563
19-6 TH E DISTR I B UTION OF M OLECU L AR SPE E DS
The total area under the plot of P(v) in Fig. 19-8a is the
total fraction of molecules (unity), and the area of the thin
gold strip (not to scale) is the fraction we seek. Let’s evaluate frac in parts:
frac ϭ (4p)(A)(v
2
)(e
B
)(⌬v),
(19-36)
where
ϭ 2.92 ϫ 10
Ϫ9
s
3
/m
3
and
Substituting A and B into Eq. 19-36 yields
frac ϭ (4p)(A)(v
2
)(e
B
)(⌬v)
ϭ (4p)(2.92 ϫ 10
Ϫ9
s
3
/m
3
)(600 m/s)
2
(e
Ϫ2.31
)(2 m/s)
ϭ 2.62 ϫ 10
Ϫ3 ϭ 0.262%.
(Answer)
ϭ Ϫ2.31.
B ϭ Ϫ
Mv
2
2RT
ϭ Ϫ
(0.0320 kg/mol)(600 m/s)
2
(2)(8.31 J/mol иK)(300 K)
A ϭ
M
2pRT
3/2
ϭ
0.0320 kg/mol
(2p)(8.31 J/molиK)(300 K)
3/2
Sample Problem 19.05 Speed distribution in a gas
In oxygen (molar mass M = 0.0320 kg/mol) at room temperature (300 K), what fraction of the molecules have speeds in
the interval 599 to 601 m/s?
KEY IDEAS
1. The speeds of the molecules are distributed over a wide
range of values, with the distribution P(v) of Eq. 19-27.
2. The fraction of molecules with speeds in a differential
interval dv is P(v) dv.
3. For a larger interval, the fraction is found by integrating
P(v) over the interval.
4. However, the interval ⌬v ϭ 2 m/s here is small compared
to the speed v ϭ 600 m/s on which it is centered.
Calculations: Because ⌬v is small, we can avoid the integration by approximating the fraction as
frac ϭ P(v) ⌬v ϭ 4p
M
2pRT
3/2
v
2
e
ϪMv 2 /2RT ⌬v.
Calculation: We end up with Eq. 19-34, which gives us
(Answer)
This result, plotted in Fig. 19-8a, is greater than v avg because
the greater speed values influence the calculation more when
we integrate the v
2
values than when we integrate the v values.
(c) What is the most probable speed v P at 300 K?
KEY IDEA
Speed v P corresponds to the maximum of the distribution
function P(v), which we obtain by setting the derivative
dP/dv ϭ 0 and solving the result for v.
Calculation: We end up with Eq. 19-35, which gives us
(Answer)
This result is also plotted in Fig. 19-8a.
ϭ 395 m/s.
ϭ A
2(8.31 J/mol иK)(300 K)
0.0320 kg/mol
v P ϭ A
2RT
M
ϭ 483 m/s.
ϭ A
3(8.31 J/mol иK)(300 K)
0.0320 kg/mol
v rms ϭ A
3RT
M
Sample Problem 19.06 Average speed, rms speed, most probable speed
The molar mass M of oxygen is 0.0320 kg/mol.
(a) What is the average speed v avg of oxygen gas molecules
at T ϭ 300 K?
KEY IDEA
To find the average speed, we must weight speed v with the
distribution function P(v) of Eq. 19-27 and then integrate
the resulting expression over the range of possible speeds
(from zero to the limit of an infinite speed).
Calculation: We end up with Eq. 19-31, which gives us
(Answer)
This result is plotted in Fig. 19-8a.
(b) What is the root-mean-square speed v rms at 300 K?
KEY IDEA
To find v rms , we must first find (v
2
) avg by weighting v
2
with
the distribution function P(v) of Eq. 19-27 and then integrating the expression over the range of possible speeds.
Then we must take the square root of the result.
ϭ 445 m/s.
