(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

(i) The specific heat of oone mole of a monoatomic gas `C_V=3/2R`.
For `mu_1 ` mole, `C-V=3/2mu_1R " " C_P=5/2mu_1R`
The specific heat of one mole of a diatomic gas
`C_V=5/2R`
For `mu_2` mole , `C_v=5/2mu_2R" " C_P=7/2mu_2R`
The specific heat of the mixture at constant pressure `C_P = 5/2mu_1R +7/2mu_2R`
The adiabatic exponent `gamma=C_P/C_V=(5mu_1+7mu_2)/(3mu_1+5mu_2)`
(ii) Form equation,
`lamda=1/(sqrt2pind^2)`
We have to find the number density n
By using ideal gas law
`n=N/V=P/(KT)` `=(101.3xx10^3)/(1.381xx10^(-23)xx300)`
`=2.449xx10^(25) "molecules"//m^3`
`lamda=1/(sqrt2xxpixx2.449xx10^(25)xx(1.2xx10^(-10)^2))`
`=1/(15.65xx10^5)`
`lamda=0.63xx10^(-6)m`