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According to this Reddit thread, the answer is no, vapor pressure can't be zero when temperature is above absolute zero. I suspect the answer might actually be yes according to a precise definition I will give later before I state what I'm really asking.

Because of the existence of gravity and the cosmological constant and because temperature might not be well defined because the zeroth law of thermodynamics hasn't been proven to be an absolute law according to this answer, we can't even define the vapour pressure of any substance sufficiently close to absolute zero because the bigger an object is, the closer to absolute zero it can get but any solid spherical object in its stable form with no pores in it of sufficient size will collapse into a black hole. I believe we fully understand a simplification of the theory of the universe that doesn't include gravity, the cosmological constant, or dark matter but I'm not sure of that because people seem to be using the Hadron collider to make new observations and learn more about the theory. However, I'm pretty sure that in that theory, we can define temperature in such a way that two substances in thermal equilibrium don't differ in Kelivn temperature by more than 8% and the intensity of blackbody radiation at any wavelength with more than 96% absorption is within 8% of what it's predicted to be.

I believe that theory almost always simulates an even simpler 3-dimensional quantum mechanical theory where electrons and nuclei are point charges and there's no nuclear chemistry. That 3-dimensional quantum mechanical theory can in turn be approximated by a simple nonrelativistic quantum mechanical theory because the fine structure constant is so small.

In nonrelativistic quantum mechanics, there are no photons and no blackbody radiation. Mercury is severely affected by the relativistic effects but it can probably be easily shown that none of the light elements are. However, for any specific substance, we can define the temperature of that specific substance in that simple theory to be the limit of what's it's defined to be in a simplified relativistic quantum mechanical theory when you set the speed of light close enough to infinite while keeping the proton mass; electron mass; proton charge; Coulumb's constant, and Planck's constant the same. For any substance, if you set the speed of light close enough to infinity, temperature is probably defined in such a way that substance's greybody radiation at any wavelength will be within 8% of prediction. According to that definition, the larger an object is, the closer to absolute zero it can get. Also according to this question, a surface of a given solid might even have a different vapour pressure depending on its orientation with respect to its crystal lattice in the simple theory where the vapour pressure of the surface whose orientation has the highest vapour pressure is at the highest pressure for its internal energy that gives a zero rate of homogeneous nucleation of the solid phase, as long as a shape whose faces are all at the orientation with the highest vapour pressure exists. I define the vapour pressure of any surface of a stable solid at any temperature above absolute zero to be the limit as its size approaches infinity of its vapour pressure at that temperature and orientation. My question is

According to the simple nonrelativistic theory, do you know how to invent a precise definition of temperature and prove entirely mathematically that the theory predicts that it has certain properties such as approximately following the zeroth law of thermodynamics on the Kelvin scale, and then prove or disprove entirely mathematically that that theory predicts that vapour pressure of any surface of a solid at any orientation above absolute zero must be nonzero without using the assumption that since something was observed in experiment, the theory predicts it?

I don't see why a substance must always have nonzero vapour pressure above absolute zero. If a solid has a microscopically thin layer of hexane on it, we could observe water to have a different contact angle with it than the theory predicts it to have. Also according to this answer, it hasn't been proven to be consistent with the simple nonrelativistic theory that a lot of the statements we were taught such as the second law of thermodynamics are all true. If you assume that a substance above absolute zero is in equilibrium and its equibrium state has certain chatoic properties where heat diffuses normally and its equilibrium state is time-invertable, then the substance must have nonzero vapour pressure. Since heat can diffuse, the reverse process can also occurr where a lot of heat randomly concentrates in a small area at the surface and knocks a surface atom loose.

I don't see why if it's in equilibrium, its equilibrium state must be time-invertable. Maybe there is a substance that at any sufficiently low temperature has a tendency to exponentially approach a non-time-invertible state where all the atoms are in phase at all times and repeat actions at regular time intervals and therefore has zero vapour pressure.

