Mean free path

Derive mean free path, collision time and collision frequency for a dilute gas, checking units and limiting cases.

University Physics Year 1

On this page

Molecules do not travel a fixed spacing between collisions. Their mean free path depends on the number of potential collision partners and the area over which they can collide. You will build that model, connect distance to collision rate and compare the collision length with the size of an apparatus.

At a glance

Prerequisites: number density, cross-sectional area, average speed.
Outcomes: derive λ = 1/(square root of 2 nσ) and related collision rates.
Key result: in dilute gases, mean free path is inversely proportional to number density.
Common trap: forgetting the square root of 2 factor from relative motion.

Define the collision geometry

Define:

Number density n = N/V.
Molecular diameter d.
Collision cross section σ = π d².
Mean free path λ: average distance between collisions.

Assumptions:

One species of identical hard-sphere molecules with constant diameter and independent pre-collision velocities. Air below is an effective single-species estimate.
Dilute gas (binary collisions dominate).
Statistical isotropy.

Physical interpretation:

λ is the mean path length between intermolecular collisions. One collision can change direction without completely randomising it; transport relaxation lengths need not equal this collision length.
Transport depends on both flight length and how many molecules carry momentum or energy. At fixed temperature in a dilute continuum gas, increasing λ by reducing density can be offset by fewer carriers. Do not infer conductivity from λ alone.
λ is not the same as the mean intermolecular spacing (∼ n^(-1/3)); it depends strongly on the collision cross-section too.

Allow both collision partners to move

If one molecule moves through stationary targets, collisions in time t are

N_coll = nσv bar t.

Distance traveled is v bar t, so

λ = (v bar t)/N_coll = 1/nσ.

For independent identical molecules at the same Maxwell–Boltzmann temperature, the mean relative speed is ⟨|v₁-v₂|⟩ = square root of 2 v bar. The mean collision rate is therefore nσ square root of 2 v bar. Each relative velocity component has twice the single-molecule variance: ⟨(v₁ₓ-v₂ₓ)²⟩ = 2kBT/m. The relative-speed Maxwell distribution therefore has a speed scale larger by square root of 2; this scales its mean speed by square root of 2 too. See Oxford kinetic theory notes, section 4.4.

Dividing the mean distance travelled per time by the collision rate gives:

λ = 1/(square root of 2 nσ) = 1/(square root of 2 π d² n).

Using the ideal-gas equation of state n = N/V = P/kBT (see UY1: equation-of-state - ideal gas law), you can also write

λ = kBT/(square root of 2 π d² P).

These are ensemble collision scales for this model, not a guarantee that each flight has that length or duration. Do not substitute rms speed into a formula defined using mean speed. Different-species collisions require their relative-speed distribution and cross section; the same square root of 2 correction is not universal.

Mean collision time:

τ = λ/(v bar).

Collision frequency:

f_coll = 1/τ = (v bar)/λ = square root of 2 nσv bar.

Units check:

[nσ] = m⁻³ · m² = m⁻¹ ⇒ [λ] = m.

Physical sanity check:

Increasing pressure (thus n at fixed T) decreases λ.
Larger molecules (larger d) collide more frequently, so λ decreases.

Limit checks:

As n → 0 (very low density), λ → ∞ (collisions become rare).
In dense gases/liquids, the dilute hard-sphere model breaks down; correlations and many-body effects matter. If λ is comparable with an apparatus gap L, wall interactions can dominate transport even while intermolecular collisions remain dilute. A local continuum model requires the collision length to be small compared with its spatial scale.

The square root of 2 factor here comes from relative motion of identical molecules with independent Maxwell–Boltzmann velocities at the same temperature.
Here n is number density in m⁻³, rather than the amount of gas in moles.
Use mean speed v bar for the stated collision rate. Replacing it with rms speed changes the rate.

Estimating the mean free path in air

Problem

Estimate

λ for air at

300 K and

1.01 × 10⁵ Pa using a dilute hard-sphere model with effective diameter

3.7 × 10⁻¹⁰ m. Use

kB = 1.38 × 10⁻²³ J K⁻¹.

Show full solution

Convert pressure to number density
Method
Use n = P/(kBT).
Reason
The collision-length formula needs molecules per unit volume, not the number of moles.
Working
Answer
n = P/kBT = (1.01 × 10⁵)/((1.38 × 10⁻²³)(300)) ≈ 2.44 × 10²⁵ m⁻³.
Include collision geometry and relative motion
Method
Substitute σ = π d² into λ = 1/(square root of 2 nσ).
Reason
Two molecules collide when their centres approach within one diameter, and both molecules are moving.
Working
Answer
λ = 1/(square root of 2 π d² n) ≈ 1/(1.414π(3.7 × 10⁻¹⁰)²(2.44 × 10²⁵)) ≈ 6.7 × 10⁻⁸ m.
Express the collision length
Working
Answer
λ ≈ 6.7 × 10⁻⁸ m = 67 nm. The inverse of number density times cross section has units m, as a path length must.

