Two moles of $H_2$ and one mole of He are mixed in a closed container whose walls are made of insulating material. If 50% of $H_2$ molecules are dissociated into atomic hydrogen, $C_P$ / $C_V$ of the mixture is 

After dissociation, number of moles of monoatomic gas in the vessel, $n_1=1+1=2$. ${\gamma }_1=\frac{3+2}{3}=\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$3$}\right.$ and number of moles of diatomic gas, $n_2=1.{\gamma }_2=\frac{5+2}{5}=\raisebox{1ex}{$7$}\!\left/ \!\raisebox{-1ex}{$5$}\right.$

Now $\frac{n_1+n_2}{\gamma -1}=\frac{n_1}{{\gamma }_1-1}+\frac{n_2}{{\gamma }_2-1}$

$\begin{array}{l}\Rightarrow \frac{2+1}{\gamma -1}=\frac{2}{{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}^{-1}}+\frac{1}{{\raisebox{1ex}{$7$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}^{-1}}=3+\frac{5}{2}=\frac{11}{2}\\ \Rightarrow \gamma =1+\raisebox{1ex}{$6$}\!\left/ \!\raisebox{-1ex}{$11$}\right.=\frac{17}{11}\end{array}$