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 $\lambda_1, \lambda_2 \ldots \ldots \lambda_n$ 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 -

$$
\lambda=\frac{\lambda_1+\lambda_2+\lambda_3+\ldots+\lambda_n}{n}
$$

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 $=\frac{n}{V}=\frac{P}{R T}$
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 $N=\frac{P N_A}{R T}$

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

If all the other molecules are not at rest then,

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

So,

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