1. For each gas, use $\gamma = \frac{C_p}{C_v}, \qquad C_p - C_v = R$ Hence, $C_v = \frac{R}{\gamma - 1}, \qquad C_p = \frac{\gamma R}{\gamma - 1}$

2. For gas 1, given $\gamma_1 = \frac{7}{5}$ So, $C_{v1} = \frac{R}{\frac{7}{5}-1} = \frac{R}{\frac{2}{5}} = \frac{5R}{2}$ $C_{p1} = C_{v1}+R = \frac{5R}{2}+R = \frac{7R}{2}$

3. For gas 2, given $\gamma_2 = \frac{5}{3}$ So, $C_{v2} = \frac{R}{\frac{5}{3}-1} = \frac{R}{\frac{2}{3}} = \frac{3R}{2}$ $C_{p2} = C_{v2}+R = \frac{3R}{2}+R = \frac{5R}{2}$

4. Since 1 mole of each gas is mixed, total heat capacities of the mixture are additive: $C_v^{(\text{mix})} = C_{v1}+C_{v2} = \frac{5R}{2}+\frac{3R}{2} = 4R$ $C_p^{(\text{mix})} = C_{p1}+C_{p2} = \frac{7R}{2}+\frac{5R}{2} = 6R$

5. Therefore, for the mixture, $\gamma_{\text{mix}} = \frac{C_p^{(\text{mix})}}{C_v^{(\text{mix})}} = \frac{6R}{4R} = \frac{3}{2}$

6. Checking options:

   • A: $\frac{7}{5}$ ❌
   • B: $\frac{2}{5}$ ❌
   • C: $\frac{3}{2}$ ✅
   • D: $\frac{12}{7}$ ❌

Hence the correct answer is Option C.