Question

The ratio of translational to rotational kinetic energies at $100 \,K$ temperature is $3: 2$. The internal energy (in $J$) of one mole gas at that temperature is $\left(R=8.3 \,J \,mol ^{-1}\, K ^{-1}\right)$

Answer 1

According to law of equipartition of energy, energies equally distributes among its degree of freedom.
Let translational and rotational degree of freedom be $f_{1}$ and $f_{2}$, respectively.
$\therefore \frac{K_{T}}{K_{R}}=\frac{3}{2}$ and $K_{T}+K_{R}=U$
Hence, the ratio of translational to rotational degrees of freedom is $3: 2$. Since translational degrees of freedom is $3$ , the rotational degrees of freedom must be $2$ .
$\therefore $ Internal energy $(U)=1 \times\left(f_{1}+f_{2}\right) \times \frac{1}{2} R T$
$U=\frac{1 \times 5 \times 8.3 \times 100}{2} $
$\Rightarrow U=2075\, J$