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Vapor pressure of a substance, in a given temperature, is the pressure exerted the vapor in a system where the gaseous and liquid phases of the pure substance are in equilibrium.

But, if for a temperature where a compound is obligatorily a gas and cannot be condensed into a liquid at any pressure (like a permanent gas at room temperature), is vapor pressure definable?

And if it isn't definable, does this imply that any mixture between the gas and a liquid (the gas being dissolved in the liquid) is obligatorily non-ideal?

I'm saying that because, for a system where the liquid phase is an ideal mixture and the vapor phase is an ideal gas, Raoult's Law is valid for every component for any proportion between the components, and Henry's Law volatility constant (the variant using molar fraction) for any component becomes simply the vapor pressure of that component at a given temperature.

But, if its vapor pressure isn't definable at that temperature, then it can't be equal.

Perhaps the very fact that one compound can be condensed at that temperature while the other cannot already means that their interactions in the liquid phase cannot be possibly similar?

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  • $\begingroup$ Vapor pressure .... are in equilibrium. That is saturated vapor pressure. Vapor pressure does not have equilibrium requirement. There is water vapor pressure in atmosphere without being saturated. // We say a gaseous, not vapor phase, as a vapor is a gas. $\endgroup$
    – Poutnik
    Commented Feb 14, 2022 at 5:39
  • $\begingroup$ In this case you would not say that Henry's law constant is equal to the non-existent partial pressure of the permanent gas. Take oxygen solubility in water, for instance (which is a permanent gas when T is above critical). But if the gas observes Henry's law then the solvent observes Raoult's law, and the mixture is considered ideal. $\endgroup$
    – Buck Thorn ♦
    Commented Feb 14, 2022 at 6:33
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    $\begingroup$ @all Please stop perpetuating this terrible misnomer. Vapour pressure is a property of the liquid. There IS NO vapour pressure of anything in the gas phase, because what you mean is partial pressure!! The vapour pressure of the liquid is identical to the equillibrium partial pressure of it's gaseous form in the free space above the liquid. $\endgroup$
    – Karl
    Commented Feb 14, 2022 at 9:17
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    $\begingroup$ @Karl You may be right, I may be a victim of a different naming due meteorological history, where vapor pressure and saturated vapor pressure is used in context of absolute and relative humidity as shortcut for vapor partial pressure and saturated vapor partial pressure. :-) So vapor pressure is partial pressure of vapor. $\endgroup$
    – Poutnik
    Commented Feb 14, 2022 at 9:23
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    $\begingroup$ Even if a gas cannot be liquefied above a certain temperature, that doesn't mean it can't be directly solidified with enough pressure. For example, even hundreds of degrees above the critical temperature of water, you can directly crush the supercritical fluid into exotic ices. In principle all you need to establish a "vapor pressure" is a condensed phase, be it liquid, solid or perhaps even something more exotic. $\endgroup$
    – Nicolau Saker Neto
    Commented Feb 14, 2022 at 9:26

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I think you are quite right, the behaviour cannot be ideal:

If you look at the solubility of a permanent gas in a liquid, you are limited to the liquid's side of Raults mixture phase diagram. It doesn't matter there's no vapour pressure on the side of the gas, because you can never get there, until you actually condense your previously permanent gas, perhaps into a solid. (Spoiler: there is probably a miscibility gap. Nonideal.)

If your gas was actually not condensable at any pressure, the curve must verge towards infinity at 100% gas, and thus obviously cannot be linear. Ergo, the mixture cannot be ideal with respect to Raults law.

qed.

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