One mole of a van der waals gas undergoes reversible isothermal transformation from an initial volume V₁ to a… · ABC26PH0838

Subject: Physical Chemistry · Chapter: Thermodynamics · Topic: Thermodynamics – General · Exam: JAM 2001-2022 · Marks: 2 · Difficulty: Medium

One mole of a van der waals gas undergoes reversible isothermal transformation from an initial volume $\mathrm{V}_{1}$ to a final volume $\mathrm{V}_{2}$. The expression for the work done is:
(a)$\mathrm{RT} \ln \frac{\mathrm{V}_{2}}{\mathrm{V}_{1}}+\mathrm{a}\left(\mathrm{V}_{2}-\mathrm{V}_{1}\right)$
(b)$-\mathrm{RT} \ln \frac{\mathrm{V}_{2}-\mathrm{b}}{\mathrm{V}_{1}-\mathrm{a}}+\mathrm{a}\left(\frac{1}{\mathrm{V}_{1}}-\frac{1}{\mathrm{V}_{2}}\right)$
(c)$R T \ln \frac{P_{2}}{P_{1}}$
(d)$\mathrm{RT} \ln \frac{\mathrm{V}_{2}-\mathrm{b}}{\mathrm{V}_{1}-\mathrm{a}}-\mathrm{a}\left(\frac{1}{\mathrm{V}_{1}}-\frac{1}{\mathrm{V}_{2}}\right)$
Answer
Answer: B ✓ checked by 4AB · confidence medium

The source book printed no answer; this one was worked out and checked — see the explanation.

Explanation
w = −∫P dV, with P = RT/(V − b) − a/V². This gives w = −RT ln[(V₂ − b)/(V₁ − b)] + a(1/V₁ − 1/V₂). Option (b) matches, with 'a' misprinted for 'b' in V₁ − b.

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