Say I have an infinite volume of water with some level of dissolved air in it. I lower the pressure to below the gas' saturation pressure. Does the air come out of solution if there are no interfaces (nuclei) for it to begin? If so, what is the driving force behind the formation of the bubbles? Are there any estimations of the time that it would take to form a bubble of a given size, or an estimation of the equilibrium size of said bubble? Presumably it would be a balance between the surface tension, radius and the water pressure (like $P + 2S/R$)...

To give some motivation, I have a cavitation tunnel where I can apply a pressure field to a large volume of water. In the literature, people speak of homogeneous nucleation where fluids below saturation pressure start to release their gas as bubbles, and I'm struggling to understand how that occurs.

  • $\begingroup$ First of all, welcome! I would try to clarify your question a bit. For example you seem to be talking about both a hypothetical "infinite volume" and the observation of a real phenomenon in the description. It may just be me but I'm having a hard time picturing that. Also you could expand on the equation you present and how you think that might explain the observation of bubbles. I think it may be an interesting question just in need of some tweaking. $\endgroup$
    – airhuff
    Jan 25, 2017 at 4:07
  • $\begingroup$ Thanks for the welcome. I added a bit of 'motivation' if that helps. $\endgroup$
    – James
    Jan 25, 2017 at 4:25
  • $\begingroup$ There will be no balance. Fusion of two bubbles into a larger bubble is always energetically favorable. The equilibrium state is one huge bubble amidst the liquid, which means your situation is essentially non-equilibrium. As for the gist of your question, you are dealing with cavitation. $\endgroup$ Jan 25, 2017 at 6:24
  • $\begingroup$ Thanks @IvanNeretin, I guess I'm wondering where these two bubbles come from in the first place. My understanding is that cavitation is driven by fluids below the vapour pressure - whereas this is taking place above that. $\endgroup$
    – James
    Jan 25, 2017 at 20:53
  • $\begingroup$ Be it vapor pressure or the pressure of dissolved gas, the phenomenon is pretty much the same. As for nucleation, well, that's a hard problem indeed. $\endgroup$ Jan 25, 2017 at 22:04

1 Answer 1


According to this abstract from a paper titled "Why Is It Much Easier To Nucleate Gas Bubbles than Theory Predicts?":

Bubble nucleation from supersaturated gas solutions generally takes place at much lower supersaturations than are expected from the theory. Furthermore, the same theory predicts that the threshold concentration of gas needed to cause nucleation should be essentially independent of the gas species used, a finding contradicted by experiment. There are two general explanations:  first, that the theory is wrong, or second, that there is a previously unidentified factor which is influencing the results of the experiments. Given the success of the fundamental theory in other areas, the second explanation is preferred. The previously unrecognized factor is identified here as being the surface activity of the gases which form the bubbles.

Although this is discussing conditions of low supersaturation and not subsaturation, I assume the same concepts apply, though it would cost me $40 to get the full article text and find out for sure.

I'm pretty sure what's happening here is that through the random motion of dissolved gas molecules, occasionally enough of them adhere to one another to form the nucleus of a bubble. Particularly at very small radii, the high surface tension of water dominates otherwise destructive forces, and a micro-bubble begins to form. But by the definition of saturation with respect to bubble formation, it would seem that you cannot have "macro-scale" bubble formation and growth persisting under conditions of subsaturation as equilibrium is reached.

If I can paraphrase your question as:

"If I lower the gas pressure to below its saturation pressure, does the gas come out of solution if there are no interfaces (nuclei) for bubbles to form?"

then the answer is yes, regardless of the presence or absence of bubbles. The dissolved gas molecules will partition at the gas-liquid interface until equilibrium concentrations are achieved in each phase. If some gas bubble began to form, it would quickly redissolve under conditions of subsaturation with respect to the gas solubility.

Unless I'm missing something about the importance of bubbles in your question, this would pretty clearly be the bottom line answer. Please comment if I've missed the crux of your question.

  • $\begingroup$ Your answer is very helpful. I'll get a hold of that paper and read it myself. I guess the bit that gets me is "through the random motion of dissolved gas molecules, occasionally enough of them adhere to one another to form the nucleus of a bubble". If there was some quantitative analysis of that then I'd be more comfortable. $\endgroup$
    – James
    Jan 25, 2017 at 20:50
  • $\begingroup$ "it would quickly redissolve under conditions of subsaturation" Is this a 'hard cut-off' phenomena at the saturation pressure or is this a gradual phenomenon? $\endgroup$
    – James
    Jan 25, 2017 at 20:51
  • $\begingroup$ I think you are mixing kinetics and thermodynamics in a way that is causing you confusion in both your question and comments. For example, when you talk about "where fluids below saturation pressure start to release their gas as bubbles", you are addressing the thermodynamic property of "saturation (implying thermodynamic equilibrium) vapor pressure". My choice of the word quickly may have confused the issue more for you, but when you ask "is this a gradual phenomenon?", then you are talking kinetics. It may help you to remember those concepts are separate from one another. $\endgroup$
    – airhuff
    Jan 25, 2017 at 21:09

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