JEE MAIN - Physics (2019 - 9th April Morning Slot - No. 27)
An HCl molecule has rotational, translational
and vibrational motions. If the rms velocity of
HCl molecules in its gaseous phase is $$\overline v $$ , m is
its mass and kB is Boltzmann constant, then its
temperature will be :
$${{m{{\overline v }^2}} \over {5{k_B}}}$$
$${{m{{\overline v }^2}} \over {6{k_B}}}$$
$${{m{{\overline v }^2}} \over {7{k_B}}}$$
$${{m{{\overline v }^2}} \over {3{k_B}}}$$
Explanation
An HCl molecule, being diatomic, has:
- 3 translational degrees of freedom
- 2 rotational degrees of freedom
- 2 vibrational degrees of freedom
The total number of degrees of freedom is $3 + 2 + 2 = 7$.
According to the equipartition theorem, each degree of freedom contributes $\frac{1}{2} k_B T$ to the total energy. So the total energy is given by: $\frac{7}{2} k_B T$
The translational kinetic energy is related to the root-mean-square (rms) speed $\overline{v}$ by: $\frac{1}{2} m \overline{v}^2 = \frac{7}{2} k_B T$
Rearranging to solve for the temperature, we find: $T = \frac{m \overline{v}^2}{7k_B}$
Comments (0)
