The value of Cv for 1 mol of N₂ gas predicted from the principle of equipartition of energy, ignoring… · ABC26PH0856

Subject: Physical Chemistry · Chapter: Thermodynamics · Topic: Thermodynamics – General · Exam: JAM 2001-2022 · Marks: 2 · Difficulty: Medium

The value of $\mathrm{C}_{\mathrm{v}}$ for 1 mol of $\mathrm{N}_{2}$ gas predicted from the principle of equipartition of energy, ignoring vibrational contribution, is ____ $\mathrm{J} \mathrm{K}^{-1} \mathrm{mol}^{-1}$ (rounded up to two decimal places).
\[ \left[\mathrm{R}=8.3 \mathrm{JK}^{-1} \mathrm{mol}^{-1}\right] \]
Answer
SELF-PRACTICE — the source book printed no answer.

Nothing is invented here, so this question has no answer on record.

Explanation
Answer: 20.75 J K⁻¹ mol⁻¹ — For a linear molecule without vibration: 3 translational + 2 rotational quadratic terms, so C_v = (5/2)R = 2.5 × 8.3 = 20.75 J K⁻¹ mol⁻¹.

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