Practice question

**Molar specific heat** from equipartition C_v = f/2 R, C_p = f/2 R + R, γ = C_p/C_v =1+2/f, for f=3 γ=1.67, f=5 γ=1.4, f=6 γ=1.33, explaining specific heat variation with molecular structure, degrees of freedom determine heat capacity. Monatomic gas: C_v = (3)/(2) R.Q = μ C_v Δ T = 0.1 × (3)/(2) × 8.31 × 30 = 37.395 J ≈ 37.4 J. Substituting values gives 37.4 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.