Practice set

Check your understanding 1

For the same dilute hard-sphere gas, what happens to λ when pressure doubles at fixed temperature?

Show hint

First relate pressure to number density.

Show answer

n = P/(kBT) doubles, so λ = 1/(square root of 2 nσ) halves. More targets per unit volume shorten the average flight.

Check your understanding 2

Write the hard-sphere collision frequency in terms of number density n, molecular diameter d and mean speed v bar.

Show hint

Use

f_coll = v bar/λ and replace the cross section.

Show answer

f_coll = square root of 2 nσv bar = square root of 2 π d²nv bar. The units are

m²m⁻³m s⁻¹ = s⁻¹.

Check your understanding 3

A dilute hard-sphere gas has n = 1.0 × 10²⁴ m⁻³ and d = 2.0 × 10⁻¹⁰ m. Find its mean free path.

Show hint

Square the diameter before multiplying by density.

Show answer

λ = 1/[square root of 2 π(2.0 × 10⁻¹⁰)²(1.0 × 10²⁴)] ≈ 5.6 × 10⁻⁶ m. This is 5.6 μm, not the intermolecular spacing.

Check the collision scale of a narrow gap

Problem

A single-species hard-sphere gas at 300 K and 200 Pa has diameter 3.70 × 10⁻¹⁰ m and molar mass 0.0280 kg mol⁻¹. It occupies a gap L = 50.0 μ m. Use kB = 1.38 × 10⁻²³ J K⁻¹ and R = 8.314 J mol⁻¹ K⁻¹. Estimate mean intermolecular free path in micrometres, mean intermolecular collision frequency and λ/L. Explain whether a local continuum transport model is automatically justified.

Work through the model, then compare your reasoning

Mean free path

Unit: µm

Mean intermolecular collision frequency

Unit: s⁻¹

Mean free path divided by the gap

Unit: dimensionless

Your explanation (self-review against the solution; paper working is also fine)

Hints

Hint 1

Check the model

Find the gas collision scale first. The gap is a separate length with which to compare it.

Show solution step by step

Calculate the intermolecular scales
Method
Use the hard-sphere mean path and Maxwell mean speed.
Reason
The correction describes collisions between moving identical molecules.
Working
Answer
λ = kBT/(square root of 2 π d²P) ≈ 34.0 μ m. The mean speed is v bar = square root of (8RT/(π M)) ≈ 476 m s⁻¹, giving f_coll = v bar/λ ≈ 1.40 × 10⁷ s⁻¹.
Check the apparatus scale
Method
Compare λ with L.
Reason
A dilute gas need not provide local equilibration over a small gap.
Working
Answer
λ/L ≈ 0.681, which is not small. Wall interactions matter, so a bulk local continuum transport model cannot be assumed adequate. The frequency above counts intermolecular collisions; it does not include wall encounters.

Change pressure, temperature and apparatus together

Check your understanding 4

The same dilute hard-sphere gas is in chamber A at 300 K and 1.00 × 10⁵ Pa with a 1 mm gap, and chamber B at 600 K and 2.00 × 10⁵ Pa with a 50 nm gap. Its diameter stays 3.70 × 10⁻¹⁰ m. Compare number density, mean intermolecular free path, mean speed and mean intermolecular collision time. A solution halves the free path in B because pressure doubles, then applies the same local continuum heat-transfer model in both gaps. Evaluate it.

Your reasoning (self-review; paper working is also fine)

Show answer

P/T is unchanged, so number density and intermolecular mean free path are unchanged: λ ≈ 68.1 nm using the stated model and kB above. Mean speed increases by square root of 2, so the mean intermolecular collision time decreases by 1/square root of 2. Halving free path from pressure alone incorrectly holds temperature fixed. In A, λ/L ≈ 6.8 × 10⁻⁵; in B it is about 1.36. A has a separation of collision and gap scales, while B needs wall-sensitive transport rather than assuming the same local continuum model.

What to keep

Specify the species, collision geometry and speed convention. Pressure changes determine number density only with temperature accounted for. A mean intermolecular free path and a wall gap answer different questions; compare them before choosing a transport model.

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Syllabus and review details

No official syllabus alignment is listed for this lesson.

Mean free path

Reading settings

At a glance

Define the collision geometry

Allow both collision partners to move

Estimating the mean free path in air

Mean free path of room-temperature air

Convert pressure to number density

Include collision geometry and relative motion

Express the collision length

Practice set

Check your understanding 1

Check your understanding 2

Check your understanding 3

Check the collision scale of a narrow gap

Compare gas collisions with an apparatus gap

Calculate the intermolecular scales

Check the apparatus scale

Change pressure, temperature and apparatus together

Check your understanding 4

What to keep