Question

The ratio of average translational kinetic energy to rotational kinetic energy of a diatomic molecule at temperature T is:

Answer 1

Hint: Recall the concepts of the chapter kinetic theory of gases. Find the degrees of freedom for a diatomic molecule. Now note down the degrees of freedom for translational and rotational energy. Multiply the degrees of freedom with the energy for one degree of freedom to obtain the values of translational kinetic energy and rotational kinetic energy. Now take the ratio of these to obtain the answer.

Complete step-by-step answer:
The kinetic theory of gases is a simple model of the behavior of gases, based on which many thermodynamic concepts have been established.
The model describes a gas as a large number of identical microscopic particles (for atoms or molecules), all of which are in constant, Brownian motion. Their size is assumed to be much smaller than the average distance between the colliding particles.
The particles undergo random elastic collisions and with the walls of the container.

The three main conclusions of the kinetic theory of gases are:
-No energy is gained or lost when molecules collide as the collisions are assumed to be completely elastic.
-The individual particles in a gas take up a negligible amount of space with respect to the container they occupy.
-The molecules of gas are in constant linear motion.
The degrees of freedom for a molecule or atom can be found by the formula given below:
$\text{f = 3N}$
Where,
N stands for the number of atoms present in one gas molecule.
${{O}_{2}}$ has 6 degrees of freedom as it is diatomic.
Out of the 6 degrees of freedom, 3 are for translational motion, 2 for rotational motion and 1 for vibrational motion.
We will now calculate the translational kinetic energy and rotational kinetic energy.
The kinetic energy of a molecule can be calculated from the formula given below.
$\text{K}\text{.E = }\dfrac{\text{f}}{2}{{\text{K}}_{b}}\text{ x T}$
Where,
K.E stands for kinetic energy of the molecule,
f stands for degrees of freedom,
T stands for temperature of the container,
${{\text{K}}_{b}}$ stands for the Boltzmann constant


Translational kinetic energy = $\dfrac{3}{2}{{\text{K}}_{b}}\text{ x T}$
Rotational kinetic energy = $\dfrac{2}{2}{{\text{K}}_{b}}\text{ x T}$
The ratio of translational kinetic energy and rotational kinetic energy is 3/2.

Therefore, the correct answer is option (D).


Note: The kinetic theory of gases is defined only for ideal gases. The equations of kinetic theory of gases do not hold good for real gases. The assumptions made to apply the above equations are:
- The size of individual particles is very small compared to the size occupied by the gas.
- There is no interaction between the particles.
- The attractive and repulsive forces are non-existent.
- The average kinetic theory of the gas is proportional to the temperature.