Timeline for In a perfect vacuum, shouldn't every solid be above its sublimation point, since its vapor pressure must exceed the atmospheric pressure?
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| Oct 1 at 23:34 | comment | added | Paul Kolk | 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. | |
| Oct 1 at 11:54 | comment | added | Buck Thorn♦ | @PaulKolk In retrospect I realize my answer was not very good. My take on the question was a bit off. Your comment on phonons etc might be the start of a better answer. The answer reduces to what fraction of surface molecules (ignoring QT) have sufficient energy to dissociate, and what the dissociation rate might be to generate a measurable particle stream, so include how ultralow pressure is measured. | |
| Oct 1 at 8:55 | comment | added | Paul Kolk | It is exceptional among molecules. Sure, many other things exhibit tunnelling, but still remain solid at absolute zero. Electron tunnelling is another story since electrons do not form a (size independent) phase in vacuum. | |
| Oct 1 at 8:20 | comment | added | Buck Thorn♦ | @PaulKolk Yes, $\ce{He}$ is interesting, but exceptional? link.springer.com/chapter/10.1007/978-3-662-05900-5_5 | |
| Oct 1 at 8:01 | comment | added | Paul Kolk | Not liberation from a surface! Only from a lattice. That is a well established fact. | |
| Oct 1 at 7:19 | comment | added | Buck Thorn♦ | @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. | |
| Sep 30 at 20:15 | comment | added | Paul Kolk | "...sufficient energy to form a disordered phase or detach from a lattice. " Detachment from a lattice needs no energy for helium. | |
| Sep 30 at 9:05 | history | edited | Buck Thorn♦ | CC BY-SA 4.0 |
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| Sep 30 at 8:59 | history | edited | Buck Thorn♦ | CC BY-SA 4.0 |
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| Sep 30 at 8:06 | history | answered | Buck Thorn♦ | CC BY-SA 4.0 |