Practice question
Question
What is the average translational kinetic energy of a nitrogen molecule at 450 K? (k_B = 1.38 × 10⁻²³ J K⁻¹)
Explanation
**Temperature dependence of RMS speed** v_rms ∝ √T, doubling T increases v_rms by √2≈1.414, e.g., at 300 K v_rms for O₂ ≈483 m/s, at 600 K ≈683 m/s, illustrating kinetic theory relation between temperature and molecular motion, average kinetic energy ½ m v_rms² =3/2 k_B T. Average translational KE = (3)/(2) k_B T.(3)/(2) × 1.38 × 10⁻²/³ × 450 = 9.315 × 10⁻²¹ J. Substituting values gives 9.31 × 10⁻²¹ 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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