MixedJEE Physics · Original learning card5 original chapter questions

Mean Free Path

For hard-sphere molecules in a dilute equilibrium gas, lambda = 1/(sqrt(2) pi d^2 n_number) = k_BT/(sqrt(2) pi d^2 P).

Why this shows up in the exam

Estimating molecular collision spacing · Comparing gases at different pressure and temperature · Relating transport scales to molecular size

Learn the idea

Mean free path is the average distance a molecule travels between successive intermolecular collisions. Crowding molecules or making their collision diameter larger shortens the unobstructed distance available to each molecule.

🧠 Memory hook: Mean path shrinks with crowding and with diameter squared.

Get this one clearly and it pays off every single time it shows up in the paper. 🎯

Formulas & facts to keep ready

  • lambda = 1/(sqrt(2) pi d² n_number) — hard-sphere mean free path at molecular number density n_number
  • lambda = k_B T/(sqrt(2) pi d² P) — ideal-gas pressure-temperature form
  • lambda proportional to T/P — scaling for fixed molecular diameter

How to approach it

  1. 1Choose number-density or P-T form
  2. 2Convert diameter to metres
  3. 3Check that the result has length units

Common slip-ups that cost marks

  • •Using molar density in the molecular formula
  • •Forgetting the square on diameter
  • •Omitting the relative-motion factor sqrt(2)

🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.

Original chapter practice

Original questions for this chapter, not past-paper questions or an exact mapping to this individual concept.

Question 1 of 5

A gas has rms molecular speed 300 m/s at 300 K. What is its rms speed at 1200 K, assuming ideal behavior?

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