Molecular Speeds & Mean Free Path

Resolve a molecule's velocity into components. Each component is Gaussian and peaked at zero, so the single most likely velocity vector is the one with no motion at all. Speed is a different question, because many different velocity vectors share one speed. Every vector with speed between v and v + dv lies in a thin shell of velocity space whose area is 4πv², so the number of ways to have speed v grows as v². Multiply the Boltzmann factor e^{(−mv²/(2k_BT))}, which falls as the kinetic energy rises, by that growing v² and you get f(v) = 4π (m/(2πk_BT))³⁄² v² e^{(−mv²/(2k_BT))}. The v² wins at small v, so f(0) = 0: almost nothing is at rest, even though zero is the most probable velocity. The exponential wins at large v. The clumsy prefactor is fixed by one demand, that ∫₀^∞ f(v) dv = 1.

Write a = m/(2k_BT), so f ∝ v² e^(−av²). Then df/dv ∝ 2v e^(−av²)(1 − av²), which vanishes at av² = 1: the peak sits at vₚ = √(2k_BT/m). Two integrals over the same curve supply the other two averages, v̄ = ∫₀^∞ v f(v) dv = √(8k_BT/(πm)) and vᵣₘₛ = √(∫₀^∞ v² f(v) dv) = √(3k_BT/m). All three carry the same factor √(k_BT/m), so their ratio never changes: vₚ : v̄ : vᵣₘₛ = √2 : √(8/π) : √3 = 1 : 1.128 : 1.225, for every gas at every temperature. The ordering is forced by the tail — the long high-speed side drags the mean above the peak, and squaring before averaging weights that side more heavily still. For nitrogen at 300 K the three are 422, 476 and 517 m s⁻¹. Use vᵣₘₛ for energy, v̄ for collision and effusion rates, and vₚ only when you mean the peak.

Every speed on the curve scales as √(T/M), so temperature and molar mass are not two knobs but one. Nitrogen at 300 K peaks at 422 m s⁻¹; heat it to 1200 K, four times the temperature, and the peak moves only to 844 m s⁻¹. Or leave the temperature alone and change the gas: helium at 300 K has M smaller by 28.0/4.00 = 7.00, so its peak sits at √7.00 × 422 = 1116 m s⁻¹. What is easy to miss is the height. The area under f(v) is pinned at 1, so a curve that stretches sideways must sag — the peak value is 4/(√π e vₚ), inversely proportional to vₚ, so the helium peak is 2.6 times further right and 2.6 times lower than the nitrogen one. The tail answers far more sharply than the peak does. Measured against the peak height, f has fallen to 4/e³ = 0.199 at 2vₚ and to 9/e⁸ = 3.0 × 10⁻³ at 3vₚ, which is why a modest change in vₚ can swing the population of fast molecules by orders of magnitude, and why reaction rates are so temperature-sensitive.

Treat the molecules as hard spheres of diameter d. Two of them touch when their centres come within d, so a molecule sweeps a cylinder of cross-section σ = πd² and hits every molecule whose centre falls inside it; for nitrogen d ≈ 3.7 × 10⁻¹⁰ m gives σ = 4.30 × 10⁻¹⁹ m². If one molecule moved and the rest stood still, in time t it would sweep volume σ v̄ t, meet n σ v̄ t of them, and average 1/(nσ) between hits. But the targets move too, and what counts is the speed of approach. Averaging |v₁ − v₂| over two independent Maxwell–Boltzmann distributions gives √2 v̄, so the number of hits rises by √2 while the distance travelled does not. Hence λ = 1/(√2 π d² n), and the collision rate is z = v̄/λ = √2 π d² n v̄. Notice which average appears: v̄, because collisions count molecules, not energy.

Put numbers in. At 1.013 × 10⁵ Pa and 300 K the number density is n = p/(k_BT) = 1.013 × 10⁵ ÷ (4.143 × 10⁻²¹) = 2.45 × 10²⁵ m⁻³. With σ = 4.30 × 10⁻¹⁹ m², λ = 1/(√2 × 2.45 × 10²⁵ × 4.30 × 10⁻¹⁹) = 6.7 × 10⁻⁸ m, or 67 nm. That is 182 molecular diameters and about 20 times the mean spacing between neighbours, 3.4 nm, so the gas really is dilute and the hard-sphere picture is safe. Dividing by v̄ = 476 m s⁻¹ gives a mean free time of 0.14 ns and a collision rate of 7.1 × 10⁹ s⁻¹. Substituting n = p/(k_BT) turns the result into λ = k_BT/(√2 π d² p), the form to reach for in the laboratory: at fixed temperature λ is inversely proportional to pressure, so pumping from 1 atm down to 1.0 × 10⁻³ Pa stretches λ from 67 nm to 6.8 m.

Four assumptions carry the whole chapter. Equilibrium: f(v) describes a gas at one uniform temperature with no net flow, so it does not apply inside a shock, a jet, or a steep thermal gradient. Classical statistics: the Boltzmann factor stands in for the true quantum statistics, which is safe while the thermal de Broglie wavelength stays far below the mean spacing — true for air at 300 K, false for conduction electrons in a metal or for helium near 1 K. Hard spheres of one fixed diameter: real molecules attract at long range and soften at short range, so the effective d drifts with temperature and tabulated values disagree by a few per cent. Diluteness: the derivation assumes uncorrelated neighbours and λ ≫ d, which fails in a dense gas or a liquid. One further caution — the molecules crossing a hole are not sampled from f(v) but from f(v) weighted by an extra factor of v, so an effusing beam is faster than the gas behind it and peaks at vᵣₘₛ, not vₚ.