Use the formula for internal energy of an ideal gas mixture

For an ideal gas, the internal energy is $U = n C_V T$ where $C_V = \dfrac{f}{2}R$ and $f$ is the number of active degrees of freedom.

Internal energy of argon

Argon is a monatomic gas, so it has $f = 3$ Hence, $C_V = \frac{3}{2}R$

Given $8$ moles of argon, $U_{\text{Ar}} = 8 \cdot \frac{3}{2}RT = 12RT$

Internal energy of oxygen

Oxygen is a diatomic gas. Neglecting vibrational modes, the active degrees of freedom are:

• 3 translational
• 2 rotational

So, $f = 5$ Therefore, $C_V = \frac{5}{2}R$

Given $6$ moles of oxygen, $U_{\text{O}_2} = 6 \cdot \frac{5}{2}RT = 15RT$

Total internal energy of the mixture

$U_{\text{total}} = U_{\text{Ar}} + U_{\text{O}_2}$ $U_{\text{total}} = 12RT + 15RT = 27RT$

Match with the options

$\boxed{27RT}$

So the correct option is B.