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
 A: 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
A: 
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.
