Question

The molecule of an ideal gas has $6$ degrees of freedom. The temperature of the gas is $T$.The average translational kinetic energy of its molecules is:
(A) ${\displaystyle \frac{3}{2}}kT$
(B) ${\displaystyle \frac{6}{2}}kT$
(C) $kT$
(D) ${\displaystyle \frac{1}{2}}kT$

Answer 1

Hint Here we know the gas molecule and the gas temperature, we can know the molecular kinetic energy and the conversion of kinetic energy accordingly. A body's translational kinetic energy is equal to one-half the product of its mass, so by using the formula we determine the average kinetic energy.
Useful formula:
The equation for Kinetic Energy
 $KE=1/2m{v}^2$
 where $m$ is the mass, and $v$ is the velocity.

Complete step by step answer
Given by,
Temperature is $T$,
We find the average translational kinetic energy,
As a function of the natural thermal movements of matter, translational energy relates to the displacement of molecules in space.
As you have read, kinetic energy is the energy of movement. The movement to which we refer is its velocity, which refers to our speed and the direction of that speed in this case.
In kinematic theory of gases, macroscopic quantities such as press and temperature are explained by considering microscopic random motion of molecules.
Hence,
The average translational energy of a molecule is given by the equipartition theorem as,
$E={\displaystyle \frac{3}{2}}kT$
Where,
 $k$ is the Boltzmann constant and $T$ is the absolute temperature.

Thus, option A is the correct answer.

Note Here If we compress a gas without altering its temperature, the gas particles' average kinetic energy stays the same. Any increase in the frequency of wall collisions should result in an increase in gas pressure. Thus, as the gas volume becomes lower, the pressure of a gas becomes higher.