Physics · Ch 12 — Kinetic Theory

Mean Free Path

12.7

Mean Free Path

The Concept of Mean Free Path

In a gas, molecules are in constant, random motion, colliding with one another. Between two successive collisions, a molecule travels freely in a straight line. The distance it covers during this free journey is not constant — it varies randomly from one free path to the next. The mean free path is the average distance a molecule travels between two successive collisions.

Note

The mean free path is a statistical average over a large number of molecules and over many collisions for a single molecule. It is not the distance a molecule travels in a fixed time — that would be its average speed multiplied by time.

Dependence on Molecular Size and Number Density

The mean free path depends on two factors:

The size of the molecules — larger molecules collide more frequently, so the mean free path is shorter.
The number density of molecules — more molecules per unit volume means more frequent collisions, again shortening the mean free path.

Let us derive an expression for the mean free path. Consider a gas of identical molecules, each treated as a hard sphere of diameter $d$ . Focus on a single molecule that is moving with an average speed $\bar{v}$ . All other molecules are assumed to be stationary for the moment — we will correct this assumption later.

The moving molecule sweeps out a cylindrical volume as it travels. In time $\Delta t$ , it covers a distance $\bar{v} \Delta t$ . Any other molecule whose centre lies within a cylinder of radius $d$ (the collision diameter) around the path of the moving molecule will be hit. The cross-sectional area of this cylinder is $\pi d^2$ , called the collision cross-section.

The volume swept out in time $\Delta t$ is $(\pi d^2)(\bar{v} \Delta t)$ . If $n$ is the number density (number of molecules per unit volume), then the number of collisions in time $\Delta t$ is:

\text{Number of collisions} = n \times (\pi d^2)(\bar{v} \Delta t)

The average time between collisions, $\tau$ , is the time in which exactly one collision occurs. Setting the number of collisions equal to 1:

1 = n \pi d^2 \bar{v} \tau

Therefore:

\tau = \frac{1}{n \pi d^2 \bar{v}}

The mean free path $\lambda$ is the average distance travelled between collisions, which is $\bar{v} \tau$ :

\lambda = \bar{v} \tau = \frac{1}{n \pi d^2}

Watch out

This derivation assumed all other molecules are stationary. In reality, all molecules are moving. When we account for the relative motion of molecules, the effective collision rate increases. For a gas in thermal equilibrium, the correct expression uses the average relative speed $\bar{v}_{\text{rel}} = \sqrt{2} \, \bar{v}$ .

The Correct Expression for Mean Free Path

When all molecules are moving, the number of collisions per unit time is proportional to the average relative speed rather than the average speed of a single molecule. The average relative speed between two molecules in a gas at the same temperature is $\sqrt{2}$ times the average speed of a molecule. Therefore, the collision rate increases by a factor of $\sqrt{2}$ , and the mean free path decreases by the same factor:

\lambda = \frac{1}{\sqrt{2} \, n \pi d^2}

This is the standard expression for the mean free path of a molecule in a gas.

Expressing Mean Free Path in Terms of Macroscopic Quantities

The number density $n$ can be expressed in terms of pressure $P$ and temperature $T$ using the ideal gas law. For $N$ molecules in volume $V$ :

P V = N k_B T

where $k_B$ is Boltzmann's constant. Therefore:

n = \frac{N}{V} = \frac{P}{k_B T}

Substituting into the expression for $\lambda$ :

\lambda = \frac{1}{\sqrt{2} \, \pi d^2} \cdot \frac{k_B T}{P}

Important

At a fixed temperature, the mean free path is inversely proportional to pressure. At a fixed pressure, it is directly proportional to temperature. This makes physical sense: higher pressure means more molecules per unit volume, so collisions are more frequent and the mean free path is shorter.

Key Properties of Mean Free Path

The textbook lists three important properties of the mean free path. Each is derived from the expression $\lambda = \frac{1}{\sqrt{2} \, n \pi d^2}$ .

›Proof

Property (I): The mean free path is independent of the speed of the molecule.

The expression $\lambda = 1/(\sqrt{2} \, n \pi d^2)$ contains no speed term. The average speed $\bar{v}$ cancels out when we account for relative motion. This is because a faster molecule covers more distance per unit time, but it also sweeps through more volume and thus collides more frequently — the two effects exactly balance.

›Proof

Property (II): The mean free path is inversely proportional to the number density $n$ .

From $\lambda = 1/(\sqrt{2} \, n \pi d^2)$ , it is clear that doubling $n$ halves $\lambda$ . More molecules per unit volume means more frequent collisions, so the average free path decreases proportionally.

›Proof

Property (III): The mean free path is inversely proportional to the square of the molecular diameter $d$ . …

Figure 12.7The volume swept by a molecule in time Δt in which any molecule will collide with it.

Fig. 12.7 — The volume swept by a molecule in time Δt in which any molecule will collide with it.

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

The figure shows a single gas molecule moving through space over a short time interval $\Delta t$ . The molecule is drawn as a sphere of diameter $d$ . Its path is represented by a long, tilted cylinder — the swept volume — whose axis is the straight line the molecule travels along. The length of this cylinder is labelled $\bar{v} \Delta t$ , where $\bar{v}$ is the average speed of the molecule. The diameter of the cylinder is labelled $d$ , the same as the molecule's diameter. The two circular end-caps of the cylinder are drawn with dashed lines, indicating that they are imaginary boundaries. Three other molecules (also spheres) are shown lying inside this cylindrical tube.

The physical idea is this: any other molecule whose centre lies inside this swept cylinder will collide with the moving molecule during the time $\Delta t$ . Why? Because the moving molecule sweeps out a tube of radius $d/2$ around its path — any molecule whose centre comes within a distance $d$ of the moving molecule's centre will be hit. The dashed end-caps simply close the volume; they do not represent physical surfaces.

Note

The cylinder is not a real object. It is a geometric construction that lets us count collisions. The moving molecule is treated as a point-like projectile, and every other molecule is treated as a stationary target of diameter $d$ . This is the standard "billiard ball" model of a gas.

From this picture, the textbook derives the mean free path $\lambda$ , the average distance a molecule travels between collisions. The number of other molecules in the swept volume is the number density $n$ (molecules per unit volume) times the volume of the cylinder:

\text{Number of collisions in } \Delta t = n \times (\pi d^2) \times (\bar{v} \Delta t)

The factor $\pi d^2$ is the collision cross-section — the effective area the moving molecule presents to the stationary ones. The total distance travelled in $\Delta t$ is $\bar{v} \Delta t$ . So the average distance between collisions is:

\lambda = \frac{\text{distance travelled}}{\text{number of collisions}} = \frac{\bar{v} \Delta t}{n \pi d^2 \bar{v} \Delta t} = \frac{1}{n \pi d^2}

This is the textbook's first result for the mean free path. A more careful treatment (accounting for the fact that all molecules move, not just one) introduces a factor of $\sqrt{2}$ :

\lambda = \frac{1}{\sqrt{2} \, n \pi d^2}

Each symbol has a clear meaning: $\lambda$ is the mean free path, $n$ is the number density of molecules (number per unit volume), and $d$ is the molecular diameter. The figure itself does not show the $\sqrt{2}$ correction — that comes from a more detailed kinetic argument — but the swept-volume picture is the essential geometric foundation. …

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