.
Express the rms speed of an atom as
a function of the temperature:
vrms =
(a) Substitute numerical values and
evaluate vrms for H2:
v rms,H2 =
3RT
M
3(8.314 J/mol ⋅ K )(123 K )
2.02 × 10 −3 kg/mol
= 1.23 km/s
(b) Evaluate vrms for O2:
v rms,O2 =
3(8.314 J/mol ⋅ K )(123 K )
32.0 × 10 −3 kg/mol
= 310 m/s
(c) Evaluate vrms for CO2:
v rms,CO2 =
3(8.314 J/mol ⋅ K )(123 K )
44.0 ×10 −3 kg/mol
= 264 m/s
(d) Calculate 20% of vesc for Jupiter:
v = 15 vesc =
1
5
(60 km/s) = 12 km/s
Because ve is greater than vrms for O2, CO2 and H2, O2, all three gasses should be
found on Jupiter.
28 ••
Estimate the average pressure on the front wall of a racquetball court,
due to the collisions of the ball with the wall during a game. Use any reasonable
numbers for the mass of the ball, its typical speed, and the dimensions of the
court. Is the average pressure from the ball significant compared to that from the
air?
Picture the Problem The average pressure exerted by the ball on the wall is the
ratio of the average force it exerts on the wall to the area of the wall. The average
force, in turn, is the rate at which the momentum of the ball changes during each
collision with the wall. Assume that the mass of a racquetball is 100 g, that the
court measures 5 m by 5 m, and that the speed of the racquetball is 10 m/s. We’ll
Temperature and the Kinetic Theory of Gases 1725
also assume that the interval Δt between collisions with the wall is 1 s.
The average pressure exerted on the
wall by the racquetball is given by:
Fav Fav
=
A
wh
where w and h are the width and height
of the wall, respectively.
Assuming head-on collisions with
the wall, the change in momentum of
the ball during each collision is 2mv
and the average force exerted by the
ball on the wall is:
Δp 2mv
=
Δt
Δt
where Δt is the elapsed time between
collisions of the ball with the wall.
Substituting for Fav yields:
Substitute numerical values and
evaluate Pav:
Pav =
Fav =
Pav =
2mv
whΔt
Pav =
2(.1 kg )(10 m/s )
= 0.08 N/m 2
(5 m )(5 m )(1 s )
≈ 0.1 Pa
Express the ratio of Pav and Patm to
obtain:
Pav
0.1 Pa
≈ 5
= 10 − 6
Patm 10 Pa
or
Pav ≈ 10 −6 Patm
and the average pressure from the ball
is not significant compared to
atmospheric pressure.
To a first approximation, the Sun consists of a gas of equal numbers of
29 ••
protons and electrons. (The masses of these particles can be found in Appendix
B.) The temperature at the center of the Sun is about 1 × 107 K, and the density
of the Sun is about 1 × 105 kg/m3. Because the temperature is so high, the protons
and electrons are separate particles (rather than being joined together to form
hydrogen atoms). (a) Estimate the pressure at the center of the Sun. (b) Estimate
the rms speeds of the protons and the electrons at the center of the Sun.
Picture the Problem We’ll assume that we can model the plasma as an ideal gas,
at least to a first approximation. Then we can apply the ideal gas law to an
arbitrary volume of the material, say one cubic meter.
(a) To the extent that the plasma acts
like an ideal gas, the pressure at the
center of the Sun is given by:
P=
nRT
V
1726
Chapter 17
The mass of our 1 m3 volume of
matter is 105 kg and this mass
consists mostly of protons. One mole
of protons has a mass of 1 g, so in
105 kg, the number of moles of
protons would be:
105 kg
nprotons =
= −3
= 108 mol
M protons 10 kg
or, counting electrons,
n = 2nprotons = 2 × 108 mol
Substitute numerical values and
evaluate P:
(2 ×10
P=
mprotons
)
(
mol (8.314 J/mol ⋅ K ) 10 7 K
1 m3
1 atm
≈ 2 ×1016 Pa ×
101.325 kPa
8
≈ 2 ×1011 atm
(b) To one significant figure, the
average speed of the protons is the
same as the rms speed:
vrms, protons =
3RT
M protons
Substitute numerical values and
evaluate v rms, protons :
v rms, protons =
3(8.314 J/mol ⋅ K ) 107 K
0.001 kg
(
)
= 5 × 105 m/s
The rms speed of an electron at the
center of the Sun is given by:
vrms, electrons =
3RT
=
M electrons
Substitute numerical values and
evaluate v rms, electrons :
vrms, electrons =
3(8.314 J/mol ⋅ K ) 10 7 K
1
2000 (0.001 kg )
1
2000
3RT
M protons
(
)
= 2 × 10 7 m/s
Remarks: The huge pressure we calculated in (a) is required to support the
tremendous weight of the rest of the sun. The very high speed we calculated
for the plasma electrons is still an order-of-magnitude smaller than the speed
of light.
You are designing a vacuum chamber for fabricating reflective
30 ••
coatings. Inside this chamber, a small sample of metal will be vaporized so that
its atoms travel in straight lines (the effects of gravity are negligible during the
brief time of flight) to a surface where they land to form a very thin film. The
sample of metal is 30 cm from the surface to which the metal atoms will adhere.
How low must the pressure in the chamber be so that the metal atoms only rarely
collide with air molecules before they land on the surface?
)
Temperature and the Kinetic Theory of Gases 1727
Picture the Problem We can use the expression for the mean free path of a
molecule to eliminate the number of molecules per unit volume nV from the idealgas law. Assume that the air in the chamber is at room temperature (300 K) and
that the average diameter of an air molecule is 4 × 10−10 m.
Apply the ideal gas law to express
the pressure in terms of the number
of molecules per unit volume and the
temperature:
The mean free path of an air
molecule is given by:
Substitute for nV to obtain:
Substitute numerical values and
evaluate P:
P=
λ=
P=
P=
NkT
= nV kT
V
1
2nV π d
2
1
⇒ nV =
2λπ d 2
kT
2λπ d 2
(1.381×10
−23
(
)
J/K (300 K )
2 (30 cm )π 4 × 10 −10 m
)
2
≈ 20 mPa
31 ••• [SSM] In normal breathing conditions, approximately 5 percent of
each exhaled breath is carbon dioxide. Given this information and neglecting any
difference in water-vapor content, estimate the typical difference in mass between
an inhaled breath and an exhaled breath.
Picture the Problem One breath (one’s lung capacity) is about half a liter. The
only thing that occurs in breathing is that oxygen is exchanged for carbon dioxide.
Let’s estimate that of the 20% of the air that is breathed in as oxygen, ¼ is
exchanged for carbon dioxide. Then the mass difference between breaths will be
5% of a breath multiplied by the molar mass difference between oxygen and
carbon dioxide and by the number of moles in a breath. Because this is an
estimation problem, we’ll use 32 g/mol as an approximation for the molar mass of
oxygen and 44 g/mol as an approximation for the molar mass of carbon dioxide.
Express the difference in mass
between an inhaled breath and an
exhaled breath:
Δm = mO 2 − mCO 2
(
)
= f CO 2 M O 2 − M CO 2 nbreath
where f CO 2 is the fraction of the air
breathed in that is exchanged for carbon
dioxide.
The number of moles per breath is
given by:
nbreath =
Vbreath
22.4 L/mol
1728
Chapter 17
Substituting for nbreath yields:
(
)
⎛ Vbreath ⎞
Δm = f CO 2 M O 2 − M CO 2 ⎜
⎟
⎝ 22.4 L/mol ⎠
Substitute numerical values and evaluate Δm:
Δm = (0.05)(44 g/mol − 32 g/mol)
(0.5 L )
22.4 L/mol
≈ 1× 10 −4 g
Temperature Scales
32 •
A certain ski wax is rated for use between –12 and –7.0 ºC. What is
this temperature range on the Fahrenheit scale?
Picture the Problem We can use the fact that 5 C° = 9 F° to set up a proportion
that allows us to make easy interval conversions from either the Celsius or
Fahrenheit scale to the other.
The proportion relating a temperature
range on the Fahrenheit scale of a
temperature range on the Celsius
scale is:
Δt F 9 F°
⎛ 9 F° ⎞
=
⇒ Δt F = ⎜
⎟Δt C
Δt C 5 C°
⎝ 5 C° ⎠
Substitute numerical values and
evaluate ΔtF:
⎛ 9 F° ⎞
Δt F = ⎜
⎟(− 7°C − (− 12°C )) = 9 F°
⎝ 5 C° ⎠
Remarks: An equivalent but slightly longer solution involves converting the
two temperatures to their Fahrenheit equivalents and then subtracting these
temperatures.
[SSM] The melting point of gold is 1945.4 ºF. Express this
33 •
temperature in degrees Celsius.
Picture the Problem We can use the Fahrenheit-Celsius conversion equation to
find this temperature on the Celsius scale.
Convert 1945.4°F to the
equivalent Celsius temperature:
tC =
5
9
(tF − 32°) = 95 (1945.4° − 32°)
= 1063°C
34 •
A weather report indicates that the temperature is expected to drop by
15.0°C over the next four hours. By how many degrees on the Fahrenheit scale
will the temperature drop?
Temperature and the Kinetic Theory of Gases 1729
Picture the Problem We can use the fact that 5 C° = 9 F° to set up a proportion
that allows us to make easy interval conversions from either the Celsius or
Fahrenheit scale to the other.
The proportion relating a temperature
range on the Fahrenheit scale of a
temperature range on the Celsius
scale is:
Δt F 9 F°
⎛ 9 F° ⎞
=
⇒ Δt F = ⎜
⎟Δt C
Δt C 5 C°
⎝ 5 C° ⎠
Substitute numerical values and
evaluate ΔtF:
⎛ 9 F° ⎞
Δt F = ⎜
⎟(15.0 C°) = 27.0 F°
⎝ 5 C° ⎠
The length of the column of mercury in a thermometer is 4.00 cm
35 •
when the thermometer is immersed in ice water at 1 atm of pressure, and 24.0 cm
when the thermometer is immersed in boiling water at 1 atm of pressure. Assume
that the length of the mercury column varies linearly with temperature. (a) Sketch
a graph of the length of the mercury column versus temperature (in degrees
Celsius). (b) What is the length of the column at room temperature (22.0ºC)?
(c) If the mercury column is 25.4 cm long when the thermometer is immersed in a
chemical solution, what is the temperature of the solution?
Picture the Problem We can use the equation of the graph plotted in (a) to (b)
find the length of the mercury column at room temperature and (c) the
temperature of the solution when the height of the mercury column is 25.4 cm.
