To solve this problem, we need to calculate the total internal energy of the gas mixture.$\\$The gas mixture consists of:$\\$• 2 moles of $O_2 \\$• 4 moles of $Ar \\$The internal energy of a gas is given by:$\\ U = n \cdot C_v \cdot T \\$where $n$ is the number of moles, $C_v$ is the molar heat capacity at constant volume, and $T$ is the temperature.$\\$For a diatomic gas like $O_2,$ the degrees of freedom are 5 (3 translational and 2 rotational).$\\$Thus, $C_v$ for $O_2$ is:$\\ C_v(O_2) = \frac{5}{2} R \\$For a monoatomic gas like $Ar,$ the degrees of freedom are 3 (all translational).$\\$Thus, $C_v$ for $Ar$ is:$\\ C_v(Ar) = \frac{3}{2} R \\$Now, calculate the internal energy for each component:$\\$• Internal energy of $O_2: \\ U(O_2) = n(O_2) \cdot C_v(O_2) \cdot T \\ U(O_2) = 2 \cdot \frac{5}{2} R \cdot T \\ U(O_2) = 5RT \\$• Internal energy of $Ar: \\ U(Ar) = n(Ar) \cdot C_v(Ar) \cdot T \\ U(Ar) = 4 \cdot \frac{3}{2} R \cdot T \\ U(Ar) = 6RT \\$The total internal energy of the system is the sum of the internal energies of $O_2$ and $Ar: \\ U_{total} = U(O_2) + U(Ar) \\ U_{total} = 5RT + 6RT \\ U_{total} = 11RT \\$Therefore, the total internal energy of the system is $11RT. \\$This corresponds to Option 3.