An ideal gas expands reversibly and isothermally from state ‘A’ (2 lit, 10 atm) to state ‘B’ (20 lit, 1 atm). Work done during the process is

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-9.3 kJ

-6.975 kJ

-2.325 kJ

-4.65 kJ

answer is A.

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Detailed Solution

$W_{rev} = - 2.303 nRTlog \frac{V_{f}}{V_{i}}$
But PV = nRT
$\begin{array}{l} W_{rev} = - 2.303 PVlog \frac{V_{f}}{V_{i}} \\ = - 2.303 (20) \log \frac{20}{2} \end{array}$
=-46.06 liter atm-46.06(0.101kJ)=-4.65kJ

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