Calculate the mean free path of nitogen at `27^(@)C` when pressure is 1.0 atm. Given, diameter of nitogen molecule = `1.5 Å, k = 1.38 xx 10^(-23) JK^(-1)`. If the average speed of nitrogen molecules is `675 ms^(-1)`, find the time taken by the molecule between two successive collsions and the frequency of collisions.

Explanation

Here, `T = 27^(@) C = 27 + 273 = 300 K, P = 1 atm = 1.01 xx 10^(5) N//m^(2)`
`d = 1.5 Å = 1.5 xx 10^(-10) m, k_(B) = 1.38 xx 10^(-23) JK^(-1) , lambda = ?`
from `lambda = (k_(B)t)/(sqrt(2) pi d^(2)P) = (1.38 xx 10^(-23) xx 300)/(1.414 xx 3.14(1.5 xx 10^(-10))^(2) xx 1.01 xx 10^(5)) = 4.1 xx 10^(-7)m`
Time interval between two successive collisions
`t = (dis t a n ce)/(speed) = (lambda)/(upsilon) = (4.1 xx 10^(-7))/(675) = 0.006xx10^(-7)s`
Collision frequency `= 1/t = 1/(0.006 xx 10^(-7)) = 1.67 xx 10^(9) s^(-1)`.