Attached is a phase diagram for carbon dioxide. At P = 0, it appears that the substance will exist as a gas regardless of temperature.

My questions are:

  1. For any liquid at a fixed temperature, is it possible to vaporize it completely by simply reducing its pressure isothermally?

  2. For any solid at a fixed temperature, is it possible to sublime it completely by simply reducing its pressure isothermally?

  3. What if the solid is at T = ~0K?

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    $\begingroup$ In outer space you get a very good vacuum (albeit the temperature is also very low) and we still get asteroids in solid state and space shuttles come down without disintegrating in between. $\endgroup$ – tschoppi Oct 5 '14 at 1:16

In general a substance will vaporize if some fraction of its particles have enough energy (due to their temperature) to exceed their binding energy. Binding energy is the energy contained in the intermolecular bonds holding a substance together. It is definitely possible for the particles of a substance to have a binding energy that is substantially higher than thermal energy will readily provide.

  1. No. While most liquids have fairly low binding energies (and thus, relatively high vapor pressures) there are liquids that will not evaporate in a vacuum, one example being ionic liquids. As I wrote above, the attraction between the ions of these liquids is strong enough that they do not evaporate over time.

    If you know the Periodic Videos YouTube channel, they've had a couple of videos about these fluids over the past couple of years. (Here's a link to one, if you're interested.) One of them featured the behavior of liquids in a vacuum, but I can't seem to find the link; it may have been on a sister channel.

  2. No, consider all of the asteroids, space ships, satellites, and such floating around in space. For all intents and purposes they will float around until the end of time without evaporating. Again, the reasoning is as I listed above. We also don't see the vast majority of solid objects here on Earth slowly evaporating away, which we would expect if sublimation were a concern in a vacuum. The reasoning is the same as above.

  3. If the solid is very, very cold, it is less likely to experience sublimation since its particles' average energies would be very low. Even dry ice would not vaporize if its temperature were very close to 0 K.

Here is a phase diagram of both water and carbon dioxide that might help illustrate the situation. You can see that as the temperature of the substance decreases the vapor pressure of the substance also decreases. At a sufficiently low temperature the vapor pressure is effectively zero (below the x-axis of this chart.) There is always the possibility of a single atom or molecule gaining enough energy to vaporize or doing something weird like tunneling off of the surface of the substance, but the process is so slow as to be unmeasurable. Again, asteroids in the solar system have been hanging around for several billion years without subliming.

Phase Diagram of CO2 and H2O

Picture from Wikipedia.

Note: Be careful if you look up phase diagrams of common substances. They often involve only a few points that are actually referenced and the rest is drawn with quite a bit of artistic license. Compare the phase diagrams of water here and here to see what I mean. They are often drawn without scales (or with minimal scales) and are sloppy at best.

  • $\begingroup$ How does one generally predict whether a substance is a solid/liquid/gas at absolute zero or at zero pressure? Do we need a phase diagram for the substance to know for certain? For CO2 and H2O, per the above phase diagram it looks like both are solid at absolute zero? $\endgroup$ – Yandle Oct 9 '14 at 23:04
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    $\begingroup$ @Yandle A phase diagram would be easiest, but a physical chemist might calculate the intermolecular binding energy and compare it to the thermal energy in the system. Based on the diagram I posted both $CO_2$ and $H_2 O$ are solid at 0K. If you look at this diagram, you can see how water behaves at significantly lower pressures still. If you're still curious, I'd encourage you to post a new, clear question. Reference this one or so that people know you've been thinking about it but that this doesn't answer your question. $\endgroup$ – Jason Patterson Oct 10 '14 at 13:49

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