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Another way to put it: Is evaporation possible when only one substance is involved?

Consider a closed vessel with water in gas and liquid state, in equilibrium. At the interface, the same amount of molecules migrate from liquid to gas and the inverse (dynamic equilibrium). Now, if we remove X molecules from the gas phase, molecules from the liquid phase will pass to the gas phase until equilibrium is achieved again.

What is the process describing this phase transition of the liquid molecules to gas: evaporation or boiling?

These are two types of vaporization. The definition from wikipedia (en.wikipedia.org/wiki/Vaporization) are rephrased below in order to mark their difference.

  • Evaporation (surface) occurs when the equilibrium vapor pressure is higher than the partial pressure of vapor of the substance.

  • Boiling (volume) occurs when the equilibrium vapor pressure is higher than the pressure of the "environmental pressure" (which I intepret as the total pressure surrounding gas).

I understand that in a single substance, the two cases are the same. Is this correct?

In other words, how would it be possible to achieve only evaporation in the single substance case. If by removing the X gas molecules will result in a pressure decrease. This suits best at the boiling definition.

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    $\begingroup$ In my opinion, the difference between evaporation and boiling is only a matter of degree. In other words, boiling is just very fast evaporation. $\endgroup$
    – David White
    Commented Dec 12, 2020 at 17:38
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    $\begingroup$ I don't think that it is a quantitative issue. If we consider a system of atmospheric air and water - there is a qualitative difference between evaporation and boiling. Namely, evaporation takes place under specific conditions (not saturated air, any temperature) and boiling under different specific conditions (boiling temperature). My problem is that I cannot see this distinction in a single-substance system. To my understanding, in a single-substance system, both evaporation and boiling occurs under the same conditions. $\endgroup$
    – Jore
    Commented Dec 12, 2020 at 23:31
  • $\begingroup$ When the vapor pressure of liquid water becomes slightly higher than the partial pressure of water vapor above that water, it boils, whether there is air and water vapor above the liquid water or only water vapor above the liquid water. For "evaporation", the rate of boiling is very low, it is constrained by how fast the liquid water can absorb heat from the environment, and it occurs at the liquid water surface. "Boiling" occurs on the bottom of a container at the heat source, and it is a very fast process due to the high rate of heat transfer. The processes are equivalent. $\endgroup$
    – David White
    Commented Dec 13, 2020 at 0:25
  • $\begingroup$ Related: physics.stackexchange.com/a/557826/44126 $\endgroup$
    – rob ♦
    Commented Nov 1 at 2:33

2 Answers 2

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One approach is to encapsulate the water in a suitably stiff container that occasionally opens briefly to expose the water to vacuum. The pressure exerted by the container suppresses bubble nucleation in the same way a covering atmosphere would. You’re left with only surface evaporation, not bulk boiling, during the brief open periods because of kinetic limitations. Is this what you’re asking about?

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  • $\begingroup$ Maybe I have not expressed myself well. What I want to understand is: in a vessel containing only one substance in liquid-gas equilibrium, if I remove part of the gas molecules, then the gas pressure will decrease and vaporization will be triggered. What is the mechanism of this vaporization - evaporation or boiling? $\endgroup$
    – Jore
    Commented Dec 12, 2020 at 17:11
  • $\begingroup$ Both are spontaneous in this case. $\endgroup$
    – Chemomechanics
    Commented Dec 12, 2020 at 19:07
  • $\begingroup$ So they both happen when the gas pressure decreases below the equilibrium vapor pressure? $\endgroup$
    – Jore
    Commented Dec 12, 2020 at 23:32
  • $\begingroup$ Jore, look at the Antoine equation: en.wikipedia.org/wiki/Antoine_equation. When the calculated vapor pressure exceeds the ambient pressure, the liquid boils. $\endgroup$
    – David White
    Commented Dec 13, 2020 at 0:29
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The difference is, as one commenter suggested, really a matter of degree. But this difference is also a rather specific one. Evaporation is any process whereby a liquid goes into a vapor state. Because every substance has a finite, if sometimes tiny, vapor pressure, there will be some degree of evaporation when the particles break free as vapor and inevitably diffuse away from the condensed phase. This leads to the observable disappearance of the condensed phase substance. This, of course, also happens in boiling, just much more noticeably.

Boiling is uniquely characterized by the vapor pressure of the system becoming equivalent to the ambient pressure of the surrounding atmosphere. Under most typical conditions on Earth, a liquid boils when it’s vapor pressure reaches 1 atm. For a really fun demonstration, you can show liquids boiling instantly at room temperature when you pull a vacuum over them, as the vapor pressure does not need to be high at all to boil in a vacuum chamber! The specific requirement of the vapor pressure matching the ambient pressure for boiling does demarcate it from simple evaporation, but only because it is so specific, not because the chemical equilibrium and diffusion processes involved are any different.

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