19-5 Mean Free Path#

Prompts

Define the mean free path \(\lambda\). Why does it depend on the number density \(N/V\) and the molecular diameter \(d\)?
Write the mean free path formula \(\lambda = 1/(\sqrt{2}\,\pi d^2\,N/V)\). Why is there a factor \(\sqrt{2}\)? Why does \(\lambda \propto 1/d^2\)?
Gas molecules move at hundreds of m/s. Why does perfume take seconds or minutes to diffuse across a room?
For an ideal gas at pressure \(p\) and temperature \(T\), express \(N/V\) in terms of \(p\) and \(T\). How does \(\lambda\) change if you double the pressure?

Lecture Notes#

Overview#

The mean free path \(\lambda\) is the average distance a molecule travels between collisions with other molecules.
\(\lambda \propto 1/(N/V)\)—higher number density → more collisions → shorter \(\lambda\).
\(\lambda \propto 1/d^2\)—larger molecules (larger collision cross section) → shorter \(\lambda\).
Despite high molecular speeds (~500 m/s for air), diffusion is slow because collisions cause frequent direction changes.

The mean free path formula#

(179)#\[ \lambda = \frac{1}{\sqrt{2}\,\pi d^2\,(N/V)} \]

\(d\): molecular diameter (effective collision size).
\(N/V\): number of molecules per unit volume (number density).
\(\pi d^2\): collision cross section—effective target area for a collision.

Physical picture: A molecule sweeps a cylinder of cross-sectional area \(\pi d^2\) as it moves. Collisions occur with molecules whose centers lie in that cylinder. The factor \(\sqrt{2}\) arises because all molecules move—the relative speed is \(\sqrt{2}\,v_{\text{avg}}\).

Dependence on density and pressure#

For an ideal gas, \(N/V = p/(kT)\). So

(180)#\[ \lambda \propto \frac{T}{p} \]

Higher pressure → more molecules per volume → shorter \(\lambda\).
Higher temperature (at fixed \(p\)) → lower density → longer \(\lambda\).

Condition	Typical \(\lambda\)
Air at sea level (1 atm)	~0.1 mm
High altitude (~100 km)	~16 cm
Very low density	meters to km

Why diffusion is slow

Molecules move fast (~500 m/s), but they collide billions of times per second. Each collision changes direction, so the net displacement from the source (diffusion) is much slower than the molecular speed.

Poll: Mean free path

You compress a gas to half its volume at constant temperature. The mean free path \(\lambda\):

(A) Doubles
(B) Halves
(C) Stays the same

Summary#

\(\lambda = 1/(\sqrt{2}\,\pi d^2\,N/V)\)—mean free path; average distance between collisions.
\(\lambda \propto 1/(N/V)\) and \(\lambda \propto 1/d^2\).
For ideal gas: \(\lambda \propto T/p\); higher pressure → shorter \(\lambda\).
Collisions cause frequent direction changes → diffusion is slow despite high molecular speeds.