The internal energy of one gram of Helium at $100 K$ and one atmospheric pressure is 

The Helium molecule is monoatomic and hence its internal energy per molecule is $\frac{3}{2}{k}_{B}T$ (where $k_B$ is Boltzmann constant). The internal energy per mole is thereafter is $\frac{3}{2}{k}_{B}T$. One gram of the helium is one fourth mole and hence its internal energy is $\frac{1}{4}x\frac{3}{2}R x 100=300 J$, taking the value of $R$ to be approximately $8 J mo{l}^{-1} K^{-1}$.