Modern Engineering Thermodynamics

738 CHAPTER 18: Introduction to Statistical Thermodynamics
EXAMPLE 18.3 (Continued )
Consider x = V/V mp = 10:00 then,
NðV ! ∞Þ
N
≈ p 2 ffiffiffi ð10:05Þe −100 = 4:22 × 10 −43
π
Thus, only one molecule in about 10 20 moles of a gas has a velocity ten times greater than V mp . Now, the velocity of
light c is 3:00 × 10 8 m/s, and for neon at 273 K. we have m = 3.35 × 10 −26 kg (see Example 18.2), and
p
V mp =
ffiffiffiffiffiffiffiffiffiffiffiffiffi
2kT/m = 474 m/s: Thus,
so that
x =
V =
c = 3:00 × 108 m/s
= 6:33 × 10 5
V mp V mp 474 m/s
Nðc ! ∞Þ
N
≈ p 2 ffiffiffi ð6:33 × 10 5 Þe −4:0×1011 ≈ 0
π
Even though we allow molecules to move faster than the speed of light in our mathematical model, we find that, for all
practical purposes, this model predicts that virtually no molecules have velocities this fast at ordinary temperatures.
Exercises
7. Plot the actual value and the four-term series expansion of the error function given in Table 18.2 for 0.1 ≤ x ≤ 2.0 and
compare the results. Partial answer: The series expansion is valid only for x < 1.0, see Figure 18.6.
8. Determine the percentage of molecules in Example 18.3 that have velocities greater than twice the most probable
velocity. Answer: N(2V mp → ∞)/N = 0.046 = 4.6%.
9. Determine the mean velocity in Example 18.3 (i.e., find the velocity V for which half the molecules have a velocity
greater than V and half have a velocity less than V) as defined by N(V → ∞)/N = 0.5 = 50%. Answer: V = 1.09 × V mp .
1.20
1.20
1.00
1.00
Exact error function
0.80
0.60
0.40
0.20
0.00
Exact error function
Approximate error function
0.80
0.60
0.40
0.20
0.00
Approximate error function
−0.20
−0.20
0.0 0.5 1.0 1.5 2.0 2.5
x
FIGURE 18.6
Example 18.3, Exercise 7.
18.6 EQUIPARTITION OF ENERGY
Equation (18.14) gives the total translational kinetic energy of a system of N molecules as 3 2
NkT. The principle
of equipartition of energy requires that the translational kinetic energy of an unrestricted molecule be equally
divided among the three translational degrees of freedom (one for each independent coordinate direction).
Therefore, the translational total internal energy in each of the x, y, andz coordinate directions must be one
third of that given in Eq. (18.14), or
ðU trans Þ x
= ðU trans Þ y
= ðU trans Þ z
= U trans /3 = 1 2 NkT