If the sublimation point of a substance is the point where its equilibrium vapor pressure exceeds the pressure of the atmosphere upon it, and if every solid above absolute zero has some vapor pressure (albeit minimal), then it would seem that in a vacuum, where the pressure of the atmosphere is zero, the sublimation point would always be reached. Yet on a phase diagram sublimation never begins until well above absolute zero.
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10$\begingroup$ The intuition behind this question is interesting but misses a critical issue. Whatever the equilibrium point is, how fast would sublimation happen? Even at room temperature most substances would sublime so slowly that the equilibrium is irrelevant on human timescales (or even on the lifetime of the universe). $\endgroup$– matt_blackCommented Sep 30 at 10:26
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4$\begingroup$ Phase diagrams are often on a log(pressure) scale, which don't show a perfect vacuum by definition... Perhaps some examples would help. $\endgroup$– SanchisesCommented Oct 1 at 17:24
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3 Answers
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Yes, but no vacuum is perfect. Intergalactic space has a matter density of about 1 atom per cubic meter, and apparently has a temperature of millions of degrees and a pressue of ~$10^{-17}$ Pa. Let's suppose we somehow managed to cool down the interstellar medium to 2.7 K. According to the ideal gas law, this would then make the pressure ~$10^{-23}$ Pa. This is certainly a very low pressure, but it isn't zero. And although I was unable to find data for the vapor pressure of iron metal at 2.7 K, I'd bet it's much, much lower than that.
It's worth noting that these astronomical extremes, are, well, extreme. Let's say some hypothetical solid had an equilibrium vapor pressure under intergalactic conditions of $1 \times 10^{-28}$ Pa. That would mean that the equilibrium vapor would have roughly one atom per 300 cubic meters. So if you don't have a 300-cubic-meter-sized system, then you can't have a meaningful equilibrium.
And it's actually much worse than that. For hydrogen at 2.7 K, at intergalactic pressures the mean free path is about 4 million meters, and the mean time between collisions is about 30 days. So there isn't a meaningful equilibrium, and therefore there is no well defined pressure, if you don't have a system that is at least millions of meters in linear size, and even if you had that, you'd have to wait at least 30 days for pressures to re-equilibrate if you change conditions.
And by the way, the lowest pressure achieved by humans is around the same not as low as that in intergalactic space.
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5$\begingroup$ I would say: yes, and that's why no vacuum is perfect. $\endgroup$ Commented Sep 30 at 7:42
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2$\begingroup$ I'm curious; if the matter density of intergalactic space is 1 atom per cubic meter... if one were to seal a 0.5m³ box at a random location in space, would this not imply that you have a 7 in 8 probability of having just created a perfect vacuum within that box? $\endgroup$ Commented Sep 30 at 9:30
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9$\begingroup$ @ConnieMnemonic Pedantic notation point: I think you mean a (0.5 m)$^3$ box, not a 0.5 m$^3$ box, but yes, if it was an imaginary box. Any real box would outgas/sublime an enormous number of atoms compared to that scale $\endgroup$– llamaCommented Sep 30 at 14:28
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4$\begingroup$ @llama Except a decent sized diamond monocrystal in an intergalactic supervoid would need proton decay to be gone before heat death of universe ;> $\endgroup$– MithoronCommented Sep 30 at 15:10
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1$\begingroup$ @llama Think about how many low energy phonons it takes to collide for a single carbon atom to evaporate at 2.7 K. Now compare to the time it takes for gaseous carbon atom to travel 1 m. Your suspicion is wrong by many orders of magnitude. $\endgroup$ Commented Sep 30 at 20:00
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Yet on a phase diagram sublimation never begins [highlight mine] until well above absolute zero.
At (theoretical) absolute zero substances lack the energy to undergo a phase transition. The structure is a perfect lattice and the only energy is zero-point energy.
In addition, points on phase diagrams represent equilibrium points. A solid in a vacuum is in non-equilibrium. Vapor pressure is a value associated with an equilibrium state without beginning or end.
