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The equation of a state of a gas is given by $p (V - b) = n R T$ . If 1 mole of a gas is isothermally expanded from volume $V$ and $2 V$ , the work done during the process is

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$R T \ln |\frac{2 V - b}{V - b}|$

$R T \ln |\frac{V - b}{2 V - b}|$

$R T \ln |\frac{V - b}{V}|$

$R T \ln |\frac{V}{V - b}|$

answer is A.

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Detailed Solution

$W = \int_{V}^{2 V} P d V$

$= \int_{V}^{2 V} (\frac{n R T}{V - b}) d V$

$= n R T \ln (\frac{2 V - b}{V - b})$

Put $n = 1$

$∴ W = R T \ln (\frac{2 V - b}{V - b})$

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