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PhysicsKinetic Theory of GasesJEE · NEET · NSEP · INPhO
  1. 1. Postulates
  2. 2. Gas Laws
  3. 3. Real Gases
  4. 4. Pressure of Air
  5. 5. Molecular Speeds
  6. 6. Pressure & KE
  7. 7. Degrees of Freedom
  8. 8. Internal Energy
  9. 9. Mean Free Path
Kinetic Theory of Gases · Part 9 of 9

Mean Free Path and Diffusion

Air molecules move at about 500 m/s, yet a smell takes seconds or minutes to cross a room. The reason is collisions: every molecule zigzags, covering only a tiny distance between hits.

Builds on: Part 5 · Distribution of Molecular Speeds.

Mean Free Path and DiffusionVideo coming soon
Mean free path

Short straight runs

Because molecules have size, they keep colliding, and each collision sends a molecule in a new direction. The average straight run between collisions is the mean free path λ. For air at STP it is about 10⁻⁷ m: a molecule collides billions of times a second.

A zigzag path through a box of molecules.
One molecule's zigzag among the others.
A molecule sweeping a translucent cylinder; molecules inside it are highlighted.
Centres inside the cylinder are hit.
Collision cylinder

λ = 1/(√2 π d² n₀)

A molecule of diameter d hits any molecule whose centre comes within d of its path: it sweeps a cylinder of cross-section πd². Allowing for the motion of the others gives , and with , . (It's π: some books misprint it as m.)

Dependence

Bigger molecules or a denser gas: shorter λ. Hotter gas at the same pressure: longer λ. At constant temperature, doubling the pressure halves λ.

Diffusion: random motion gives a net flow from high to low concentration. Fick's law: , where D depends on the substances, temperature and pressure. The net flow stops when the concentrations are equal, though the molecules keep moving.

A 1/P curve of mean free path against pressure.
Air at 273 K: λ against pressure.
Summary

Key formulas

Mean free path
Mean free path
Diffusion and Fick's law
JEE-style question

Your turn

At constant temperature, the pressure of a gas is doubled. What happens to its mean free path?

1)Halves
2)Doubles
3)Unchanged
4)Quarters
Show the answer and the traps

λ = kT/(√2 π d² P) ∝ 1/P: it halves. Option 1.

'Doubles' inverts the relationship. 'Unchanged' forgets that n₀ rises. 'Quarters' squares the pressure, confusing it with d².

Watch out

Common mistakes

Writing m for π in .It's a geometric factor from the cross-section πd².
"Doubling P at constant T leaves λ unchanged."n₀ = P/kT doubles, so λ halves.
Practice

Try these

1. Mean free path of a gas at 300 K and 1 atm, with molecular diameter 3 × 10⁻¹⁰ m?

≈ 1.0 × 10⁻⁷ m.

2. The temperature of a gas doubles (a) at constant pressure, (b) in a sealed rigid vessel. What happens to λ?

(a) , so it doubles. (b) n₀ is fixed, so is unchanged.

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