Final Equilibrium Temperature of Ideal Monoatomic Gas Mixture

To find the final equilibrium temperature of the gas mixture, we apply the principle of conservation of internal energy. Since both gases are ideal monoatomic gases, their molar heat capacities at constant volume $C_v$ are equal.

The total internal energy before mixing is equal to the total internal energy after mixing: $n_1 C_v T_1 + n_2 C_v T_2 = (n_1 + n_2) C_v T_{\text{mix}}$

Canceling $C_v$ from both sides gives: $n_1 T_1 + n_2 T_2 = (n_1 + n_2) T_{\text{mix}}$

Substituting the given values $n_1 = 2\text{ mole}$, $T_1 = T$, $n_2 = 6\text{ mole}$, and $T_2 = 2T$: $(2)(T) + (6)(2T) = (2 + 6) T_{\text{mix}}$

$2T + 12T = 8 T_{\text{mix}}$

$14T = 8 T_{\text{mix}}$

$T_{\text{mix}} = \frac{14}{8}T = \frac{7}{4}T$

Thus, the final equilibrium temperature of the mixture is $\frac{7}{4}T$, which corresponds to Option D.