Modern Engineering Thermodynamics

12.9 Mixtures of Real Gases 437
Solution
a. For ideal gas mixture behavior,
and
v m = RT m
p m
b. From Table C.12a, we find
= ½1545:35 ft .lbf/ðlbmole.RÞŠð−100: + 459:67 RÞ
ð1500: lbf=in 2 Þð144 in 2 /ft 2 Þ
2:57 − 1:315
% error = ð100Þ = 95:4% high
1:315
= 2:57 ft 3 /lbmole
ðp c Þ N2
= 492 psia
ðT c Þ N2
= 227:1R
and
ðP c Þ CH4 = 673 psia
ðT c Þ CH4
= 343:9R
Since the mixture specific volume is the unknown here, it must be determined by a trial and error method, using the
reduced pseudospecific volume v′ Ri , which, as was shown earlier, can be written in a variety of forms, for example, as
v′ Ri = v m p ci /ðx i RT ci Þ
We assume values for v m ; find (Z Dm ) N ,(Z D ) CH ,andZ Dm , then check the assumption with v m ¼ Z Dm =RT m /p m . Assume
v m ¼ 1:51ft 3 / lbmole, then
and
Then,
and
ðv′ R Þ N2
=
ðv′ R Þ CH4 =
From Figure 7.6 in Chapter 7, we find that, for these values,
and
Then, from Eq. (12.36), we have
Now, checking the v m assumption,
ð1:51Þð492Þð144Þ
ð0:300Þð1545:35Þð227:1Þ = 1:02
ð1:51Þð673Þð144Þ
ð0:700Þð1545:35Þð343:9Þ = 0:393
ðT R Þ N2
= ð −100: + 459:67Þ/227:1 = 1:58
ðT R Þ CH4
= ð −100: + 459:67Þ/343:9 = 1:05
ðZ D Þ N2
= 0:91
ðZ D Þ CH4
= 0:39
Z Dm = 0:300ð0:91Þ + 0:700ð0:39Þ = 0:59
v m = Z Dm =RT m /p m =
= 1:52 ft 3 /lbmole
0:59ð1545:35Þð−100: + 459:67Þ
ð1500:Þð144Þ
which is close enough to our original assumption. Then,
1:52 − 1:315
% error = ð100Þ = 15:6 % high
1:315
(Continued )