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JEE Main 2022 · 28 Jun · Shift 1, Physics question 58

The total internal energy of two mole monoatomic ideal gas at temperature T = 300 K will be ______ J. (Given R = 8.31 J/mol.K)

Answer and worked solution

Official answer: 7479

Internal energy. U=n⋅32RT=2×1.5×8.31×300=7479U=n\cdot\frac32RT=2\times1.5\times8.31\times300=7479 J.

Answer: 7479

Topics:Degrees of freedom and equipartition
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