Show that each degree of freedom of a gas particle adds 1/2 k_B T to the average

Question

Show that each degree of freedom of a gas particle adds 1/2 k_B T to the average kinetic energy of the gas, K_a upsilon = 1/2 m upsilon^2, where upsilon^2 = upsilon_x^2 + upsilon_y^2 + upsilon_z^2. In other words, show that = = = 1/2 k_B T where K_a upsilon = (K_x) + (K_y) + (K_z) and K_x = 1/2 m upsilon_x^2, K_y = 1/2 m upsilon_y^2 and K_z = 1/2 m upsilon_z^2. The angle brackets denote averaging along a single direction, e.g. (K_x) = integral 1/2 m upsilon_x^2 middot f V_x (upsilon_x) d upsilon_x.

Explanation / Answer

Ans. K.E = 1/2 mv2 (in one direction say x direction)

total K.E= 3/2 mv2 ( in all three direction x, y and z )

and we know that K.E = 1/2 kT (in one direction direction)

and hence Total K.E = 3/2 kT

so, (K.E)x = (K.E)y = (K.E)z = 1/2 kT

It is law of equipartition law of energy

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