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Q & A: kinetic energies and equipartition

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Most recent answer: 01/08/2015
Q:
If a liquid and a gas are at the same temp,,then do their constituent particles have the same kinetic energies??if not then why??
- Jenniffer (age 18)
india
A:

To a good approximation, the kinetic energies of the molecules are the same in the liquid and the gas at the same temperature. The reason is the Equipartition Theorem. It says that for some mode (like velocity vx in some direction or angular velocity around some axis) that obeys some conditions, in thermal equilibrium the average energy is a universal constant times the absolute temperature. The conditions are:

0) The temperature is high enough for the particle to explore many quantum states with different values of the parameter, e.g. vx for velocity in the x direction.

1) The density of quantum states doesn't depend on the value of the parameter, e.g. there are the same number of quantum states around vx=1 m/s as around vx=-2 m/s.

2) The energy is quadratic in the parameter, e.g. mvx2/2.

When these conditions hold, the thermal probability distribution for the parameter (e.g. vx) looks like like a Gaussian whose width is just right to make the average of the energy (e.g. mvx2/2) equal to kT/2, where k is Boltzmann's constant. (This probability distribution is derived from the Boltzmann probability for occupying states with different energies at some temperature. That in turn is derived from entropy maximization, which is in effect a first principle.) These conditions usually hold well enough for the translational kinetic energy of gases and liquids. Condition (0) doesn't necessarily hold for rotations, and I suppose the deviations from equipartiton will be somewhat different in the liquid and gas, since the rotational states are different. For cold solids it often doesn't hold for translations either.

Mike W.


(published on 01/08/2015)

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