Practice question
Question
What is the average translational kinetic energy of an argon atom at 900 K? (k_B = 1.38 × 10⁻²³ J K⁻¹)
Explanation
**Collision frequency** Z = √2 n π d² v_avg, n number density, d molecular diameter, v_avg average speed, proportional to n and v_avg, mean free path λ = v_avg/Z =1/(√2 n π d²), inversely proportional to n, so λ ∝1/P at constant T because n ∝ P, collision frequency increases with pressure, λ decreases. Average translational KE = (3)/(2) k_B T.(3)/(2) × 1.38 × 10⁻²/³ × 900 = 1.863 × 10⁻²⁰ J. Substituting values gives 1.863 × 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
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