Question

A polyatomic gas $(\gamma =\frac{4}{3})$ is compressed to $\frac{1}{27}$ of its volume adiabatically and reversibly. If its initial pressure is $P_0$ its new pressure will be

• A. $64P_0$
• B. $21P_0$
• C. $27P_0$
• D. $81P_0$

Answer 1

The correct option is D $81P_0$

For adiabatic process,

$\frac{P_2}{P_1}={\left(\frac{V_1}{V_2}\right)}^{\gamma }$
where $P_1=P_0$ initial pressure
$P_2=$ final pressure
$V_1=$ initial volume
$V_2=$ final volume $=\frac{1}{27}{V}_1$
$\gamma =4/3$
so,
$P_2=P_1{\left(\frac{V_1}{V_2}\right)}^{\gamma }=P_0(27{)}^{4/3}=81P_0$