A polyatomic gas $(\gamma=4 / 3)$ is compressed to $\left(\frac{1}{8}\right)^{\text {th }}$ of its volume…
A polyatomic gas $(\gamma=4 / 3)$ is compressed to $\left(\frac{1}{8}\right)^{\text {th }}$ of its volume adiabatically.If its initial pressure is $\mathrm{P}_0$, its new pressure will be
$2 \mathrm{P}_0$
$8 \mathrm{P}_0$
$6 \mathrm{P}_0$
$16 \mathrm{P}_0$
Solution
For adiabatic expression, we have
$\begin{aligned}
& \mathrm{P}_2 \mathrm{~V}_2^\gamma=\mathrm{P}_1 \mathrm{~V}_1^\gamma \\
& \therefore \frac{\mathrm{P}_2}{\mathrm{P}_1}=\left(\frac{\mathrm{V}_1}{\mathrm{~V}_2}\right)^\gamma=(8)^{\frac{4}{3}}=16 \\
& \therefore \mathrm{P}_2=16 \mathrm{P}_1=16 \mathrm{P}_0
\end{aligned}$
*