Only when T>0 K will molecules have sufficient energy to form a disordered phase or detach from a lattice. A disordered interfacial phase on a solid surface does not represent a minimum free energy state relative to (a perfect) vacuum. Surface molecules with enough energy will tend to detach and escape into a persistent vacuum. Note they will tend to take energy resulting in evaporative cooling. A similar principle is used to generate super-cold atoms, see for instance this Scientific American article:
Further cooling is done by evaporative cooling, by selective removal of the most energetic atoms from the system. The same process cools a cup of coffee when the most energetic molecules escape as steam, thus lowering the average energy and therefore the temperature of the remaining molecules. In a magnetic trap, the most energetic atoms can move farther against the pull of the magnetic forces, and can therefore reach regions with higher magnetic fields than the colder atoms can. At those high magnetic fields, they get into resonance with radio waves or microwaves, which changes the magnetic moment in such a way that the atoms fly away and escape from the trap. [ Source: How are temperatures close to absolute zero achieved and measured? Scientific American, Jan 19, 2004 ]
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1$\begingroup$ "...sufficient energy to form a disordered phase or detach from a lattice. " Detachment from a lattice needs no energy for helium. $\endgroup$ Commented Sep 30 at 20:15
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$\begingroup$ @PaulKolk Quantum tunneling could account for liberation of some atoms from a surface? I was thinking classically about thermal equilibria. But I would not go so far as saying "needs no energy". Energy is conserved. $\endgroup$– Buck Thorn ♦Commented Oct 1 at 7:19
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$\begingroup$ Not liberation from a surface! Only from a lattice. That is a well established fact. $\endgroup$ Commented Oct 1 at 8:01
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$\begingroup$ @PaulKolk Yes, $\ce{He}$ is interesting, but exceptional? link.springer.com/chapter/10.1007/978-3-662-05900-5_5 $\endgroup$– Buck Thorn ♦Commented Oct 1 at 8:20
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2$\begingroup$ I don't want to answer this incomplete question. The answer depends on whether there is a source of heat behind the solid, to keep the temperature from dropping, because perfect vacuum provides no thermal radiation to compensate the evaporative cooling heat loss. Without a heat source, it doesn't even matter what the initial dissociation rate is. The rate goes to zero and some solid remains. QT doesn't change that. BTW, "zero-point energy" is not something like thermal energy: it is just a correction on top of a classical calculation. I think, it is worth explicitly mentioning in your answer. $\endgroup$ Commented Oct 1 at 23:34
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The sublimation curve describes pressure/temperature when a molecule/atom moves from a solid into the vapour phase, and as you see in a P/T plot the vapour pressure increases, effectively from zero, with temperature as sublimation occurs. (Atmospheric pressure in not relevant here.) Past the triple point there is also liquid so now we call the process vapourisation/evaporation instead of sublimation but its basically the same process.
The molecules in the solid or liquid have many intermolecular interactions and to enter the gas phase these have to be overcome, obviously. These interactions are electronic in nature and so do not depend on temperature therefore to overcome these intermolecular interactions energy, such as thermal energy, is needed. Strong interactions, as in a metal such as iron, means that even at room temperature a few Fe atoms, but so very, very few go into the vapour phase that the vapour pressure is minuscule as described in other answers. In a solid made of planar molecules, such as naphthalene, intermolecular interactions are weaker and the solid readily sublimes. You know this because you can smell it at room temperature.
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1$\begingroup$ The relevant P is the vapor pressure not the atmosphere. I would imagine that at some T above 0K there would not be enough energy left in the lattice for even one molecule [2 or more are probably needed] to have enough KE to escape the lattice and gravity. This question should be settled when we go to the Moon again and harvest all the water ice hidden in the forever dark, forever cold, nooks and crannies. If ice does not sublimate at reasonable T rocks and metals will not. Above the triple point when vapor is present solid vanishes. $\endgroup$– jimchmstCommented Oct 5 at 21:04
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$\begingroup$ @jimchmst That's what I thought! You may like If there is actual ice on the moon, why hasn't it sublimated? $\endgroup$– uhohCommented Oct 5 at 21:59