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For an isothermal reversible expansion of an ideal gas from volume $V_1$ to $V_2$, the work done **by** the gas is:
A$-2.303\,nRT\,\log\dfrac{V_2}{V_1}$
B$+2.303\,nRT\,\log\dfrac{V_2}{V_1}$
C$0$
D$-p_\text{ext}(V_2 - V_1)$
Answer & Solution
Correct answer: B. $+2.303\,nRT\,\log\dfrac{V_2}{V_1}$
For an isothermal reversible expansion, $W_\text{by gas} = +nRT\ln(V_2/V_1) = +2.303\,nRT\log(V_2/V_1)$. In the IUPAC convention $W_\text{on system} = -nRT\ln(V_2/V_1)$ is negative, but the question asks for work done *by* the gas, which is the positive of this.
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