Find `gamma` for a mixture of gases containing `n_(1)` moles of a monoatomic gas and `n_(2)` moles of a diatomic gas. Assuming diatomic molecules to be right.

Explanation

`gamma = ?`
Internal energy of `n_(1)` of monoatomic gas
`Delta U_(1) = n_(1) (3/2 RT) DeltaT`
Internal energy of `n_(2)` moles of diatomic gas
`Delta U - Delta U_(1) +Delta U_(2)`
`=R/2 (3n_(1)+5n_(2))Delta T`
If `C_(V)` is specific heat at constant volume for the mixture containing `(n_(1)+n_(2))` moles,
then `C_(V) = (Delta U)/((n_(1)+n_(2))Delta T)`
`C_(V) = R/2 ((3n_(1)+5n_(2))Delta T)/((n_(1)+n_(2))Delta T) = ((3n_(1)+5n_(2)))/((n_(1)+n_(2))) R/2`
Now, `C_(p) = C_(upsilon) +R`
`=R/2 ((3n_(1)+5n_(2)))/(n_(1)+n_(2)) +R = R/2 [(3n_(1)+5n_(2))/((n_(1)+n_(2)))+2]`
`C_(p) = R/2 ((5n_(1)+7n_(2))/(n_(1)+n_(2)))`
`gamma = (C_p)/(C_(upsilon)) = (5n_(1)+7n_(2))/(3n_(1)+5n_(2))`.