If a box of ideal gas and another box of diatomic gas are in thermal equilibrium,

  1. does it mean that the average translational energy of ideal gas particle (A) is the same as that of diatomic gas particle (B)?

  2. or does it mean that A is equal to the sum of the average translational energy (B) + average vibrational energy (C) + average rotational energy (D) of diatomic gas particle?

  3. Equilibrium is achieved when A = B = C = D?

  4. or is it A = B + C + D?


The Equipartition theorem states that each degree of freedom has an average energy of 1/2KT.

This is valid at large enough temperatures where quantum mechanics does not play a role.

A = 3/2 KT(3 degrees of freesom)

B = 3/2 KT

C = 1/2 KT (1 vibrational degree of freedom in a di-atomic molecule)

D = KT (2 axis of rotation, the third has very low Moment of Inertia, and will not be excited)

  • $\begingroup$ I'm confused about D. If you narrowed it down to one degree of freedom, why wouldn't it be 1/2kT according to your prior argument? $\endgroup$ – Alan Rominger Oct 11 '12 at 13:05
  • $\begingroup$ No, Lets picture of the diatomic molecule as a dumbbell. There are 3 axis of rotation. One along the axis of the dumbbell. And 2 Perpendicular to it. The one along the axis of the dumbbell we say does not contribute to the internal energy of the Gas. The molecule is free to rotate about the 2 axis perpendicular to the dumbbell. These are 2 degrees of freedom, so it must be KT. $\endgroup$ – Prathyush Oct 12 '12 at 8:13
  • $\begingroup$ That's right, I had it wrong. I was thinking about the fact that angles are defined with only two variables $(\theta,\psi)$, for instance. But with that you also have magnitude, the moment of inertia vector (right hand rule) has fully 3-degrees of freedom. $\endgroup$ – Alan Rominger Oct 12 '12 at 14:06
  • $\begingroup$ You need 3 angles to specify orientation of any object.(Read Euler angles). To specify the direction of an axis you need 2 angles, but that is different from orientation. Also Moment of Inertia is not a vector, It is a 2 Tensor. In special situations It can be viewed as a scalar. $\endgroup$ – Prathyush Oct 12 '12 at 17:01

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