I've always had a hard time wrapping my head around the 2 below statements being true for vaporizing a liquid into a gas:

  1. When a liquid reaches its boiling point the temperature stops rising (and any further energy added goes into breaking bonds and changing phase)
  2. A specific amount of heat (latent heat) is required for said phase change.

I don't see how both these conditions can exist simultaneously. For example, say you hypothetically had a liquid right spot on its boiling point (and perhaps to simplify things assume there wasn't a distribution to the molecular energies, they were all essentially at the same average energy). What would happen then if you gave a molecule an amount of energy less than the latent heat required for a single molecule to vaporize? Per statement 2 above that molecule wouldn't change phase, however then I would expect the KE energy to increase and the temperature to come up which would then violate statement 1?

I think I can maybe simplify my question/issue into the below:

If a molecule is in a liquid at boiling point, if it is given the latent heat will that case it to just barely break free (i.e it needs the latent heat amount to just barely escape and go into the gas phase with zero KE)? Or will any amount of energy at this stage (no matter how small) cause it to break free (essentially it is already on the verge of breaking free and stretching the bonds to the limit), and the latent heat is actually the amount required for it to both break out of the liquid phase and have the same temp/KE as before?

  • 2
    $\begingroup$ First of all, the phase of matter never talks about a single molecule, it talks about the properties of a lot of particles. Second, when you're at a phase-change temperature, energy is no longer a function of temperature, meaning there isn't a single energy possible for that temperature. $\endgroup$ Mar 15 at 12:47
  • $\begingroup$ If the amount of heat were insufficient to vaporize the liquid it would mean, by definition, that the liquid is not saturated liquid. $\endgroup$
    – Bob D
    Mar 15 at 14:34
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    $\begingroup$ As discussed in this answer, the temperature doesn’t remain exactly the same in a phase change because of the surface energy penalty of nucleating the new phase. So your conditions (1) and (2) don’t strictly exist simultaneously because (1) is an idealization. $\endgroup$ Mar 15 at 15:23

2 Answers 2


The key to this is that molecules do not have phase transition. A molecule of water vapor is intrinsically exactly the same as a molecule of liquid water in every way. The difference between liquid water and water vapor lies in the organization of the molecules, not the molecules themselves. In liquid water, the energy is lower compared to the energy of the intermolecular bonds between the molecules. In water vapor, the energy is higher.

These statements are, of course, made statistically. In the usual statistical model behind phase transitions, each molecule's energy is described statistically. In liquid water, sometimes the molecules at the surface have an amount of energy sufficient to break the intermolecular bonds and the water evaporates. The higher the temperature is, the larger the probability that any molecule has enough energy to break free and evaporation increases.

When we talk about phase transitions, we're talking about a LOT of molecules. And in this case, the central limit theorem applies. The standard deviation of the energy of the particles gets closer and closer together on average. It starts to look more like a single spike, which is moved to higher energies or lower energies depending on temperature (or perhaps more accurately, temperature is a measure of that mean energy).

In such a case with lots of molecules, the transition from "the vast majority of molecules do not have enough energy to break free" to "the vast majority of molecules do have enough energy to break free" is tremendously sudden.

As we go through this transition, putting energy into vaporization, we don't so much increase the kinetic energy of the molecules as much as we increase their potential energy, moving them into configurations where more and more energy has been put into breaking those intermolecular bonds. This process is called "boiling." Very quickly, the random interactions of the molecules give some atoms enough energy to escape their bonds and become a gas.

If you are very clever, you can construct ensembles of molecules where their behavior is not well described by these simple concepts like vaporization and a "boiling point". But these are special cases (super-saturation is a good one to look at). But for the vast majority of compounds we come across, this statistical approach with the central limit theorem does a good job of describing how the molecules behave statistically.

