[Physics Class Notes] on Law of Equipartition of Energy Pdf for Exam

The Law of Equipartition of Energy states that the total energy in thermal equilibrium for any dynamic system gets divided equally among the degrees of freedom.

The kinetic energy along the x-axis, the y-axis, and the z-axis for a single molecule is given by-

Along x-axis  [frac {1} {2}]  mv_x²

Along y-axis   [frac {1} {2}]  mv_y²

Along z-axis   [frac {1} {2}]  mv_z²

In the case of thermal equilibrium the average kinetic energy of the gas is given by-

Along x-axis ( [frac {1} {2}]  mv_x2)

Along y-axis ( [frac {1} {2}]  mv_y²)

Along z-axis ( [frac {1} {2}]  mv_z²)

Now, the average kinetic energy of a molecule according to the kinetic theory of gases is represented as-

[frac {1} {2}]mv_rms²=32K_bT

where the root-mean-square velocity is represented by v_rms

&

the Boltzmann constant is represented by K_b

&

T is temperature

Degree of Freedom

There are three degrees of freedom in the case of the monoatomic gas. Thus, the average kinetic energy per degree of freedom is represented as-

K_Ex= [frac {1} {2}] K_bT

A molecule possesses three translational degrees of freedom, which is free to move in space and hence needs three coordinates in order to specify its location. The molecule possesses two degrees of freedom if it is constrained to move in a plane and if it is in a straight line, there is one translational degree of freedom. On the contrary in the case of a molecule that is triatomic, the translational degree of freedom is 6 and in this case, the kinetic energy of the per molecule is given by-

6 x N  x [frac {1} {2}] x KbT= 3 x  [frac {R} {N}] x NKbT=3 RT

The translational degree of freedom in the case of the molecules of mono-atomic gases such as helium and argon is one. Then, the kinetic energy per molecule is given by-

3 x N x  [frac {1} {2}] x KbT=3 x [frac {R} {N}] x NKbT= [frac {3} {2}]   RT

The translational degree of freedom in the case of diatomic gases such as oxygen and nitrogen is 3.