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JEE Main 2022 (28 Jun, Shift 1) · Degrees of Freedom & Internal Energy

Degrees of Freedom & Internal Energy question from JEE Main 2022 (28 Jun, Shift 1)

Q21JEE Main 2022 · 28 JUN · SHIFT 1Easy

Degrees of Freedom & Internal Energy

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)

Numerical answer type. Work it out, then open the answer.

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Answer: 7479

Monoatomic gas: $U = n\frac{3}{2}RT$ .
$U = 2 \times \frac{3}{2} \times 8.31 \times 300$
$U = 7479$ J

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