State and explain the law of equipartition of energy.

(N/A) Law of Equipartition of Energy: In thermal equilibrium,the total energy of a system is equally distributed among all its active degrees of freedom,and each degree of freedom is associated with an average energy equal to $\frac{1}{2} k_{B} T$,where $k_{B}$ is the Boltzmann constant and $T$ is the absolute temperature.
Consider a monoatomic gas molecule in thermal equilibrium at temperature $T$. The average kinetic energy of the molecule is given by:
$\langle E_{t} \rangle = \langle \frac{1}{2} m v_{x}^{2} \rangle + \langle \frac{1}{2} m v_{y}^{2} \rangle + \langle \frac{1}{2} m v_{z}^{2} \rangle$
From the kinetic theory of gases,we know that the average kinetic energy of a molecule is $\frac{3}{2} k_{B} T$:
$\langle E_{t} \rangle = \frac{3}{2} k_{B} T$
Since the gas is isotropic,the average kinetic energy along each axis is equal:
$\langle \frac{1}{2} m v_{x}^{2} \rangle = \langle \frac{1}{2} m v_{y}^{2} \rangle = \langle \frac{1}{2} m v_{z}^{2} \rangle$
Substituting this into the total energy equation:
$\frac{3}{2} k_{B} T = 3 \langle \frac{1}{2} m v_{x}^{2} \rangle$
Therefore,the energy associated with each degree of freedom is:
$\langle \frac{1}{2} m v_{x}^{2} \rangle = \frac{1}{2} k_{B} T$
This confirms that for every degree of freedom,the associated average energy is $\frac{1}{2} k_{B} T$. This is known as the Law of Equipartition of Energy.