  • $\begingroup$ Thanks for the answer. I suppose my question originated from me trying to visualise the molecules as they change phase with added energy. I was picturing them as escaping a potential well. This led me to think that when they reach the boiling point and are on the cusp of breaking free into vapour, shouldn't essentially any extra energy added to any number of molecules at this point cause that to happen (with higher energies meaning more leftover KE once the potential well has been escaped). This seemed to contradict the idea that a fixed amount of energy per mass was needed $\endgroup$
    – LWilkinson
    Mar 15 at 16:09
  • $\begingroup$ I don't quite see the contradiction. As you say, there's a fixed amount of energy required to break the intermolecular bonds, and any remaining kinetic energy the molecule has remains kinetic energy (because it wasn't used to break the bonds) $\endgroup$
    – Cort Ammon
    Mar 15 at 16:14
  • $\begingroup$ It's the idea that it's a fixed amount of energy required that seems off to me. As in there seems to be some amount of energy, that if less than, won't be enough to free a molecule into the vapour phase if given to it at the boiling point. It makes more sense to me that if at the boiling point, some molecules recieved a very very small (almost 0) amount of additional energy, they would still change phase, however their KE once in the vapour phase would basically be 0 (essentially them just barely escaping the well) $\endgroup$
    – LWilkinson
    Mar 15 at 16:33
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    $\begingroup$ That is true statistically. What really happens is that there's always some movement "up" and "down" the potential well, but only the tiniest fraction of molecules escape it. The rest fall back in, turning the energy back into random thermal motion. Any molecule that does escape takes with it all the energy it took to break free of the potential well (the fixed part), plus some kinetic energy. Once you hit the boiling temperature, it goes from the tiniest fraction of molecules escaping to most molecules escaping, and it happens over a very small temperature range. $\endgroup$
    – Cort Ammon
    Mar 15 at 17:53
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    $\begingroup$ And remember that this is idealized. In a real boiling situation, we don't get to assume that the temperature at the top of the liquid is exactly the same as the temperature at the bottom. So these processes which sound instantaneous in theory do take time in real life. $\endgroup$
    – Cort Ammon
    Mar 15 at 17:55

It is a common misconception that the liquid to gas transition is an abrupt process that happens suddenly at the boiling point. The liquid is actually turning into a gas (and vice versa) at all temperatures, which is why wet clothes dry if we hang them on a washing line. It's just below the boiling point we call this process evaporation instead of boiling. It's true that boiling is "special" in that it happens at a constant temperature, but as we'll see in the moment the constant temperature is just because the process happens as a constant pressure and it doesn't reflect anything fundamental.

As a rough approximation the rate at which molecules escape from the liquid into the gas phase is proportional to the temperature of the liquid. The rate at which molecules in the gas phase recondense into the liquid depends on the partial pressure of the gaseous molecules.

Suppose we hang out wet clothes to dry in a closed room with initially dry air in it. Water will start to evaporate from the clothes and the partial pressure of water vapour in the room will increase. As the partial pressure of the water vapour increases the rate at which water vapour recondenses increases, and we reach a dynamic equilibrium at which the rate the water evaporates from the clothes is equal to the rate at which the water recondenses onto them, and the drying process stops. One way of describing this is to say the air in the room has reached 100% relative humidity.

If we increase the temperature the rate of evaporation increases, and as a result the partial pressure of water increases until we reach equilibrium again.

And so on, we can keep increasing the temperature and the partial pressure of water vapour increases to maintain equilibrium. We can keep doing this to arbitrarily high temperatures, well above 100°C, because the pressure can rise without limit as well. In this system the water never boils so there is no boiling point.

But, if we attempt this in the open air there is a limit to the pressure because the ambient pressure is 1 atmosphere. So when the partial pressure of the water vapour has increased to 1 atmosphere it cannot get any higher, and this happens at 100°C. If we increase the temperature of the water the increased rate of evaporation is no longer balanced by an increased rate of condensation and that's why we see boiling.

The point of all this is that you cannot understand the process by looking at a single molecule because single molecules are continuously undergoing the liquid to gas (and back) transition at all temperatures. The reason boiling happens at constant temperature is nothing to do with the properties of single molecules. It's because we have limited the partial pressure of the gas phase to a constant value of 1 atm.

Response to comment

In your question you say:

and perhaps to simplify things assume there wasn't a distribution to the molecular energies, they were all essentially at the same average energy

but you cannot make this simplification as a thermal distribution of energies is fundamental to the process. The energy required to remove an atom from the liquid to the gas phase is roughly independent of temperature, and the reason the evaporation rate increases with temperature is because the probability of any one molecule getting that much energy increases with temperature.

You ask:

if you gave a molecule an amount of energy less than the latent heat required for a single molecule to vaporize?

and the answer is that within a few microseconds it would have thermalised with the molecules around it and its energy would randomly increase, decrease or remain roughly constant. If it's the first of these it could escape to the gas, or if it's the second it could transfer enough energy to another molecule for that molecule to escape to the gas.

You really cannot understand boiling by considering just a single molecule. It is an ensemble process.

  • $\begingroup$ Hi John, thanks for the answer. I think I feel farely comfortable with the above. I suppose my question is more about latent heat of vaporisation and my single molecule example was me trying to create a hypothetical scenario to highlight my issue. Perhaps I could frame my issue as so: If a liquid is at boiling why cant I vaporise it with an amount of energy less than the stated latent heat for that mass? I feel if most the molecules were on the verge of escaping their potential well, then shouldnt just a little bit of energy given to the liquid be enough for them to escape (albeit with low KE) $\endgroup$
    – LWilkinson
    Mar 16 at 12:24
  • 1
    $\begingroup$ @LWilkinson I have extended my answer to address your comment. $\endgroup$ Mar 16 at 13:48

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