For an ideal gas expanding isothermally from volume $V_1$ to $V_2$ at temperature $T$, the work done by the gas is:
A$nRT (V_2 - V_1)$
B$nRT \ln\dfrac{V_2}{V_1}$
C$\dfrac{nRT}{V_2 - V_1}$
D$nRT \ln\dfrac{V_1}{V_2}$
Answer & Solution
Correct answer: B. $nRT \ln\dfrac{V_2}{V_1}$
For an isothermal process, $P = \dfrac{nRT}{V}$. Work done by the gas equals $\int_{V_1}^{V_2} P\,dV = nRT \int_{V_1}^{V_2} \dfrac{dV}{V} = nRT \ln\dfrac{V_2}{V_1}$.
If $V_2 > V_1$ (expansion) the log is positive and the gas does positive work, as it should. Option D flips the limits and gives the wrong sign.
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