Question

Explain : 

Each translational and rotational degree of freedom contributes only one quadratic term to the energy but one vibrational mode contributes two quadratic terms.

Answer 1

With three translational degrees of freedom, the average translational energy per molecule of a gas is

where m is the mass of the molecule and v_x , v_y and v_z are the x-, y- and z-components of the molecular velocity.

A diatomic molecule has two rotational degrees of freedom. If ω₁ and ω₂ are the angular speeds about the two axes and I₁ and I₂ are the corresponding moments of inertia, the rotational energy of a diatomic molecule,

A diatomic molecule is regarded to have two degrees of vibrational freedom for the vibrational mode in which the two atoms vibrate relative to, and without affecting, the centre of mass of the molecule. Comparing this system with a vibrating body of mass m connected to a spring of force constant k, the vibrational energy has two terms corresponding to the kinetic and potential energies :

where x is the displacement from the mean position.

From Eqs. (1), (2) and (3), each translational and rotational degree of freedom contributes only one quadratic term to the average energy of a gas molecule while one vibrational mode contributes two quadratic terms.

Note : For a gas at an absolute temperature T, the average kinetic energy molecule, <KE> = <\(\frac{1}2\)mv_x²> + <\(\frac{1}2\)mv_y²> + <\(\frac{1}2\)mv_z²> = \(\frac{3}2\)k_BT

As there is no preffered direction x or y or z, we have,

 <\(\frac{1}2\)mv_x²> = <\(\frac{1}2\)mv_y²> = <\(\frac{1}2\)mv_z²> = \(\frac{3}2\)k_BT