Chemistry · Thermodynamics & Thermochemistry

JEE Main 2026 — 28 January, Morning Shift — Question 44

$20.0 \mathrm{dm}^{3}$ of an ideal gas ' X ' at 600 K and 0.5 MPa undergoes isothermal reversible expansion until pressure of the gas is 0.2 MPa . Which of the following option is correct? (Given: $\log 2=0.3010$ and $\log 5=0.6989$ )

Option A:
$\mathrm{w}=-9.1 \mathrm{~kJ}, \Delta \mathrm{U}=0, \Delta \mathrm{H}=0, \mathrm{q}=9.1 \mathrm{~kJ}$
Correct
Option B:
$\mathrm{w}=9.1 \mathrm{~J}, \Delta \mathrm{U}=9.1 \mathrm{~J}, \Delta \mathrm{H}=0 ; q=0$
Option C:
$\mathrm{w}=+4.1 \mathrm{~kJ}, \Delta \mathrm{U}=0, \Delta \mathrm{H}=0 ; q=-4.1 \mathrm{~kJ}$
Option D:
$\mathrm{w}=-3.9 \mathrm{~kJ}, \Delta \mathrm{U}=0, \Delta \mathrm{H}=0 ; \mathrm{q}=3.9 \mathrm{~kJ}$

Answer: A

Step-by-step solution

For isothermal reversible process $\Delta \mathrm{U}=\Delta \mathrm{H}=0$

\begin{aligned} & \mathrm{W}_{\text {iso }}=-\mathrm{p}_{1} \mathrm{v}_{1} \ln \frac{\mathrm{P}_{1}}{\mathrm{P}_{2}} & \mathrm{~W}_{\text {iso }}=-0.5 \times 10^{6} \times 20 \times 10^{-3} \ln \frac{0.5}{0.2} & \mathrm{~W}_{\text {iso }}=-0.5 \times 10^{6} \times 20 \times 10^{-3} \times 2.303 \times(.6989-.3010) & \mathrm{W} \simeq-9.1 \mathrm{~kJ} & \mathrm{q}=-\mathrm{W}=9.1 \mathrm{~kJ} \end{aligned}

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 28 January, Morning Shift
Subject: Chemistry
Chapter: Thermodynamics & Thermochemistry
Topic: Work Done in Different Cases