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I have been wondering about it all day as I heard that particles of bodies at absolute zero have no kinetic energy. So, is it true that the total kinetic energy of an ideal gas is equal to its thermal energy?

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  • $\begingroup$ Kinetic energy is not temperature. At low temperatures, particles have kinetic energy due to quantum physics. $\endgroup$ – Pieter Jan 19 '18 at 20:01
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Absolutely not except in the case of an ideal, classical, monatomic gas, with no further degrees of freedom.

A counter-example would be a completely degenerate gas. The degenerate particles have a range of non-zero kinetic energies, right up to the Fermi energy, but none of this can be extracted as heat. The ideal degenerate gas therefore has zero thermal energy and much less thermal than kinetic energy even when not completely degenerate.

For an ideal classical, monatomic gas, all of the kinetic energy can be extracted as heat and indeed the thermal energy and kinetic energy are the same. However, if you start thinking about classical gases of molecules then there may be further degrees of rotational and oscillatory freedom that also store thermal energy, so the translational kinetic energy and thermal energy are not equivalent.

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  • $\begingroup$ About the 1st paragraph, it seems you focused on a Fermi gas, which is, according to wikipedia, the QM equivalent of a classical ideal gas. The OP didn't state whether he considered a QM or CM ideal gas. So you could precise that the answer is no in the QM case in the 1st paragraph. $\endgroup$ – thermomagnetic condensed boson Jan 20 '18 at 15:04
  • $\begingroup$ @no_choice99 I think my second paragraph does make this clear. The answer is also no for a classical molecular gas, not just quantum gases. $\endgroup$ – Rob Jeffries Jan 20 '18 at 15:44
  • $\begingroup$ What about the Bose gas? $\endgroup$ – thermomagnetic condensed boson Jan 20 '18 at 19:16
  • $\begingroup$ Thanks for the Answer. By an Ideal Gas I meant in Classical Mechanics. $\endgroup$ – Star Platinum ZA WARUDO Feb 7 '18 at 3:36
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In short, yes. Although in thermal physics it is often more useful to define "internal energy" of a system (rather than thermal energy) as being the sum of both kinetic and potential energies of the constituent particles, since separating the kinetic and potential components can be difficult.

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  • $\begingroup$ This claim is only for true for a select set of simple models and fails on most physically realistic models. $\endgroup$ – dmckee Jan 19 '18 at 22:05

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