Chemistry · Thermodynamics & Thermochemistry

JEE Main 2024 — 8 April, Shift 1 — Question 74

Consider the figure provided.

1 mol of an ideal gas is kept in a cylinder, fitted with a piston, at the position A , at $18^{\circ} \mathrm{C}$ . If the piston is moved to position B ,

keeping the temperature unchanged, then ' $x$ ' L atm work is done in this reversible process. $\mathrm{x}=$ $\qquad$

L atm. (nearest integer) [Given : Absolute temperature $={ }^{\circ} \mathrm{C}+$ 273.15, $\left.\mathrm{R}=0.08206{\mathrm{~L} \mathrm{~atm} \mathrm{~mol}^{-1}}^{-1} \mathrm{~K}^{-1}\right]$

Answer: 55

Numerical answer — enter this value.

Step-by-step solution

$\omega=-\mathrm{nRT} \ln \left(\frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}\right)$

$=-1 \times .08206 \times 291 \cdot 15 \ln \left(\frac{100}{10}\right)$

$=-55.0128$

Work done by system $\approx 55$ atm lit.

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 8 April, Shift 1
Subject: Chemistry
Chapter: Thermodynamics & Thermochemistry
Topic: Work Done in Different Cases