Briefly discuss the law of equipartition of energy.

Explanation

The energy of molecule corresponding to translational mode is given by
`E_(r)=(1)/(2)mbarv_(x)^(2)+(1)/(2)mbarv_(y)^(2)+(1)/(2)mbarv_(x)^(2)=(3)/(2)KT`
so that along x,y and z average translational may be taken indpependently as
`<<(1)/(2)mbarv_(x)^(2)>ge<<(1)/(2)mbarv_(y)^(2)>ge<<(1)/(2)mbarv_(x)^(2)>ge(1)/(2)KT.`
However for rotational mode of a molecule
`E_(r)=(1)/(2)1,omega_(1)^(2)+(1)/(2)I,omega_(2)^(2)`
For vibratory motion of a molecule
`E_(v)=(1)/(2)m((dy)/(dt))^(2)+(1)/(2)k_(t)y^(3)` where `K_(t)` is force constant of the atomic oscillator and y the vibrational coordinate. Since for each mode of a absorption of energy, `(1)/(2)K_(B)T` is associated, the net vibrational energy per degree of freedom `=2xx(1)/(2)K_(B)T=K_(B)T` We easily note that `(1)/(2)K_(t)Y^(2)` corresponds to potential energy and `(1)/(2)m((dy)/(dt))^(2)` corresponds to kinetic energy.
For polyatomic gases, one mole of gas has energy.
`U=[(3)/(2)K_(B)T+(3)/(2)K_(B)T+K_(B)T]N_(A) "where" 'N_(A)'` is Avogdro's number.