Name: Introduction to Brownian Motion
Uploaded: 2019-07-19
Duration: 3 min 35 s
Description: The study of Brownian motion is important in the study of the kinetic theory of gases. But what is meant by the Brownian motion? Let's understand it in this...

Question

Derive an expression for the mean free path of a molecule.

Accepted Answer

We will derive the equation using the following assumptions, let’s assume that the molecule is spherical, and the collision occurs when one molecule hits the other, and only the molecule we are going to study will be in motion and rest molecules will be stationary.

Let’s consider our single-molecule to have a diameter of $d$ And all the other molecules to be points this does not change our criteria for collision, as our single-molecule moves through the gas, it sweeps out a short cylinder of cross-section area ${πd}^{2}$ Between successive collisions, for a small-time $t$ It will move a distance of $vt$ where $v$ Is the velocity of the molecule, now if we sweep this cylinder we will get a volume of ${πd}^{2} vt$ So the number of point molecules inside this volume will give us the number of collisions the molecule might have,

Since $N / V$ Is the number of molecules per unit volume, the number of molecule in the cylinder will be $N / V$ Multiplied by the volume of cylinder i.e. ${πd}^{2} vt$ . The mean free path can be derived as follows,

$λ = \frac{length of path during the time t}{number of collision in time r}$ $\approx \frac{vt}{{πd}^{2} vt \frac{N}{V}} = \frac{1}{{πd}^{2} \frac{N}{V}}$

We have cancelled two velocities in the above equation but actually the $v$ In the numerator is the average velocity.And $v$ In the denominator is relative velocity hence they both differ from each other with a factor $\sqrt{2}$

Therefore, the final equation would be, $λ = \frac{1}{\sqrt{2} {πd}^{2} \frac{N}{V}}$

Derive an expression for the mean free path of a molecule.

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