One mole of an ideal gas at temperature $T_{1}$ expands according to the law $\frac{P}{V^{2}} = a$ (constant). The work done by the gas till temperature of gas becomes $T_{2}$ is

\frac{1}{5} R (T_{2} - T_{1})

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\frac{1}{4} R (T_{2} - T_{1})

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\frac{1}{3} R (T_{2} - T_{1})

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\frac{1}{2} R (T_{2} - T_{1})

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Solution

The correct option is C $\frac{1}{3} R (T_{2} - T_{1})$
Work done by the gas is given by
$\int d W = \int P d V$

$W = a \int_{V_{1}}^{V_{2}} V^{2} d V = \frac{a}{3} (V_{2}^{3} - V_{1}^{3})$

$W = \frac{R}{3} (T_{2} - T_{1})$

Alternate solution:
Given:
$\frac{P}{V^{2}} = a \Rightarrow P V^{- 2} = a$
So, gas is expanding via polytropic process where polytropic exponent $η = - 2$
Now, work done in polytropic process is given by
$W = \frac{- n R Δ T}{η - 1} = \frac{- R (T_{2} - T_{1})}{- 2 - 1}$
or, $W = \frac{1}{3} R (T_{2} - T_{1})$
Hence, option $(B)$ is correct.

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