Physics · Thermodynamics

JEE Main 2026 — 21 January, Evening Shift — Question 47

A diatomic gas $(\gamma=1.4)$ does 100 J of work when it is expanded isobarically. Then the heat given to the gas $\_\_\_\_$ J.

Answer: 350

Numerical answer — enter this value.

Step-by-step solution

$\mathrm{w}=100 \mathrm{~J}=\mathrm{nR} \Delta \mathrm{T}$ for isobaric process. $\mathrm{Q}=\mathrm{nC}_{\mathrm{p}} \Delta \mathrm{T}=\left(\frac{\mathrm{f}}{2}+1\right) \mathrm{nR} \Delta \mathrm{T}$ $=\frac{7}{2} .(100)=350$ Joule.

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 21 January, Evening Shift
Subject: Physics
Chapter: Thermodynamics
Topic: Calculation of Work, Heat and Internal Energy