In open air, when vapour pressure reaches 1 atm, boiling takes place.

I read that if we add two immiscible liquids together, the total vapour pressure of the 'mixture' is close to $p = p^*_A + p^*_b$, means that the vapour pressure of the 'mixture' is higher than the vapour pressure of both the constituent liquids, subsequently means that the 'mixture' boils at lower temperature than both of the constituent liquids.

If we have many liquids which are not immiscible with each other, and we add them together to form a 'mixture' such that $p = p^*_A + p^*_B + ... > \pu{1atm}$ , will boiling results at room temperature?

I suspect we have many liquids which are not miscible with each other, since we classify solvents as organic solvent and inorganic solvent only, but let's say we indeed have that (or we have two immiscible liquids volatile enough to make $p>\pu{1atm}$), what will be observed? Boiling at room temperature? In that case, we can use the vapour to do work (e.g. rotate turbine), but when we mix immiscible liquid together there should be no energy transaction right? The particles do not react with each other. If all these are true, we are actually doing work without any input which violates the first law, so there must be something wrong with my reasoning above.

  • 2
    $\begingroup$ This brings interesting question how many non miscible liquid phases can coexist in equilibrium. $\endgroup$
    – Poutnik
    Apr 16, 2019 at 16:13
  • $\begingroup$ I guess a good answer here should make clear what exactly is going to happen, and how we can predict the final state of such a system. $\endgroup$
    – Karl
    Oct 25, 2020 at 20:06
  • $\begingroup$ @Karl IMHO AKs answer describes the situation pretty well. The OP is apparently confused about when something will boil, as AK explains. The steam distillation technique does not rely on changes in boiling points, that is a misunderstanding, but if the volatile compounds are soluble in and form an azeotrope with water then the boiling point will be lowered. However the question asks about immiscible solvents, so azeotropes are essentially irrelevant. Steam distillation relies on transport of the volatile compound, not changes in b.p. $\endgroup$
    – Buck Thorn
    Oct 27, 2020 at 17:59
  • $\begingroup$ @BuckThorn Steam distillation is done with nonmixing azeotropes. Some high-boiling unpolar organic substance and water. Toluene and water would be a perfectly good example. since the partial pressure of each of the phases is less than ambient, no boiling will occur. is of course the correct answer here, no doubt. But what exactly will happen? $\endgroup$
    – Karl
    Oct 27, 2020 at 18:15
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    $\begingroup$ @Poutnik: I got it up to six at room temperature and minimal hazard. Gallium, Fluorinert (liquid breathing), Silicone oil, mineral oil (has to be heavy enough or else miscible with silicone), rubbing alcohol, salt water (salt forces out the isopropal phase). Shake the whole thing and all 6 liquids will re-form. $\endgroup$ May 7 at 17:49

4 Answers 4


Yes they will boil all right. Sure, there might be some kinetic impediment to it if you let the liquids to settle in layers, but if you stir them so as to expose their surfaces, they will boil. This is what steam distillation is all about.

As for the first law, it will hold just fine. You burn your firewood, you get the heat, but it is not for free: the firewood is gone. Same thing here. Your liquids are gone. What you have now is a vapor mixture. Sure, you may cool it down, and it will separate back into the liquids, which you may then heat up and repeat... Congratulations, you've just invented the steam engine. Pity it's been done before by one James Watt. Also, you don't really need two liquids for it.

So it goes.

(†) There seem to be some ambiguity as to what exactly "boiling" entails.

  • Will vapor bubbles form in the whole bulk of both liquids? No.
  • Will both liquids eventually evaporate completely, if left in a cylinder under a piston and maintained at these pressure and temperature until they reach equilibrium? Yes.

For a single pure liquid these two questions are equivalent, hence the confusion.

  • $\begingroup$ Found a link where the same is explained from a different perspective: en.wikipedia.org/wiki/… $\endgroup$ Apr 17, 2019 at 19:20
  • $\begingroup$ What happens when one of the components has evaporated completely? $\endgroup$
    – Karl
    Oct 25, 2020 at 11:55
  • $\begingroup$ Then the other will stop evaporating too. $\endgroup$ Oct 25, 2020 at 13:18
  • $\begingroup$ Vapor pressure of A and B, obviously. $\endgroup$ Oct 25, 2020 at 14:04
  • 1
    $\begingroup$ Both A and B will evaporate until they reach their respective saturated pressures. There is nothing ambiguous about it. $\endgroup$ Oct 26, 2020 at 6:43

Nothing special would happen, immiscible liquids would just form layers. As for the expression, $$p_T=\Sigma p^o_i$$

I suggest you read this answer. Quoting Ivan Neretin:

This is not just some vapor pressure. This is the equilibrium vapor pressure. Thermodynamics is all about equilibrium, you know. And equilibrium, roughly speaking, is what takes place in a closed container after a billion years.

