Write a note on mean free path.

Explanation

Molecules of a gas have finite size and undego collisions. The path of these molecules gets deflected. If a molecule is assumed to spherical in diameter "d" then it colide with another molecule within a distance of between their centres. The path travlled by a molecule in a time interval of `Deltat` with an average spped will be `Deltat`. A molecule will sweeep a volume of `((pid^(2))`<>`Deltat)/(Deltat)=npid^(2)`<>
Number of collisions suffered `=pid^(2)`<>Deltatn therefore(1)/(pid^(2)<>)=Deltat`
The average time taken b/w two successive collisions = `=pid^(2)<>Deltatn therefore(1)/(pid^(2)<>)=Deltat`
The mean free path is the average distance b/w two successive collisions.
ie., `l= pi = (1)/(npid^(2))` the other molecules are assumed to be at rest. However the average relative velocity of the molecular determine the actual rate of collision.
It can be shown that `l=((1)/(sqrt2npid^(2))).`
We note that n is of the order `10^(23) "to" 10^(25)`, diameter of the order of a few `overset(**)(A)` and mean free path `10^(-6)m - 10^(-s)m` and time b/w the succissive collisions `10^(-7)s "to" 10^(-6)s`.