Q.3 Derive an expression for maximum work obtainable during isothermal reversible expansion of an ideal gas from initial volume (V1) to final volume (V2).
Best answer
Answer: \(W_{max} = -2.303\,nRT\log_{10}\dfrac{V_2}{V_1}\)
For an infinitesimal reversible expansion, the external pressure differs from the gas pressure only by an infinitesimal amount, so the work done is \(dW = -P\,dV\).
Total work for expansion from \(V_1\) to \(V_2\): \(W_{max} = -\int_{V_1}^{V_2} P\,dV\)
For n mol of an ideal gas, \(P = \dfrac{nRT}{V}\). At constant temperature:
\(W_{max} = -nRT\int_{V_1}^{V_2}\dfrac{dV}{V} = -nRT\ln\dfrac{V_2}{V_1} = -2.303\,nRT\log_{10}\dfrac{V_2}{V_1}\)
Since at constant T, \(P_1V_1 = P_2V_2\), i.e. \(\dfrac{V_2}{V_1} = \dfrac{P_1}{P_2}\), we can also write \(W_{max} = -2.303\,nRT\log_{10}\dfrac{P_1}{P_2}\).
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