Work done by a system under isothermal change from a volume $V_{1}$ to $V_{2}$ for a gas, which obeys vander Waals equation $(V - β n) (P + \frac{a n^{2}}{V}) = n R T$ is

n R T l o g_{e} (\frac{V_{2} - n β}{V_{1} - n β}) + a n^{2} (\frac{V_{1} - V_{2}}{V_{1} V_{2}})

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n R T l o g_{10} (\frac{V_{2} - α β}{V_{1} - α β}) + α n^{2} (\frac{V_{1} - V_{2}}{V_{1} V_{2}})

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n R T l o g_{e} (\frac{V_{2} - n α}{V_{1} - n α}) + β n^{2} (\frac{V_{1} - V_{2}}{V_{1} V_{2}})

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n R T l o g_{e} (\frac{V_{2} - n β}{V_{1} - n β}) + α^{2} (\frac{V_{1} V_{2}}{V_{1} - V_{2}})

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Solution

The correct option is A $n R T l o g_{e} (\frac{V_{2} - n β}{V_{1} - n β}) + a n^{2} (\frac{V_{1} - V_{2}}{V_{1} V_{2}})$
Given Vander Waals equation $(V - β n)$ ( $P$ $+$ $\frac{a n^{2}}{V^{2}}$ ) $=$ $n R T$
Work done by the system( $W$ ) = $-$ $\int_{V_{1}}^{V_{2}} P d V$
$=$ $\int_{V_{2}}^{V_{1}} P d V$
From the Vander Waals equation:
$P$ $=$ $\frac{n R T}{(V - β n)}$ $-$ $\frac{a n^{2}}{V^{2}}$
Substituting $^{'} P^{'}$ in the Work done, we get
$W$ $=$ $\int_{V_{2}}^{V_{1}}$ ( $\frac{n R T}{(V - β n)}$ $-$ $\frac{a n^{2}}{V^{2}}$ ) $d V$
By integrating we get,
$W$ $=$ $[$ $n R T$ ${log}_{e} (V - β n)$ $-$ ( $(-) \frac{a n^{2}}{V}$ ) ] $_{V_{2} \to V_{1}}$
$W$ $=$ $n R T$ ${log}_{e}$ $($ $\frac{V_{2} - β n}{V_{1} - β n}$ $)$ $+$ $a n^{2}$ $($ $\frac{V_{1} - V_{2}}{V_{1} V_{2}}$ $)$
Hence, the Correct Option is $^{'} A^{'}$ .

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