Isentropic Process for Ideal Gases

Recall that the TdS Equations state that for ideal gases, we have:

$$s_2-s_1={\overline{c}}_v\ln \frac{T_2}{T_1}+R\ln \frac{v_2}{v_1}$$

For an isentropic process, $s_2=s_1$, so that $s_2-s_1=0$. This means

$$\begin{array}{r}\ln \left(\frac{T_2}{T_1}\right)=-\frac{R}{{\overline{c}}_v}\ln \left(\frac{v_2}{v_1}\right)\\ \ln \left(\frac{T_2}{T_1}\right)=\ln {\left(\frac{v_1}{v_2}\right)}^{R/{\bar{c}}_v}\end{array}$$

Recall that the specific heat ratio is given as $\frac{{\bar{c}}_p}{{\bar{c}}_v}=k$, and $R={\bar{c}}_p-{\bar{c}}_v$. Thus, we have:

$$\frac{R}{{\bar{c}}_v}=\frac{{\bar{c}}_p-{\bar{c}}_v}{{\bar{c}}_v}=k-1$$

So:

$$\begin{array}{rl}\frac{T_2}{T_1}& ={\left(\frac{v_1}{v_2}\right)}^{k-1}\\ T{v}^{k-1}& =\text{const}\end{array}$$

By the ideal gas equation, we also have:

$$P{v}^k=\text{const}$$

which is in the form of a polytropic process. Thus, the polytropic process $P{v}^k$ is isentropic (reversible and adiabatic) if $n=k$, where $k=\frac{{\bar{c}}_p}{{\bar{c}}_v}$.