Question 10.1: A pressure vessel contains a 50/50 mixture of He and Ar by v...
A pressure vessel contains a 50/50 mixture of He and Ar by volume at 500 K and 10 bar. Assuming ideal gas behavior, calculate the specific entropy of the mixture (J/K mol). Note that, for all inert gases, the first excited electronic state is at least 10 eV above the ground electronic state.
From Eq. (10.28), the specific entropy for a 50/50 mixture of He and Ar by volume is
S = \sum\limits_{i=1}^{M}{n_{i}}s_{i}. (10.28)
\frac{s}{R} = \sum\limits_{i=1}^{2}{x_{i}}\left(\frac{s_{i}}{R}\right) = 0.5\left(\frac{s_{1}}{R}\right) + 0.5\left(\frac{s_{2}}{R}\right),
where s_{1} is the specific entropy of He and s_{2} is the specific entropy of Ar. To calculate the specific entropy for these two monatomic gases, we require their translational and electronic contributions, as given by Eqs (9.12) and (9.20):
\left(\frac{s}{R}\right)_{tr} = \frac{5}{2}\ln T+\frac{3}{2}\ln M-\ln P-1.1516
\left(\frac{s}{R}\right)_{el} = \frac{Z^{\prime}_{el}}{Z_{el}}+\ln{Z_{el}} = \ln g_{0}.
From Appendix J.1, the ground-state term symbol for all noble gases is ¹S_{0}, from which we conclude that the ground-state degeneracy is g_{0} = 1. Therefore, the only effective contribution to the entropy comes from the translational mode. Now, because each component of this ideal gas mixture acts as if it alone occupies the pressure vessel at 500 K, the pressure used in the Sackur–Tetrode equation to determine the translational contribution of each component must be the partial pressure, which is P_{1} = P_{2} = 5 bar. Hence, we have
\frac{s_{1}}{R} = \frac{5}{2} ln(500) + \frac{3}{2}ln(4.0026) − ln (5) − 1.1516
\frac{s_{2}}{R} = \frac{5}{2} ln(500) + \frac{3}{2}ln(39.948) − ln (5) − 1.1516
for helium and argon, respectively. On this basis, the dimensionless entropy for the mixture becomes
\frac{s}{R} = \frac{5}{2}ln (500) + \frac{3}{4}[ln (4.0026) + ln (39.948)] − ln(5) − 1.1516 = 16.581 .
Hence, evaluating the specific entropy for the mixture, we obtain
s = 16.581 (8.3145 J/K · mol) = 137.865 J/K · mol.