Given one mole each of two ideal gases:

• Monatomic gas: $\gamma_1 = \dfrac{5}{3}$
• Diatomic gas: $\gamma_2 = \dfrac{7}{5}$

We need $\gamma$ for the mixture.

Use the relation $\gamma = \frac{C_p}{C_v}, \qquad C_p - C_v = R$ Hence, $C_v = \frac{R}{\gamma - 1}, \qquad C_p = \frac{\gamma R}{\gamma - 1}$

For the monatomic gas $(1$ mole$)$: $C_{v1} = \frac{R}{\frac{5}{3}-1} = \frac{R}{\frac{2}{3}} = \frac{3R}{2}$ $C_{p1} = C_{v1}+R = \frac{3R}{2}+R = \frac{5R}{2}$

For the diatomic gas $(1$ mole$)$: $C_{v2} = \frac{R}{\frac{7}{5}-1} = \frac{R}{\frac{2}{5}} = \frac{5R}{2}$ $C_{p2} = C_{v2}+R = \frac{5R}{2}+R = \frac{7R}{2}$

For the mixture of 1 mole + 1 mole: Total heat capacities are additive: $C_{v,\text{mix}} = C_{v1}+C_{v2} = \frac{3R}{2}+\frac{5R}{2} = 4R$ $C_{p,\text{mix}} = C_{p1}+C_{p2} = \frac{5R}{2}+\frac{7R}{2} = 6R$

Therefore, $\gamma_{\text{mix}} = \frac{C_{p,\text{mix}}}{C_{v,\text{mix}}} = \frac{6R}{4R} = \frac{3}{2}$

Match with options: $\boxed{\gamma = \frac{3}{2}}$ So the correct option is C.