Physics · Thermodynamics

JEE Main 2024 — 1 February, Shift 2 — Question 34

A diatomic gas $(\gamma=1.4)$ does 200 J of work when it is expanded isobarically. The heat given to the gas in the process is :

Option A:
850 J
Option B:
800 J
Option C:
600 J
Option D:
700 J
Correct

Answer: D

Step-by-step solution

$\quad \gamma=1+\frac{2}{\mathrm{f}}=1.4 \Rightarrow \frac{2}{\mathrm{f}}=0.4$

$\Rightarrow \mathrm{f}=5$ $\mathrm{W}=\mathrm{nR} \Delta \mathrm{T}=200 \mathrm{~J}$ $\mathrm{Q}=\left(\frac{\mathrm{f}+2}{2}\right) \mathrm{nR} \Delta \mathrm{T}$

$=\frac{7}{2} \times 200=700 \mathrm{~J}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 1 February, Shift 2
Subject: Physics
Chapter: Thermodynamics
Topic: Calculation of Work, Heat and Internal Energy