(a) A graph of the length of the
mercury column versus temperature
(in degrees Celsius) is shown to the
right. The equation of the line is:
cm ⎞
⎛
L = ⎜ 0.200
⎟ t C + 4.00 cm (1)
°C ⎠
⎝
L, cm
24.0
4.00
0
(b) Evaluate L(22.0°C )
t C , °C
cm ⎞
⎛
L = ⎜ 0.200
⎟ (22.0°C ) + 4.00 cm
°C ⎠
⎝
= 8.40 cm
(c) Solve equation (1) for tC to
obtain:
100
tC =
L − 4.00 cm
cm
0.200
°C
1730
Chapter 17
Substitute numerical values and
evaluate t C (25.4 cm ) :
t C (25.4 cm ) =
25.4 cm − 4.00 cm
cm
0.200
°C
= 107°C
36 •
The temperature of the interior of the Sun is about 1.0 × 107 K. What is
this temperature in (a) Celsius degrees, (b) kelvins, and (c) Fahrenheit degrees?
Picture the Problem We can use the temperature conversion equations
tF = 95 tC + 32° and tC = T − 273.15 K to convert 107 K to the Fahrenheit and
Celsius temperatures.
Express the kelvin temperature in
terms of the Celsius temperature:
T = tC + 273.15 K
(a) Solve for and evaluate tC:
t C = T − 273.15 K
= 1.0 × 10 7 K − 273.15 K
≈ 1.0 × 10 7 °C
(b) Use the Celsius to Fahrenheit
conversion equation to evaluate tF:
(
)
t F = 95 1.0 × 107°C + 32°
≈ 1.8 × 107°F
37 •
The boiling point of nitrogen, N2, is 77.35 K. Express this temperature
in degrees Fahrenheit.
Picture the Problem While we could
convert 77.35 K to a Celsius
temperature and then convert the
Celsius temperature to a Fahrenheit
temperature, an alternative solution is
to use the diagram to the right to set up
a proportion for the direct conversion
of the kelvin temperature to its
Fahrenheit equivalent.
Use the diagram to set up the
proportion:
F
K
212
373.15
32
273.15
tF
77.35
32°F − t F
273.15 K − 77.35 K
=
212°F − 32°F 373.15 K − 273.15 K
or
32°F − t F 195.8
=
180 F°
100
Temperature and the Kinetic Theory of Gases 1731
Solving for tF yields:
t F = 32°F −
195.8
× 180 F° = − 320 °F
100
38 •
The pressure of a constant-volume gas thermometer is 0.400 atm at the
ice point and 0.546 atm at the steam point. (a) Sketch a graph of pressure versus
Celsius temperature for this thermometer. (b) When the pressure is 0.100 atm,
what is the temperature? (c) What is the pressure at 444.6 ºC (the boiling point of
sulfur)?
Picture the Problem We can use the equation of the graph plotted in (a) to (b)
find the temperature when the pressure is 0.100 atm and (c) the pressure when the
temperature is 444.6ºC.
(a) A graph of pressure (in atm)
versus temperature (in degrees
Celsius) for this thermometer is
shown to the right. The equation
of this graph is:
atm ⎞
⎛
P = ⎜1.46 × 10−3
⎟t C + 0.400 atm
°C ⎠
⎝
P , atm
0.546
0.400
0
100
t C , °C
P − 0.400 atm
atm
1.46 × 10 −3
°C
(b) Solving for tC yields:
tC =
Substitute numerical values and
evaluate t C (0.100 atm ) :
t C (0.100 atm ) =
0.100 atm − 0.400 atm
atm
1.46 × 10−3
°C
= − 205°C
Substitute numerical values and evaluate P (444.6°C ) :
atm ⎞
⎛
P (444.6°C ) = ⎜1.46 × 10−3
⎟ (444.6°C ) + 0.400 atm = 1.05 atm
°C ⎠
⎝
39 •
[SSM] A constant-volume gas thermometer reads 50.0 torr at the
triple point of water. (a) Sketch a graph of pressure vs. absolute temperature for
this thermometer. (b) What will be the pressure when the thermometer measures a
temperature of 300 K? (c) What ideal-gas temperature corresponds to a pressure
of 678 torr?
1732
Chapter 17
Picture the Problem We can use the equation of the graph plotted in (a) to (b)
find the pressure when the temperature is 300 K and (c) the ideal-gas temperature
when the pressure is 678 torr.
(a) A graph of pressure (in torr)
versus temperature (in kelvins)
for this thermometer is shown to
the right. The equation of this
graph is:
⎛ 50.0 torr ⎞
P =⎜
⎟T
⎝ 273 K ⎠
(1)
(b) Evaluate P when T = 300 K:
P , torr
50.0
0
273
0
T, K
⎛ 50.0 torr ⎞
P (300 K ) = ⎜
⎟ (300 K )
⎝ 273 K ⎠
= 54.9 torr
(c) Solve equation (1) for T to obtain:
⎛ 273 K ⎞
T =⎜
⎟P
⎝ 50.0 torr ⎠
Evaluate T (678 torr ) :
⎛ 273 K ⎞
T (678 torr ) = ⎜
⎟ (678 torr )
⎝ 50.0 torr ⎠
= 3.70 × 103 K
40 •
A constant-volume gas thermometer has a pressure of 30.0 torr when it
reads a temperature of 373 K. (a) Sketch a graph of pressure vs. absolute
temperature for this thermometer. (b) What is its triple-point pressure P3?
(c) What temperature corresponds to a pressure of 0.175 torr?
Picture the Problem We can use the equation of the graph plotted in (a) to (b)
find the triple-point pressure P3 and (c) the temperature when the pressure is
0.175 torr.
Temperature and the Kinetic Theory of Gases 1733
(a) A graph of pressure versus
absolute temperature for this
thermometer is shown to the right.
The equation of this graph is:
⎛ 30.0 torr ⎞
P =⎜
⎟T
⎝ 373 K ⎠
(1)
P , torr
30.0
0
T, K
0
373
(b) Solve for and evaluate the
thermometer’s triple-point
(273.16 K) pressure:
⎛ 30.0 torr ⎞
P (273.16 K ) = ⎜
⎟ (273.16 K )
⎝ 373 K ⎠
(c) Solve equation (1) for T to
obtain:
⎛ 373 K ⎞
T =⎜
⎟P
⎝ 30.0 torr ⎠
Evaluate T (0.175 torr ) :
⎛ 373 K ⎞
T (0.175 torr ) = ⎜
⎟ (0.175 torr )
⎝ 30.0 torr ⎠
= 22.0 torr
= 2.18 K
41 •
At what temperature do the Fahrenheit and Celsius temperature scales
give the same reading?
Picture the Problem We can find the temperature at which the Fahrenheit and
Celsius scales give the same reading by setting tF = tC in the temperatureconversion equation.
Set tF = tC in tC =
5
9
(tF − 32°) :
Solve for and evaluate tF:
tF =
5
9
(tF − 32°)
t C = t F = − 40.0°C = − 40.0°F
Remarks: If you’ve not already thought of doing so, you might use your
graphing calculator to plot tC versus tF and tF = tC (a straight line at 45°) on
the same graph. Their intersection is at (−40, −40).
42 •
Sodium melts at 371 K. What is the melting point of sodium on the
Celsius and Fahrenheit temperature scales?
1734
Chapter 17
Picture the Problem We can use the Celsius-to-absolute conversion equation to
find 371 K on the Celsius scale and the Celsius-to-Fahrenheit conversion equation
to find the Fahrenheit temperature corresponding to 371 K.
Express the absolute temperature as a
function of the Celsius temperature:
T = tC + 273.15 K
Solve for and evaluate tC:
t C = T − 273.15 K
= 371 K − 273.15 K = 98°C
Use the Celsius-to-Fahrenheit
conversion equation to find tF:
tF = 95 tC + 32° =
9
5
(97.9°) + 32°
= 208°F
43 •
The boiling point of oxygen at 1.00 atm is 90.2 K. What is the boiling
point of oxygen at 1.00 atm on the Celsius and Fahrenheit scales?
Picture the Problem We can use the Celsius-to-absolute conversion equation to
find 90.2 K on the Celsius scale and the Celsius-to-Fahrenheit conversion equation
to find the Fahrenheit temperature corresponding to 90.2 K.
Express the absolute temperature as a
function of the Celsius temperature:
T = tC + 273.15 K
Solve for and evaluate tC:
t C = T − 273.15 K
= 90.2 K − 273.15 K = − 183°C
Use the Celsius-to-Fahrenheit
conversion equation to find tF:
tF = 95 tC + 32° =
9
5
(− 183°) + 32°
= − 297°F
44 •• On the Réaumur temperature scale, the melting point of ice is 0ºR and
the boiling point of water is 80ºR. Derive expressions for converting temperatures
on the Réaumur scale to the Celsius and Fahrenheit scales.
Picture the Problem We can use the following diagram to set up proportions that
will allow us to convert temperatures on the Réaumur scale to Celsius and
Fahrenheit temperatures.
Temperature and the Kinetic Theory of Gases 1735
100
tC
0
R
F
C
212
80
tF
tR
32
0
Referring to the diagram, set up a
proportion to convert temperatures
on the Réaumur scale to Celsius
temperatures:
tC − 0°C
t − 0°R
= R
100°C − 0°C 80°R − 0°R
Simplify to obtain:
tC
t
= R ⇒ tC =
100 80
Referring to the diagram, set up a
proportion to convert temperatures
on the Réaumur scale to Fahrenheit
temperatures:
t F − 32°F
t − 0°R
= R
212°F − 32°F 80°R − 0°R
Simplify to obtain:
t F − 32 t R
=
⇒ tF =
180
80
5
4 R
t
t + 32
9
4 R
45
••• [SSM] A thermistor is a solid-state device widely used in a variety of
engineering applications. Its primary characteristic is that its electrical resistance
varies greatly with temperature. Its temperature dependence is given
approximately by R = R0eB/T, where R is in ohms (Ω), T is in kelvins, and R0 and B
are constants that can be determined by measuring R at calibration points such as
the ice point and the steam point. (a) If R = 7360 Ω at the ice point and 153 Ω at
the steam point, find R0 and B. (b) What is the resistance of the thermistor at
t = 98.6 ºF? (c) What is the rate of change of the resistance with temperature
(dR/dT) at the ice point and the steam point? (d) At which temperature is the
thermistor most sensitive?
Picture the Problem We can use the temperature dependence of the resistance of
the thermistor and the given data to determine R0 and B. Once we know these
quantities, we can use the temperature-dependence equation to find the resistance
at any temperature in the calibration range. Differentiation of R with respect to T
will allow us to express the rate of change of resistance with temperature at both
the ice point and the steam point temperatures.
