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At $\mathrm{T}(\mathrm{K}) 2$ mole of an ideal gas is allowed to expand reversibly and isothermally from a pressure of 10 atmospheres to 1 atmosphere. The work done (in $\mathrm{kJ}$ ) is $\left(\mathrm{R}=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right.$ )
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The correct answer is:
$-3.82 \times 10^{-2} \times \mathrm{T}$
$\mathrm{W}=-\mathrm{nRT} \ln \frac{\mathrm{P}_2}{\mathrm{P}_1}=-2.303 \mathrm{nRT} \log \frac{\mathrm{P}_2}{\mathrm{P}_1}$
$\begin{aligned} & \mathrm{W}=-2.303 \times 2 \times 8.314 \times \mathrm{T} \log \frac{10}{1} \\ & \mathrm{~W}=38.29 \mathrm{~J} \\ & \mathrm{~W}=3.82 \times 10^{-2} \mathrm{~kJ}\end{aligned}$
$\begin{aligned} & \mathrm{W}=-2.303 \times 2 \times 8.314 \times \mathrm{T} \log \frac{10}{1} \\ & \mathrm{~W}=38.29 \mathrm{~J} \\ & \mathrm{~W}=3.82 \times 10^{-2} \mathrm{~kJ}\end{aligned}$
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