Question

The expression for mean free path is:
A. $\lambda ={\displaystyle \frac{KT}{\sqrt{2}\pi d^{2}P}}$
B. $\lambda ={\displaystyle \frac{\pi dP}{kT}}$
C. $\lambda ={\displaystyle \frac{\pi d^{2}P}{kT}}$
D. $\lambda ={\displaystyle \frac{kT}{\pi dP}}$

Answer 1

Hint: The actual distance a molecule travels before a collision is called free path. The average of such distances is known as mean free path. As the size of molecules gets larger the mean free path gets shorter.
The mean free path is given as $\lambda ={\displaystyle \frac{1}{\sqrt{2}\pi d^2{\displaystyle \frac{N}{V}}}}$

Complete step by step answer:
 The motion of a gas molecule inside an ideal gas changes its direction and speed as it collides elastically with other molecules of the same gas. Between the collisions the molecules move in a straight line at some constant speed. Measuring the mean free path is difficult as it depends on a number of factors.
Let us derive the equation for the mean free path. Consider $\lambda$ to be the average distance travelled by a molecule between collisions. As the number of molecules per unit volume increases, the number of collisions increases hence the distance travelled by molecule decreases.
Thus, we have:
$\lambda \alpha {\displaystyle \frac{1}{\left({\displaystyle \frac{N}{V}}\right)}}$ -- equation $1$
Here, ${\displaystyle \frac{N}{V}}$ is the number of molecules per unit volume.
Also, if we consider molecules to be point masses then there is no collision as their point mass would not have any surface. Thus, if a molecule has a larger surface that is larger in diameter than the probability of collision will increase and thus reduce the mean free path. Thus, the mean free path must be proportional to $\pi$ times the square of the diameter $d$ as the collision takes place on the surface and not only at the diameter line
$\lambda \alpha {\displaystyle \frac{1}{\pi d^2}}$
Thus, the number of collisions in time $t$ having when the particles is having velocity $v$ will be given as
$\pi d^{2}vt{\displaystyle \frac{N}{V}}$ -- equation $2$
The particle covers a path of length $vt$ in time $t$ .
Now, mean free length will be the ratio of the length of path covered in time $t$ divided by the number of collisions.
$\lambda ={\displaystyle \frac{v_{avg}t}{\pi d^{2}vt{\displaystyle \frac{N}{V}}}}$
Here $v_{avg}$ is the average velocity.
$v$ is the relative velocity.
The ratio ${\displaystyle \frac{v}{v_{avg}}}=\sqrt{2}$ thus
$\Rightarrow \lambda ={\displaystyle \frac{1}{\sqrt{2}\pi d^2{\displaystyle \frac{N}{V}}}}$ --equation $3$
Using gas equation, we have ${\displaystyle \frac{N}{V}}={\displaystyle \frac{P}{KT}}$
Here, $P$ is the pressure inside the cylinder.
$K$ is the Boltzmann constant
$T$ is the temperature.
Substituting this value in equation $3$ we get
$\lambda ={\displaystyle \frac{KT}{\sqrt{2}\pi d^{2}P}}$
This is the expression for mean free path.

So, the correct answer is “Option A”.

Note:
Two velocities are considered in the derivation; average velocity and relative velocity. It is difficult to measure the path between collisions as the collisions take place in a random manner. As the size of the atom increases, the number of collisions increases and thus the free path decreases. Mean free path is the average of the sum of free paths.