Since equal volumes of gases are taken at the same temperature and pressure, their number of moles will be equal, let it be $n$. For monoatomic gases, $C_v = \frac{3}{2}R$ and $C_p = \frac{5}{2}R$. The molar heat capacity at constant volume for the mixture is given by: $C_{v(\text{mix})} = \frac{n_1C_{v1} + n_2C_{v2}}{n_1 + n_2} = \frac{n(\frac{3}{2}R) + n(\frac{3}{2}R)}{n + n} = \frac{3}{2}R$ The molar heat capacity at constant pressure for the mixture is given by: $C_{p(\text{mix})} = \frac{n_1C_{p1} + n_2C_{p2}}{n_1 + n_2} = \frac{n(\frac{5}{2}R) + n(\frac{5}{2}R)}{n + n} = \frac{5}{2}R$ Therefore, the ratio of specific heats for the mixture is: $\gamma_{\text{mix}} = \frac{C_{p(\text{mix})}}{C_{v(\text{mix})}} = \frac{\frac{5}{2}R}{\frac{3}{2}R} = \frac{5}{3} \approx 1.67$ Alternatively, since both are monoatomic gases, the mixture will also behave as a monoatomic gas, so $\gamma = \frac{5}{3} \approx 1.67$.