What is the average speed of a molecule if its mean free path is 4 × 10⁻⁷ m and time between collisions is 8 × 10⁻¹⁰ s?
**Mean free path variation** λ ∝1/n ∝1/P at constant T, λ ∝ T/P, temperature increase increases λ because n decreases at constant P, but also v increases, overall λ ∝ T/P, for gas at 2 atm λ=4×10⁻⁷ m, at 4 atm λ=2×10⁻⁷ m halves when pressure doubles, as n doubles. Mean free path l =tau.= (l)/(tau) = 4 × 10⁻⁷⁸ × 10⁻¹⁰ = 500 m/s . Substituting values gives 500 m/s, 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 > Collision Frequency and Mean Free Path Variation
