When one mole of monoatomic gas is mixed with 3 moles of diatomic gas, then find `C_(p), C_(v), f and gamma` for this mixture. Here, `f` stands for degrees of freedom.

Explanation

Here, `n_(1)=1, f_(1)= 3` for monoatomic gas
`n_(2)=3, f_(2)=5`, for diatomic gas.
`f_(mix)= (n_(1)f_(1)+n_(2)f_(2))/(n_(1)+n_(2))=(1xx3+3xx5)/(1+2)`
`= (18)/3 = 6`
`(C_(v))_(mix.)= f_(mix.)xxR/2=6xxR/2= 3R`
`(C_(p))_(mix.)= (C_(v))_(mix.)+R=3R+R=4R`
`gamma_(mix.)= 1+(2)/(f_(mix.))=1+2/6= 4/3`