# Mean Free Path of Air

There is a simple approximation for the mean free path, λmfp, of particles in air at room temperature (T = 25 oC) given by,

$\lambda_\text{mfp} \approx \frac{5 \times 10^{-3}}{P}$

where λmfp is in units of centimeters and the pressure, P, is in units of Torr. While this is a commonly used approximation, it is not clear how it is derived. The following is one way to arrive at this result.

The mean free path of a single species of gas is given by,

$\lambda_\text{mfp} = \frac{1}{n \sigma}$

where n is the gas density and σ is the collisional cross section. While this expression for a single type of gas (i.e. not a mixture such as air), we will apply it for the derivation of our approximate solution. The composition of air is more than 70% nitrogen so it is “approximately” correct to solve for the mean free path of nitrogen.

Solve for the density using the ideal gas law,

$PV = NkT \\ P = \frac{N}{V}kT = nkT \\ n = \frac{P}{kT}$

where V is the volume occupied by the gas, N is the total number of gas particles, and k is Boltzmann’s constant. The term N/V is the density, n. The result of this step is that the density is written in terms of the pressure and temperature, which are the variable given in the final solution. Inserting the density expression into the general form of &lambdamfp gives:

$\lambda_\text{mfp} = \frac{1}{\frac{P}{kT} \sigma} = \frac{kT}{P\sigma}$

It remains to solve for the collisional cross section before a numerical result may be obtained. The collisional cross section is given by,

$\sigma = \pi d^2$

where d is the particle diameter. Recall that this is a measure of the area (centered on the center of mass of one of the particles) through which the particles cannot pass each other without colliding. This leads to a discussion of the proper value to use for the atomic diameter of nitrogen. Since this is air, the nitrogen is actually diatomic, so we need to determine the diameter of a nitrogen molecule.

Atomic diameters can be difficult to determine. The nature of atomic structure can make the very definition of size a challenging prospect. Again, in this approximate treatment it is acceptable to take the more direct route of treating the particles like hard spheres. In this case the diameter of a single nitrogen atom is roughly 3.1 Å (3.1 × 10-10 m). The diameter of diatomic nitrogen is then 6.2 Å†. The collisional cross section is σ = π × (6.2 × 10-10)2 = 1.21 × 10-18 [m2].

The expression for the mean free path can be simplified further by inserting some of the numerical values now available. Boltzmann’s constant is k = 1.38 × 10-23 [J]/[K] and the temperature is T = 25 oC ~ 298 [K].

$\lambda_\text{mfp} = \frac{kT}{P\sigma} \\ = \frac{411.24 \times 10^{-23}}{P \left(1.21 \times 10^{-18} \right)} \\ = \frac{3.40 \times 10^{-3}}{P}$

This is already close to the desired expression. It remains to convert this into the correct units. The original expression is in SI units, giving the mean free path in meters and the pressure in Pascals. The necessary conversions are 1 m = 100 cm and 7.5 × 10-3 Torr = 1 Pa. Inserting these conversions will lead to the desired form of this expression (units are enclosed in square brackets),

$\lambda_\text{cm} = 100 \frac{\text{[cm]}}{\text{[m]} \cdot \lambda_\text{m}} \\ = 100 \cdot \left( \frac{3.4 \times 10^{-3}}{P_\text{Pa}} \right) \\ \\ = \frac{3.4 \times 10^{-3}}{P_\text{Pa}}$

$P_\text{Torr} = 7.5\times 10^{-3} \frac{\text{[Torr]}}{\text{[Pa]}} \cdot P_\text{Pa}$

$P_\text{Pa} = \frac{3.4 \times 10^{-1}}{\lambda_\text{cm}}$

$P_\text{Torr} = 7.5 \times 10^{-3} \cdot \frac{3.4 \times 10^{-1}}{\lambda_\text{cm}} \\ = \frac{2.6 \times 10^{-3}}{\lambda_\text{cm}}$

$\lambda_\text{cm} = \frac{2.6 \times 10^{-3}}{P_\text{Torr}}$

where this is the expression we have been trying to verifyE. We are off by a factor of two, though that seems pretty good considering we only used the molecular size of nitrogen.

Having this approximately correct expression, what use is it? Consider the mean free path of an air particle. Atmospheric pressure (sea level) is about 760 Torr. Plugging this into the final expression gives a mean free path of λmfp = 3.4 × 10-6 cm. This is 34 nanometers, which is roughly half of the commonly reported value of 65 nm. Particles in air do not travel very far before they collide with other particles.

† – I should use 6 Å for the diameter because that would make the calculation easier without sacrificing the low level of accuracy needed for this problem. The first time I solved this problem I used 6.2, however, so I’m keeping that now in order to use the other numbers for which I have already solved.

E – Thanks to Eric Lawrence from UCLA for helping out with this.

## One thought on “Mean Free Path of Air”

1. sohail mumtaz says:

Nicely explained. Thanks