Timothy
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    Wood? Tungsten? Perhaps from a theoretical standpoint, the answer is "no". However, there are many objects that have been in the vacuum of space for eons, whose temperature is definitely NOT absolute zero, that have not evaporated due to their non-zero vapor pressure (e.g., asteroids). – David White Aug 26 '18 at 01:54
  • @DavidWhite People have thought wrong things in the past. Maybe there was confusion between different people of what they meant and they derived the incorrect conclusion, and no researchers did the math to double check whether the theory is consistent and predicts that a substance can never have zero vapour pressure above absolute zero. Maybe the mistake was never noticed because of the bystander effect and because people stubbornly insisted on the truth of what they learned and built opon it figuring out new stuff they claimed must be true. – Timothy Aug 26 '18 at 02:13
  • At which temperature do you propose that the vapor pressure drops from a nonzero value to zero? – Chemomechanics Aug 26 '18 at 02:37
  • @Chemomechanics I'm not sure. I don't claim to have proven that some substances can have zero vapour pressure above absolute zero. I'm just not sure whether it can happen or not. – Timothy Aug 26 '18 at 02:55
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    @DavidWhite. Just because objects in the vacuum of space haven't evaporated over eons does not mean that their vapor pressure is zero. All it means is that the vapor pressure is very low. Have you actually calculated the evaporation rate (mass flux per unit area) for such a substance/object to estimate how rapidly (or slowly) its mass would actually decrease over time (at typical space temperatures of only a few degrees K)? – Chet Miller Aug 26 '18 at 16:44
  • @ChesterMiller, I recognize that the theoretical vapor pressure may be greater than zero. However, if such a vapor pressure indicates that an object will evaporate in a time frame that is on the order of or greater than the age of the universe, I consider that to be effectively zero. – David White Aug 26 '18 at 17:49
  • That’s fair enough – Chet Miller Aug 26 '18 at 17:58
  • @ChesterMiller I'm not claiming that objects a bit above absolute zero must have zero vapour pressure because they existed for eons. I'm just being careful and refusing to claim with certainty that the opposite is true. I described in my question why I'm not sure a substance above absolute zero can't have zero vapour pressure, not why I'm sure one can. Maybe some people have knowledge I don't have from which they can prove that no substance can. – Timothy Aug 27 '18 at 00:49
  • Well, we do know from thermodynamics what the relationship is between vapor pressure and temperature. This is the Clapeyron equation, which predicts that the vapor pressure at any temperature is non-zero (even at very low temperatures). – Chet Miller Aug 27 '18 at 01:42
  • @ChesterMiller I learned in University chemistry the relationship between vapour pressure and temperature. I think it has been derived from the formula for entropy of a gas which too hasn't been proven to be consistent with the properties of all those other defined concepts according to quantum theory. The book "The selfish gene" taught me so many things that I'm not absolutely sure are true in all situations probably because stating that certain statements have been by researchers to make them more than 98% sure that they're a simplified model of reality can still help with research. – Timothy Aug 27 '18 at 02:21
  • How do you feel about quantitative experimental confirmation of the predictions from the Clapeyron equation? Another conspiracy? – Chet Miller Aug 27 '18 at 02:38
  • @ChesterMiller I'd still not be sure there wasn't a mistake. I don't understand the laws of quantum theory but if I did, I would try to do the mental math do double check whether the theory was consistent and there existed a way of simultaneously defining all those concepts that approximately satisfies all those assumed properties and if yes, then also verify with my math ability that the claimed experimental result was consistent with those definitions, and would only upvote an answer that I could verify with my own math ability was correct. – Timothy Aug 27 '18 at 02:47
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    Well, since I haven’t provided an answer, there is nothing for you to upvote. But please, enjoy yourself trying to prove or disprove whatever it is you are trying to prove or disprove. As for me, i’m outta here. – Chet Miller Aug 27 '18 at 02:55
  • Just to note it, this is one of those hyper-theoretical questions. By which I mean, the experimental distinction between extremely-low and truly-zero values doesn't really exist. – Nat Sep 04 '18 at 03:23
  • @Nat Did you mean the distinction between near zero vapour pressure and zero vapour pressure? We already have a simple nonrelativistic quantum theory. My question was not what occurrs in the real universe but whether somebody knew how to mathematically prove that according to the theory, the vapour pressure is always nonzero above absolute zero without using the assumption that since something was observed, the theory predicts it. I fixed up the question to make it clear that I'm asking that. – Timothy Sep 04 '18 at 03:39
  • Was it because I linked another question that hasn't been answered that that question has been deleted? That's a reason to close this question, not a reason to delete that question. That question already was worthy of closing before but probably not worthy of deleting and I didn't want to wait a really long time until after it got answered to ask this question. I'm not supposed to ask that question and this question in one question and that question probably has to exist and be answered in order for this question not to be worthy of closing. – Timothy Sep 04 '18 at 04:14
  • related: https://physics.stackexchange.com/q/729698/226902 https://physics.stackexchange.com/q/234018/226902 https://physics.stackexchange.com/q/404555/226902 https://physics.stackexchange.com/q/489231/226902 https://physics.stackexchange.com/q/708104/226902 – Quillo Sep 28 '22 at 16:30

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