ThermodynamicsMedium
Question
The magnitude of work done by one mole of a Van der Waals gas during its isothermal reversible expansion from volume V1 to V2 at temperature T K, is
Options
A.$RT\ln\left( \frac{V_{2}}{V_{1}} \right)$
B.$RT\ln\left( \frac{V_{2} - b}{V_{1} - b} \right)$
C.$RT\ln\left( \frac{V_{2} - b}{V_{1} - b} \right) + a\left( \frac{1}{V_{2}} - \frac{1}{V_{1}} \right)$
D.$RT\ln\left( \frac{V_{2} - b}{V_{1} - b} \right) - a\left( \frac{1}{V_{2}} - \frac{1}{V_{1}} \right)$
Solution
$W = - \int_{V_{1}}^{V_{2}}{P_{ext}.dv} = - \int_{V_{1}}^{V_{2}}{P.dv}$ (For reversible process)
$= - \int_{V_{1}}^{V_{2}}\left( \frac{nRT}{V - nb} - \frac{an^{2}}{V^{2}} \right).dV = - \left\lbrack nRT.\ln\frac{V_{2} - nb}{V_{1} - nb} + an^{2}\left( \frac{1}{V_{2}} - \frac{1}{V_{1}} \right) \right\rbrack$
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