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Gas molecules move at hundreds of metres per second, yet a smell takes a while to cross a room. The reason is the mean free path: molecules keep hitting each other, so each one travels only a tiny distance before changing direction. This post explains where the formula comes from, what it depends on, a fully worked problem and the slips that cost marks.
What is the mean free path, and is it in the syllabus?
The mean free path is the average distance a molecule covers between two successive collisions. The Syllabus for JEE (Main) 2026 lists "mean free path" in Unit 9, Kinetic Theory of Gases. It is Section 12.7 of the NCERT Class 11 Physics chapter on kinetic theory. Check the syllabus for your own exam year, since NTA reissues it each cycle.
Where does the formula come from?
NCERT's derivation treats molecules as spheres of diameter d.
A molecule hits any other molecule whose centre comes within a distance d of its own centre.
Moving with average speed v for a time Δt, it sweeps out a cylinder of volume πd²vΔt.
If there are n molecules per unit volume, it makes nπd²vΔt collisions in that time.
Distance travelled ÷ number of collisions gives l = 1/(nπd²).
That version imagines the other molecules standing still. They are not, and the collision rate depends on the average relative speed. Allowing for this adds a factor of √2:
l = 1/(√2 nπd²)
For an ideal gas, n = p/(kT), where k is the Boltzmann constant. Substituting gives l = kT/(√2 πd²p). OpenStax University Physics writes the same result using the radius r, as kT/(4√2 πr²p), since d = 2r.
What does the mean free path depend on?
Number density: l ∝ 1/n. Fewer molecules per cubic metre, longer paths.
Molecular size: l ∝ 1/d². Double the diameter and l falls to a quarter.
Pressure and temperature: l ∝ T/p. At fixed temperature, halving the pressure doubles l. At fixed pressure, raising the temperature spreads molecules out and increases l.
Not speed directly. At a fixed number density, faster molecules collide more often but travel the same average distance between hits.
NCERT notes that in a highly evacuated tube the mean free path can be as long as the tube itself.
Worked example: nitrogen at 2.0 atm and 17 °C
This is NCERT Exercise 12.10. Find the mean free path and collision frequency of a nitrogen molecule at 2.0 atm and 17 °C. Take the radius of a nitrogen molecule as 1.0 Å and its molecular mass as 28.0 u. Then compare the time between collisions with the time a collision takes.
Convert units. T = 17 + 273 = 290 K. p = 2.0 × 1.013 × 10⁵ = 2.026 × 10⁵ Pa. d = 2r = 2.0 × 10⁻¹⁰ m.
Number density. n = p/(kT) = 2.026 × 10⁵ ÷ (1.38 × 10⁻²³ × 290) = 2.026 × 10⁵ ÷ 4.00 × 10⁻²¹ ≈ 5.06 × 10²⁵ m⁻³.
Mean free path. √2 × π × d² × n = 1.414 × 3.142 × (4.0 × 10⁻²⁰) × (5.06 × 10²⁵) ≈ 9.0 × 10⁶ m⁻¹. So l ≈ 1/(9.0 × 10⁶) ≈ 1.1 × 10⁻⁷ m. That is about 550 molecular diameters.
Speed. v_rms = √(3RT/M) = √(3 × 8.314 × 290 ÷ 0.028) ≈ √(2.58 × 10⁵) ≈ 508 m/s.
Collision frequency. v/l = 508 ÷ (1.11 × 10⁻⁷) ≈ 4.6 × 10⁹ collisions per second, so the time between collisions is about 2.2 × 10⁻¹⁰ s.
Time for one collision. Roughly the time to cover one diameter: d/v = 2.0 × 10⁻¹⁰ ÷ 508 ≈ 3.9 × 10⁻¹³ s.
Compare. 2.2 × 10⁻¹⁰ ÷ 3.9 × 10⁻¹³ ≈ 560. A molecule spends roughly 500 times longer flying freely than colliding, which is why the ideal gas picture works so well.
A note on NCERT's estimate for air
In the chapter text, NCERT estimates l ≈ 2.9 × 10⁻⁷ m for air at STP, using n = 2.7 × 10²⁵ m⁻³ and d = 2 × 10⁻¹⁰ m. That number matches the simpler 1/(nπd²). Including the √2 gives about 2.1 × 10⁻⁷ m. Both are a few tenths of a micrometre. In exams, use the formula with √2, which is the one in NCERT's summary, unless the question tells you otherwise.
Which mistakes cost marks?
Radius for diameter. Gives an answer four times too big.
Pressure in atm. Convert to pascals before using k.
Celsius instead of kelvin. 17 °C is 290 K.
Mixing moles and molecules. n here is molecules per cubic metre, not moles.
Thinking faster molecules have longer free paths. At fixed density they do not.
Dropping the √2 and then picking a wrong option built from that slip.
A practice routine for this topic
Re-derive l = 1/(nπd²) from the swept cylinder without notes, then explain the √2 in one sentence.
Scale, do not recompute. Using the worked example, find l for the same nitrogen at 1.0 atm and 290 K. You should get about 2.2 × 10⁻⁷ m.
Push to a vacuum. At 1.0 × 10⁻³ Pa and 290 K, show that l is about 22 m, longer than most vacuum chambers.
Mix it with rms speed. Pair one mean free path question with one rms speed question from our kinetic theory guide.
When a step goes wrong and you cannot see why, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer. For how real gases differ from the ideal model, see our post on real gases and the van der Waals equation.
Tonight, redo the nitrogen example with the pressure halved and check that the mean free path doubles.
