Modern Engineering Thermodynamics

756 CHAPTER 18: Introduction to Statistical Thermodynamics
EXAMPLE 18.9 (Continued )
Exercises
25. Determine the values of c v and the ratio of c v /R for the nitrous oxide in Example 18.9 at 2000. K. Answer: (c v ) NO =
0.930 kJ/kg·K and (c v /R) NO = 3.36.
26. Using Eqs. (18.51a–d), compute the values for the specific enthalpy and the ratio of h/R for the nitrous oxide in
Example 18.9 at 20.0°C. Answer: h NO = 948 kJ/kg (h trans = 203 kJ/kg, h rot = 81.2 kJ/kg, h vib = 664 kJ/kg), and
(h/R) NO = 3420 K.
27. Using Eqs. (18.54a–d), compute the values of the specific entropy and the ratio s/R for the nitrous oxide in Example
18.9 at 20.0°C and a pressure of 1 atm. Answer: s NO = 6.63 kJ/kg·K (s trans = 5.03 kJ/kg·K, s rot = 1.60 kJ/kg·K,
s vib = 0.0003 kJ/kg·K) and (s/R) NO = 23.9.
18.13 POLYATOMIC MAXWELL-BOLTZMANN GASES
Polyatomic gases are divided into two categories of molecular geometry: linear and nonlinear. Linear polyatomic
molecules have only 2 degrees of rotational freedom, as in the case of diatomic molecules. However, they have
3b−5 degrees of vibrational freedom, where b is the number of atoms in the molecule. Therefore, the equations
used to calculate the translational and rotational contributions to the molecular energy are the same as those
used for the diatomic molecule (i.e., parts a and b of Eqs. (18.49), and (18.51) through (18.54)). The equations
for the vibrational contribution to the molecular energy of a linear polyatomic Maxwell-Boltzmann gas are
u vib = ðu o Þ vib
+ R∑ 3b−5
i=1
h vib = u vib = ðu o Þ vib + R∑ 3b−5
Θ vi
½expðΘ vi /TÞ− 1Š
i=1
Θ vi
½expðΘ vi /TÞ− 1Š
(18.55a)
(18.55b)
ðc v Þ vib = R∑ 3b−5
i=1
ðΘ vi /TÞ 2 ½expðΘ vi /TÞŠ
½expðΘ vi /TÞ− 1Š 2 (18.55c)
and
ðc p Þ vib = ðc v Þ vib = R∑ 3b−5
i=1
ðΘ vi /TÞ 2 ½expðΘ vi /TÞŠ
½expðΘ vi /TÞ− 1Š 2 (18.55d)
s vib = R∑ 3b−5
i=1
ln ½1 − expð−Θ vi /TÞŠ −1 + ðΘ vi /TÞ/ ½expðΘ vi /TÞ− 1Š
where the vibrational internal energy at absolute zero temperature is now found from
(18.55e)
since there are now 3b−5 characteristic vibrational temperatures Θ vi.
3b−5
ðu o Þ vib = ∑ RΘ vi /2 (18.56)
i=1
EXAMPLE 18.10
Carbon dioxide is a linear triatomic molecule that has the following characteristic temperatures
Θ r = 0:562 K
Θ v1 = 1932 K
Θ v2 = Θ v3 = 960: K
Θ v4 = 3380 K