Question

The ratio of translational kinetic energies at 100 K temperature is 3:2. Then the internal energy of one mole gas at that temperature is [R = 8.3 J/mol-K ]
$
  {\text{A}}{\text{. 1175 J}} \\
  {\text{B}}{\text{. 1037}}{\text{.5 J}} \\
  {\text{C}}{\text{. 2075 J}} \\
  {\text{D}}{\text{. 4150 J}} \\
 $

Answer 1

Hint: From the ratio of the kinetic energies given, get the degrees of freedom and then use them in the total energy formula. Then the internal energy of one mole gas can be found.
Formula Used:
$U = \dfrac{{nRT({f_t} + {f_R})}}{2}$
$U = {K_t} + {K_r}$

Complete Step-by-Step solution:
From the ratio:
$\dfrac{{{K_t}}}{{{K_r}}} = \dfrac{3}{2}$
We know that the translational degree of freedom is 3. So, from the ratio it can be concluded that the rotational degree of freedom will be 2.
Now, from the formula for total internal energy:
$
  U = \dfrac{{1 \times (5) \times 8.3 \times 100}}{2} \\
  U = 2075J \\
 $
The correct option is (C).

Note: From the equipartition theorem, the average kinetic energies are equally stored in their degrees of freedom.
The translational degree of freedom is always 3.
The total internal energy is the sum of translational and rotational kinetic energy.