Practice question

**Molecular diameter from mean free path** uses λ =1/(√2 n π d²), solving d = √(1/(√2 n π λ)). At higher pressure n ∝ P, λ ∝1/P, so doubling P halves λ, illustrating pressure dependence of collision distance. Triatomic gas: 6 degrees of freedom (3 translational + 3 rotational).U = 3 R T = 3 × 8.31 × 300 = 7479 J ≈ 7.48 kJ. Substituting values gives 7.48 kJ, 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.