(i) Find the adiabatic exponent `gamma` for mixture of `mu_1` moles of monoatomic gas and `mu_2` moles of a diatomic gas at normal temperature.
(ii) An oxygen molecule is travelling in air at 300 K and 1 atm , and the diameter of oxygen molecule is `1.2xx10^(-10)` m . Calculate the mean free path of oxygen molecule.

Explanation

The specific heat of one mole of a monoatomic gas `C_(V)=3/2R`
For `mu` mole, `C_(V)=3/2mu_(1)R,C_(P)=5/2mu_(1)R`
The Specific heat of one mole of a diatomic gas `C_(V)=5/2R`
For `mu_(2)` mole, `C_(V)=5/2mu_(2)R , C_(P)=7/2mu_(2)R`
The specific heat of the mixture at constant volume `C_(V)=3/2mu_(1)R+5/2mu_(2)R`
The specific heat of the mixture at constant pressure `C_(P)=5/2mu_(1)R+7/2mu_(2)R`
The adiabatic exponent `gamma =C_(P)/C_(V)=(5mu_(1)+7mu_(2))/(3mu_(1)+5mu_(2))`