Physics - Heat & Thermo Question with Solution | TestHub

Change in entropy is given by $dS=\frac{dQ}{T}$. Since, the process is $dS=0$, $dQ=0$. Therefore, constant entropy means the process is adiabatic.

$P{V}^{\gamma }=$constant

$\Rightarrow P_1{V}_1^{\gamma }=P_2{V}_2^{\gamma }$

For a monatomic gas $\gamma =\frac{5}{3}$

$\Rightarrow P_1{V}_1^{\frac{5}{3}}=P_2{\left(\frac{V_1}{8}\right)}^{\frac{5}{3}}$

$\Rightarrow P_1{V}_1^{\frac{5}{3}}=\frac{P_2 V_1^{{\displaystyle \frac{5}{3}}}}{32}$

$\Rightarrow P_2=32P_1$