Mean Free Path MCQ - Practice Questions with Answers

Mean Free Path - 

On the basis of the kinetic theory of gases, it is assumed that the molecules of a gas are continuously colliding against each other. So, the distance traveled by a gas molecule between any two successive collisions is known as the free path.

There are assumptions for this theory that during two successive collisions, a molecule of a gas moves in a straight line with constant velocity. Now, let us discuss the formula of mean free path - 

Let ${\lambda }_1,{\lambda }_2\dots \dots {\lambda }_n$ be the distance traveled by a gas molecule during $\mathbf{n}$ collisions respectively, then the mean free path of a gas molecule is defined as -

$$\lambda =\frac{\text{ Total distance travelled by a gas molecule between successive collisions }}{\text{ Total number of collisions }}$$

Here, $\lambda$ is the mean free path.

It can also be written as -

$$\lambda =\frac{{\lambda }_1+{\lambda }_2+{\lambda }_3+\dots +{\lambda }_n}{n}$$

Now, let us take d = Diameter of the molecule,
$N=$ Number of molecules per unit volume.

Also, we know that, $\mathrm{PV}=\mathrm{nR}T$
So, Number of moles per unit volume $=\frac{n}{V}=\frac{P}{RT}$
Also, we know that the number of molecules per unit mole $=N_A=6.023\times {10}^{23}$
So, the number of molecules in ' $n$ ' moles $=n{N}_A$
So the number of molecules per unit volume is $N=\frac{P{N}_A}{RT}$

$$\begin{array}{rl}\lambda & =\frac{\mathrm{RT}}{\pi {\mathrm{d}}^2{\mathrm{PN}}_{\mathrm{A}}}=\frac{\mathrm{kT}}{\pi {\mathrm{d}}^2\mathrm{P}}\\ \lambda & =\frac{1}{\sqrt{2}\pi N{d}^2}=\frac{\mathrm{RT}}{\sqrt{2}\lambda {\mathrm{d}}^2{\mathrm{PN}}_{\mathrm{A}}}=\frac{\mathrm{kT}}{\sqrt{2}\pi {\mathrm{d}}^2\mathrm{P}}\end{array}$$

If all the other molecules are not at rest then,

$$\begin{array}{rl}& \mathit{\lambda }=\frac{1}{\sqrt{2}\pi \mathit{N}{\mathit{d}}^2}\text{ and }\mathrm{m}=\text{ mass of each }\mathrm{r}\\ & \text{ Now, if }\\ & \lambda =\frac{1}{\sqrt{2}\pi N{d}^2}=\frac{m}{\sqrt{2}\pi (mN)d^2}=\frac{m}{\sqrt{2}\pi d^2\rho }\end{array}$$

So,

$$\lambda \propto \frac{1}{\rho }\text{ and }\lambda \propto m$$