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1 mole of an ideal gas expands isothermally and reversibly by decreasing pressure form $210 \mathrm{kPa}$ to $105 \mathrm{kPa}$ at $300 \mathrm{~K}$. What is the work done? $\left(\mathrm{R}=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\right)$
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The correct answer is:
1729 J
For Isothermal reversible process,
$$
\begin{aligned}
& \mathrm{W}=-2.303 \mathrm{nRT} \log \frac{\mathrm{P}_1}{\mathrm{P}_2} \\
& =-2.303 \times 1 \times 8.314 \times 300 \log \frac{210}{105} \\
& =-2.303 \times 8.314 \times 300 \times \log 2 \\
& =-1729 \mathrm{~J}(\log 2=0.3010) \\
& |\mathrm{W}|=1729 \mathrm{~J}
\end{aligned}
$$
$$
\begin{aligned}
& \mathrm{W}=-2.303 \mathrm{nRT} \log \frac{\mathrm{P}_1}{\mathrm{P}_2} \\
& =-2.303 \times 1 \times 8.314 \times 300 \log \frac{210}{105} \\
& =-2.303 \times 8.314 \times 300 \times \log 2 \\
& =-1729 \mathrm{~J}(\log 2=0.3010) \\
& |\mathrm{W}|=1729 \mathrm{~J}
\end{aligned}
$$
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