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Chemistry · Thermodynamics & Thermochemistry

JEE Main 2025 — 4 April, Morning Shift — Question 3

One mole of an ideal gas expands isothermally and reversibly from 10dm310 \mathrm{dm}^{3} to 20dm320 \mathrm{dm}^{3} at 300 K.ΔU,q300 \mathrm{~K} . \Delta \mathrm{U}, \mathrm{q} and work done in the process respectively are Given : R=8.3 J K−1 mol−1\mathrm{R}=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}

In 10=2.310=2.3

log⁡2=0.30\log 2=0.30

log⁡3=0.48\log 3=0.48

  1. Option A:

    0,21.84 kJ,21.84 kJ0,21.84 \mathrm{~kJ}, 21.84 \mathrm{~kJ}

  2. Option B:

    0,21.84 kJ,−1.726 J0,21.84 \mathrm{~kJ},-1.726 \mathrm{~J}

  3. Option C:

    0,−17.18 kJ,1.718 J0,-17.18 \mathrm{~kJ}, 1.718 \mathrm{~J}

  4. Option D:

    0,1.718 kJ,−1.718 kJ0,1.718 \mathrm{~kJ},-1.718 \mathrm{~kJ}

    Correct

Answer: D

Step-by-step solution

ΔU=0\Delta \mathrm{U}=0, for isothermal process

∴q=−w\therefore \mathrm{q}=-\mathrm{w}

w=−nRTln⁡(V2 V1)\mathrm{w}=-\mathrm{nRT} \ln \left(\frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}\right)

w=−1×8.3×300×ln⁡(2010)w=-1 \times 8.3 \times 300 \times \ln \left(\frac{20}{10}\right)

w=−8.3×300×2.3×log⁡2w=-8.3 \times 300 \times 2.3 \times \log 2

w=−1.718 kJw=−1.718 kJw=-1.718 \mathrm{~kJ} \quad \mathrm{w}=-1.718 \mathrm{~kJ}

∴q=+1.718 kJ\therefore \mathrm{q}=+1.718 \mathrm{~kJ}

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
4 April, Morning Shift
Subject
Chemistry
Chapter
Thermodynamics & Thermochemistry
Topic
Work Done in Different Cases

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