Change in Entropy — Question 2

Question 2

Pressure-temperature data give an equivalent entropy formula

For a fixed amount of ideal gas with constant molar capacities, use $\Delta S=nC_V\ln(T_f/T_i)+nR\ln(V_f/V_i)$ and $C_p=C_V+R$, with $R=8.314$ J/(mol K). One mole has $C_V=5R/2$. Its temperature doubles while its pressure quadruples.

1. Eliminate the volume ratio and derive $\Delta S=nC_p\ln(T_f/T_i)-nR\ln(p_f/p_i)$. Evaluate it.

2. For a temperature doubling, find the pressure ratio giving zero entropy change. Can pressure increase while entropy also increases?

Question 2 - Solution

Pressure-temperature data give an equivalent entropy formula

(a) Method and calculation

$$V_f/V_i=(T_f/T_i)/(p_f/p_i).$$$$\Delta S=n(C_V+R)\ln(T_f/T_i)-nR\ln(p_f/p_i)=R[\tfrac72\ln2-\ln4]\approx8.64424\,\mathrm{J/K}.$$

(b) Interpretation and further calculation

$$\Delta S=0\ \Rightarrow\ p_f/p_i=(T_f/T_i)^{C_p/R}=2^{7/2}\approx11.3137.$$$$1<p_f/p_i<2^{7/2}\ \Rightarrow\Delta S>0\quad(T_f/T_i=2).$$Compression and heating contribute oppositely; pressure increase alone cannot determine the entropy sign.

Physical check

The use of $C_p$ follows algebraically from the ideal equation of state; it does not imply a constant-pressure process.