A gas mixture consists of 2 moles of oxygen and 4 moles of neon at temperature $$T$$. Neglecting all vibrational modes, the total internal energy of the system will be:

We need to find the total internal energy of a gas mixture consisting of 2 moles of oxygen ($$O_2$$) and 4 moles of neon ($$Ne$$) at temperature $$T$$, neglecting vibrational modes.

We recall that the internal energy of an ideal gas is given by the equipartition theorem:

$$U = n \cdot \frac{f}{2} \cdot RT$$

where $$n$$ is the number of moles, $$f$$ is the number of degrees of freedom per molecule, $$R$$ is the universal gas constant, and $$T$$ is the absolute temperature.

We begin by determining the degrees of freedom for oxygen ($$O_2$$, a diatomic gas). It has the following degrees of freedom when vibrational modes are neglected:
- 3 translational degrees of freedom (motion along the x, y, and z axes)
- 2 rotational degrees of freedom (rotation about two axes perpendicular to the bond axis; rotation about the bond axis itself contributes negligibly because the moment of inertia about that axis is extremely small)
Total for $$O_2$$: $$f_{O_2} = 3 + 2 = 5$$

Next, we determine the degrees of freedom for neon ($$Ne$$, a monoatomic gas). A monoatomic gas consists of single atoms and therefore can only translate; it has no rotational or vibrational modes:
- 3 translational degrees of freedom only
Total for $$Ne$$: $$f_{Ne} = 3$$

Substituting into the equipartition formula gives the internal energy of oxygen as:

$$U_{O_2} = n_{O_2} \cdot \frac{f_{O_2}}{2} \cdot RT = 2 \cdot \frac{5}{2} \cdot RT = 5RT$$

Similarly, for neon we obtain:

$$U_{Ne} = n_{Ne} \cdot \frac{f_{Ne}}{2} \cdot RT = 4 \cdot \frac{3}{2} \cdot RT = 6RT$$

Therefore, since internal energy is an extensive property, the total internal energy of the mixture is the sum of the individual contributions:

$$U_{total} = U_{O_2} + U_{Ne} = 5RT + 6RT = 11RT$$

The correct answer is Option 1: $$11RT$$.