Mean Free Path

Definition

The molecules of a gas are in continuous random motion. When one molecule approaches another they repel each other, so the molecules behave as small hard spheres and the collisions between them are elastic; both speed and direction change abruptly at each collision. Between two successive collisions the molecules move along straight paths with constant velocity, because the molecules exert force on each other only during collisions.

The mean free path $\lambda$ is defined as the average distance traversed by a molecule with constant velocity between two successive collisions.

Diagram in words

A typical molecule in a gas traces a zig-zag path: a sequence of straight segments of unequal length, each ending in a sharp change of direction where a collision occurs. The mean free path is the average length of these straight segments.

The factors on which the mean free path depends

• Number density. The mean free path varies inversely with the density of the gas, expressed through $\dfrac{N}{V}$, the number of molecules $N$ enclosed in a volume $V$. The higher the density, the more collisions there are and the smaller is $\lambda$.

• Size of the molecule. $\lambda$ is inversely proportional to the diameter $d$ of the molecule: the smaller the molecule, the less the chance of collision and the larger the mean free path. Further, $\lambda$ is inversely proportional to $d^{2}$ and not merely to $d$, because collision depends on the cross section of a molecule, which goes as $d^{2}$.

Expression

$$\lambda = \dfrac{1}{\sqrt{2}\,\pi d^{2}\left(\dfrac{N}{V}\right)}$$

Using $PV = Nk_{B}T$, so that $\dfrac{N}{V} = \dfrac{P}{k_{B}T}$, the same relation becomes

$$\lambda = \dfrac{k_{B}T}{\sqrt{2}\,\pi d^{2}P}$$

which is the form to use whenever the pressure and temperature are given instead of the number density. It shows directly that $\lambda \propto T$ at constant pressure and $\lambda \propto \dfrac{1}{P}$ at constant temperature.

Worked value from the textbook

For nitrogen at $T = 0\,^{\circ}$C $= 273$ K, $P = 1$ atm $= 1.01 \times 10^{5}$ Pa and molecular diameter $d = 324$ pm $= 324 \times 10^{-12}$ m,

$$\lambda = \dfrac{\left(1.38 \times 10^{-23}\right)(273)}{\sqrt{2}\,\pi\left(324 \times 10^{-12}\right)^{2}\left(1.01 \times 10^{5}\right)} = 8.0 \times 10^{-8}\ \text{m}$$

which is about 247 times the molecular diameter.

Application

If the pressure of a gas in an enclosure is reduced by evacuating it, the density of the gas decreases and the mean free path increases. This is used in coating articles with metal films: the metal is heated and evaporated inside an enclosure whose pressure is reduced until the mean free path of the air molecules is larger than the dimensions of the enclosure. The atoms in the metal vapour then do not collide with air molecules, so they travel straight to the target and get deposited on it.

Standard results

• $\lambda = \dfrac{1}{\sqrt{2}\pi d^{2}(N/V)} = \dfrac{k_{B}T}{\sqrt{2}\pi d^{2}P}$

• $\lambda \propto \dfrac{1}{N/V}$, $\lambda \propto \dfrac{1}{d^{2}}$, $\lambda \propto T$ at constant $P$, $\lambda \propto \dfrac{1}{P}$ at constant $T$.

• For nitrogen at S.T.P., $\lambda \approx 8 \times 10^{-8}$ m.