Answer to Question #103757 in Molecular Physics | Thermodynamics for Ajay

Question

Obtain an expression for mean free path of the molecules in a gas in the first order
approximation. Also, show that mean free path of the molecules in a gas decreases
when all molecules are moving

Answer 1

It is the average distance traveled by molecule between the successive collisions, which modify it's direction or energy or other particle property.

Let the duration between the collision is t and the distance covered by the molecule is $\bar{v}t$ and the diameter of the cylinder is 2d.

so, volume of the cylinder $=\pi d^2 vt$

Average number of collisions z=$n_v\pi d^2 vt$

where $n_v$ is the number of collisions.

So, mean free path(l) =$\dfrac{vt}{n_v\pi d^2 vt}=\dfrac{1}{n_v\pi d^2}$

we know that $n_v=\dfrac{p}{K_bT}$

So, $l=$ $\dfrac{1}{n_v\pi d^2}=\dfrac{K_bT}{p\pi d^2}$

we know that, $v\rightarrow\sqrt{2}v$

So, $z=n_v\pi d^2 vt=n_v\pi d^2 \sqrt{2}vt$

Hence, mean free path (l)$=\dfrac{vt}{n_v\pi d^2 \sqrt{2}vt}=\dfrac{1}{\sqrt{2}n_v\pi d^2}$

$l=$ $\dfrac{K_b T}{\sqrt{2}p\pi d^2}$

As the number of molecules are getting increase, the molecules will become closer to each other, therefore they more likely to run each other so mean free path decreases.