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I had always thought that temperature of a substance was a measure of the average kinetic energy of the particles in that substance:

$E_k = (3/2) k_bT $

where $E_k$ is the average kinetic energy of a molecule, $k_b$ Boltzmann's constant, and $T$ the temperature. (I'm not sure of the 3/2 coefficient.) Then I heard from several folks that this is a simplistic notion, not strictly true, but they didn't explained what they thought was flawed with this idea. I'd like to know what (if anything) is objectionable about this idea? Is it that the system must be macroscopically at rest? Is it that it ignores the quantum mechanically required motion of particles that persists at low temperatures? When is it not valid?

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There are a number of ways of defining temperature, for example using:

$$ {\partial S \over \partial E} = {1\over T} $$

This definition is the basis of negative temperatures. This is a case where the temperature is not a measure of average kinetic energy.

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The expression your wrote down for the energy is the expression for the ensemble average kinetic energy of a monatomic ideal gas. Therefore, we see that for this system, the average energy of the system is simply proportional to the temperature.

For a general statistical mechanical system, however, it might not even make sense to talk about the "average kinetic energy" of the system. For example, take a quantum system consisting of a single spin $1/2$ particle interacting with a magnetic field. It is possible to define a temperature for this system when it is in contact with a heat bath even if the particle is not moving around. In such cases, one appeals to more general definitions of temperature such as that to which John Rennie refers in his answer.

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temperature of a substance was a measure of the average kinetic energy of the particles in that substance

This is only true for a monatomic gas. Even without quantum mechanics, a classical diatomic gas has three more degress of freedom than a monatomic gas—two rotational and one vibrational—for a total of 6. The equipartition theorem tells us that the kinetic energy will be divided equally among these. Thus the kinetic energy of the gas particles in this case represents only 1/2 of the energy represented by the temperature.

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  • $\begingroup$ Based on measurements of specific heat of multiatom molecular substances, equipartition theorem isn't really valid for those other degress of freedom. The only degrees where it is reliable in practice are translational degrees of freedom for the whole molecule. So translational energy of molecules is on average $3/2k_B T$ but other energies (rotational, vibrational) are not so simple linear function of temperature. $\endgroup$ – Ján Lalinský Aug 25 '19 at 21:51
  • $\begingroup$ @JánLalinský ar you saying that the classical equipartition theorem is wrong for classical systems? $\endgroup$ – abalter Aug 25 '19 at 22:38
  • $\begingroup$ No, it is wrong for real gases. $\endgroup$ – Ján Lalinský Aug 25 '19 at 22:40
  • $\begingroup$ Of course it is. That's why a specifically said "a classical diatomic gas." I believe the OP is trying to understand the issue at a more basic level and did not want to make the issue more complex than necessary. At the most simplistic and classical level, the molecular KE is only one part of the thermodynamic energy. $\endgroup$ – abalter Aug 25 '19 at 22:44

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