Mean Free Path

The mean free path $\lambda$ is the average distance a gas molecule travels between successive collisions. It connects microscopic molecular properties (size, density) to transport phenomena like viscosity, thermal conductivity, and diffusion.

Concept

Imagine a single molecule (radius $r$) flying through a gas of similar molecules at rest. It collides with any molecule whose centre lies within a cylinder of radius $2r$ around its path (because two spheres of radius $r$ touch when their centres are $2r$ apart).

The cross-section for collision is

$\sigma = \pi (2r)^2 = 4\pi r^2$

(often also called the "scattering cross-section").

Derivation

Step 1 — All other molecules at rest. In time $t$, the moving molecule sweeps out a cylindrical volume $V_{\text{swept}} = \sigma v t$, where $v$ is its speed.

If $n$ is the number density of molecules, the number of collisions in this time is $N_{\text{coll}} = n \sigma v t$.

Mean free path = (distance travelled) / (number of collisions):

$\lambda = \frac{vt}{n \sigma v t} = \frac{1}{n \sigma}$

This is a first estimate, valid only if other molecules are truly stationary.

Step 2 — All molecules in motion. In reality, target molecules are also moving. The relative speed between two molecules averages to $\sqrt{2}$ times the average speed of a single molecule:

$\langle v_{\text{rel}}\rangle = \sqrt{2}\,\langle v\rangle$

This is because for independent random velocities $\vec v_1, \vec v_2$, the magnitude of $\vec v_1 - \vec v_2$ has mean square $\langle v_1^2\rangle + \langle v_2^2\rangle = 2 \langle v^2\rangle$, so the typical relative speed is $\sqrt{2}$ times higher.

Thus the effective collision rate is $\sqrt{2}$ times larger, and the mean free path is reduced by $\sqrt{2}$:

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

where $d = 2r$ is the molecular diameter.

Pressure dependence

Using $n = P / (k_B T)$:

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

So $\lambda$ increases with $T$ at fixed $P$, and decreases with $P$ at fixed $T$.

Typical values

For air at STP, $d \approx 3.7 \times 10^{-10}$ m, $n \approx 2.7 \times 10^{25}$ m$^{-3}$, giving

$\lambda \approx \frac{1}{\sqrt 2 \cdot 2.7\times 10^{25} \cdot \pi (3.7 \times 10^{-10})^2} \approx 6 \times 10^{-8} \text{ m}$

That is, about 60 nm — roughly 200 molecular diameters between collisions.

Collision frequency

The number of collisions per unit time is

$\nu = \frac{\langle v\rangle}{\lambda} = \sqrt 2\, n \sigma\,\langle v\rangle$

For air molecules at room temperature, $\nu \sim 10^9$ s$^{-1}$.

Worked Example

Q: A gas with diameter $d = 4 \times 10^{-10}$ m has number density $n = 3 \times 10^{25}$ m$^{-3}$. Find the mean free path.

A: $\sigma = \pi d^2 = \pi (4\times 10^{-10})^2 = 5.03 \times 10^{-19}$ m².

$\lambda = 1/(\sqrt 2 \cdot 3\times 10^{25} \cdot 5.03 \times 10^{-19}) \approx 1/(2.13 \times 10^7) \approx 4.7 \times 10^{-8}$ m.

So $\lambda \approx 47$ nm.

Common Confusions

• Don't forget the $\sqrt 2$ — beginners often write $\lambda = 1/(n\sigma)$, which is incorrect because target molecules also move.
• The cross-section is $\sigma = \pi d^2$ where $d$ is the molecular diameter, not the radius. ($\sigma = 4\pi r^2$ is the same thing.)
• Mean free path is not the same as average distance to a nearest neighbour.

Key Takeaways

• $\lambda = 1/(\sqrt 2\, n\, \pi d^2)$ — explicitly involves the relative-speed factor $\sqrt 2$.
• $\lambda \propto T/P$ — inverse with pressure, direct with temperature.
• Typical values for air at STP: $\lambda \sim 10^{-7}$ m, collision frequency $\sim 10^9$ Hz.