Question

The pressure of an ideal gas varies with volume as $P=\alpha V$, where $\alpha$ is a constant. One mole of the gas is allowed to undergo expansion such that its volume becomes $\text{m}$ times its initial volume. The work done by the gas in the process is

• A. $\frac{\alpha V^2}{2}(m^2-1)$
• B. $\frac{\alpha }{2}(m^2-1)$
• C. $\frac{{\alpha }^2{V}^2}{2}(m^2-1)$
• D. $\frac{\alpha V}{2}(m^2-1)$

Answer 1

The correct option is A $\frac{\alpha V^2}{2}(m^2-1)$

The work done by a gas is given by
$W=\underset{V_1}{\overset{V_2}{\int }}PdV$

In this case it is given that $P=\alpha V$. Also if the initial volume $V_1$ is $V$ then the final volume $V_2$ is $mV$

So, $W=\underset{V}{\overset{mV}{\int }}\alpha VdV=\alpha {\left[\frac{V^2}{2}\right]}_V^{mV}$

$E=\frac{\alpha }{2}[m^2{V}^2-V^2]$

$W=\frac{\alpha V^2}{2}(m^2-1)$

So, option $\left(a\right)$ is correct.