Q.A gas made of a mixture of 2 moles of oxygen and 4 moles of argon at temperature T. Calculate the energy of the gas in terms of RT. Neglect the vibrational modes.

Statement. The average kinetic energy of a system of molecules in thermal equilibrium at temperature $T$ is shared out completely uniformly across every one of its available degrees of freedom, with each individual degree of freedom receiving on average exactly $\tfrac12kT$ of energy, regardless of what kind of motion (translational, rotational, or vibrational) that degree of freedom actually corresponds to.

Where it comes from. The starting point is the translational case: kinetic theory already shows $\tfrac12m\overline{v_x^2}=\tfrac12kT$ for motion along x alone, and by the assumed isotropy of molecular motion, the identical result $\tfrac12m\overline{v_y^2}=\tfrac12kT$ and $\tfrac12m\overline{v_z^2}=\tfrac12kT$ holds for y and z as well -- so each of the three translational degrees of freedom independently carries $\tfrac12kT$. The law of equipartition simply generalises this even sharing to every kind of degree of freedom a molecule might have, not just translational ones. …