For the case of nitrogen under standard conditions find :
(a) the mean number of collisions experienced by each molecule per second ,
(b) the total number of collisions occurring between the molecules within `1 cm^3` of nitrogen per second.

Explanation

(a) `v = (1)/(tau) = (1)/(lamda//lt v gt) = (lt v gt)/(lamda)`
=`sqrt(2) pi d^2 n lt v gt = .74 xx 10^10 s^-1 (sec 2.223)`
(b) Total number of collisions is
`(1)/(2) nv ~~ 1.0 xx 10^29 s cm^-3`
Note, the factor `(1)/(2)`. When two molecules collide we must not count it twice.