ϭ A
8(8.31 J/mol иK)(300 K)
p(0.0320 kg/mol)
v avg ϭ A
8RT
pM
Additional examples, video, and practice available at WileyPLUS
19-6 TH E DISTR I B UTION OF M OLECU L AR SPE E DS
The total area under the plot of P(v) in Fig. 19-8a is the
total fraction of molecules (unity), and the area of the thin
gold strip (not to scale) is the fraction we seek. Let’s evaluate frac in parts:
frac ϭ (4p)(A)(v
2
)(e
B
)(⌬v),
(19-36)
where
ϭ 2.92 ϫ 10
Ϫ9
s
3
/m
3
and
Substituting A and B into Eq. 19-36 yields
frac ϭ (4p)(A)(v
2
)(e
B
)(⌬v)
ϭ (4p)(2.92 ϫ 10
Ϫ9
s
3
/m
3
)(600 m/s)
2
(e
Ϫ2.31
)(2 m/s)
ϭ 2.62 ϫ 10
Ϫ3 ϭ 0.262%.
(Answer)
ϭ Ϫ2.31.
B ϭ Ϫ
Mv
2
2RT
ϭ Ϫ
(0.0320 kg/mol)(600 m/s)
2
(2)(8.31 J/mol иK)(300 K)
A ϭ
M
2pRT
3/2
ϭ
0.0320 kg/mol
(2p)(8.31 J/molиK)(300 K)
3/2
Sample Problem 19.05 Speed distribution in a gas
In oxygen (molar mass M = 0.0320 kg/mol) at room temperature (300 K), what fraction of the molecules have speeds in
the interval 599 to 601 m/s?
KEY IDEAS
1. The speeds of the molecules are distributed over a wide
range of values, with the distribution P(v) of Eq. 19-27.
2. The fraction of molecules with speeds in a differential
interval dv is P(v) dv.
3. For a larger interval, the fraction is found by integrating
P(v) over the interval.
4. However, the interval ⌬v ϭ 2 m/s here is small compared
to the speed v ϭ 600 m/s on which it is centered.
Calculations: Because ⌬v is small, we can avoid the integration by approximating the fraction as
frac ϭ P(v) ⌬v ϭ 4p
M
2pRT
3/2
v
2
e
ϪMv 2 /2RT ⌬v.
Calculation: We end up with Eq. 19-34, which gives us
(Answer)
This result, plotted in Fig. 19-8a, is greater than v avg because
the greater speed values influence the calculation more when
we integrate the v
2
values than when we integrate the v values.
(c) What is the most probable speed v P at 300 K?
KEY IDEA
Speed v P corresponds to the maximum of the distribution
function P(v), which we obtain by setting the derivative
dP/dv ϭ 0 and solving the result for v.
Calculation: We end up with Eq. 19-35, which gives us
(Answer)
This result is also plotted in Fig. 19-8a.
ϭ 395 m/s.
ϭ A
2(8.31 J/mol иK)(300 K)
0.0320 kg/mol
v P ϭ A
2RT
M
ϭ 483 m/s.
ϭ A
3(8.31 J/mol иK)(300 K)
0.0320 kg/mol
v rms ϭ A
3RT
M
Sample Problem 19.06 Average speed, rms speed, most probable speed
The molar mass M of oxygen is 0.0320 kg/mol.
(a) What is the average speed v avg of oxygen gas molecules
at T ϭ 300 K?
KEY IDEA
To find the average speed, we must weight speed v with the
distribution function P(v) of Eq. 19-27 and then integrate
the resulting expression over the range of possible speeds
(from zero to the limit of an infinite speed).
Calculation: We end up with Eq. 19-31, which gives us
(Answer)
This result is plotted in Fig. 19-8a.
(b) What is the root-mean-square speed v rms at 300 K?
KEY IDEA
To find v rms , we must first find (v
2
) avg by weighting v
2
with
the distribution function P(v) of Eq. 19-27 and then integrating the expression over the range of possible speeds.
Then we must take the square root of the result.
ϭ 445 m/s.
ϭ A
8(8.31 J/mol иK)(300 K)
p(0.0320 kg/mol)
v avg ϭ A
8RT
pM
Additional examples, video, and practice available at WileyPLUS