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