How can sublimation occur at much lower temperatures than 0 degrees celsius provided that the pressure is sufficiently low? The melting point of water is relatively unaffected by the external pressure compared to the boiling point. Therefore it seems reasonable to assume that below 0 the molecules wouldn't have enough kinetic energy to overcome the binding forces in ice no matter how low the external pressure is.

If the boiling point is below the melting point, shouldn't sublimate happen at the melting point (which is at a higher temperature)?

My question is: Why does the phase diagram of water look like this enter image description here

and not like this enter image description here

  • 2
    $\begingroup$ All solids have a vapor pressure - I'm unclear exactly what the question is here. How many atoms may or may not have the energy to sublime is a kinetics question, not a thermodynamic question. $\endgroup$
    – Jon Custer
    Jul 1 '19 at 21:53

Your first misconception is the thing about kinetic energy of particles: Particle energies follow a distribution that does not have a cutoff. I.e. there will always be particles present with enough energy to go into the gaseous phase. If you lower the temperature, the count of these particles decreases, but it never reaches zero. Consequently, you get sublimation whenever you have an ice surface exposed to perfect vacuum. The count of sufficiently energetic particles dictates the maximum rate of the sublimation, and there is always a vapor pressure at which this rate of sublimation equals the rate at which gaseous particles hit the ice, dissipate their energy, and enter its phase.

The melting point is largely invariant of pressure because melting/freezing does not change volume much. Volume grows a bit when freezing at room temperatures, so very high pressures can discourage water from entering the solid state, lowering the melting point a bit.

Of course, the same volume dependency is present with the gaseous phase: Water takes much more space when it's low-pressure vapor than when it's solid or fluid. As such, it's natural that both the sublimation and the boiling point show a strong pressure dependence: The more pressure, the more energy (= temperature) you need to put a particle into the gaseous phase. The volume dependency between vapor and other phases is much, much stronger at the triple point than the volume dependency between solid and fluid phases, so you have to expect the sublimation point to qualitatively continue the path of the boiling point to lower temperatures and pressures. There is a slight angle between the two curves at the triple point, though, as far as I remember.


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