1736
Chapter 17
(a) Express the resistance at the ice
point as a function of temperature of
the ice point:
7360 Ω = R0e B 273 K
(1)
Express the resistance at the steam
point as a function of temperature of
the steam point:
153 Ω = R0e B 373 K
(2)
Divide equation (1) by equation (2)
to obtain:
7360 Ω
= 48.10 = e B 273 K − B 373 K
153 Ω
Solve for B by taking the logarithm
of both sides of the equation:
1 ⎞ −1
⎛ 1
−
ln 48.1 = B⎜
⎟K
⎝ 273 373 ⎠
and
ln 48.1
= 3.944 × 10 3 K
B=
1 ⎞ −1
⎛ 1
−
⎜
⎟K
⎝ 273 373 ⎠
= 3.94 × 10 3 K
Solve equation (1) for R0 and
substitute for B:
R0 =
7360 Ω
= (7360 Ω )e − B 273 K
B 273 K
e
= (7360 Ω )e −3.944×10
3
K 273 K
= 3.913 ×10 −3 Ω
= 3.91×10 −3 Ω
(
)
3
(b) From (a) we have:
R = 3.913 × 10 −3 Ω e 3.944×10
Convert 98.6 °F to kelvins to obtain:
T = 310 K
Substitute for T to obtain:
R(310 K ) = 3.913 × 10 −3 Ω e 3.944×10
(
K T
)
3
K 310 K
= 1.31 kΩ
(c) Differentiate R with respect to T
to obtain:
dR
d
d ⎛B⎞
R0e B T ) = R0e B T
=
(
⎜ ⎟
dT dT
dT ⎝ T ⎠
RB
−B
= 2 R0e B T = − 2
T
T
Temperature and the Kinetic Theory of Gases 1737
Evaluate dR/dT at the ice point:
(
(7360 Ω ) 3.943 ×103 K
⎛ dR ⎞
=
−
⎜
⎟
(273.16 K )2
⎝ dT ⎠ice point
)
= − 389 Ω / K
Evaluate dR/dT at the steam point:
(
(153 Ω ) 3.943 ×103 K
⎛ dR ⎞
=
−
⎜
⎟
(373.16 K )2
⎝ dT ⎠steam point
)
= − 4.33 Ω / K
(d) The thermistor is more sensitive (has greater sensitivity) at lower
temperatures.
The Ideal-Gas Law
46 •
An ideal gas in a cylinder fitted with a piston (Figure 17-20) is held at
fixed pressure. If the temperature of the gas increases from 50° to 100°C, by what
factor does the volume change?
Picture the Problem Let the subscript 50 refer to the gas at 50°C and the
subscript 100 to the gas at 100°C. We can apply the ideal-gas law for a fixed
amount of gas to find the ratio of the final and initial volumes.
Apply the ideal-gas law for a fixed
amount of gas:
P100V100 P50V50
=
T100
T50
or, because P100 = P50,
V100 T100
=
V50
T50
Substitute numerical values and
evaluate V100/V50:
V100 (273.15 + 100 )K
=
= 1.15
(273.15 + 50)K
V50
or a 15% increase in volume.
47 •
[SSM] A 10.0-L vessel contains gas at a temperature of 0.00ºC and a
pressure of 4.00 atm. How many moles of gas are in the vessel? How many
molecules?
Picture the Problem We can use the ideal-gas law to find the number of moles of
gas in the vessel and the definition of Avogadro’s number to find the number of
molecules.
1738
Chapter 17
Apply the ideal-gas law to the gas:
PV = nRT ⇒ n =
Substitute numerical values and
evaluate n:
n=
PV
RT
(4.00 atm )(10.0 L )
(8.206 ×10
−2
)
L ⋅ atm/mol ⋅ K (273 K )
= 1.786 mol = 1.79 mol
N = nN A
Relate the number of molecules N in
the gas in terms of the number of
moles n:
Substitute numerical values and evaluate N:
(
)
N = (1.786 mol) 6.022 × 10 23 molecules/mol = 1.08 × 10 24 molecules
48 ••
A pressure as low as 1.00 × 10–8 torr can be achieved using an oil
diffusion pump. How many molecules are there in 1.00 cm3 of a gas at this
pressure if its temperature is 300 K?
Picture the Problem We can use the ideal-gas law to relate the number of
molecules in the gas to its pressure, volume, and temperature.
Solve the ideal-gas law for the
number of molecules in a gas as a
function of its pressure, volume, and
temperature:
N=
PV
kT
Substitute numerical values and evaluate N:
N=
(1.00 ×10
−8
)
(
)
torr (133.32 Pa/torr ) 1.00 × 10 −6 m 3
= 3.22 × 108
1.381× 10−23 J/K (300 K )
(
)
You copy the following paragraph from a Martian physics textbook:
49 ••
″1 snorf of an ideal gas occupies a volume of 1.35 zaks. At a temperature of
22 glips, the gas has a pressure of 12.5 klads. At a temperature of –10 glips, the
same gas now has a pressure of 8.7 klads.″ Determine the temperature of absolute
zero in glips.
Picture the Problem Because the gas is ideal, its pressure is directly proportional
to its temperature. Hence, a graph of P versus T will be linear and the linear
equation relating P and T can be solved for the temperature corresponding to zero
pressure. We’ll assume that the data was taken at constant volume.
Temperature and the Kinetic Theory of Gases 1739
A graph of P, in klads, as function of T,
in glips, is shown to the right. The
equation of this graph is:
P, klads
12.5
⎛ 3.8 klads ⎞
⎟⎟T + 9.9 klads
P = ⎜⎜
⎝ 32 glips ⎠
8.7
T0
−10
0
22
When P = 0:
⎛ 3.8 klads ⎞
⎟⎟T0 + 9.9 klads
0 = ⎜⎜
⎝ 32 glips ⎠
Solve for T0 to obtain:
T0 = − 83 glips
T , glips
50 ••
A motorist inflates the tires of her car to a gauge pressure of 180 kPa
on a day when the temperature is –8.0ºC. When she arrives at her destination, the
tire pressure has increased to 245 kPa. What is the temperature of the tires if we
assume that (a) the tires do not expand or (b) that the tires expand so the volume
of the enclosed air increases by 7 percent?
Picture the Problem Let the subscript 1 refer to the tires when their gauge
pressure is 180 kPa and the subscript 2 to conditions when their gauge pressure is
245 kPa. Assume that the air in the tires behaves as an ideal gas. Then, we can
apply the ideal-gas law for a fixed amount of gas to relate the temperatures to the
pressures and volumes of the tires.
(a) Apply the ideal-gas law for a
fixed amount of gas to the air in the
tires:
P2V2 P1V1
=
T2
T1
(1)
where the temperatures and pressures
are absolute.
Solve for T2:
T2 = T1
Substitute numerical values to
obtain:
T2 = (265 K )
P2
because V1 = V2.
P1
245 kPa + 101 kPa
180 kPa + 101 kPa
= 326.3 K = 53°C
1740
Chapter 17
(b) Use equation (1) with
V2 = 1.07 V1. Solve for T2:
Substitute numerical values and
evaluate T2:
T2 =
⎛P ⎞
P2V2
T1 = (1.07 ) ⎜⎜ 2 T1 ⎟⎟
P1V1
⎝ P1 ⎠
T2 = (1.07 )(326.3 K ) = 349.1 K
= 76°C
51 ••
A room is 6.0 m by 5.0 m by 3.0 m. (a) If the air pressure in the room
is 1.0 atm and the temperature is 300 K, find the number of moles of air in the
room. (b) If the temperature increases by 5.0 K and the pressure remains constant,
how many moles of air leave the room?
Picture the Problem We can apply the ideal-gas law to find the number of moles
of air in the room as a function of the temperature.
(a) Use the ideal-gas law to relate the
number of moles of air in the room
to the pressure, volume, and
temperature of the air:
n=
PV
RT
Substitute numerical values and
evaluate n:
n=
(101.325 kPa )(6.0 m )(5.0 m )(3.0 m )
(8.314 J/mol ⋅ K )(300 K )
(1)
= 3.66 ×10 3 mol = 3.7 ×10 3 mol
(b) Letting n′ represent the number
of moles in the room when the
temperature rises by 5 K, express the
number of moles of air that leave the
room:
Δn = n − n'
Apply the ideal-gas law to obtain:
n' =
Divide equation (2) by equation (1)
to obtain:
n' T
T
and n' = n
=
n T'
T'
Substitute for n′ to obtain:
PV
RT '
Δn = n − n
(2)
T
⎛ T⎞
= n⎜1 − ⎟
T'
⎝ T' ⎠
Temperature and the Kinetic Theory of Gases 1741
Substitute numerical values and
evaluate Δn:
⎛ 300 K ⎞
⎟⎟
Δn = 3.66 × 103 mol ⎜⎜1 −
⎝ 305 K ⎠
(
)
= 60 mol
52 •• Image that 10.0 g of liquid helium, initially at 4.20 K, evaporate into an
empty balloon that is kept at 1.00 atm pressure. What is the volume of the
balloon at (a) 25.0 K and (b) 293 K?
Picture the Problem Let the subscript 1 refer to helium gas at 4.2 K and the
subscript 2 to the gas at 293 K. We can apply the ideal-gas law to find the volume
of the gas at 4.2 K and a fixed amount of gas to find its volume at 293 K. See
Appendix C for the molar mass of helium.
(a) Apply the ideal-gas law to the
helium gas to express its volume:
m
RT
nRT1 M 1 mRT1
V1 =
=
=
P1
P1
MP1
Substitute numerical values and evaluate V1:
V1 =
(10.0 g )(0.08206 L ⋅ atm/mol ⋅ K )(25.0 K ) = 5.125 L =
(4.003 g/mol)(1.00 atm)
(b) Apply the ideal-gas law for a
fixed amount of gas and solve for the
volume of the helium gas at
293 K:
Substitute numerical values and
evaluate V2:
5.13 L
P2V2 P1V1
=
T2
T1
and, because P1 = P2,
T
V2 = 2 V1
T1
V2 =
293 K
(5.125 L ) = 60.1L
25.0 K
53 •• A closed container with a volume of 6.00 L holds 10.0 g of liquid
helium at 25.0 K and enough air to fill the rest of its volume at a pressure of 1.00
atm. The helium then evaporates and the container warms to room temperature
(293 K). What is the final pressure inside the container?
Picture the Problem We can apply the law of partial pressures to find the final
pressure inside the container. See Appendix C for the molar mass of helium and of
air.
