Modern Engineering Thermodynamics

18.13 Polyatomic Maxwell-Boltzmann Gases 757
Determine the specific internal energy, specific enthalpy, and specific entropy of CO 2 at a temperature of 1000. K and a pressure
of 1.00 atm.
Solution
The mass of the CO 2 molecule is
and the gas constant for CO 2 is
m = M/N o = 44:01/6:023 × 10 26 = 7:31 × 10 −26 kg/molecule
R = R/M = 8:3143/44:01 = 0:1889 KJ/ ðkg⋅KÞ
Equation (18.56) gives the vibrational specific internal energy at absolute zero temperature as
ðu o Þ vib = ð0:1889Þð1932 + 960: + 960: + 3380Þ/2 = 683 kJ/kg
and Eq. (18.55a) gives the vibrational component of the specific internal energy as
u vib = 683 + ð0:1889Þfð1932Þ½expð1:932Þ−1
+ 2ð960:
Þ½expð0:960:
Þ−1
Š −1
Š −1
+ ð3380Þ½expð3:380Þ−1Š −1 g = 992 kJ/kg
The translational and rotational components are given by Eqs. (18.49a) and (18.49b) as
Then,
The specific enthalpy is now given simply by
u trans = 3 2 RT = 3 2 ð0:1889Þð1000:
Þ = 283:4 kJ/kg
u rot = RT = ð0:1889Þð1000:
Þ = 188:9 kJ/kg
u = u trans + u rot + u vib = 283:4 + 188:9 + 992 = 1465 kJ/kg
h = u + RT = 1465 + ð0:1889Þð1000:
Þ = 1654 kJ/kg
The translational and rotational specific entropy values are calculated from Eqs. (18.54a) and (18.54b). First, we calculate
2πm/ħ 2 h i
3/2
ðkTÞ
5/2 /p = 2πð7:31 × 10 –26 Þ/6:626 ð × 10 –34 Þ 2 3/2
× ½ð1:38 × 10 –23 Þð1000ÞŠ 5/2 /101,325
= 2:36 × 10 8 per molecule
and then Eq. (18.54a) gives
h
s trans = ð0:1889Þ 1n 2:36 × 10 8 + 5 i
2
= 4:11 kJ/ ðkg⋅KÞ
and Eq. (18.54b) with σ = 2 from Table 18.9 gives
s rot = ð0:1889Þfln ½1000/ ð2Þð0:562ÞŠ+ 1g = 1:47 kJ/ðkg⋅KÞ
Equation (18.55e) is then used to find the vibrational component of the specific entropy as
Then, the specific entropy is
s vib = ð0:1889Þfln ½1 − expð− 1:932ÞŠ −1 + ð1:932Þ½expð1:932Þ−1
+ ln½l − expð− 0:960ÞŠ −1 + ð0:960Þ½expð0:960Þ−1Š − 1
+ ln½l − expð− 0:960ÞŠ −1 + ð0:960Þ½expð0:960Þ−1
+ ln½l − expð− 3:380ÞŠ −1 + ð3:380Þ½expð3:380Þ−1
= 0:527 kJ/ ðkg⋅KÞ
s = s trans + s rot + s vib = 4:11 + 1:47 + 0:527 = 6:11 kJ/ ðkg ⋅ KÞ
Š −1
Š −1
Š −1
(Continued )