Physics · Thermodynamics

JEE Main 2024 — 6 April, Shift 2 — Question 32

A total of 48 J heat is given to one mole of helium kept in a cylinder. The temperature of helium increases by $2^{\circ} \mathrm{C}$ . The work done by the gas is : (Given, $\mathrm{R}=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$ .)

Option A:
72.9 J
Option B:
24.9 J
Option C:
48 J
Option D:
23.1 J
Correct

Answer: D

Step-by-step solution

$1^{\text {st }}$ law of thermodynamics

$\Delta \mathrm{Q}=\Delta \mathrm{U}+\mathrm{W}$

$\Rightarrow+48=\mathrm{nC}_{\mathrm{v}} \Delta \mathrm{T}+\mathrm{W}$

$\Rightarrow 48=(1)\left(\frac{3 \mathrm{R}}{2}\right)(2)+\mathrm{W}$

$\Rightarrow \mathrm{W}=48-3 \times \mathrm{R}$ $\Rightarrow \mathrm{W}=48-3 \times(8.3)$

$\Rightarrow \mathrm{W}=23.1$ Joule

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 6 April, Shift 2
Subject: Physics
Chapter: Thermodynamics
Topic: Calculation of Work, Heat and Internal Energy