The pressure of an ideal gas varies with volume as $P = αV$ , where $α$ is a constant. One mole of the gas is allowed to undergo expansion such that its volume becomes ‘m’ times its initial volume. The work done by the gas in the process is:

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$\frac{αV}{2} (m^{2} - 1)$

$\frac{{αV}^{2}}{2} (m^{2} - 1)$

$\frac{α^{2} V^{2}}{2} (m^{2} - 1)$

$\frac{α}{2 V} (m^{2} - 1)$

answer is B.

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Detailed Solution

$\begin{array}{l} W = \int PdV = \int_{V_{1} = V}^{V_{f} = mV} αVdV \\ W = {[\frac{{αV}^{2}}{2}]}_{V}^{mV} \\ W = \frac{{αV}^{2}}{2} (m^{2} - 1) \end{array}$

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