Question

The ratio of translational and rotational kinetic energies at 100 K temperature is 3 : 2. Then the internal energy of one mole gas at that temperature is $[R=8.3J/mol-K]$

• A. $1175J$
• B. $1037.5J$
• C. $2075J$
• D. $4150J$

Answer 1

Answer: (C)

$2075J$

According to law of equipartition of energy, energies
equally distributed among its degree of freedom,
Let translational and rotational degree of freedom be $f_1$ and $f_2$.
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.
Internal energy $U=1\times (f_1+f_2)\times \frac{1}{2}RT$

$=\frac{5}{2}RT=2075J$