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Chemistry · Thermodynamics & Thermochemistry

NEET (UG) 2024 — Question 87

The work done during reversible isothermal expansion of one mole of hydrogen gas at 25∘C25^{\circ} \mathrm{C} from pressure of 20 atmosphere to 10 atmosphere is (Given R=2.0calK−1 mol−1\mathrm{R}=2.0 \mathrm{cal} \mathrm{K}^{-1} \mathrm{~mol}^{-1} )

  1. Option A:

    0 calorie

  2. Option B:
    • 413.14 calories
    Correct
  3. Option C:

    413.14 calories

  4. Option D:

    100 calories

Answer: B

Step-by-step solution

Wrev, iso =−2.303nRTlog⁡PiPfW_{\text {rev, iso }}=-2.303 n R T \log \frac{P_{i}}{P_{f}}

=−2.303×1×2×298×log⁡2=-2.303 \times 1 \times 2 \times 298 \times \log 2

=−2.303×1×2×298×0.3=-2.303 \times 1 \times 2 \times 298 \times 0.3

=−413.14 calories =-413.14 \text { calories }

Answer key and solution verified before publishing.

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Exam
NEET (UG) 2024
Paper
2024 paper
Subject
Chemistry
Chapter
Thermodynamics & Thermochemistry
Topic
Work Done in Different Cases

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