What is the triple point pressure for Gallium?

I have seen various discussions about the triple point of Gallium determined to a very precise value, so precise that it is used as a reference for NIST scales and measurements.

However, these reference related documents mean 'temperature' only when talking about triple point. Triple-point is defined as a (temperature, pressure) pair, so I fail to understand why despite so many discussions on the triple point of Ga, the pressure value is not easy to find.

I did find a value of $$10^{-38} \mathrm{atm}$$ (if I'm not wrong) sometime in the past but I can't find the source. However I have two questions related to this value:

• Does it even makes sense to talk about such a minuscule value of pressure? Wouldn't this mean one particle every cubic light-year or something? (edit: More like one particle every cubic kilometer, depending on temperature, but still).
• Is it safe to say that Gallium melts at absolute zero pressure (perfect vacuum), and never sublimates or desorbs (outgassing of Ga atoms/clusters from solid Ga surface)?

In general, are there other metals that have such a small triple-point pressure that they are guaranteed to never exhibit the process of sublimation/desorption, even in a perfect vacuum? (i.e., they will always melt first before evaporating or boiling).

• pubs.acs.org/doi/pdf/10.1021/ja01102a057 has the vapor pressure of gallium. Commented Dec 29, 2019 at 3:16
• @JonCuster Thanks. Very helpful resource. I calculated the pressure using eq. 8 and temperature of $302.9166$ K. I get $6.77 \times 10^{-34}$ atm. Since this is pretty much as minuscule as $10^{-38}$ atm, posted questions are still relevant. Commented Dec 29, 2019 at 4:03

At the very least your pressure has to be consistent with a whole number of molecules or atoms in your vapor phase. Suppose you have a pressure of $$7×10^{-34}$$ atmosphere or (roughly) $$7×10^{-29}$$ Pa at $$300$$ Kelvins. The average volume per atom/molecule is computed by the molecular level version of the Ideal Gas Law, in which the gas constant $$R$$ per mole is replaced by Boltzmann's constant $$k=1.38×20^{-23}$$ J/K per unit entity. Thus

$$V=kT/P=(1.38×10^{-23})×300/(7×10^{-29})=6.9×10^7$$

cubic meters per vaporized entity.

So unless you are equilibrating your gallium over that much vapor volume, with high probability you do not have in practice even one atom or molecule of vaporized gallium. In realistic laboratory or even industrual settings the triple point content of gallium vapor is most probably zero.