Two ideal gases have same value of `(C_p / C_v = gamma)`. What will be the value of this ratio for a mixture of the two gases in the ratio `(1:2)`?

Explanation

Considering gases, in gas (1) we have, `C_(p_1)`(Sp. heat at constant P), `C_(v_1)` (Sp. heat at constant V), `n_1` (No. of moles)
`(C_(p_1))/(C_(p_1)) = gamma` and ` C_(p_1) - C_(v_1) =R`
implies `gamma C_(v_1) - C_(v_1) =R`
implies `C_(v_1)(gamma-1) =R`
implies `C_(v_1) = (R)/(gamma-1)` and `C_(p_1) = gamma (R)/((gamma-1))`
In gas (2), we have
`C_(p_2)` (Sp. heat at constant P), `C_(v_2)` (Sp. heat at constant V) , `n_2` (No. of moles)
`(C_(p_2))/(C_(v_2)) = gamma` and ` C_(p_2)-C_(v_2)=R`
` implies `gamma C_(v_2)-C_(v_2) =R`
`C_(v_2)(gamma -1) =R`
implies `C_(v_2) = (R)/((gamma-1))` and
`C_(p_2) = gamma (R)/((gamma-1))`
Given, `n_1 = n_2 = 1:2`
`dU_1=nC_(v_1)dt` and `dU_2`
`=2nC_(v_2)dT`
When gases are mixed,
`nC_(v_1)dT + 2nC_(v_2)dT = 3nC_vdT`
`C_v = (C_(v_1) + 2C_(v_2))/(3)`
` = ((R)/(gamma-1)+(2R)/(gamma-1))/(3)`
`(3R)/((gamma-1)3) = (R)/(gamma-1)`
Hence `C_p//C_v` in the mixture `=gamma`.