Modern Engineering Thermodynamics

758 CHAPTER 18: Introduction to Statistical Thermodynamics
EXAMPLE 18.10 (Continued )
Exercises
28. Note that the only property of a polytropic Maxwell-Boltzmann gas that depends on pressure is entropy. Compute the
entropy for the carbon dioxide in Example 18.10 if the pressure is increased from 1.00 to 10.0 atm. Assume all other
variables remain unchanged. Answer: s = 5.68 kJ/kg·K (s trans = 3.68 kJ/kg·K, s rot = 1.47 kJ/kg·K, s vib = 0.528 kJ/kg·K).
29. Recompute the specific internal energy, enthalpy, and entropy of the carbon dioxide in Example 18.10 when the
temperature is lowered from 1000. K to 300. K but the pressure remains constant at 1.00 atm. Answer: u = 841 kJ/kg
(u trans = 85.0 kJ/kg, u rot = 56.7 kJ/kg, u vib = 699 kJ/kg); h = 898 kJ/kg; s = 4.86 kJ/kg·K (s trans = 3.55 kJ/kg·K,
s rot = 1.24 kJ/kg·K, s vib = 0.0694 kJ/kg·K).
30. Determine the specific internal energy, enthalpy, and entropy of the carbon dioxide in Example 18.10 for a temperature
of 5000. K and a pressure of 0.100 atmosphere. Answer: u = 6190 kJ/kg (u trans = 1420 kJ/kg, u rot = 945 kJ/kg, u vib =
3832 kJ/kg); h = 7138 kJ/kg; s = 8.72 kJ/kg·K (s trans = 5.31 kJ/kg·K, s rot = 1.78 kJ/kg·K, s vib = 1.64 kJ/kg·K).
The nonlinear polyatomic molecule has 3 translational degrees of freedom, 3 rotational degrees of freedom, and
3b−6 degrees of vibrational freedom. Thus, it has the same equations for the translational molecular energy as
in the linear polyatomic case, but the rotational and vibrational contribution equations are different. In this
case,
u rot = h rot = 3 2 RT
(18.57a)
ðc v Þ rot = ðc p Þ rot = 3 2 R
n
s rot = R ln½T/ðσ Θ r ÞŠ+ 3 o
2
(18.57b)
(18.57c)
and
u vib = ðu o Þ vib
+ R∑ 3b−6
i=1
h vib = u vib = ðu o Þ vib
+ R∑ 3b−6
ðc v Þ vib = R∑ 3b−6
i=1
Θ vi / ½expðΘ vi /TÞ−1Š (18.58a)
i=1
Θ vi / ½expðΘ vi /TÞ−1Š (18.58b)
ðΘ vi /TÞ 2 ½expðΘ vi /TÞŠ/ ½expðΘ vi /TÞ−1Š 2 (18.58c)
and
ðc p Þ vib
= ðc v Þ vib
+ R∑ 3b−6
i=1
ðΘ vi /TÞ 2 ½expðΘ vi /TÞŠ/ ½expðΘ vi /TÞ−1Š 2 (18.58d)
s vib = R∑ 3b−6
i=1
ln ½1 − expð−Θ vi /TÞŠ −1 + ðΘ vi /TÞ½expðΘ vi /TÞ−1
where the vibrational internal energy at absolute temperature is now found from
Š −1
(18.58e)
3b−6
ðu o Þ vib = ∑ RΘ vi /2 (18.59)
since the nonlinear polyatomic molecule has 3b−6 characteristic vibrational temperatures Θ vi .
SUMMARY
The subject of statistical thermodynamics is inherently mathematically complex and conceptually difficult. It is
often the subject of an entire advanced engineering course, usually at the graduate level. The material presented
in this chapter is intended only as an introduction to this subject.
i=1