As far as I know, for all substances other than helium, if you extrapolate the solid-gas line on the phase diagram, it passes through the origin. That is, no matter how low the temperature is, you can still vaporize the substance by lowering the pressure, and no matter how low the pressure is, you can still freeze the substance by lowering the temperature.

This makes sense to me on thermodynamic grounds. If we keep lowering the pressure at a given temperature, the entropy of vaporization increases without bound (although only logarithmically) and at some point the gas phase is favoured, whereas if we keep lowering the temperature at a given pressure, then the entropy term in free energy becomes negligible and the solid phase is favoured by enthalpy.

The phase diagrams I've found online for helium-4, however, don't follow this rule. They seem to show that helium-4 is exclusively a liquid in a neighbourhood of the origin. Example here: http://ltl.tkk.fi/research/theory/helium.html

I wonder whether this is true, and if so, how it can be possible. How can it be that no matter how low the pressure is, helium-4 will not boil? I understand that superfluidity is a quantum phenomenon but I can't see how my previous reasoning could not apply.

Edit since some people think this is a duplicate: Again, my question is whether helium-4 will not boil at zero pressure, and if so, how that can be; it concerns the liquid-gas transition and not the solid-liquid one.

  • $\begingroup$ possible duplicate of Why can't helium be solidified at 'ordinary' pressures? $\endgroup$
    – user15489
    May 23 '15 at 4:10
  • $\begingroup$ @santiago I don't think this is a duplicate. My question asks why helium-4 will not boil at zero pressure, whereas every other liquid apparently will. Not why it will not freeze. $\endgroup$
    – Brian
    May 23 '15 at 4:36
  • $\begingroup$ Interesting question and preamble. I'd never considered the content in your second paragraph. $\endgroup$ May 23 '15 at 11:01
  • $\begingroup$ en.wikipedia.org/wiki/Lambda_point $\endgroup$
    – Mithoron
    May 23 '15 at 12:07
  • $\begingroup$ It will boil all right. $\endgroup$ Jul 2 '17 at 6:49

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