Question

State and explain law of equipartition of energy.

Answer 1

Hint: State the law of equipartition of energy and then prove the law of equipartition for monoatomic gas in the thermal equilibrium any degree of freedom contributing potential energy will have another ${\displaystyle \frac{1}{2}}KT$ associated with it.

Complete answer:
We know that the law of the equipartition of energy states that for a dynamical system in thermal equilibrium that the total energy of the system is shared equally by all the degrees of freedom. The energy associated with each degree of freedom per molecule is ${\displaystyle \frac{1}{2}}KT$ where, K is the Boltzmann constant. In this law an equal amount of energy will be associated with each independent energy state. Let us consider one mole of a monatomic gas in thermal equilibrium at temperature T. each gas molecule has 3 degrees of freedom due to its translational motion.
We know that according to kinetic theory of gases, and then mean kinetic energy of translational motion of a gas molecule is given by,
${\displaystyle \frac{1}{2}}m{v}^2={\displaystyle \frac{3}{2}}KT....\left(1\right)$
Where $v^2$ is the mean square velocity of the gas molecule of mass ‘m’, ‘K’ is Boltzmann constant and ‘T’ is the temperature of gas. The average kinetic energy is corresponding to each degree of freedom is the same that is
${\displaystyle \frac{1}{2}}m{v_x}^2={\displaystyle \frac{1}{2}}m{v_y}^2={\displaystyle \frac{1}{2}}m{v_z}^2$
Using (1) equation, we get,
${\displaystyle \frac{1}{2}}m{v_x}^2={\displaystyle \frac{1}{2}}m{v_y}^2={\displaystyle \frac{1}{2}}m{v_z}^2={\displaystyle \frac{1}{2}}KT$
Thus, the mean kinetic energy per molecule per degree of freedom is the${\displaystyle \frac{1}{2}}KT$. This result was first deduced by Boltzmann.

Additional information:
The degrees of freedom are the minimum number of the coordinates required to specify the state of a system completely.

Note:
We know that if a molecule is free to move in space, it needs three coordinates to specify its location. Thus it possesses three translational degrees of freedom. It is noted that total energy has an average energy of ${\displaystyle \frac{1}{2}}KBT$ in thermal equilibrium and contributes ${\displaystyle \frac{1}{2}}KB$ to the system’s heat capacity.