Question

The ratio of rms speed to the rms angular speed of a diatomic gas at certain temperature is: (assume m =mass of 1 molecule, M = molecular mass, I = moment of inertia of the molecules)
A. $\sqrt{\dfrac{3}{2}}$
B. $\sqrt{\dfrac{3I}{2M}}$
C. $\sqrt{\dfrac{3I}{2m}}$
D. $1$

Answer 1

Hint: Begin with the number of degrees of freedom possessed by a diatomic gas and classify them. In other words, a diatomic gas has 3 translational and 2 rotational degrees of freedom. Then, attribute equal fractions of the total energy of the diatomic molecule to each degree of freedom. Following this, equate the average translational kinetic energy in terms of rms velocity with the total energy associated with the translational degrees of freedom. Do the same for the rotational degrees of freedom as well. To this end, just by dividing the two relations and rearranging them, you should arrive at the demanded ratio.

Formula used: Average kinetic energy $(KE)_{avg} = \dfrac{1}{2}kT$, where k is Boltzmann’s constant and T is gas temperature.
Average translational kinetic energy $(KE)_t = \dfrac{1}{2}mv_{rms}^2$, where m is the mass of the molecule and $v_{rms}$ is the root mean square velocity of the molecule constituting the diatomic gas.
Average rotational kinetic energy $(KE)_r = \dfrac{1}{2}I\omega_{rms}^2$, where I is the moment of inertia of the molecule, and $\omega_{rms}$ is the root mean squared angular velocity of a molecule of the diatomic gas.

Complete step by step answer:
Let us begin with the foundational principle behind this problem, which is the equipartition theorem.
Let us first understand what degrees of freedom mean.
If you consider a single atom in a three-dimensional space, it is free to move along the x, y and z direction. Thus, it can execute motion in 3 directions. We can then say that the single atom has 3 degrees of freedom. Since the atom moves from one point to another as it traverses along the axes, we call this type of motion a translational movement. Thus, a single atom executes translational motion via its 3 degrees of freedom.
If you now consider a diatomic molecule (where two atoms are bound to each other), they are also free to move from one point to another along the axes and execute translational motion, but additionally, as a system, they are able to rotate about their centre of mass in two ways: either along the axis that joins the two atoms or the axis perpendicular to the axis joining them. Thus, they have an additional 2 degrees of freedom which arise from rotational motion. Therefore, a diatomic molecule has in total 5 degrees of freedom: 3 translational and 2 rotational.
Now, let us look at the equipartition theorem.
The law of equipartition of energy suggests that in thermal equilibrium, the total energy of a molecule is divided equally among all its degrees of freedom, and each degree of freedom thus possesses an energy of $\dfrac{1}{2}kT$ in accordance with the kinetic theory of gases, where k is Boltzmann’s constant and T is gas temperature.
This means that, for a diatomic molecule, the total energy of the molecule is divided equally among its 5 degrees of freedom, and each degree of freedom is attributed with an energy of $\dfrac{1}{2}kT$ to carry out the motion of the atom within that degree of freedom.
Remember that the kinetic energy given by $\dfrac{1}{2}kT$ is the average kinetic energy. Therefore, any translational or rotational velocities associated with it will be the rms velocities.
Since translational motion ensures 3 degrees of freedom $(KE)_t = 3 \times \dfrac{1}{2}kT$
Now, we know that for a molecule, translational kinetic energy $(KE)_t = \dfrac{1}{2}mv_{rms}^2$
Equating the two we get: A: $ \dfrac{1}{2}mv_{rms}^2 = 3 \times \dfrac{1}{2}kT$
Similarly, since rotational motion ensures 2 degrees of freedom $(KE)_r = 2 \times \dfrac{1}{2}kT$
And the rotational kinetic energy $(KE)_r = \dfrac{1}{2}I\omega_{rms}^2$
Equating the two we get B: $\dfrac{1}{2}I\omega_{rms}^2 = 2 \times \dfrac{1}{2}kT $
Dividing A by B, we get:
$\dfrac{mv_{rms}^2}{I\omega_{rms}^2} = \dfrac{3}{2} \Rightarrow \dfrac{v_{rms}^2}{\omega_{rms}^2} = \dfrac{3I}{2m} \Rightarrow \dfrac{v_{rms}}{\omega_{rms}} = \sqrt{\dfrac{3I}{2m}}$.

So, the correct answer is “Option C”.

Note: The root mean square velocity is the square root of the average of the square of n individual velocities, i.e.,
$v_{rms} = \sqrt{\dfrac{1}{n}(v_1^2 + v_2^2 + ….. + v_n^2)}$
Whereas, $v_{average} = \dfrac{1}{n}(v_1 + v_2 +….. +v_n)$
The reason we use rms velocity instead of the average velocity is due to the fact that for a typical gas sample, the net velocity is zero since the constituent particles are moving in all directions. Therefore, the average velocity would yield no quantifiable result. This is why sometimes the rms velocity is referred to as rms speed since it does not indicate the resultant velocity direction. In fact, the rms velocity is independent of the direction of any of its individual velocities and considers only the magnitude.