Question

An ideal gas whose adiabatic exponent $\gamma$ is expanded according to the law $P=\alpha V$, where $\alpha$ is a constant. The initial volume of the gas is equal to $V_0$. As a result of expansion, the volume increases 4 times. Select the correct option for above information.

• A. Molar heat capacity of gas in the process is $\frac{R(\gamma +1)}{2(\gamma -1)}$
• B. Work done by the gas is $-15V_0^2\frac{\alpha }{2}$
• C. Molar heat capacity of gas in the process is $\frac{R(\gamma -1)}{2(\gamma +1)}$
• D. Work done by the gas is $+15V_0^2\frac{\alpha }{2}$

Answer 1

Answer: (A), (B)

The correct options are
A Molar heat capacity of gas in the process is $\frac{R(\gamma +1)}{2(\gamma -1)}$
B Work done by the gas is $-15V_0^2\frac{\alpha }{2}$

$P=\alpha V$
$P{V}^{-1}=\alpha$
We know for polytropic process
$P{V}^x=\text{constant}$
$\therefore x=-1$
We know $C=C_v-\frac{R}{x-1}$
$C=C_v+\frac{R}{2}$
$C=\frac{R}{\gamma -1}+\frac{R}{2}$
$C=\frac{R(\gamma +1)}{2(\gamma -1)}$

$dW=-PdV$
$dW=-\alpha VdV$
On integrating above equation
$W=-\alpha [\frac{(4V_0{)}^2}{2}-\frac{V_0^2}{2}]$
$W=-\frac{\alpha }{2}[16V_0^2-V_0^2]$
$W=-15V_0^2\frac{\alpha }{2}$