Two ideal gases have the same value of  $\frac{{\text{C}}_{\text{p}}}{{\text{C}}_{\text{v}}}=\gamma$ . The value of this ratio for the mixture of two gases in the ratio$1:2$ is

given that two ideal gasses are having same $\gamma$ value they are mixed in 1 : 2

ratio of their moles

${\gamma }_{\text{mixture}}=?$

We know

$\frac{{\text{C}}_{\text{p}}}{{\text{C}}_{\text{v}}}=\gamma$

${\text{C}}_{\text{P}}-{\text{C}}_{\text{V}}=\text{R}$

${\text{C}}_{\text{V}}\left(\frac{{\text{C}}_{\text{P}}}{{\text{C}}_{\text{V}}}-1\right)=\text{R}$

${\text{C}}_{\text{V}}=\frac{\text{R}}{\gamma -1}$

${\text{C}}_{\text{P}}=\frac{\gamma \text{R}}{\gamma -1}$

  $\frac{{\text{C}}_{\text{P}}}{{\text{C}}_{\text{V}}}=\gamma ,\frac{{\text{n}}_{\text{1}}}{{\text{n}}_{\text{2}}}=\frac{1}{2}$

${\gamma }_{\text{mixture}}=\frac{{\left({\text{C}}_{\text{P}}\right)}_{\text{mix}}}{{\left({\text{C}}_{\text{V}}\right)}_{\text{mix}}}=\frac{{\text{n}}_{\text{1}}{\left({\text{C}}_{\text{P}}\right)}_{\text{m}}+{\text{n}}_{\text{2}}{\left({\text{C}}_{\text{P}}\right)}_{\text{d}}}{{\text{n}}_{\text{1}}{\left({\text{C}}_{\text{V}}\right)}_{\text{m}}+{\text{n}}_{\text{2}}{\left({\text{C}}_{\text{V}}\right)}_{\text{d}}}$

${\gamma }_{\text{mixture}}=\frac{{\text{n}}_1\left(\frac{\gamma \text{R}}{\gamma -1}\right)+2{\text{n}}_1\left(\frac{\gamma \text{R}}{\gamma -1}\right)}{{\text{n}}_1\left(\frac{\text{R}}{\gamma -1}\right)+2{\text{n}}_1\left(\frac{\text{R}}{\gamma -1}\right)}$                                        

$\Rightarrow \frac{{\left(C_P\right)}_{mix}}{{\left(C_V\right)}_{mix}}=\frac{n_1\left(\frac{R}{\gamma -1}\right)\left(\gamma +2\gamma \right)}{n_1\left(\frac{R}{\gamma -1}\right)\left(1+2\right)}$

$=\frac{3\gamma }{3}=\gamma$

$\therefore {\gamma }_{\text{mixture}}=\gamma$