State and prove law of equipartition of energy.

Answer

It states that for a dynamical system in thermal equilibrium the energy of the system is equally distributed amongst the varies degree of freeom and the energy associated with each degree of freedom per molecule is `(1)/(2)kT` where `k` is the Boltzmann's constant. Let us consider one mole of a monoatomic gas in thermal equilibrium at temperature `T`. Each gas molecule has `3` degrees of freedon due to its translational motion. Accodring to kinetic theory of gases, the mean kinetic energy of translational motion of a gas molecules is given by
`(1)/(2) mv^(2) = (3)/(2) kT ..............(i)`
where `bar(v^(2))` is mean square velocity of the gas molecule of mass m.
If `bar(v_(x)^(2)),bar(v_(y)^(2))` and `bar(v_(z)^(2))` are the components of mean square velocity of the gas molecules along the three axes, then average energy of a gas molecule
`(1)/(2) mv^(bar2) = (1)/(2)mv_(x)^(bar2)+ (1)/(2) bar(mv_(y)^(2)) +(1)/(2) bar(mv_(z)^(2)) .......(ii)`
From the equations (i) and (ii) we have
`(1)/(2) mv_(x)^(bar2) +(1)/(2)mv_(x)^(bar2) +(1)/(2) bar(mv_(z)^(2)) = (3)/(2)kT ....(ii)`
The molecular motion is random in nature and no direction of motion is preferred one. Therefore, the average kinetic energy correspondig to each degree of preferred one. Therefore, the average kinetic energy coresponding to each degree of freedom is the same i.e.,
`(1)/(2) mv_(x)^(bar2) =(1)/(2)mv_(y)^(bar2)=(1)/(2)mv_(z)^(bar2)`
Hence, the equation (iii) gives
`(1)/(2)mv_(x)^(bar2)=(1)/(2)mv_(y)^(bar2)=(1)/(2)mv_(z)^(bar2)=(1)/(2)kT ......(iv)`
Thus, the mean kinetic energy per molecule per degree of freedom is `(1)/(2)kT`. This result was first deduced by Boltzmann.