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When we look at the phase diagram of a material, at low enough temperature and pressure, it tends to sublimate:

enter image description here So can it be true that no matter how strong a solid is, if at a low enough temperature, I maintain pressure below a certain level, it is bound to sublimate.

But is this true for all solids, no matter how strong the bonding is and no matter how much energy has been lowered by formation of solid?? on Wikipedia page, even common metals seem to sublimate (https://en.wikipedia.org/wiki/Materials_for_use_in_vacuum)

Suppose I consider a simple chain with nearest neighbour hopping, the electrons occupy the valence band with energy as $-2t_0(1-\cos(ka))$ where $t_0$ is nearest neighbour hopping, this total energy is negative hence shouldn't the solid be always energetically favourable than separate atoms roaming in space (if phonons are not energetic enough to break the chains)??

I am thinking taking into account the thermodynamics of photons and phonons is also important but I am not able to think of their role in this

Kutsit
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    I am not aware that helium has a conventional triple point. I don't believe there is a solid-gas transition at all. I might be wrong. The phase diagrams that I am aware of do not extend to extremely high pressures so there might be new phases similar to metallic hydrogen and there might be sublimation for those after all. – FlatterMann Sep 28 '22 at 08:16
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    There are an awful lot of different materials that we have on Earth, let alone in the Universe. I rather doubt we have generated phase diagrams for a significant percentage thereof. – Carl Witthoft Sep 28 '22 at 13:11
  • Well, all materials have an equilibrium vapor pressure of it or it’s constituents at all temperatures. – Jon Custer Sep 28 '22 at 13:36
  • related: https://physics.stackexchange.com/q/489231/226902 and https://physics.stackexchange.com/q/234018/226902 – Quillo Sep 28 '22 at 15:54
  • Also: https://chemistry.stackexchange.com/q/17066 Regarding the fact that no solids are thermodynamically stable at zero pressure, regardless of what the temperature is: https://physics.stackexchange.com/q/404555/226902 – Quillo Sep 28 '22 at 16:02

1 Answers1

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Yes, all condensed matter is expected to spontaneously sublimate/evaporate at any pressure and temperature. (Put another way, all materials have a nonzero vapor pressure.)

The standard argument is that the first molecule that detaches and flies off into the void increases the entropy of the universe by a tremendous amount due to the new positional opportunities. In Gibbs free energy terms, this is more than enough to offset any enthalpy penalty arising from bond breaking during detachment. Minimization of the Gibbs free energy can in turn be traced back to the Second Law (derivation sketch).

The equilibrium vapor pressure increases somewhat with increasing total pressure because the strain energy stored in the condensed matter increases the relative stability of the gas phase. This dependence is typically small and often ignored.

More importantly, the vapor pressure is a strong—exponential—function of the temperature and decreases to essentially zero (but not exactly zero) as the temperature decreases toward 0 K:

enter image description here (Larger version here.)

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    Is there a general theoretical proof from first principles that is valid to any substance? Do you have a reference? I vaguely remember that solids with covalent bonds can sublime, while ionic ones no, but I may be wrong. – Quillo Sep 28 '22 at 16:32
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    The Second Law favors entropy maximization. The entropy of the material decreases because it cools down when the first atom breaks its bonds, which is endothermic. But the total entropy in the universe increases by a larger amount because that first atom can now go anywhere. So we always see spontaneous initial sublimation/evaporation, regardless of the material or material class. – Chemomechanics Sep 28 '22 at 17:14
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    OK, but.... What-If ( :-) ) the entropy/enthalpy relationship is radically different for unobtanium molecules going liquid such that sublimation is much less statistically probable? – Carl Witthoft Sep 28 '22 at 17:14
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    I don't see how "much less" could take you to exactly zero. – Chemomechanics Sep 28 '22 at 17:16
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    Although this answer is technically correct, it could use a caveat that with many materials the sublimation is so slow as to be irrelevant for all practical purposes. The reason being that you're far out in the tail of a Boltzmann distribution, i.e. each particle does have a nonzero probability of breaking free but it is exponentially small. If it takes longer than the age of the universe for a measurable quantity to sublimate, you might as well say it doesn't happen at all... – leftaroundabout Sep 29 '22 at 08:59
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    However, there is a limit to sublimation: If there's not enough energy available to break the bonds of a single atom/molecule, energy preservation forbids the detachment. Of course, since we are comparing the total free energy of the condensed matter to the energy required to break a single atom/molecule free, there will always be enough energy available for sufficiently large collectives at achievable temperatures. And below that threshold, we simply don't talk about condensed matter with a temperature, we talk about a bunch of atoms/molecules with some small amount of kinetic energy. – cmaster - reinstate monica Sep 29 '22 at 09:24
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    @cmaster-reinstatemonica There is always the Cosmic microwave background to deliver some energy (if everything else fails) – fraxinus Sep 29 '22 at 14:42
  • @fraxinus Not inside a metal box. – cmaster - reinstate monica Sep 29 '22 at 18:32
  • @fraxinus I just looked it up: Temperatures of less than 1nK have been achieved to create Bose Einstein Condensates. Compared to that, the CMB is positively searingly hot at 2.7K. So, obviously, those people managed to ban the CMB from their experiments. – cmaster - reinstate monica Sep 29 '22 at 18:43
  • They don't ban it, they just limit the heat exchange to some acceptable level. – fraxinus Sep 29 '22 at 19:37