1742
Chapter 17
The final pressure inside the
container is the sum of the partial
pressures of helium gas and air:
Pfinal = PHe gas + Pair
The pressure exerted by the air
molecules at room temperature is
given by the ideal-gas law:
Pair =
=
(1)
nair RT mair RT ρ airVRT
=
=
V
M airV
M airV
ρ air RT
M air
The pressure exerted by the helium
molecules at room temperature is
also given by the ideal-gas law:
PHe gas =
Substituting in equation (1) yields:
Pfinal =
nHe RT mHe RT
=
V
M HeV
mHe RT ρ air RT
+
M HeV
M air
⎛ m
ρ
= ⎜⎜ He + air
⎝ M HeV M air
⎞
⎟⎟ RT
⎠
Substitute numerical values and evaluate P2:
Pfinal
⎛
kg ⎞⎟
⎜
1.293 3
10.0 g
⎜
m ⎟ ⎛⎜ 8.314 J ⎞⎟ (293 K )
=⎜
+
3
−3
g ⎟⎝
mol ⋅ K ⎠
⎛
⎞
⎜ ⎛⎜ 4.003 g ⎞⎟ ⎜ 6.00 L × 10 m ⎟ 28.81
⎟
⎜⎝
mol ⎟
mol ⎠ ⎜⎝
L ⎟⎠
⎝
⎠
1 atm
= 1.124 × 106 Pa ×
= 11.1 atm
101.325 kPa
54 •• An automobile tire is filled to a gauge pressure of 200 kPa when its
temperature is 20ºC. (Gauge pressure is the difference between the actual pressure
and atmospheric pressure.) After the car has been driven at high speeds, the tire
temperature increases to 50ºC. (a) Assuming that the volume of the tire does not
change and that air behaves as an ideal gas, find the gauge pressure of the air in
the tire. (b) Calculate the gauge pressure if the tire expands so the volume of the
enclosed air increases by 10 percent.
Picture the Problem Let the subscript 1 refer to the tire when its temperature is
20°C and the subscript 2 to conditions when its temperature is 50°C. We can
apply the ideal-gas law for a fixed amount of gas to relate the temperatures to the
pressures of the air in the tire.
Temperature and the Kinetic Theory of Gases 1743
(a) Apply the ideal-gas law for a
fixed amount of gas and solve for
pressure at the higher
temperature:
P2V2 P1V1
=
T2
T1
(1)
and
P2 =
T2
P1
T1
because V1 = V2
Substitute numerical values to
obtain:
323 K
(200 kPa + 101 kPa )
293 K
= 332 kPa
P2 =
and
P2,gauge = 332 kPa − 101 kPa = 231 kPa
(b) Solve equation (1) for P2:
P2 =
V1T2
P1
V2T1
Because V2 = 1.10 V1:
P2 =
T2
V1T2
P1 =
P1
1.10V1T1
1.10 T1
Substitute numerical values and
evaluate P2 and P2,gauge:
P2 =
323 K
(200 kPa + 101kPa )
(1.10)(293 K )
= 302 kPa
and
P2,gauge = 302 kPa − 101 kPa = 201 kPa
55 ••
[SSM] After nitrogen (N2) and oxygen (O2), the most abundant
molecule in Earth's atmosphere is water, H2O. However, the fraction of H2O
molecules in a given volume of air varies dramatically, from practically zero
percent under the driest conditions to as high as 4 percent where it is very humid.
(a) At a given temperature and pressure, would air be denser when its water vapor
content is large or small? (b) What is the difference in mass, at room temperature
and atmospheric pressure, between a cubic meter of air with no water vapor
molecules, and a cubic meter of air in which 4 percent of the molecules are water
vapor molecules?
Picture the Problem (a) At a given temperature and pressure, the ideal-gas law
tells us that the total number of molecules per unit volume, N/V, is constant. The
denser gas will, therefore, be the one in which the average mass per molecule is
greater. (b) We can apply the ideal-gas law and use the relationship between the
masses of the dry and humid air and their molar masses to find the difference in
1744
Chapter 17
mass, at room temperature and atmospheric pressure, between a cubic meter of air
containing no water vapor, and a cubic meter of air containing 4% water vapor.
See Appendix C for the molar masses of N2, O2, and H2O.
(a) The molecular mass of N2 is 28.014 amu (see Appendix C), that of O2 is
31.999 amu, and that of H2O is 18.0152 amu. Because H2O molecules are lighter
than the predominant molecules in air, a given volume of air will be less when its
water vapor content is lower.
(b) Express the difference in mass
between a cubic meter of air
containing no water vapor, and a
cubic meter of air containing 4%
water vapor:
Δm = mdry − mhumid
The masses of the dry air and humid
air are related to their molar masses
according to:
mdry = 0.04nM dry
and
m humid = 0.04nM humid
Substitute for mdry and mhumid and
simplify to obtain:
Δm = 0.04nM dry − 0.04nM humid
From the ideal-gas law we have:
PatmV
RT
and so Δm can be written as
0.04 PatmV
(M dry − M humid )
Δm =
RT
= 0.04n(M dry − M humid )
n=
Substitute numerical values (28.811 g/mol is an 80% nitrogen and 20% oxygen
weighted average) and evaluate Δm:
(
0.04)(101.325 kPa ) (1.00 m 3 )
(28.811 g/mol − 18.0152 g/mol) ≈
Δm =
(8.314 J/mol ⋅ K )(300 K )
18 g
56 ••
A scuba diver is 40 m below the surface of a lake, where the
temperature is 5.0ºC. He releases an air bubble that has a volume of 15 cm3. The
bubble rises to the surface, where the temperature is 25ºC. Assume that the air in
the bubble is always in thermal equilibrium with the surrounding water, and
assume that there is no exchange of molecules between the bubble and the
surrounding water. What is the volume of the bubble right before it breaks the
surface? Hint: Remember that the pressure also changes.
Temperature and the Kinetic Theory of Gases 1745
Picture the Problem Let the subscript 1 refer to the conditions at the bottom of
the lake and the subscript 2 to the surface of the lake and apply the ideal-gas law
for a fixed amount of gas.
Apply the ideal-gas law for a fixed
amount of gas:
P2V2 P1V1
VT P
=
⇒ V2 = 1 2 1
T1 P2
T2
T1
The pressure at the bottom of the
lake is the sum of the pressure at its
surface (atmospheric) and the
pressure due to the depth of the lake:
P1 = Patm + ρgh
Substituting for P1 yields:
V2 =
V1T2 (Patm + ρgh )
T1 P2
Substitute numerical values and evaluate V2:
V2 =
(15 cm )(298 K )[101.325 kPa + (1.00 ×10
3
)(
]
)
kg/m 3 9.81 m/s 2 (40 m )
(278 K )(101.325 kPa )
3
= 78 cm3
57 ••
[SSM] A hot-air balloon is open at the bottom. The balloon has a
volume of 446 m3 is filled with air with an average temperature of 100°C. The air
outside the balloon has a temperature of 20.0°C and a pressure of 1.00 atm. How
large a payload (including the envelope of the balloon itself) can the balloon lift?
Use 29.0 g/mol for the molar mass of air. (Neglect the volume of both the
payload and the envelope of the balloon.)
Picture the Problem Assume that the volume of the balloon is not changing.
Then the air inside and outside the balloon must be at the same pressure of about
1.00 atm. The contents of the balloon are the air molecules inside it. We can use
Archimedes principle to express the buoyant force on the balloon and we can find
the weight of the air molecules inside the balloon. You’ll need to determine the
molar mass of air. See Appendix C for the molar masses of oxygen and nitrogen.
Express the net force on the balloon
and its contents:
Fnet = B − wair inside the balloon
Using Archimedes principle, express
the buoyant force on the balloon:
B = wdisplaced fluid = mdisplaced fluid g
(1)
or
B = ρoVballoon g
where ρo is the density of the air
outside the balloon.
1746
Chapter 17
Express the weight of the air inside
the balloon:
wair inside the balloon = ρiVballoon g
where ρi is the density of the air inside
the balloon.
Substitute in equation (1) for B and
wair inside the balloon to obtain:
Fnet = ρ oVballoon g − ρiVballoon g
Express the densities of the air
molecules in terms of their number
densities, molecular mass, and
Avogadro’s number:
ρ=
Using the ideal-gas law, relate
the number density of air N/V to
its temperature and pressure:
Substitute for
(2)
= (ρ o − ρ i )Vballoon g
M ⎛N⎞
⎜ ⎟
NA ⎝ V ⎠
PV = NkT and
N
to obtain:
V
ρ=
Substitute in equation (2) and
simplify to obtain:
N
P
=
V kT
M ⎛ P ⎞
⎜ ⎟
N A ⎝ kT ⎠
MP ⎛ 1 1 ⎞
⎜ − ⎟Vballoon g
N A k ⎜⎝ To Ti ⎟⎠
MP ⎛ 1 1 ⎞ 1
⎜⎜ − ⎟⎟ 6 π d 3 g
=
N A k ⎝ To Ti ⎠
Fnet =
(
)
Assuming that the average molar mass of air is 28.81 g/mol, substitute numerical
values and evaluate Fnet:
[
⎛
Fnet =
](
1
1 ⎞ 1
⎟⎟ 6 π (15.0 m )3 9.81 m/s 2
−
⎝ 297 K 348 K ⎠
23
6.022 × 10 particles/mol 1.381 × 10 −23 J/K
(28.81g/mol)(101.325 kPa )⎜⎜
(
)(
)
)
= 3.0 kN
58 ••• A helium balloon is used to lift a load of 110 N. The weight of the
envelope of the balloon is 50.0 N and the volume of the helium when the balloon
is fully inflated is 32.0 m3. The temperature of the air is 0ºC and the atmospheric
pressure is 1.00 atm. The balloon is inflated with a sufficient amount of helium
gas that the net upward force on the balloon and its load is 30.0 N. Neglect any
effects due to the changes of temperature as the altitude changes. (a) How many
moles of helium gas are contained in the balloon? (b) At what altitude will the
balloon be fully inflated? (c) Does the balloon ever reach the altitude at which it is
fully inflated? (d) If the answer to (c) is ″Yes,″ what is the maximum altitude
attained by the balloon?
Temperature and the Kinetic Theory of Gases 1747
Picture the Problem (a) We can find the number of moles of helium gas in the
balloon by applying the ideal-gas law to relate n to the pressure, volume, and
temperature of the helium and Archimedes principle to find the volume of the
helium. In Part (b), we can apply the result of Problem 13-91 to relate
atmospheric pressure to altitude and use the ideal-gas law to determine the
pressure of the gas when the balloon is fully inflated. In Part (c), we’ll find the net
force acting on the balloon at the altitude at which it is fully inflated in order to
decide whether it can rise to that altitude.
