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Physics · Thermodynamics

JEE Main 2026 — 21 January, Morning Shift — Question 46

10 mole of oxygen is heated at constant volume from 30∘C30^{\circ} \mathrm{C} to 40∘C40^{\circ} \mathrm{C}. The change in the internal energy of the gas is ____\_\_\_\_ cal. (The molecular specific heat of oxygen at constant pressure, Cp=7cal./mol∘C\mathrm{C}_{\mathrm{p}}=7 \mathrm{cal} . / \mathrm{mol}^{\circ} \mathrm{C} and R=2cal./mol∘C\mathrm{R}=2 \mathrm{cal} . / \mathrm{mol}^{\circ} \mathrm{C}.)

Answer: 500

Numerical answer — enter this value.

Step-by-step solution

ΔU=nCvΔT\Delta \mathrm{U}=\mathrm{nC}_{\mathrm{v}} \Delta \mathrm{T}

=n(Cp−R)ΔT=10(7−2)(40−30)ΔU=500\begin{aligned} & =n\left(C_{p}-R\right) \Delta T & =10(7-2)(40-30) \Delta U & =500 \end{aligned}

Answer key and solution verified before publishing.

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