Mean Free Path Calculator (gas molecules: λ = kT/(√2·π·d²·p))

Understanding the mean free path calculator — concept and principle

This calculator finds the mean free path of gas molecules, the average distance a molecule travels between collisions, from temperature, pressure, molecular diameter and molar mass. It also gives the number density, RMS speed, mean free time and collision rate. It is used in vacuum technology, kinetic theory courses and when deciding whether a gas behaves as a continuum.

Each atom travels about 213 of its own diameters between hits, billions of times a second.

How it is calculated

λ = k_B T ÷ (√2 π d² p)

Number density n = p ÷ (k_B T)

v_rms = √(3 R T ÷ M); mean free time τ = λ ÷ v_rms; collision rate = 1 ÷ τ

k_B = 1.380649 × 10⁻²³ J/K; pressure in atm, Pa, Torr or mbar (1 atm = 101,325 Pa); d in nm; M in g/mol.

The principle

Treat each molecule as a hard sphere of diameter d. A molecule sweeps out a cylinder of cross-section πd² as it moves, and collides with any other molecule whose centre lies inside it. The number of collisions per unit length is that area times the number density, so the mean free path is 1 ÷ (n π d²). The factor √2 corrects for the fact that the other molecules are also moving. Since n = p/(k_B T), the path grows with temperature and is inversely proportional to pressure.

Worked example

Default inputs: T = 273 K, p = 1 atm, d = 0.34 nm, M = 39.9 g/mol (argon).

n = 101,325 ÷ (1.380649 × 10⁻²³ × 273) = 2.688 × 10²⁵ m⁻³.
λ = 1.380649 × 10⁻²³ × 273 ÷ (√2 × π × (0.34 × 10⁻⁹)² × 101,325) = 7.243 × 10⁻⁸ m = 72.43 nm.
v_rms = √(3 × 8.314 × 273 ÷ 0.0399) = 413 m/s.
τ = 72.43 × 10⁻⁹ ÷ 413 = 1.753 × 10⁻¹⁰ s; about 5,704,000,000 collisions per second.
λ ÷ d = 213.

Each point is the calculator re-run with only “Molecular diameter d (nm)” changed; the gold point is the default example. Change the inputs above and this chart follows your numbers.

Molecular diameter d (nm)	Mean free path (m)
0.085	0.000001159
0.17	2.897e-7
0.255	1.288e-7
0.34	7.243e-8
0.425	4.635e-8
0.51	3.219e-8
0.68	1.811e-8

How to read the result

At atmospheric pressure the mean free path is tens of nanometres, a couple of hundred molecular diameters. Each tenfold drop in pressure makes it ten times longer, so in a high vacuum it reaches metres or kilometres and molecules fly from wall to wall without colliding. Comparing λ with the size of a channel or particle tells you whether gas flow is continuum or molecular.

Common mistakes

Using the radius instead of the diameter, or vice versa.
Forgetting the √2 factor.
Mixing pressure units.
Applying the hard-sphere value to gases where effective diameters vary with temperature.

Keep in mind: Molecular diameters are effective values from kinetic theory and vary between sources. Treat the result as an estimate.

Frequently asked questions

How long is the mean free path in air at sea level?

Around 70 nm, comparable to the argon example.

Why does it get longer in a vacuum?

There are fewer molecules to hit, so the path is inversely proportional to pressure.

What is the √2 factor for?

It accounts for the motion of the target molecules.

What is mean free time?

The average time between collisions, λ divided by the molecular speed.

Sources: OpenStax, University Physics Volume 2 (2016), section 2.2 "Pressure, Temperature, and RMS Speed" · E. Al-Dmour, "Fundamentals of Vacuum Physics and Technology", CERN Accelerator School (2017)

Checked 9 Nov 2026. OpenStax, University Physics Volume 2 (2016), §2.2 “Pressure, Temperature, and RMS Speed” (eqs. 2.10–2.12: λ = k_BT/(4√2πr²p), τ = λ/v_rms; Example 2.7: argon, r = 1.70 × 10⁻¹⁰ m, 273 K, 1.00 atm → τ = 1.76 × 10⁻¹⁰ s); E. Al-Dmour, “Fundamentals of Vacuum Physics and Technology”, CERN Accelerator School 2017, Table 3 (air at room temperature: 7 × 10⁻⁶ cm at 1,000 mbar). Constants: c = 299,792,458 m/s, e = 1.602176634 × 10⁻¹⁹ C, k_B = 1.380649 × 10⁻²³ J/K, h = 6.62607015 × 10⁻³⁴ J·s and N_A = 6.02214076 × 10²³ mol⁻¹ are exact in the 2019 SI.

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