(a) Apply the ideal-gas law to the
helium in the balloon and solve
for n:
n=
Relate the net force on the balloon to
its weight:
FB − wskin − wload − wHe = 30 N
Use Archimedes principle to express
the buoyant force on the balloon in
terms of the volume of the balloon:
FB = wdisplaced air
Substitute to obtain:
ρ airVg − wskin − wload − ρ HeVg = 30.0 N
Solve for the volume of the helium:
V =
30.0 N + wskin + wload
(ρ air − ρ He )g
Substituting for V in equation (1)
yields:
n=
P(30.0 N + wskin + wload )
RT (ρ air − ρ He )g
PV
RT
(1)
= ρ airVg
Substitute numerical values and evaluate n:
(1.00 atm )(30.0 N + 50.0 N + 110 N )⎛⎜⎜
1L ⎞
⎟
10 −3 m 3 ⎟⎠
⎝
n=
8.206 × 10−2 L ⋅ atm/mol ⋅ K (273 K ) 1.293 kg/m 3 − 0.179 kg/m 3 9.81 m/s 2
(
)
(
= 776 mol
(b) Using the result of Problem
13-91, express the variation in
atmospheric pressure with
altitude:
P(h ) = P0 e −Ch
where C = 0.13 km −1 .
)(
)
1748
Chapter 17
h=
1 ⎡ P0 ⎤
ln
C ⎢⎣ P(h ) ⎥⎦
Using the ideal-gas law, express P:
P=
nRT
V
Substitute for P in equation (2) and
simplify to obtain:
⎡
1 ⎢ P
h = ln ⎢ 0
C ⎢ nRT
⎣ V
Solving for h yields:
(2)
⎤
⎥ 1 ⎡ P0V ⎤
⎥ = ln ⎢
⎥
⎥ C ⎣ nRT ⎦
⎦
Substitute numerical values and evaluate h:
⎡
⎤
⎛
⎞
(1.00 atm )⎜⎜ 32.0 m 3 × 1−3L 3 ⎟⎟
⎢
⎥
10 m ⎠
1
⎛
⎞ ⎢
⎝
⎥
h=⎜
ln
−1 ⎟
−2
⎢
⎥
(
)
(
)
×
⋅
⋅
0
.
13
km
776
mol
8.206
10
L
atm/mol
K
273
K
⎝
⎠
⎢
⎥
⎢⎣
⎥⎦
(
)
= 4.84 km = 4.8 km
(c) Express the condition that must
be satisfied if the balloon is to reach
its fully inflated altitude:
Fnet = FB − wtot ≥ 0
The total weight is the sum of the
weights of the load, balloon skin, and
helium:
w tot = wload + wskin + wHe
or, because wHe = ρ HeVg ,
w tot = w load + w skin + ρ HeVg
Express the buoyant force on the
balloon at h = 4.84 km:
FB = ρ air, hVg
Express the dependence of the
density of the air on atmospheric
pressure:
P
P ρ air, h
=
⇒ ρair, h = ρ air (5)
P0
ρ air
P0
Substitute for ρair,h in equation (4) to
obtain:
FB =
P
ρ airVg
P0
(3)
(4)
Temperature and the Kinetic Theory of Gases 1749
Substitute for wtot and FB in equation
(3) and simplify to obtain:
Fnet =
P
Vg (ρ air − ρ He ) − wload − wskin
P0
or, because P(h ) = P0 e −Ch ,
Fnet = e −ChVg (ρ air − ρ He ) − wload − wskin
Substitute numerical values and evaluate Fnet:
−1
Fnet = e − (0.13 km )(4.84 km ) (32.0 m 3 )(9.81 m/s 2 )(1.293 kg/m 3 − 0.179 kg/m 3 )
− 110 N − 50 N = 30 N
Because Fnet = 30 N > 0, the balloon will rise higher than the altitude at which it
is fully inflated.
(d) The balloon will rise until the net force acting on it is zero. Because the
buoyant force depends on the density of the air, the balloon will rise until the
density of the air has decreased sufficiently for the buoyant force to just equal the
total weight of the balloon.
ρ
1
ln air
C ρ air, h
Substitute equation (5) in equation
(2) to obtain:
h=
Using equation (4), express the
density of the air in terms of FB:
ρ air, h =
Substituting for ρ air, h and
⎡
⎤
⎢
ρ ⎥ 1 ⎡Vgρ air ⎤
1
h = ln ⎢ air ⎥ = ln ⎢
⎥
C ⎢ FB ⎥ C ⎣ FB ⎦
⎢⎣ Vg ⎥⎦
simplifying yields:
FB
Vg
Substitute numerical values and evaluate h:
(
)(
)(
)
3
2
3
1
⎛
⎞ ⎡ 32.0 m 9.81 m/s 1.293 kg/m ⎤
h=⎜
ln
⎢
⎥ = 6.0 km
−1 ⎟
190.5 N
⎝ 0.13 km ⎠ ⎣
⎦
Kinetic Theory of Gases
59 •
[SSM] (a) One mole of argon gas is confined to a 1.0-liter container
at a pressure of 10 atm. What is the rms speed of the argon atoms? (b) Compare
your answer to the rms speed for helium atoms under the same conditions.
1750
Chapter 17
Picture the Problem We can express the rms speeds of argon and helium atoms
by combining PV = nRT and vrms = 3RT M to obtain an expression for vrms in
terms of P, V, and M. See Appendix C for the molar masses of argon and helium.
Express the rms speed of an atom as
a function of the temperature and its
molar mass:
From the ideal-gas law we have:
RT =
Substitute for RT to obtain:
3RT
M
vrms =
PV
n
3PV
nM
vrms =
(a) Substitute numerical values and evaluate vrms for argon atoms:
v rms, Ar =
(
)
3(10 atm )(101.325 kPa/atm ) 1.0 × 10 −3 m 3
= 0.28 km/s
(1mol) 39.948 ×10−3 kg/mol
(
)
(b) Substitute numerical values and evaluate vrms for helium atoms:
v rms, He =
(
)
3(10 atm )(101.325 kPa/atm ) 1.0 × 10−3 m 3
= 0.87 km/s
(1mol) 4.003 ×10−3 kg/mol
(
)
The rms speed of argon atoms is slightly less than one third the rms speed of
helium atoms.
60 •
Find the total translational kinetic energy of the molecules of 1.0 L of
oxygen gas at a temperature of 0.0ºC and a pressure of 1.0 atm.
Picture the Problem We can express the total translational kinetic energy of the
oxygen gas by combining K = 32 nRT and the ideal-gas law to obtain an expression
for K in terms of the pressure and volume of the gas.
Relate the total translational kinetic
energy of translation to the
temperature of the gas:
K = 32 nRT
Using the ideal-gas law, substitute
for nRT to obtain:
K = 32 PV
Temperature and the Kinetic Theory of Gases 1751
Substitute numerical values and
evaluate K:
K=
3
2
⎛
(101.325 kPa )⎜⎜1.0 L × 10
⎝
m3 ⎞
⎟
L ⎟⎠
−3
= 0.15 kJ
61 •
Estimate the rms speed and the average kinetic energy of a hydrogen
atom in a gas at a temperature of 1.0 × 107 K. (At this temperature, which is
approximately the temperature in the interior of a star, hydrogen atoms are
ionized and become protons.)
Picture the Problem Because we’re given the temperature of the hydrogen atom
and know its molar mass, we can find its rms speed using vrms = 3RT M and its
average kinetic energy from K av = 32 kT . See Appendix C for the molar mass of
hydrogen.
Relate the rms speed of a hydrogen
atom to its temperature and molar
mass:
Substitute numerical values and
evaluate vrms:
vrms =
3RT
MH
v rms =
3(8.314 J/mol ⋅ K ) 1.0 × 107 K
1.0079 × 10−3 kg/mol
(
)
)(
)
= 5.0 × 105 m/s
Express the average kinetic energy of
the hydrogen atom as a function of
its temperature:
K av = 32 kT
Substitute numerical values and
evaluate Kav:
K av = 32 1.381×10 −23 J/K 1.0 × 10 7 K
(
= 2.1×10 −16 J
62 •
Liquid helium has a temperature of only 4.20 K and is in equilibrium
with its vapor at atmospheric pressure. Calculate the rms speed of a helium atom
in the vapor at this temperature, and comment on the result.
Picture the Problem The rms speed of helium atoms is given by vrms =
See Appendix C for the molar mass of helium.
3RT
.
M He
1752
Chapter 17
The rms speed of helium atoms is
given by:
v rms =
3 RT
M He
Substitute numerical values and
evaluate vrms:
v rms =
3(8.314 J/mol ⋅ K )(4.20 K )
4.003 g/mol
= 162 m/s
Thermal speeds, even at temperatures as low as 4.20 K, are very large compared
to most of the speeds we experience directly.
63 •
Show that the mean free path for a molecule in an ideal gas at
temperature T and pressure P is given by λ = kT 2 Pπ d 2 .
Picture the Problem We can combine λ =
1
and PV = nRT to express
2nvπ d 2
the mean free path for a molecule in an ideal gas in terms of the pressure and
temperature.
Express the mean free path of a
molecule in an ideal gas:
λ=
1
2nvπ d 2
(1)
where
N nN A
nv = =
V
V
Solve the ideal-gas law for the
volume of the gas:
V =
nRT
P
Substitute for V in the expression for
nv to obtain:
nv =
nN A
P
P=
nRT
kT
Substitute for nv in equation (1) to
obtain:
λ=
kT
2 Pπ d 2
64 ••
State-of-the-art vacuum equipment can attain pressures as low as
–11
7.0 × 10 Pa. Suppose that a chamber contains helium at this pressure and at
room temperature (300 K). Estimate the mean free path and the collision time for
helium in the chamber. Assume the diameter of a helium atom is 1.0 × 10–10 m.
Picture the Problem We can find the collision time from the mean free path and
the average (rms) speed of the helium molecules. We can use the result of Problem
Temperature and the Kinetic Theory of Gases 1753
63 to find the mean free path of the molecules and vrms = 3RT M to find the
average speed of the molecules. See Appendix C for the molar mass of helium.
