Q24 / 75 · JEE Main 2025 (4 Apr Shift 1) · Chemistry
One mole of an ideal gas expands isothermally and reversibly from $10 \mathrm{dm}^3$ to $20 \mathrm{dm}^3$ at $300 \mathrm{~K}. \Delta \mathrm{U}$, q and work done in the process respectively are :
Given : $\mathrm{R}=8.3 \mathrm{JK}^{-1}$ and $\mathrm{mol}^{-1}$
$\begin{aligned}<br />& \text { In } 10=2.3 \\<br>& \log 2=0.30 \\<br>& \log 3=0.48<br />\end{aligned}$
Given : $\mathrm{R}=8.3 \mathrm{JK}^{-1}$ and $\mathrm{mol}^{-1}$
$\begin{aligned}<br />& \text { In } 10=2.3 \\<br>& \log 2=0.30 \\<br>& \log 3=0.48<br />\end{aligned}$
- A$0,21.84 \mathrm{~kJ},-1.26 \mathrm{~kJ}$
- B$0,-17.18 \mathrm{~kJ}, 1.718 \mathrm{~J}$
- C$0,21.84 \mathrm{~kJ}, 21,84 \mathrm{~kJ}$
- Correct answerD$0,178 \mathrm{~kJ},-1.718 \mathrm{~kJ}$
Solution
$\begin{aligned} & (10 \mathrm{~L}, 300 \mathrm{~K}) \xrightarrow{\mathrm{n}=1}(20 \mathrm{~L}, 300 \mathrm{~K}) \\ & -\mathrm{q}=\mathrm{w}=-\mathrm{nRT} \ln \frac{\mathrm{V}_2}{\mathrm{~V}_1} \\ & =-8.3 \times 300 \times \ln \left(\frac{20}{10}\right) \\ & =-1.718 \mathrm{~kJ} \\ & \Rightarrow \mathrm{q}=1.718 \mathrm{~kJ} \\ & \mathrm{w}=-1.718 \mathrm{~kJ} \\ & \Delta \mathrm{U}=0(\because \Delta \mathrm{~T}=0)\end{aligned}$
Hence, the correct answer is option: 'D'.