1 mole of a gas with $$\gamma = \frac{7}{5}$$  is mixed with 1 mole of a gas with $$\gamma = \frac{5}{3},$$  then the value of $$\gamma $$ for the resulting mixture is

Solution :
If $${n_1}$$ moles of adiabatic exponent $${\gamma _1}$$ is mixed with $${n_2}$$ moles of adiabatic exponent $${\gamma _2}$$ then the adiabatic component of the resulting mixture is given by
$$\eqalign{ & \frac{{{n_1} + {n_2}}}{{\gamma - 1}} = \frac{{{n_1}}}{{{\gamma _1} - 1}} + \frac{{{n_2}}}{{{\gamma _2} - 1}} \cr & \frac{{1 + 1}}{{\gamma - 1}} = \frac{1}{{\frac{7}{5} - 1}} + \frac{1}{{\frac{5}{3} - 1}} \cr & \therefore \,\,\frac{2}{{\gamma - 1}} = \frac{5}{2} + \frac{3}{2} \cr & = 4 \cr & \therefore 2 = 4\gamma - 4 \cr & \Rightarrow \,\,\gamma = \frac{6}{4} \cr & = \frac{3}{2} \cr} $$