Express the collision time in terms of
the mean free path for and the
average speed of a helium molecule:
τ=
From Problem 63, the mean free
path of the gas is given by:
λ=
Substitute numerical values and
evaluate the mean free path λ:
λ
vav
λ=
≈
λ
(1)
vrms
kT
2 Pπ d 2
(
(1.381×10
2 7.0 ×10
−11
−23
)
J/K (300 K )
) (
Pa π 1.0 ×10 −10 m
)
2
= 1.332 ×10 9 m = 1.3 ×10 9 m
Express the rms speed of the helium
molecules:
vrms =
Substitute for vrms in equation (1)
and simplify to obtain:
τ=
3RT
M He
λ
3RT
M He
=λ
M He
3RT
Substitute numerical values and evaluate τ :
τ = (1.332 × 109 m )
4.007 g/mol
= 9.7 × 10 5 s
3(8.314 J/mol ⋅ K )(300 K )
65 ••
[SSM] Oxygen (O2) is confined to a cube-shaped container 15 cm on
an edge at a temperature of 300 K. Compare the average kinetic energy of a
molecule of the gas to the change in its gravitational potential energy if it falls
15 cm (the height of the container).
Picture the Problem We can use K = 32 kT and ΔU = mgh = Mgh N A to express
the ratio of the average kinetic energy of a molecule of the gas to the change in its
gravitational potential energy if it falls from the top of the container to the bottom.
See Appendix C for the molar mass of oxygen.
Express the average kinetic energy of
a molecule of the gas as a function of
its temperature:
K av = 32 kT
1754
Chapter 17
Letting h represent the height of the
container, express the change in the
potential energy of a molecule as it
falls from the top of the container to
the bottom:
ΔU = mgh =
Express the ratio of Kav to ΔU and
simplify to obtain:
M O 2 gh
NA
3
kT
K av
3N A kT
= 2
=
ΔU M O 2 gh 2M O 2 gh
NA
Substitute numerical values and evaluate Kav/ΔU:
(
)(
)(
)
K av 3 6.022 ×10 23 particles/mol 1.381×10 −23 J/K (300 K )
=
= 7.9 ×10 4
2
−3
ΔU
2 32.0 × 10 kg/mol 9.81 m/s (0.15 m )
(
)
*The Distribution of Molecular Speeds
66 ••
Use calculus to show that f(v), given by Equation 17-36, has its
maximum value at a speed v = 2 kT / m .
Picture the Problem Equation 17-36 gives the Maxwell-Boltzmann speed
distribution. Setting its derivative with respect to v equal to zero will tell us where
the function’s extreme values lie.
Differentiate Equation 17-36with
respect to v:
32
df
d ⎡ 4 ⎛ m ⎞ 2 − mv 2
=
⎜
⎟ ve
⎢
dv dv ⎢⎣ π ⎝ 2kT ⎠
4 ⎛ m ⎞
=
⎜
⎟
π ⎝ 2kT ⎠
Set df/dv = 0 for extrema and solve
for v:
2v −
32
2 kT
⎤
⎥
⎦⎥
⎛
mv3 ⎞ − mv 2
⎜⎜ 2v −
⎟e
kT ⎟⎠
⎝
mv3
=0⇒ v=
kT
2 kT
2kT
m
Examination of the graph of f(v) makes it clear that this extreme value is, in fact, a
maximum. See Figure 17-17 and note that it is concave downward at v = 2kT m .
Remarks: An alternative to the examination of f(v) in order to conclude that
v = 2kT m maximizes the Maxwell-Boltzmann speed distribution function
is to show that d2f/dv2 < 0 at v = 2kT m .
Temperature and the Kinetic Theory of Gases 1755
67 ••
[SSM] The fractional distribution function f(v) is defined in Equation
17-36. Because f(v) dv gives the fraction of molecules that have speeds in the
range between v and v + dv, the integral of f(v) dv over all the possible ranges of
speeds must equal 1. Given that the integral
∫
∞
0
2
v 2 e− av dv =
π
4
a−3/ 2 , show that
∫ f (v )dv = 1 , where f(v) is given by Equation 17-36.
∞
0
Picture the Problem We can show that f(v) is normalized by using the given
integral to integrate it over all possible speeds.
Express the integral of Equation 1736:
∞
Let a = m 2kT to obtain:
∞
∫
0
∫
4 ⎛ m ⎞
f (v )dv =
⎜
⎟
π ⎝ 2kT ⎠
f (v )dv =
0
π
∞
∫v e
2 − mv 2 2 kT
dv
0
a 3 2 ∫ v 2e − av dv
2
0
⎛ π −3 2 ⎞
a 3 2 ⎜⎜
a ⎟⎟ = 1
π
0
⎝ 4
⎠
That is, f(v) is normalized.
Use the given integral to obtain:
∞
∫ f (v )dv =
∞ 3 − av 2
∫0 v
4
1
dv =
2 , calculate the average
2a
speed vav of molecules in a gas using the Maxwell–Boltzmann distribution
function.
68
••
Given that the integral
4
3 2∞
e
Picture the Problem In Problem 67 we showed that f(v) is normalized. Hence we
∞
can evaluate vav using ∫ vf (v )dv .
0
The average speed of the molecules
in the gas is given by:
∞
vav = ∫ vf (v )dv
0
=
Substitute a = m 2kT :
vav =
4 ⎛ m ⎞
⎜
⎟
π ⎝ 2kT ⎠
4
π
∞
a
32
3 2∞
∫v e
3 − mv 2 2 kT
0
3 − av
∫ v e dv
0
2
dv
1756
Chapter 17
Use the given integral to obtain:
⎛ a −2 ⎞
2 1
⎟⎟ =
a 3 2 ⎜⎜
π
π a
⎝ 2 ⎠
4
vav =
2
=
2kT
m
π
The translational kinetic energies of the molecules of a gas are
69 ••
distributed according to the Maxwell-Boltzmann energy distribution, Equation
17-38. (a) Determine the most probable value of the translational kinetic energy
(in terms of the temperature T) and compare this value to the average value.
(b) Sketch a graph of the translational kinetic energy distribution [f(E) versus E]
and label the most probable energy and the average energy. (Do not worry about
calibrating the vertical scale of the graph.) (c) Your teacher says, ″Just looking at
the graph f(E) versus E of allows you to see that the average translational kinetic
energy is considerably greater than the most probable translational kinetic
energy.″ What feature(s) of the graph support her claim?
Picture the Problem We can set the derivative of f(E) with respect to E equal to
zero in order to determine the most probable value of the kinetic energy of the gas
molecules.
(a) The Maxwell-Boltzmann energy
distribution function is:
⎛ 2 ⎞⎛ 1 ⎞
f (E ) = ⎜⎜
⎟⎟ ⎜ ⎟
⎝ π ⎠ ⎝ kT ⎠
32
E e − E kT
Differentiate this expression with respect to E to obtain:
d
⎛ 2 ⎞⎛ 1 ⎞
f (E ) = ⎜⎜
⎟⎟ ⎜ ⎟
dE
⎝ π ⎠ ⎝ kT ⎠
32
⎛ 1 −1 2 − E kT
⎞
⎛ −1 ⎞
⎜⎜ 2 E e
+ E 1 2 ⎜ ⎟e − E kt ⎟⎟
⎝ kT ⎠
⎝
⎠
Set this derivative equal to zero for
extrema values and simplify to obtain:
1
2
Solving for Epeak yields:
E peak =
The average value of the energy of
the gas molecules is given by:
Eav = 32 kT
Express the ratio of Epeak to Eav:
E peak
−1 2
Epeak
e
Eav
=
− E peak kT
1
2
1
2
3
2
1 2 ⎛ − 1 ⎞ − E peak
+ Epeak
⎜ ⎟e
⎝ kT ⎠
kt
=0
kT
kT 1
= 3 ⇒ E peak =
kT
1
3
Eav
(b) The following graph of the energy distribution was plotted using a spreadsheet
program. Note that kT was set equal to 1 and that, as predicted by our result in
(a), the peak value occurs for E = 0.5.
Temperature and the Kinetic Theory of Gases 1757
0.50
0.45
0.40
0.35
f (E )
0.30
0.25
0.20
0.15
0.10
0.05
0.00
0.0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0
E
(c) The graph rises from zero to the peak much more rapidly than it falls off to the
right of the peak. Because the distribution is so strongly skewed to the right of the
peak, the outlying molecules with relatively high energies pull the average
(3kT/2) far to the right of the most probable value (kT/2).
General Problems
70 •
Find the temperature at which the rms speed of a molecule of hydrogen
gas equals 343 m/s.
Picture the Problem We can use vrms = 3RT M to relate the temperature of the
H2 molecules to their rms speed. See Appendix C for the molar mass of hydrogen.
Relate the rms speed of the hydrogen
molecules to its temperature:
vrms =
Substitute numerical values and
evaluate T:
T=
2
M H 2 vrms
3RT
⇒T =
M H2
3R
(2.016 ×10
)
kg/mol (343 m/s )
3(8.314 J/mol ⋅ K )
−3
2
= 9.51K
71 •• (a) If 1.0 mol of a gas in a cylindrical container occupies a volume of
10 L at a pressure of 1.0 atm, what is the temperature of the gas in kelvins?
(b) The cylinder is fitted with a piston so that the volume of the gas (Figure 1720) can vary. When the gas is heated at constant pressure, it expands to a volume
of 20 L. What is the temperature of the gas in kelvins? (c) Next, the volume is
fixed at 20 L, and the gas’s temperature is increased to 350 K. What is the
pressure of the gas now?
1758
Chapter 17
Picture the Problem We can use the ideal-gas law to find the initial temperature
of the gas and the ideal-gas law for a fixed amount of gas to relate the volumes,
pressures, and temperatures resulting from the given processes.
(a) Apply the ideal-gas law to
express the temperature of the gas:
T=
PV
nR
Substitute numerical values and
evaluate T:
T=
(101.325 kPa ) (10 ×10 −3 m 3 )
(1.0 mol)(8.314 J/mol ⋅ K )
= 121.9 K = 1.2 × 10 2 K
(b) Use the ideal-gas law for a fixed
amount of gas to relate the
temperatures and volumes:
Solve for and evaluate T2:
P1V1 P2V2
=
T1
T2
or, because P1 = P2,
V1 V2
=
T1 T2
T2 =
V2
T1 = 2(121.9 K ) = 243.7 K
V1
= 2.4 × 10 2 K
(c) Use the ideal-gas law for a fixed
amount of gas to relate the
temperatures and pressures:
P1V1 P2V2
=
T1
T2
or, because V1 = V2,
P1 P2
=
T1 T2
Solve for T2:
P2 =
T2
P1
T1
Substitute numerical values and
evaluate P2:
P2 =
350 K
(1.0 atm) = 1.4 atm
243.7 K
72 ••
(a) The volume per molecule of a gas is the reciprocal of the number
density (the number of molecules per unit volume). (a) Find the average volume
per molecule for dry air at room temperature and atmospheric pressure. (b) Take
the cube root of your answer to Part (a) to obtain a rough estimate of the average
distance d between air molecules. (c) Find or estimate the average diameter D of
an air molecule, and compare it to your answer to Part (b). (d) Sketch the
molecules in a cube-shaped volume of air, with the edge length of the cube equal
Temperature and the Kinetic Theory of Gases 1759
to 3d. Make your figure to scale and place the molecules in what you think is a
typical configuration. (e) Use your picture to explain why the mean free path of
an air molecule is much greater than the average distance between molecules.
