# Mean free path for gas mixture

How to derive an expression for mean free path ($\lambda$) of a gas mixture with gases of different molecular mass and diameter? What modification will be required in the general technique used by Maxwell to reach at

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

Where $n$ is number density and $d$ is molecular diameter

It's the same general idea as lies behind an ideal gas calculation, with the additional factor that there are extra types of collisions to incorporate into your estimate.

Take a mixture of 50% $\rm CO_2$ and 50% $\rm He$ (by molarity).

The available collisions that may occur for $\rm CO_2$ are ($\rm CO_2 + CO_2$) and ($\rm CO_2+He$).

The equivent collision probabilities for $\rm He$ are ($\rm He + He$) and ($\rm He + CO_2$).

Obviously the collisions of ($\rm CO_2 + He$) and ($\rm He + CO_2$) represent the same event and have the same collision rate.

Because ($\rm CO_2 + CO_2$) occurs at a slower rate than ($\rm He + He$), the total collision rate for $\rm CO_2$ will be slower than the lighter $\rm He$.

Also, the mean velocity of $\rm CO_2$ will be slower than $\rm He$.

The mean velocity divided by the collision rate, provides the mean free path.

The source for this answer is PhysicsForums

• If the mean velocity of one species is different from that of the other one then the mixture is not in thermal equilibrium, is it?
– Jeff
Commented Jul 23, 2017 at 13:41
• Now, TBH I have only started studying the nice ideal gases, but read this about a mixture of two ideal gases : physics.stackexchange.com/questions/96089/… intuitively I think you would have thermal equilibrium, it might just take longer to achieve than single species ideal gases
– user163104
Commented Jul 23, 2017 at 13:49
• Well, temperature is usually defined as part of the exponent in the Maxwell-Boltzmann distribution, and thus it uniquely determines the mean velocity of the species. That's why I don't think it's fair to use the argument that the difference in mean free path is caused by different mean velocities: if both species have the same temperature, they have the same mean velocity.
– Jeff
Commented Jul 23, 2017 at 13:58
• Oh hang on, that's not true! The mean velocity does also depend on the species mass, not only the temperature. So indeed, a heavier species WILL have a lower mean velocity. See this wiki article
– Jeff
Commented Jul 23, 2017 at 14:13
• My question, related to this, is that I know about the Maxwell distribution for the d.o.f. related to translation, but I don't know ( yet), how you modify the distribution for the range of vibrational and rotational velocities of molecules. Slightly off topic here as regards ideal gases, sorry.
– user163104
Commented Jul 23, 2017 at 14:21

It is a simple solution that I have been used.This equation may be used for all ideal gas mixtures. The mean free path is $$\lambda\frac{N}{V}\pi r^2 \approx 1$$ where $$r$$ is the radius of a molecule. This gives $$\lambda = \frac{1}{(N/V)\pi r^2}$$

So we have $$\lambda_t = \frac{1}{\Sigma (x_i / \lambda_i)} \\ \Sigma x_i =1$$ where$$\lambda_t$$ is the mean value of the multi system and $$x$$ is the mole fraction.