Collision Rate, Mean Free Path & Diffusion

Give every molecule a diameter d and freeze all of them except one. Two spheres touch when their centres are d apart, so the moving molecule collides with any centre lying within d of its track: it sweeps a cylinder of radius d, cross-section σ = πd², and in time t of length v̄t. With n centres per unit volume the cylinder contains nσv̄t of them, so the collision rate is nσv̄ and the distance per collision is 1/(nσ). The numbers already say something. Nitrogen at 300 K and 1 atm has n = p/kT = 2.45 × 10²⁵ m⁻³; with d = 0.37 nm, σ = 4.30 × 10⁻¹⁹ m² and 1/(nσ) = 95 nm. That is 260 diameters, so the gas is mostly empty space and a molecule misses almost every neighbour it passes. Check the dimensions: [nσ] = m⁻³ × m² = m⁻¹, the reciprocal of a length, and no speed appears in it. A faster molecule collides more often; it does not fly any less far between collisions.

The frozen crowd is a fiction: the targets move too, and what decides whether two molecules meet is their relative velocity vᵣₑₗ = v₁ − v₂. Square it and average: ⟨vᵣₑₗ²⟩ = ⟨v₁²⟩ + ⟨v₂²⟩ − 2⟨v₁⋅v₂⟩, and the cross term vanishes because the two velocities are independent and every direction is equally likely, leaving ⟨vᵣₑₗ²⟩ = 2⟨v²⟩. For a Maxwell-Boltzmann gas the mean magnitudes obey the same rule exactly: the relative velocity of two molecules of mass m is itself Maxwell-Boltzmann distributed with the reduced mass m/2, so ⟨|vᵣₑₗ|⟩ = √2 v̄. The swept cylinder is therefore √2 times longer per second than a lone moving molecule suggests, and Z = √2 n σ v̄, λ = v̄/Z = 1/(√2 n σ). For nitrogen at 300 K, v̄ = √(8kT/πm) = 476 m s⁻¹, so Z = √2 × 2.45 × 10²⁵ × 4.30 × 10⁻¹⁹ × 476 = 7.1 × 10⁹ s⁻¹ and λ = 67 nm: a collision every 0.14 ns. Limiting case: a fast ion or electron crossing the same gas sees targets that are effectively at rest, so its rate is nσv with no √2 and its free path is 1/(nσ).

Collisions have no memory. A molecule that has already flown a distance x is no more likely to hit something in the next dx than one that has just collided, so the fraction P(x) of molecules still uncollided obeys dP/dx = −P/λ with P(0) = 1. The solution is P(x) = e^(−x/λ), and the probability density of a free path ℓ is f(ℓ) = −dP/dℓ = (1/λ)e^(−ℓ/λ), normalised because ∫₀^∞ f dℓ = 1. Its mean, ∫₀^∞ ℓ f dℓ = λ, is what makes λ the mean free path, but the distribution is wide: 63% of flights are shorter than λ, a fraction e⁻³ = 5% are longer than 3λ, and the mean square ∫₀^∞ ℓ² f dℓ = 2λ² is twice what equal steps of length λ would give. That factor of two reappears in the diffusion coefficient. The same exponential is what an experiment measures: fire a beam of molecules through a cell of gas, fast enough that the gas can be treated as stationary, and the transmitted intensity is I = I₀ e^(−nσx), so the slope of ln(I/I₀) against nx is the cross-section itself, with no diameter assumed.

A molecule has no edge. Its electron cloud fades out, the force between two molecules is a soft repulsion that steepens only when they are close, and the distance of closest approach depends on how hard they hit. So d is not read off a picture: it is the diameter that makes πd² reproduce a measured transport property. Nitrogen's viscosity, 1.78 × 10⁻⁵ Pa s at 300 K, is reproduced with d ≈ 0.37 nm by the exact hard-sphere transport result η = (5π/32) n m v̄ λ ≈ 0.49 n m v̄ λ; the cylinder argument gives the same dependence on n, m, v̄ and λ but the prefactor ⅓, about 1.5 times too small, so the d a fit returns depends on the formula it was fitted with. Van der Waals' excluded volume for nitrogen, b = 3.91 × 10⁻⁵ m³ mol⁻¹, is four times the molecules' own volume, b = (2/3)πN_A d³, and returns d = 0.31 nm, 15% smaller, which is a factor of 1.4 in σ and therefore in λ. Neither is wrong; they are different questions asked of a soft object. Fitted diameters also fall slowly with temperature, because hotter collisions penetrate further into the repulsive wall before turning back — Sutherland's correction, and the reason measured viscosities rise faster than the √T of the rigid-sphere model. Quote d to two figures, name the property it came from, and treat λ as good to about 10%.

Follow one molecule through N free flights ℓ₁, ℓ₂, …, ℓ_N, each in a random direction. Its displacement is R = Σℓᵢ, so R² = Σℓᵢ² + Σ ℓᵢ⋅ℓⱼ over the pairs i ≠ j. Average over many molecules and every cross term vanishes, because ℓᵢ⋅ℓⱼ is as often negative as positive when directions are uncorrelated, leaving ⟨R²⟩ = N⟨ℓ²⟩. With ⟨ℓ²⟩ = 2λ² and N = Zt = v̄t/λ flights in time t, ⟨R²⟩ = 2λv̄t. The mean-square displacement grows linearly with time, so the rms distance grows as √t: not the vt of free flight, not the t² of anything accelerating. Write ⟨R²⟩ = 6Dt, two per Cartesian axis, and the model gives D = λv̄/3 = 6.7 × 10⁻⁸ × 476/3 = 1.1 × 10⁻⁵ m² s⁻¹ for nitrogen at 1 atm and 300 K, about half the measured self-diffusion coefficient of 2 × 10⁻⁵ m² s⁻¹: the scalings with T and p are right, the prefactor is not. Along one axis ⟨x²⟩ = 2Dt, so an rms spread of 1 cm takes (0.01)²/(2 × 1.1 × 10⁻⁵) ≈ 5 s and 3 m takes about 5 days. A smell that crosses a room in a minute rides on air currents; diffusion alone is a molecule flying at 476 m s⁻¹ and getting almost nowhere.

Because λ = kT/(√2πd²p), lowering the pressure stretches it without limit in the model. Nitrogen at 300 K has λ = 67 nm at 1 atm, 6.8 mm at 1 Pa and 68 m at 10⁻⁴ Pa. Once λ exceeds the size L of the container, so that the Knudsen number λ/L passes 1, molecules cross the chamber and strike the walls without meeting each other. Heat and momentum are then carried wall to wall by individual flights, viscosity and conductivity stop being properties of the gas alone, and the heat a hot filament loses becomes proportional to pressure, which is what a Pirani gauge exploits. That is the regime of molecular beams and electron tubes. At the opposite end, compress the gas until λ approaches d: the cylinder argument assumed many diameters of flight between hits, and when the spacing n^(−1/3) falls to about d there is no free flight to average; the substance is a liquid, and its correlated many-body motion is beyond this model. In between, the model is also blind to velocity persistence: a molecule keeps part of its forward motion through a collision, which is much of why D = λv̄/3 comes out low.