A gas has a C_p of 35.4 J mol⁻¹ K⁻¹. What is the ratio of specific heats (gamma)? (R = 8.31 J mol⁻¹ K⁻¹)
**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. C_v = C_p - R = 35.4 - 8.31 = 27.09 J mol⁻¹ K⁻¹.γ = (C_p)/(C_v) = (35.4)/(27.09) ≈ 1.31. Substituting values gives 1.31, 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.
Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Mean Free Path and Molecular Diameter