It would take infinite time for that expression to prove itself correct.

In speaking within an average man's lifespan, the liquids would remain in separate layers, with the pressure above liquid surface would remain approximately the pure liquid vapour pressure ($p^o$) of the uppermost layer if left undisturbed (according to your question: no energy transactions).

Don't worry about the first law: it is rarely violated these days.

This "equilibrium" can be attained sooner by agitation (stirring) of the mixture, but mechanical stirring certainly counts as work done.

From a chemguide page:

Obviously if you have two immiscible liquids in a closed flask and keep everything still, the vapour pressure you measure will simply be the vapour pressure of the one which is floating on top. There is no way that the bottom liquid can turn to vapour. The top one is sealing it in.

Agitated mixtures of immiscible liquids will boil at a temperature lower than the boiling point of either of the pure liquids. Their combined vapour pressures are bound to reach the external pressure before the vapour pressure of either of the individual components get there.

In other words, although less evident, even if you do get the mixture to boil and do work, it would be because you had done work on it earlier during agitation.

  • 2
    $\begingroup$ I object to the last point. Work done during agitation can easily be made negligible. For example, we may put the liquids into two separate open compartments, so that their vapors mix, but the liquids themselves don't. Then they both will have open surface without any work. $\endgroup$ Apr 15, 2019 at 16:08
  • $\begingroup$ @IvanNeretin I wish to respectfully mention that I do not agree. Could I get back to you in a day? I have an exam tomorrow and am short on time today. I will certainly participate in a fruitful discussion. $\endgroup$ Apr 15, 2019 at 17:04
  • $\begingroup$ Looking forward to hear your arguments. $\endgroup$ Apr 15, 2019 at 17:33
  • $\begingroup$ Ok, well, as A.K. mentions in his answer, boiling happens when the vapour pressure of the phase is greater than ambient pressure. Keeping the liquids in separate partitions will not create a single phase. $\endgroup$ Apr 16, 2019 at 1:26
  • 2
    $\begingroup$ What problem? You seemed to agree that an agitated mixture will boil below the boiling points of both liquids, and that because of exposed surface. I just proposed a setup that lets you have exposed surface of both liquids without any agitation. $\endgroup$ Apr 16, 2019 at 8:21

Is it possible to boil a liquid by just mixing many immiscible liquids together?

No*, boiling is when the vapor pressure of a phase is greater than the ambient pressure. You might create a total pressure of all of the partial pressures in excess of ambient, but it wont cause boiling since the partial pressure of each of the phases is less than ambient, no boiling will occur.

To get boiling by mixing you need two mixable substances that form a positive azeotrope when mixed and are at a sufficiently high temperature that the mixture can boil. Take for example methanol and chloroform system in the phase diagrams below. At $\pu{330K}$ neither pure methanol nor pure chloroform boil, but if you mixed $\pu{2 mol}$ of chloroform at $\pu{330K}$ with $\pu{1 mol}$ of methanol at $\pu{330K}$ together the phase would boil upon mixing as the vapor pressure of the phase is now above ambient.

enter image description here

Okay I said no, but that technically isn't correct. Grease fires are a (too) common example of mixing immiscible liquids and causing boiling as the oil temperature is well above the boiling temperature of water. This is why you can't throw water on a grease fire. For perspective on thanksgiving of 2016 there were about 1570 fires in the US related to this phenomenon.

  • $\begingroup$ Is steam distillation not a thing at all, then? $\endgroup$ Apr 16, 2019 at 5:21

Yes, but there are no many-component mixtures with all partial pressures contributing significantly to the total vapor pressure.

That said, the following can be applied to two-immiscible-component liquids: No violation of the first law: energy originates from the process that liquefied the two phases in the setup. There is no way to liquefy them without using energy.
Bubbles do form, given that they can nucleate. No stirring needed.
The mechanism that drives their growth is diffusion of one species through the liquid phase comprising mostly of another (which evaporates directly from the inner surface of the bubble). It doesn't matter for the equilibrium, whether this diffusion is slow. It does matter for the growth rate of the bubble, however.


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