Question 5.3: For the ideal-gas state and constant heat capacities, Eq. (3...
For the ideal-gas state and constant heat capacities, Eq. (3.23b) for a reversible adiabatic (and therefore isentropic) process can be written:
T P^{(1-\gamma )/\gamma}= const (3.23b)
\frac{T_2}{T_1}=\left(\frac{P_2}{P_1}\right)^{(\gamma-1) / \gamma}
Show that this same equation results from application of Eq. (5.10) with ΔS^{ig} = 0.
\frac{\Delta S^{i g}}{R} = \int_{T_0}^T \frac{C_P^{i g}}{R} \frac{d T}{T} – \ln \frac{P}{P_0} (5.10)
Because C_P^{i g} is constant, Eq. (5.10) becomes:
0=\frac{C_P^{i g}}{R} \ln \frac{T_2}{T_1}-\ln \frac{P_2}{P_1}=\ln \frac{T_2}{T_1} – \frac{R}{C_P^{i g}} \ln \frac{P_2}{P_1}
By Eq. (3.12) for the ideal-gas state, with \gamma=C_P^{i g} / C_V^{i g} :
C_P^{i g} \equiv \frac{d H^{i g}}{d T}=\frac{d U^{i g}}{d T}+R=C_V^{i g}+R (3.12)
C_P^{i g}=C_V^{i g}+R \quad \text { or } \quad \frac{R}{C_P^{i g}}=\frac{\gamma – 1}{\gamma}
Thus,
\ln \frac{T_2}{T_1}=\frac{\gamma-1}{\gamma} \ln \frac{P_2}{P_1}
Exponentiating both sides of this equation leads to the given equation.