Given: n=10 mol,
ΔT=40−30=10°C,
CP=7 cal/mol·°C,
R=2 cal/mol·°C
For ideal gas: CV=CP−R=7−2=5 cal/mol·°C
Change in internal energy: ΔU=nCVΔT=10×5×10=500 cal
JEE Main 2026 — Physics Thermodynamics
10 mole of oxygen is heated at constant volume from 30∘C to 40∘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 and R=2cal/mol.∘C.)
Held on 21 Jan 2026 · Verified 6 Jul 2026.
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