For oxygen gas at `27^(@)C`, calculate the mean free path at 1 atm pressure. The collision diameter of oxygen molecule is 361 picometre.

Explanation

`P= 1.0 atm= 1.01325 xx 10^(5) Nm^(-2), T= 27^(@)C= 300K, d= 361 pm= 361 xx 10^(-12)m = 3.61 xx 10^(-10)m`
Mean free path at 1 atm pressure,
`lambda= (kT)/(sqrt(2)pi d^(2)P)= (1.38xx 10^(-23)JK^(-1)(300K))/((1.4142)(3.1416)(3.61xx 10^(-10)m)^(2)(1.01325xx 10^(5)Nm^(-2)))`
`=7.07 xx 10^(-8) m= 70.7nm" "[J= kg m^(2)s^(-2)= Nm]`.