One mole of an ideal gas at temperature $T_{1}$ expends according to the law $P / V^{2} = a$ (constant). the work done by the gas till the temperature of gas become $T_{2}$ is:

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

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

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

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

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Solution

The correct option is B $\frac{1}{3} R (T 2 - T 1)$
The work done by the gas is given by
$W = \int P . d V$ . . . . .(1)
From the question.
$\frac{P}{V^{2}} = a$
$P = a V^{2}$
From equation (1), we get
$W = \int a V^{2} d V = a \frac{V^{3}}{3} |_{V} {_{1}}^{V}_{2}$
$W = \frac{a (V_{2}^{3} - V_{1}^{3})}{3}$ . . . . . . .(2)
From ideal gas equation, $P V = n R T$
Substitute, $n = 1$ and $P = a V^{2}$
$a V^{2} . V = 1. R T$
$a V^{3} = R T$
Now equation (2) become,
$W = \frac{R}{3} (T_{2} - T_{1})$
The correct option is B.

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