We were discussing super-critical fluids with a colleague and as were thinking about different thought-experiments, it occurred to me that you cannot really have only liquid in a piston in a real situation. I want to make sure I'm not missing something. Are there liquids which have zero vapor pressure under certain conditions? If not, am I right to think even if you would perfectly enclose a volume of liquid in an hermetic container, you would always have a tiny fraction of vapor due to evaporation and vapor-pressure equilibrium? If that is the case, ato turn this liquid into a supercritical fluid, I will only need to reduce the volume of the container by a very small amount and this can be explained in two ways:

  • from the liquid point of view, since it is almost incompressible, I need a very large force to change that volume even by a small amount. Thus the pressure in my piston is large and likely over the critical pressure of the liquid, turning it into a super-critical fluid
  • from the gas point of view, since the volume occupied by the gas is very small and since it will "feel" the compression first, it is easy to see the gas pressure increasing very rapidly for small changes in volume, once again leading to a supercritical fluid.

Thanks in advance for your answers !


1 Answer 1


Vapor pressure is something which applies at equilibrium for a liquid vaporizing into an established gas phase. If the gas phase is not established, the liquid must first cavitate. And if the gas phase is already established, the liquid can still exist in a metastable state for quite a while before reaching equilibrium.

So if you take an indestructible piston-cylinder with no nucleation sites and fill it with degassed water, the water will remain as pure liquid with no vapor phase. But this isn't really a challenge, since atmospheric pressure keeps the water pressurized at 1 bar. If you push the piston further, the water will mostly just push back. Sometime well beyond 10 000 bar, the water will solidify into a denser form of ice, maybe ice VII or ice X, and then the piston will give way a bit more. Conversely, you could also pull the piston all you want, and the water will just pull back. This way, you could even get negative pressures – one paper reports getting down to -277 bar this way, but theoretically you could pull at ten times that tension before cavitation lets the piston-cylinder expand.

If you start with a degassed saturated mixture of liquid water and water vapor, that's not a complication. Pushing the piston in that case could condense the vapor phase entirely. Conversely, pulling the piston could vaporize the liquid phase entirely.

You would see something similar if the water had air bubbles. Pushing would make water vapor in the bubbles condense into the liquid phase, while simultaneously dissolving more air into the liquid phase. Pulling would encourage water in the liquid phase to vaporize to maintain vapor pressure in the bubbles.

Some takeaways:

  • No liquids have zero vapor pressure, but they may only cavitate under tension.
  • It's quite possible to have a pure enclosed liquid with no vapor.
  • If you push an incompressible fluid, it will just push back. Its pressure is entirely determined by how hard you push. If you push harder than the fluid's critical pressure, then so be it. It's up to you whether you want to call it a "supercritical fluid" at that point; personally, I reserve that label for fluids above the critical temperature.
  • $\begingroup$ Well in my thought experiment, I cannot determine the temperature, but you are right, I probably should have called the fluid a compressed liquid (top-left quadrant of the p-T diagram) and not a supercritical fluid. Let me digest the rest... $\endgroup$ Commented Mar 5, 2012 at 19:05
  • $\begingroup$ When you say "If you push the piston further, the water will mostly just push back." that does indicate that of course my equilibrium pressure (i'm pushing very slowly) is increasing.I think my mistake was that in the case you do get a pure liquid in a given volume, the p-T conditions can be outside the boiling curve and thus there will be no evaporation right? $\endgroup$ Commented Mar 5, 2012 at 19:18
  • $\begingroup$ @FrenchKheldar There's a big difference between evaporation and boiling. Evaporation is a diffusion-driven process – it simply means that if there's a liquid and a gas phase, water molecules will constantly wander from one phase to the other such that the partial pressure of water in the gas phase tends toward vapor pressure. Boiling means that water molecules near a nucleation site will turn into a vapor bubble right there, all at once, and then try to find a gas phase. The "boiling curve" says when this may start, but metastable liquid states can last for weeks before boiling. $\endgroup$
    – Chel
    Commented Mar 5, 2012 at 20:47

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