Physics · Thermodynamics

JEE Main 2026 — 2 April, Evening Shift — Question 20

5 moles of unknown gas is heated at constant volume from 10∘C10^\circ\mathrm{C} to 20∘C20^\circ\mathrm{C}. The molar specific heat at constant pressure cp=8 cal/mol.∘Cc_p = 8\,\mathrm{cal/mol.^\circ C} and R=8.36 J/mol.∘CR = 8.36\,\mathrm{J/mol.^\circ C}. The change in internal energy of the gas is ______ calorie.

Answer: 300

Numerical answer — enter this value.

Step-by-step solution

cv=cp−R=8−2=6 cal/mol.∘Cc_v = c_p - R = 8 - 2 = 6\,\mathrm{cal/mol.^\circ C} (since R≈2 cal/mol.K), ΔU=ncvΔT=5×6×10=300 cal\Delta U = n c_v \Delta T = 5\times6\times10 = 300\,\mathrm{cal}

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Subject
Physics
Chapter
Thermodynamics
Topic
First Law of Thermodynamics
5 moles of unknown gas is heated at constant volume from 10 ° C to 20… | JEE Main 2026 PYQ with Solution · DhiX AI