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For a start, let me clarify that by "almost identical configuration" I mean same volume, temperature and number of molecules (but different pressure). One could for instance take two identical systems of 1 mole of water each and following different paths of volume and temperature bringing them in liquid and solid state on both side of the phase transition such that the volume and the temperature of both systems are the same (thus with different pressure).

My question (if it makes sense) is: "How can we understand the fact that those systems with a structure that is so different (i.e. solid vs liquid) can have, in the same volume filled with the same number of molécules, the same temperature ? Do they have the same average translational kinetic energy ?" If yes, this is strange because a liquid is way more free to move than a solid and this tought-experiment is even more confusing when thinking about gas.

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I'm not sure whether this is an answer or a stretched out comment.

Thermal energy is not really related to displacements, but instantaneous velocities; The diffusion behaviour of the different states is most certainly different. This is to say that in the solid, the particles can move at a great speeds, but immediately collide with particles next to them and change directions. Classically it does not matter what state the matter is in, the particles will follow the Maxwell-Boltzmann velocity distribution.

Quantum effects complicate this picture. For example, some degrees of freedom might be frozen out by quantum effects (breaking down of the equipartition theorem), which means, for example, that the heat capacity is not a constant.

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  • $\begingroup$ Out of curiosity: how clear is it that Maxwell-Boltmann velocity distribution also applies to liquid and solid? That's puzzling me. I'm not sure to see how the heat capacity is related to my question. Nevertheless, I gave an answer above to my question which is not redudant to yours. I would be interested in having your opinion about it. $\endgroup$
    – SMR
    Dec 1, 2014 at 20:13
  • $\begingroup$ @SMR It follows immediately from statistical mechanics with minimal assumptions: $p = \frac{e^{-\beta \mathcal{H}}}{Z}$. Integrate over all the positions and over all but one particle's velocity: $p(v_i) = \frac{1}{C}\exp\left(-\beta\frac{m_iv_i^2}{2}\right)$. Note that integrating in spherical coordinates (we want speed, not velocity) we get the $v_i^2$ in front. So we have the M-B distribution. Heat capacity is an example of where quantum effects start breaking stuff: A quantum oscillator will not obey M-B because some DOFs are frozen; This is most easily described through heat capacity. $\endgroup$
    – alarge
    Dec 1, 2014 at 20:35
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If I hold a ruler from one end and let it swing it will com to rest with one end, call it end $A$, pointing vertically down. If I start to move the point I'm holding down the ruler it will continue to hang down until suddenly, when my fingers pass the centre of mass, it will flip round and come to rest with the other end pointing down and end $A$ pointing up. This is a discontinuous change; there is in point I can hold the ruler so that it will come to rest horizontally. This happens because the system has 2 points of equilibrium; one with end $A$ pointing down and one with end $A$ pointing up and the system will try to find the state of lowest energy. The energies of both these states vary continuously as I vary the pivot point, however there is a unique point at which the energy of one state drops below the energy of the other, leading to a discontinuous jump in the behaviour of the system.

Exactly the same thing happens in phase transitions. The system will come to equilibrium in a state which minimises its free energy, however if the system has multiple minima of free energy there can be discontinuous jumps if the energy of one local minima drops below the level of another and becomes the new global minimum of the free energy.

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  • $\begingroup$ While the analogy is appropriate---the minimum free energy state is chosen by the system---I'm not sure how this answers any of the questions presented, all of which dealt with kinetic energy. $\endgroup$
    – alarge
    Nov 25, 2014 at 20:53
  • $\begingroup$ The point you raise with your analogy is pretty clear to me. My question was not about the possibility of having two different points of equilibrium in similar configuration (note that the problem I posed is in equal volume and not equal pressure so both systems are not in two states lying on both side of the phase transition interface). My question is on the interpretation of having both sytems in almost identical configuration but still with a very different structure of molecular arrangement in terms of translational kinetic energy. It's hard for me to clearly picture what's happeing there. $\endgroup$
    – SMR
    Nov 25, 2014 at 20:59
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Of course I wasn't asking this question in order to answer it myself but because I discussed about it with some other "érudits" that were puzzled. But eventually, one pointed me the figure (sorry it's in french but you get the point)

http://upload.wikimedia.org/wikipedia/commons/b/ba/P-V-T_Diagram_%28Water%29.fr.svg

where the mixed phase zone are parallel to the volume V axis.

This figure indicate that there the triple point is a line which defines the concentration on the possible state of water depending on the volume, which makes it possible to have 1 mole of water at same temperature and pressure to be purely gas, liquid and solid and this will of course go with a different volume for each configuration. This can be described as saying that the volume is discontinuous between purely liquid and purely solid water at a given pressure and temperature. The triple point was in fact the starting point hidden behind my question but it wasn't my question, I agree. What this picture makes now clear is that their is no such discontinuity along the pressure axis, i.e., what I describe in my question is just impossible. At a given volume and temperature, there is only one pressure which dictates the state of matter. This explains why I was puzzled by the possiblity of existence of such systems in almost identical configuration: it's one or the other.

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  • $\begingroup$ You described a system whose pressure you control (so the system can change its size at will). Now do the opposite: Control the volume instead and let the system relax to some pressure. The pressure has to be uniform, but the system will now phase separate: Half of the container might be liquid and the other half gas. They have different densities, but wherever you look, you will find the system in the same temperature, and thus each particle to have, on average, the same velocity. $\endgroup$
    – alarge
    Dec 1, 2014 at 20:48
  • $\begingroup$ The point of my answer above is to say that though the question is quite confusing, I claim that the situation I described is impossible because I meant to have two different systems reached through different paths of (V,T) such that the first one is fully solid and the second one fully liquid. So it was not a matter of phase separation in one system. The reason I was puzzled that both these so different systems had same volume and temperature was simply that the truth is either one system, or the other (or a coexistence of separated phases). My question was somehow a paradox that's wrong. $\endgroup$
    – SMR
    Dec 2, 2014 at 21:12

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