When 5 moles of He gas expand isothermally and reversibly at 300 K from 10 litre to 20 litre, the magnitude of the maximum work obtained is __________ J. [nearest integer] (Given : R = 8.3 J K$-$1 mol$-$1 and log 2 = 0.3010)
Answer (integer)
8630
Solution
$$
\begin{aligned}
\mathrm{W}_{\mathrm{rev}} &=-2.303 ~\mathrm{nRT} ~\log _{10}\left(\frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}\right) \\\\
&=-2.303 \times 5 \times 8.3 \times 300 \times \log _{10}\left(\frac{20}{10}\right) \\\\
& \simeq-8630 \mathrm{~J}
\end{aligned}
$$
About this question
Subject: Chemistry · Chapter: Thermodynamics · Topic: Zeroth and First Law
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