The average kinetic energy of one $\text{mole}$ of gas per degree of freedom (on the basis of theory of gases)

The correct option is D $\frac{1}{2}RT$
Equipartition of energy states that the average energy $\left(E\right)$ of a molecule in a gas associated with each degree of freedom is $\frac{1}{2}kT$ where $k$ is Boltzmann's constant and $T$ is its absolute temerature.
$\therefore$ In one mole, number of molecules $=N_A$
$\therefore E$ of one mole of gas per degree of freedom $=N_A\left(\frac{1}{2}kT\right)$
$\Rightarrow E=\frac{1}{2}\left(N_{A}k\right)T=\frac{1}{2}RT$
($\because k=\frac{R}{N_A}$)
where $N_A=$ Avogadro's number
$R=$ Gas constant.