560
CHAPTE R 19 TH E KI N ETIC TH EORY OF GASES
Additional examples, video, and practice available at WileyPLUS
19-6 THE DISTRIBUTION OF MOLECULAR SPEEDS
After reading this module, you should be able to . . .
19.24 Explain how Maxwell’s speed distribution law is used
to find the fraction of molecules with speeds in a certain
speed range.
19.25 Sketch a graph of Maxwell’s speed distribution, showing the probability distribution versus speed and indicating
the relative positions of the average speed v avg , the most
probable speed v P , and the rms speed v rms .
19.26 Explain how Maxwell’s speed distribution is used to
find the average speed, the rms speed, and the most
probable speed.
19.27 For a given temperature T and molar mass M, calculate the average speed v avg , the most probable speed v P ,
and the rms speed v rms .
Learning Objectives
● The Maxwell speed distribution P(v) is a function such that
P(v) dv gives the fraction of molecules with speeds in the
interval dv at speed v:
● Three measures of the distribution of speeds among the
molecules of a gas are
P(v) ϭ 4p
M
2pRT
3/2
v
2
e
ϪMv 2 /2RT
.
(average speed),
(most probable speed),
and
(rms speed).
v rms ϭ A
3RT
M
v P ϭ A
2RT
M
v avg ϭ A
8RT
pM
Key Ideas
Sample Problem 19.04 Mean free path, average speed, collision frequency
(a) What is the mean free path l for oxygen molecules at
temperature T ϭ 300 K and pressure p ϭ 1.0 atm? Assume
that the molecular diameter is d ϭ 290 pm and the gas is
ideal.
KEY IDEA
Each oxygen molecule moves among other moving oxygen
molecules in a zigzag path due to the resulting collisions. Thus,
we use Eq. 19-25 for the mean free path.
Calculation: We first need the number of molecules per unit
volume, N/V. Because we assume the gas is ideal, we can use
the ideal gas law of Eq. 19-9 (pV NkT) to write N/V
p/kT. Substituting this into Eq. 19-25, we find
ϭ 1.1 ϫ 10
Ϫ7
m.
(Answer)
This is about 380 molecular diameters.
(b) Assume the average speed of the oxygen molecules is
v ϭ 450 m/s. What is the average time t between successive
ϭ
(1.38 ϫ 10
Ϫ23
J/K)(300 K)
12p (2.9 ϫ 10
Ϫ10
m)
2
(1.01 ϫ 10
5
Pa)
l ϭ
1
12pd
2
N/V
ϭ
kT
12pd
2
p
ϭ
ϭ
collisions for any given molecule? At what rate does
the molecule collide; that is, what is the frequency f of its
collisions?
KEY IDEAS
(1) Between collisions, the molecule travels, on average, the
mean free path l at speed v. (2) The average rate or frequency at which the collisions occur is the inverse of the
time t between collisions.
Calculations: From the first key idea, the average time
between collisions is
ϭ 2.44 ϫ 10
Ϫ10
s Ϸ 0.24 ns.
(Answer)
This tells us that, on average, any given oxygen molecule has
less than a nanosecond between collisions.
From the second key idea, the collision frequency is
(Answer)
This tells us that, on average, any given oxygen molecule
makes about 4 billion collisions per second.
f ϭ
1
t
ϭ
1
2.44 ϫ 10
Ϫ10
s
ϭ 4.1 ϫ 10
9
s
Ϫ1
.
t ϭ
distance
speed
ϭ
l
v
ϭ
1.1 ϫ 10
Ϫ7
m
450 m/s
CHAPTE R 19 TH E KI N ETIC TH EORY OF GASES
Additional examples, video, and practice available at WileyPLUS
19-6 THE DISTRIBUTION OF MOLECULAR SPEEDS
After reading this module, you should be able to . . .
19.24 Explain how Maxwell’s speed distribution law is used
to find the fraction of molecules with speeds in a certain
speed range.
19.25 Sketch a graph of Maxwell’s speed distribution, showing the probability distribution versus speed and indicating
the relative positions of the average speed v avg , the most
probable speed v P , and the rms speed v rms .
19.26 Explain how Maxwell’s speed distribution is used to
find the average speed, the rms speed, and the most
probable speed.
19.27 For a given temperature T and molar mass M, calculate the average speed v avg , the most probable speed v P ,
and the rms speed v rms .
Learning Objectives
● The Maxwell speed distribution P(v) is a function such that
P(v) dv gives the fraction of molecules with speeds in the
interval dv at speed v:
● Three measures of the distribution of speeds among the
molecules of a gas are
P(v) ϭ 4p
M
2pRT
3/2
v
2
e
ϪMv 2 /2RT
.
(average speed),
(most probable speed),
and
(rms speed).
v rms ϭ A
3RT
M
v P ϭ A
2RT
M
v avg ϭ A
8RT
pM
Key Ideas
Sample Problem 19.04 Mean free path, average speed, collision frequency
(a) What is the mean free path l for oxygen molecules at
temperature T ϭ 300 K and pressure p ϭ 1.0 atm? Assume
that the molecular diameter is d ϭ 290 pm and the gas is
ideal.
KEY IDEA
Each oxygen molecule moves among other moving oxygen
molecules in a zigzag path due to the resulting collisions. Thus,
we use Eq. 19-25 for the mean free path.
Calculation: We first need the number of molecules per unit
volume, N/V. Because we assume the gas is ideal, we can use
the ideal gas law of Eq. 19-9 (pV NkT) to write N/V
p/kT. Substituting this into Eq. 19-25, we find
ϭ 1.1 ϫ 10
Ϫ7
m.
(Answer)
This is about 380 molecular diameters.
(b) Assume the average speed of the oxygen molecules is
v ϭ 450 m/s. What is the average time t between successive
ϭ
(1.38 ϫ 10
Ϫ23
J/K)(300 K)
12p (2.9 ϫ 10
Ϫ10
m)
2
(1.01 ϫ 10
5
Pa)
l ϭ
1
12pd
2
N/V
ϭ
kT
12pd
2
p
ϭ
ϭ
collisions for any given molecule? At what rate does
the molecule collide; that is, what is the frequency f of its
collisions?
KEY IDEAS
(1) Between collisions, the molecule travels, on average, the
mean free path l at speed v. (2) The average rate or frequency at which the collisions occur is the inverse of the
time t between collisions.
Calculations: From the first key idea, the average time
between collisions is
ϭ 2.44 ϫ 10
Ϫ10
s Ϸ 0.24 ns.
(Answer)
This tells us that, on average, any given oxygen molecule has
less than a nanosecond between collisions.
From the second key idea, the collision frequency is
(Answer)
This tells us that, on average, any given oxygen molecule
makes about 4 billion collisions per second.
f ϭ
1
t
ϭ
1
2.44 ϫ 10
Ϫ10
s
ϭ 4.1 ϫ 10
9
s
Ϫ1
.
t ϭ
distance
speed
ϭ
l
v
ϭ
1.1 ϫ 10
Ϫ7
m
450 m/s