Modern Engineering Thermodynamics

18.5 Molecular Velocity Distributions 737
Equation (18.24) can be evaluated to find the fraction of molecules whose velocities lie in the range from
0toV as
Nð0 ! VÞ
N
= erfðxÞ − p 2 ffiffiffi xe −x2 (18.25)
π
and to find the fraction of molecules whose velocities lie in the range from V to ∞ as
where x = V/V mp in each case.
NðV ! ∞Þ
N
= 1 −
Nð0 ! VÞ
N
= 1 − erfðxÞ + p 2 ffiffiffi xe −x2 (18.26)
π
EXAMPLE 18.3
Test assumption 7 at the beginning of this section by computing the fraction of neon molecules at 273 K whose velocities are
faster than (a) V mp , (b) V avg , (c) V rms , and (d) c (the speed of light). Use the molecular data for neon given in Example 18.2.
Solution
a. The fraction having velocities greater than V mp is given by Eq. (18.26) with x = V mp /V mp = 1:0 as
or
NðV mp ! ∞Þ
N
NðV mp ! ∞Þ
N
Thus, 57.24% of the molecules have velocities faster than V mp .
b. Here, Eq. (18.20) gives
= 1 − erfð1:0Þ + p 2 ffiffiffi ð1:0Þ e −1:0
π
= 1 − 0:8427 + 0:4151 = 0:5724
rffiffiffiffiffiffiffiffi
rffiffiffiffiffiffiffiffi
V avg =
8kT
, V mp =
2kT
,
πm m
and
x = V avg
V mp
=
rffiffiffiffiffi
8
= 1:128
2π
Therefore, the fraction of molecules having velocities greater than V avg is given by interpolating in Table 18.2 to find
NðV avg ! ∞Þ
N
= 1 − erfð1:128Þ + p 2 ffiffiffi ð1:128Þe −1:272
π
= 1 − 0:8893 + 0:3566 = 0:4673
Consequently, 46.73% of the molecules have velocities faster than V
p
avg.
c. Here, Eq. (18.21) gives V rms = ffiffiffiffiffiffiffiffiffiffiffiffiffi
3kT/m,so
qffiffi
x = V rms /V mp = 3 = 1:225
2
and
NðV rms ! ∞Þ
N
= 1 − erfð1:225Þ + p 2 ffiffiffi ð1:225Þe − 1:501
π
= 1 − 0:9168 + 0:3081 = 0:3913
or 39.13% of the molecules have velocities greater than V rms .
d. It can be shown (see Problem 10 at the end of this chapter) that, when V/V mp ≫ 1, the fraction of molecules with
velocities in the range from V to ∞ is approximately given by
NðV ! ∞Þ
N
≈ 2
pffiffiffi
π
x + 1
e −x2 ðfor x ≫ 1 onlyÞ
2x
(Continued )