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Work out the heat change (cal) when $40 \mathrm{~g}$ of He gas at $27^{\circ} \mathrm{C}$ undergoes isothermal and reversible compression from initial pressure of $1 \mathrm{~atm}$ to $10 \mathrm{~atm}\left(\mathrm{R}=2 \mathrm{cal} \mathrm{K}^{-1} \mathrm{~mol}^{-1}\right)$.
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The correct answer is:
$-13.818 \mathrm{kcal}$
$\begin{aligned} \mathrm{q} &=-\mathrm{w}_{\mathrm{rev}}=-\left(-2.303 \mathrm{nRT} \log \frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}\right) \\ &=2.303 \times \frac{40}{4} \times 2 \times 300 \log \frac{1}{10}=-13.82 \mathrm{kcal} \end{aligned}$
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