Modern Engineering Thermodynamics

18.12 Diatomic Maxwell-Boltzmann Gases 755
and
ðc p Þ trans = 5 2 R
(18.53a)
ðc p Þ rot = ðc v Þ rot = R
(18.53b)
Thus,
ðc p Þ vib
= ðc v Þ vib
= R ð Θ v/TÞ 2 ½expðΘ v /TÞŠ
½expðΘ v /TÞ− 1Š 2 (18.53c)
c p = ðc p Þ trans
+ ðc p Þ rot
+ ðc p Þ vib
= 7 2 R + R ð Θ v/TÞ 2 ½expðΘ v /TÞŠ
½expðΘ v /TÞ− 1Š 2 (18.53d)
Finally,
n
s trans = R ln 2πm/ħ 2 3/2
ðkTÞ 5/2 /p + 5 o
2
(18.54a)
s rot = Rfln ½T/ðσ Θ r ÞŠ+ 1g (18.54b)
n
s vib = R ln 1 − expð−Θ v /TÞ o
−1
+ ð Θv /TÞ/ ½expðΘ v /TÞ− 1Š
(18.54c)
Thus,
s = s trans + s rot + s vib
n
= R ln ð2πm/ħ 2 Þ 3/2 ðkTÞ 5/2 /p + 5 o
+ Rfln ½T/ðσ Θ r ÞŠ+ 1g
2
n
+ R ln 1 − expð−Θ v /TÞ o
−1
+ ð Θv /TÞ/ ½expðΘ v /TÞ− 1Š
(18.54d)
These equations produce reasonably accurate results for diatomic gases at moderate to high temperatures, as
illustrated by the following example.
EXAMPLE 18.9
To test the diatomic Maxwell-Boltzmann gas equations just developed, compute the value of c v /R for nitrous oxide (NO) at
20.0°C and compare it with the measured result given in Table 18.3.
Solution
For a diatomic Maxwell-Boltzmann gas, the constant volume specific heat is given by Eq. (18.52d) as
then,
From Table 18.8, we find that Θ v = 2740 K for NO, so,
c v = 5 2 R + RðΘ v/TÞ 2 expðΘ v /TÞ
½expðΘ v /TÞ − 1Š 2
c v
R = 5 2 + ðΘ v/TÞ 2 expðΘ v /TÞ
½expðΘ v /TÞ − 1Š 2
2
2
2740
2740
c v
= 5 R NO 2 + 20:0 + 273:15 exp
20:0 + 273:15
= 2:51
exp
2740
− 1
20:0 + 273:15
as in Table 18.3. Note that, since R NO = Y/M NO = 8.3143/30.01 = 0.2771 kJ/kg·K, then (c v ) NO = 0.2771 × 2.51 = 0.695 kJ/kg·K
at 20.0°C.
(Continued )