The work done in the expansion of a real gas ( $n$ moles) from volume $V_{1}$ to $V_{2}$ , obeying Van der Waals' equation at temperature T where volume occupied by the molecules may be taken as negligible in comparison to the total volume of gas is :

= - n R T log \frac{V_{1}}{V_{2}} - n^{2} a [\frac{1}{V_{2}} - \frac{1}{V_{1}}]

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= - n R T log \frac{V_{2}}{V_{1}} + n^{2} a [\frac{1}{V_{1}} - \frac{1}{V_{2}}]

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= - n R T log \frac{V_{2}}{V_{1}} - n^{2} a [\frac{1}{V_{2}} - \frac{1}{V_{1}}]

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= - n R T log \frac{V_{2}}{V_{1}} - n^{2} a [\frac{1}{V_{1}} - \frac{1}{V_{2}}]

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Solution

The correct option is C $= - n R T log \frac{V_{2}}{V_{1}} - n^{2} a [\frac{1}{V_{2}} - \frac{1}{V_{1}}]$
For a real gas, $[P + \frac{n^{2} a}{V^{2}}] [V - n b] = n R T$
if $n b << V$ ,(as given), then
$[P + \frac{n^{2} a}{V^{2}}] V = n R T$
or $P = \frac{n R T}{V} - \frac{n^{2} a}{V^{2}}$
Now work done in reversible expansion is
$w = - \int P d V = \int_{V_{1}}^{V_{2}} \frac{n R T d V}{V} + \int_{V_{1}}^{V_{2}} \frac{n^{2} a}{V^{2}} . d V$
$= - n R T log \frac{V_{2}}{V_{1}} - [\frac{n^{2} a}{V_{2}} - \frac{n^{2} a}{V_{1}}]$

$= - n R T log \frac{V_{2}}{V_{1}} - n^{2} a [\frac{1}{V_{2}} - \frac{1}{V_{1}}]$

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The work done in the expansion of a real gas (n moles) from volume V1 to V2, obeying Van der Waals' equation at temperature T where volume occupied by the molecules may be taken as negligible in comparison to the total volume of gas is :

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The work done in the expansion of a real gas ( $n$ moles) from volume $V_{1}$ to $V_{2}$ , obeying Van der Waals' equation at temperature T where volume occupied by the molecules may be taken as negligible in comparison to the total volume of gas is :