Question

According to Maxwell-Boltzmann distribution law of molecular velocities of gases, the average translational kinetic energy is:
(A) $\dfrac{KT}{2}$ per molecule
(B) $KT$ per molecule
(C) $RT$ per molecule
(D) $\dfrac{3}{2}KT$ per molecule

Answer 1

Hint: Recollect the concept of kinetic theory of gases and its postulates. Write the law of Maxwell-Boltzmann distribution for molecular velocities of gases. Take a look at all the options and then tick the correct option for translational kinetic energy of one molecule of gas.

Complete step by step solution:
- Kinetic energy is given as, K.E. = $\dfrac{1}{2}m{{v}^{2}}$ where m is the mass and v is the velocity.
- According to kinetic theory of gases, the kinetic energy of a gas is, K.E. = $\dfrac{1}{2}mv_{rms}^{2}$ where m is the mass of a molecule of gas and ${{v}_{rms}}$ is the root mean square velocity of the gas which is equal to $\sqrt{\dfrac{3KT}{m}}$ where K is the Boltzmann constant, m is the mass of a molecule of gas and T is the absolute temperature.
- If we substitute the value of root mean square velocity of a gas in kinetic energy equation we get, $K.E.=\dfrac{1}{2}mv_{rms}^{2}=\dfrac{1}{2}m\times \dfrac{3KT}{m}=\dfrac{3}{2}KT$.
- According to Maxwell-Boltzmann distribution law of molecular velocities of gases, at a given temperature, all the gas molecules irrespective of their masses, have the same average translational kinetic energy.
- Therefore, the average translational kinetic energy for a molecule is equal to $\dfrac{3}{2}KT$ where K is the Boltzmann constant and T is the absolute temperature.

Therefore, the correct answer is option (D).

Note: Remember according to Maxwell-Boltzmann distribution law of molecular velocities of gases, at a given temperature, all the gas molecules irrespective of their masses, have the same average translational kinetic energy which is equal to $\dfrac{3}{2}KT$ per molecule.