JEE MAIN - Chemistry (2025 - 4th April Morning Shift - No. 9)
One mole of an ideal gas expands isothermally and reversibly from $10 \mathrm{dm}^3$ to $20 \mathrm{dm}^3$ at 300 K . $\Delta \mathrm{U}, \mathrm{q}$ and work done in the process respectively are
Given: $\mathrm{R}=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$
$\ln 10=2.3$
$\log 2=0.30$
$$\log 3=0.48$$
$0,21.84 \mathrm{~kJ},-1.726 \mathrm{~J}$
$0,21.84 \mathrm{~kJ}, 21.84 \mathrm{~kJ}$
$0,1.718 \mathrm{~kJ},-1.718 \mathrm{~kJ}$
$0,-17.18 \mathrm{~kJ}, 1.718 \mathrm{~J}$
Explanation
$$\begin{aligned}
& (10 \mathrm{~L}, 300 \mathrm{~K}) \xrightarrow{\mathrm{n}=1}(20 \mathrm{~L}, 300 \mathrm{~K}) \\
& -\mathrm{q}=\mathrm{w}=-\mathrm{nRT} \ln \frac{\mathrm{~V}_2}{\mathrm{~V}_1} \\
& =-8.3 \times 300 \times \ln \left(\frac{20}{10}\right) \\
& =-1.718 \mathrm{~kJ} \\
& \Rightarrow \mathrm{q}=1.718 \mathrm{~kJ} \\
& \mathrm{w}=-1.718 \mathrm{~kJ} \\
& \Delta \mathrm{U}=0(\because \Delta \mathrm{~T}=0)
\end{aligned}$$
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