Question

An ideal gas with adiabatic exponent $\gamma$ , is according to the law P =$\alpha V$ where $\alpha$ is a constant. The initial volume of the gas is $V_0$. As a result increases $\eta$ times. Find the increment in internal energy and work done.

Answer 1

The process is given as $P=\alpha V$

$⟹$ $P{V}^{-1}=\alpha =constant$

Comparing with $P{V}^m=constant$ we get $m=-1$

Given : $V_1=V_o$ $V_2=\eta V_o$

$⟹$ $P_1=\alpha V_o$ and $P_2=\alpha \eta V_o$

Change in internal energy $\Delta U=\frac{R}{\gamma -1}\times n(T_2-T_1)=\frac{P_2{V}_2-P_1{V}_1}{\gamma -1}$

$⟹$ $\Delta U=\frac{\left[\alpha \eta V_o\right(\eta V_o)-\alpha V_o(V_o\left)\right]}{\gamma -1}=\frac{\alpha V_o^2({\eta }^2-1)}{\gamma -1}$

Work done $W=\frac{P_2{V}_2-P_1{V}_1}{1-m}$

$\therefore$ $W=\frac{\left[\alpha \eta V_o\right(\eta V_o)-\alpha V_o(V_o\left)\right]}{1-(-1)}=\frac{\alpha V_o^2({\eta }^2-1)}{2}$