Picture the Problem (a) We can use the ideal-gas law to find the average volume
per molecule. (b) If one were to divide a container of air into little cubes, with one
cube per molecule, then this distance would be the width of each cube or,
equivalently, the distance from the center of one cube to the centers of the
neighboring cubes. On average, we can imagine that the molecules are at the
centers of their respective cubes, so this distance is also the average distance
between neighboring molecules.
(a) The ideal-gas law relates the
number of molecules N in a gas to
the volume V they occupy:
PV = NkT ⇒
Substitute numerical values and
evaluate V/N:
1.381× 10 −23 J/K (293 K )
V
=
N
101.325 kPa
V kT
=
N
P
(
)
= 3.99 × 10 −26 m 3
(b) Taking the cube root of V/N
yields:
d = 3 3.99 × 10−26 m 3 ≈ 3.42 nm
(c) Example 17-10 gives the average
diameter of an air molecule as:
D = 3.75 × 10 −10 m = 0.375 nm
or about 1/10 the average distance
between molecules.
(d) A sketch of the molecules in a cube-shaped volume of air, with the edge
length of the cube equal to 3d, follows. The random distribution of the molecules
is a typical configuration.
3d
3d
3d
1760
Chapter 17
(e) If a particular molecule in the diagram is moving in a random direction, its
chance of colliding with a neighbor is very small because it can miss in either of
the two directions perpendicular to its motion. So the mean free path, or average
distance between collisions, should be many times larger than the distance to the
nearest neighbor.
73 ••
[SSM] The Maxwell-Boltzmann distribution applies not just to
gases, but also to the molecular motions within liquids. The fact that not all
molecules have the same speed helps us understand the process of evaporation.
(a) Explain in terms of molecular motion why a drop of water becomes cooler as
molecules evaporate from the drop’s surface. (Evaporative cooling is an
important mechanism for regulating our body temperatures, and is also used to
cool buildings in hot, dry locations.) (b) Use the Maxwell-Boltzmann distribution
to explain why even a slight increase in temperature can greatly increase the rate
at which a drop of water evaporates.
Determine the Concept
(a) To escape from the surface of a droplet of water, molecules must have enough
translational kinetic energy to overcome the attractive forces from their neighbors.
Therefore the molecules that escape will be those that are moving faster, leaving
the slower molecules behind. The slower molecules have less kinetic energy, so
the temperature of the droplet, which is proportional to the average translational
kinetic energy per molecule, decreases.
(b) As long as the temperature isn’t too high, the molecules that evaporate from a
surface will be only those with the most extreme speeds, at the high-energy ″tail″
of the Maxwell-Boltzmann distribution. Within this part of the distribution,
increasing the temperature only slightly can greatly increase the percentage of
molecules with speeds above a certain threshold. For example, suppose that we
set an initial threshold at E = 5kT1, then imagine increasing the temperature by
10% so T2 = 1.1T1. At the threshold, the ratio of the new energy distribution to
the old one is
−3 − 5
F (T2 ) ⎛ T1 ⎞ kT2 kT1
= ⎜⎜ ⎟⎟e e = (1.1) 2 e 1.1e 5 = 1.366
F (T1 ) ⎝ T2 ⎠
−E
+E
an increase of almost 37%.
74 ••
A cubic metal box that has 20-cm–long edges contains air at a pressure
of 1.0 atm and a temperature of 300 K. The box is sealed so that the enclosed
volume remains constant, and it is heated to a temperature of 400 K. Find the
force due to the internal air pressure on each wall of the box.
Temperature and the Kinetic Theory of Gases 1761
Picture the Problem We can use the definition of pressure to express the net
force on each wall of the box in terms of its area and the pressure differential
between the inside and the outside of the box. We can apply the ideal-gas law for
a fixed amount of gas to find the pressure inside the box after it has been heated.
Using the definition of pressure,
express the net force on each wall
of the box:
F = AΔP
= A(Pinside − Patm )
Use the ideal-gas law for a fixed
amount of gas to relate the initial
and final pressures of the gas:
PatmVinitial PinsideVfinal
=
Tinitial
Tfinal
(1)
or, because Vinitial = Vfinal,
Patm
P
= inside
Tinitial Tfinal
Solve for Pinside to obtain:
Pinside =
Substituting in equation (1) and
simplifying yields:
⎛T
⎞
F = A⎜⎜ final Patm − Patm ⎟⎟
⎝ Tinitial
⎠
⎛T
⎞
= A⎜⎜ final − 1⎟⎟ Patm
⎝ Tinitial ⎠
Substitute numerical values and
evaluate F:
⎞
2 ⎛ 400 K
F = (0.20 m ) ⎜
− 1⎟ (101.325 kPa )
⎝ 300 K ⎠
Tfinal
Patm
Tinitial
= 1.4 kN
75 •• [SSM] In attempting to create liquid hydrogen for fuel, one of the
proposals is to convert plain old water (H2O) into H2 and O2 gases by electrolysis.
How many moles of each of these gases result from the electrolysis of 2.0 L of
water?
Picture the Problem We can use the molar mass of water to find the number of
moles in 2.0 L of water. Because there are two hydrogen atoms in each molecule
of water, there must be as many hydrogen molecules in the gas formed by
electrolysis as there were molecules of water and, because there is one oxygen
atom in each molecule of water, there must be half as many oxygen molecules in
the gas formed by electrolysis as there were molecules of water. See Appendix C
for the molar masses of hydrogen and oxygen.
1762
Chapter 17
Express the electrolysis of water into
H2 and O2:
Express the number of moles in
2.0 L of water:
nH 2 O → nH 2 + 12 nO 2
g
L = 110 mol
=
18.02 g/mol
2.0 L × 1000
nH 2 O
Because there are two hydrogen
atoms for each water molecule:
nH 2 = 110 mol
Because there is one oxygen atom
for each water molecule:
nO 2 = 12 nH 2 O =
1
2
(110 mol) =
55 mol
76 ••
A hollow 40-cm-long cylinder of negligible mass rests on its side on a
horizontal frictionless table. The cylinder is divided into two equal sections by a
vertical non-porous membrane. One section contains nitrogen and the other
contains oxygen. The pressure of the nitrogen is twice that of the oxygen. How far
will the cylinder move if the membrane breaks?
Picture the Problem The diagram shows the cylinder before removal of the
membrane. We’ll assume that the gases are at the same temperature. The
approximate location of the center of mass (CM) is indicated. We can find the
distance the cylinder moves by finding the location of the CM after the membrane
is removed. See Appendix C for the molar masses of oxygen and nitrogen.
N2
O2
x , cm
0
CM
40
Express the distance the cylinder
will move in terms of the movement
of the center of mass when the
membrane is removed:
Δx = xcm,after − xcm,before
Apply the ideal-gas law to both
collections of molecules to obtain:
PN 2 VN 2 = nN 2 kT
and
PO 2 VO 2 = nO 2 kT
Temperature and the Kinetic Theory of Gases 1763
Divide the first of these equations by
the second to obtain:
PN 2
PO 2
=
nN 2
nO 2
or, because PN 2 = 2 PO 2 ,
2 PO 2
PO 2
=
nN 2
nO 2
⇒ nN 2 = 2nO 2
Express the mass of O2 in terms of
its molar mass and the number of
moles of oxygen:
mO 2 = nO 2 M O 2
Express the mass of N2 in terms of
its molar mass and the number of
moles of nitrogen:
mN 2 = nO 2 M N 2
Using the definition of center of mass, express the center of mass before the
membrane is removed:
∑x m
=
∑m
i
xcm,before
i
i
i
=
n(N 2 )M (N 2 )xcm, N 2 + n(O 2 )M (O 2 )xcm,O 2
n(N 2 )M (N 2 ) + n(O 2 )M (O 2 )
i
=
=
2n(O 2 )M (N 2 )xcm, N 2 + n(O 2 )M (O 2 )xcm,O 2
2n(O 2 )M (N 2 ) + n(O 2 )M (O 2 )
2 M (N 2 )xcm, N 2 + M (O 2 )xcm,O 2
2M (N 2 ) + M (O 2 )
Substitute numerical values and evaluate xcm,before:
xcm,before =
2(10 cm )(28.01g ) + (30 cm )(32.00 g )
= 17.27 cm
2(28.01g ) + 32.00 g
Locate the center of mass after the membrane is removed:
xcm,after =
Substitute to obtain:
2(20 cm )(28.01g ) + (20 cm )(32.00 g )
= 20.00 cm
2(28.01g ) + 32.00 g
Δx = 20.00 cm − 17.27 cm = 2.7 cm
1764
Chapter 17
Because momentum must be conserved during this process and the center of mass
moved to the right, the cylinder moved 2.7 cm to the left.
77 ••
A cylinder of fixed volume contains a mixture of helium gas (He) and
hydrogen gas (H2) at a temperature T1 and pressure P1. If the temperature is
doubled to T2 = 2T1, the pressure would also double, except for the fact that at
temperature the H2 is essentially 100 percent dissociated into H1. In reality, at
pressure P2 = 2P1 the temperature is T2 = 3T1. If the mass of the hydrogen in the
cylinder is m, what is the mass of the nitrogen in the cylinder?
Picture the Problem We can apply the ideal-gas law to the two processes to find
the number of moles of hydrogen in terms of the number of moles of nitrogen in
the gas. Using the definition of molar mass, we can relate the mass of each gas to
the number of moles of each gas and their molar masses. See Appendix C for the
molar masses of nitrogen gas and hydrogen gas.
[
]
Apply the ideal-gas law to the first
case:
P1V = 2nN 2 + nH 2 RT1
Apply the ideal-gas law to the
second case:
3P1V = 2nN 2 + 2nH 2 2 RT1
Divide the second of these equations
by the first and simplify to express
nH 2 in terms of nN2 :
nH 2 = 2nN 2
Relate the mN to nN2 :
mN = nN 2 M N 2
[
]
= n N 2 (28.01g/mol)
and
nN 2 =
Relate the m to nH2 :
mN 2
28.01g/mol
m = nH 2 M H 2
= nH 2 (2.016 g/mol)
and
nH 2 =
mH 2
2.016 g/mol
(1)
Temperature and the Kinetic Theory of Gases 1765
Substitute in equation (1) and solve
for mN:
2m N 2
m
=
2.016 g/mol 28.01g/mol
and
mN 2 ≈ 7m
78 ••
The mean free path for O2 molecules at a temperature of 300 K and at
1.00 atm pressure is 7.10 × 10–8 m. Use this data to estimate the size of an O2
molecule.
