A gas has a C_v of 24.93 J mol⁻¹ K⁻¹. What is the ratio of specific heats (gamma)? (R = 8.31 J mol⁻¹ K⁻¹)
**Mean free path** λ = 1/(√2 n π d²) is average distance molecule travels between collisions, n number density (m⁻³), d molecular diameter (m), π≈3.14. Inversely proportional to n and d², larger n or d reduces λ. Rearranged d² = 1/(√2 n π λ), so d = √(1/(√2 n π λ)), enabling diameter estimation from measured λ and n. C_p = C_v + R = 24.93 + 8.31 = 33.24 J mol⁻¹ K⁻¹.γ = (C_p)/(C_v) = (33.24)/(24.93) ≈ 1.33. Substituting values gives 1.33, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π d²), ideal gas law P V =
Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Mean Free Path and Molecular Diameter