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Question

What is the average translational kinetic energy of an argon atom at 500 K? (k_B = 1.38 × 10⁻²³ J K⁻¹)

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Explanation

**Degrees of freedom** f counts independent motions, monatomic 3 translational, diatomic 3 translational +2 rotational =5 at room T, vibrational adds at high T, molar specific heat at constant volume C_v = f/2 R, at constant pressure C_p = C_v + R, ratio γ = C_p/C_v = (f+2)/f, monatomic γ=5/3≈1.67, diatomic γ=7/5=1.4. Average translational KE = (3)/(2) k_B T.(3)/(2) × 1.38 × 10⁻²/³ × 500 = 1.035 × 10⁻²⁰ J. Substituting values gives 1.035 × 10⁻²⁰ J, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π d²), ideal gas law P V = n R T and v_rms = √(3 R T/M) relations.

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