Picture the Problem Because the O2 molecule resembles 2 spheres stuck
together, which in cross section look something like two circles, we can estimate
the radius of the molecule from the formula for the area of a circle. We can
express the area, and hence the radius, of the circle in terms of the mean free path
and the number density of the molecules and use the ideal-gas law to express the
number density.
Express the area of two circles of
diameter d that touch each other:
⎛ πd 2 ⎞ πd 2
⎟⎟ =
⇒d =
A = 2⎜⎜
4
2
⎝
⎠
Relate the mean free path of the
molecules to their number
density and cross-sectional area:
λ=
Substitute for A in equation (1) to
obtain:
Use the ideal-gas law to relate the
number density of the O2
molecules to their temperature
and pressure:
Substituting for nv yields:
Substitute numerical values and
evaluate d:
d=
2A
π
(1)
1
1
⇒A=
nv A
nv λ
2
πnv λ
PV = NkT or nv =
N
P
=
V kT
d=
2kT
πPλ
d=
2 1.381×10 − 23 J/K (300 K )
π (101.325 kPa ) 7.10 ×10 −8 m
(
(
)
)
= 0.605 nm
79 ••
[SSM] Current experiments in atomic trapping and cooling can
create low-density gases of rubidium and other atoms with temperatures in the
nanokelvin (10–9 K) range. These atoms are trapped and cooled using magnetic
1766
Chapter 17
fields and lasers in ultrahigh vacuum chambers. One method that is used to
measure the temperature of a trapped gas is to turn the trap off and measure the
time it takes for molecules of the gas to fall a given distance. Consider a gas of
rubidium atoms at a temperature of 120 nK. Calculate how long it would take an
atom traveling at the rms speed of the gas to fall a distance of 10.0 cm if (a) it
were initially moving directly downward and (b) if it were initially moving
directly upward. Assume that the atom doesn’t collide with any others along its
trajectory.
Picture the Problem Choose a coordinate system in which downward is the
positive direction. We can use a constant-acceleration equation to relate the fall
distance to the initial velocity of the molecule, the acceleration due to gravity, the
fall time, and vrms = 3kT m to find the initial velocity of the rubidium
molecules.
(a) Using a constant-acceleration
equation, relate the fall distance to
the initial velocity of a molecule,
the acceleration due to gravity, and
the fall time:
Relate the rms speed of rubidium
atoms to their temperature and
mass:
Substitute numerical values and
evaluate vrms:
y = v0t + 12 gt 2
(1)
vrms =
3kT
mRb
v rms =
3 1.381× 10−23 J/K (120 nK )
(85.47 u ) 1.6606 ×10−27 kg/u
(
)
(
= 5.918 × 10−3 m/s
Letting vrms = v0, substitute in equation (1) to obtain:
(
)
(
)
0.100 m = 5.918 × 10 −3 m/s t + 12 9.81m/s 2 t 2
Use the quadratic formula or your
graphing calculator to solve this
equation for its positive root:
t = 0.14218 s = 142 ms
(b) If the atom is initially moving
upward:
v rms = v0 = −5.918 ×10 −3 m/s
)
Temperature and the Kinetic Theory of Gases 1767
Substitute in equation (1) to obtain:
(
)
(
)
0.100 m = − 5.918 × 10−3 m/s t + 12 9.81 m/s 2 t 2
Use the quadratic formula or your
graphing calculator to solve this
equation for its positive root:
t = 0.1464 s = 146 ms
80 ••• A cylinder is filled with 0.10 mol of an ideal gas at standard
temperature and pressure, and a 1.4-kg piston seals the gas in the cylinder (Figure
17-21) with a frictionless seal. The trapped column of gas is 2.4-m high. The
piston and cylinder are surrounded by air, also at standard temperature and
pressure. The piston is released from rest and starts to fall. The motion of the
piston ceases after the oscillations stop with the piston and the trapped air in
thermal equilibrium with the surrounding air. (a) Find the height of the gas
column. (b) Suppose that the piston is pushed down below its equilibrium position
by a small amount and then released. Assuming that the temperature of the gas
remains constant, find the frequency of vibration of the piston.
Picture the Problem (a) Let A be the cross-sectional area of the cylinder. We can
use the ideal-gas law to find the height of the piston under equilibrium conditions.
In (b), we can apply Newton’s 2nd law and the ideal-gas law for a fixed amount of
gas to the show that, for small displacements from its equilibrium position, the
piston executes simple harmonic motion.
(a) Express the pressure inside the
cylinder:
Pin = Patm +
Apply the ideal-gas law to obtain a
second expression for the pressure
of the gas in the cylinder:
Pin =
Equating these two expressions
yields:
Patm +
Solve for h to obtain:
h=
=
mg
A
nRT nRT
=
V
hA
mg nRT
=
A
hA
(2.4 m )APatm
nRT
=
APatm + mg
APatm + mg
2.4 m
mg
1+
APatm
(1)
1768
Chapter 17
At STP, 0.10 mol of gas occupies
2.24 L. Therefore:
Substitute numerical values and
evaluate h:
(2.4 m )A = 2.24 × 10−3 m3
and
A = 9.333 × 10 −4 m 2
h=
2.4 m
(
1.4 kg ) 9.81 m/s 2
1+
9.333 ×10 − 4 m 2 (101.325 kPa )
(
(
)
)
= 2.096 m = 2.1 m
(b) Relate the frequency of vibration
of the piston to its mass and a
″stiffness″ constant:
1 k
(2)
2π m
where m is the mass of the piston and k
is a constant of proportionality.
Letting y be the displacement from
equilibrium, apply ∑ Fy = ma y to
Pin A − mg − Patm A = 0
f =
the piston in its equilibrium position:
For a small displacement y above
equilibrium:
P' in A − mg − Patm A = ma y
or
P' in A − Pin A = ma y
Using the ideal-gas law for a fixed
amount of gas and constant
temperature, relate Pin' to Pin :
P'inV' = PinV
or
Substituting for V and simplifying
yields:
P' in A = Pin A
Substitute in equation (3) to obtain:
y⎞
⎛
Pin A⎜1 + ⎟ − Pin A = ma y
⎝ h⎠
or, for y << h,
y⎞
⎛
Pin A⎜1 − ⎟ − Pin A ≈ ma y
⎝ h⎠
Simplify equation (4) to obtain:
(3)
P' in (V + Ay ) = PinV ⇒ P'in = Pin
V
V + Ay
Ah
1
= Pin A
y
Ah + Ay
1+
h
−1
− Pin A
y
≈ ma y
h
(4)
Temperature and the Kinetic Theory of Gases 1769
Substitute in equation (1) to obtain:
⎛ nRT ⎞ y
⎛ nRT ⎞
−⎜
⎟ A ≈ ma y ⇒ − ⎜ 2 ⎟ y ≈ ma y
⎝ Ah ⎠ h
⎝ h ⎠
Solving for ay yields:
ay = −
nRT
y
mh 2
or
k
y , the condition for SHM
m
k nRT
where
=
m mh 2
ay = −
Substitute in equation (2) to obtain:
f =
1
2π
nRT
mh 2
Substitute numerical values and evaluate f:
f =
1
2π
(0.10 mol)(8.314 J/mol ⋅ K )(300 K ) =
(1.4 kg )(2.096 m )2
1.0 Hz
81 ••• During this problem, you will use a spreadsheet to study the
distribution of molecular speeds in a gas. Figure 17-22 should help you get
started. (a) Enter the values for constants R, M, and T as shown. Then in column
A, enter values of speed ranging from 0 to 1200 m/s, in increments of 1 m/s.
(This spreadsheet will be long.) In cell B7, enter the formula for the MaxwellBoltzmann fraction fractional speed distribution. This formula contains
parameters v, R, M and T. Substitute A7 for v, B$1 or R, B$2 for M and B$3 for T.
Then use the FILL DOWN command to enter the formula in the cells below B7.
Create a graph of f(v) versus v using the data in columns A and B. (b) Explore
how the graph changes as you increase and decrease the temperature, and describe
the results. (c) Add a third column in which each cell contains the cumulative
sum of all f(v) values, multiplied by the interval size dv (which equals 1), in the
rows above and including the row in question. What is the physical interpretation
of the numbers in this column? (d) For nitrogen gas at 300 K, what percentage of
the molecules has speeds less than 200 m/s? (e) For nitrogen gas at 300 K, what
percentage of the molecules has speeds greater than 700 m/s?
(a) The first few rows of the spreadsheet are shown below. Note that the column
for Part (c) is included.
A
B
C
1
R= 8.31
J/mol-K
2
M= 0.028
kg/mol
3
T= 300
K
4
5
f(v)
sum f(v)dv
v
1770
Chapter 17
6
7
8
9
10
11
12
(m/s)
0
1
2
3
4
5
(s/m)
0.00E+00
3.00E−08
1.20E−07
2.70E−07
4.80E−07
7.51E−07
(unitless)
0.00E+00
3.00E−08
1.50E−07
4.20E−07
9.01E−07
1.65E−06
A graph of f(v) for nitrogen at 300 K follows:
0.0025
f (v ), s/m
0.0020
0.0015
0.0010
0.0005
0.0000
0
200
400
600
800
1000
1200
v , m/s
(b) As the temperature is increased the horizontal position of the peak moves to
the right in proportion to the square root of the temperature, while the height of
the peak drops by the same factor, preserving the total area under the graph
(which must be 1.0).
(c) Each number in column C of the spreadsheet (shown in (a)) is approximately
equal to the integral of f(v) from zero up to the corresponding v value. This
integral represents the probability of a molecule having a speed less than or equal
to this value of v.
Temperature and the Kinetic Theory of Gases 1771
1.2
1.0
f (v ) dv
0.8
0.6
0.4
0.2
0.0
0
200
400
600
800
1000
1200
v , m/s
(d) Looking in cell C207, we find that the probability (at 300 K) of a nitrogen
molecule having a speed less than 200 m/s is 0.0706 , or about 7%. Note that
this value is consistent with the graph of f (v )dv shown immediately above.
(e) Looking in cell C707, we find that the probability of a nitrogen molecule
having a speed less than 700 m/s is approximately 0.862, so the probability of it
having a speed greater than this would be 1 – 0.862 or 0.138 , or a little under
14%.
1772
Chapter 17