On new year's eve, after a splendid red and an assortment of sumptuous repasts, I made a bold remark which, on further consideration, may turn out to be incorrect. Unless! Unless I can concoct an impressive scientific explanation - for which I will need your input


  • Temperature is the average kinetic energy of a "bulk" liquid
  • Particles within the liquid will have a range of kinetic energies according to kinetic molecular theory (Maxwell-Boltzmann curve etc.)
  • Water boils at 100 deg Celsius (101.3kpa etc)
  • Evaporation can happen at any temperature (>273K)
  • Individual particles don't have temperature but rather kinetic energy


Assuming we could measure the kinetic energy of multiple evaporating water molecules just as they left the surface of liquid water (at standard temp and pressure) over time, would the average kinetic energy of the sample of evaporating water molecules equal the average kinetic energy of boiling water (at 101.3kpa)? That is, if the evaporating water molecules had a temperature - would it be approximately one hundred degrees. Or, as I may have put it at the time, at a molecular level, do my drying undies effectively boil? (Note: I understand that the bulk water is at room temperature)


If I knew the speed of evaporating water molecules I could calculate their energy and compare this to the average energy of boiling water and see if they were similar - but I can't see how to estimate the speed of evaporating water molecules?


...I have read "Is it true that an evaporating molecule has the same kinetic energy as a molecule in a pot of boiling water?" on this site. I don't think it answers this question.

Diagram for discussion

enter image description here

If I assume that Boltzmann's distribution works for liquids (I get that it's meant for ideal gases) and assume 9 degrees of freedom for water molecules then:

Rough and dirty

This suggests (if it's even remotely correct) that a very small number of molecules in room temperature water are moving very fast and therefore are at very high temperatures? It's not a relationship per se but, if correct, it does affirm the initial idea that evaporating water molecules are "hot"...even boiling?

Final Comments?

Thanks for the input and careful consideration. The chart below is my attempt to summarise my thoughts inspired by your comments. It's rendered in excel from the equation shown and accords well with the chart for water included in Boltzmann distribution for water which didn't extend far enough on the x axis for what we're trying to show here. Thanks for the heads up re energy distribution rather than speed - much easier to understand.

Boltzmann Energy Distribution in Excel

The equation comes from BC Campus Molecular Speeds I have ignored degrees of freedom effects in this and the other equations on the chart.

Clearly the energy of evaporating water molecules (let's say liquid immediately before take off), at room temperature (25 deg) is significantly higher than the average energy of boiling water.

What can be said about the little molecules about to liberate themselves from my drying undies then? The original question precipitated from the idea that molecules evaporating from washing 'boiled'.

  • They are at similar energy, a little greater in fact, than the energy of molecules about to jump from a pot of boiling water (44kJ vs 41kJ, 7% difference), and massively more energetic than the average energy of boiling water (4.7kJ)
  • If we could measure the temperature of a bunch of them they would be at an energy equivalent temperature some ten fold greater (3527K/373K) than the average temperature of boiling water (T = 2E/3Nk, bc campus university physics internal energy), but close to the equivalent (theoretical) temperature of the 'boiling' molecules. I know we can't measure their temperature as it is related to the average energy of the bulk liquid - but theoretically... (there's a case of beer in this)
  • The proportion of molecules ready to let loose from my y-fronts is much smaller per unit of room temperature water than it would be for boiling water (see the area under the curve right of the 41kJ and 44kJ points)

So...to all intents and purposes, the undie vaporizing molecules are doing what the boiling water vaporizing molecules are doing ....boiling (but they're not at 100 degrees, and we can't really measure their temperature - maybe half a case of beer then?). We don't feel the heat of the highly excited little blighters because it's a (relatively) small number of individual molecules within a bulk liquid at an average temperature of 25 degrees.

Other Stuff

In coming to the answer from a kinetic energy perspective, I found this paper Velocity of a droplet evaporated from waterwhich measures and models the speed of protonated water molecules evaporated from nano droplets. Speeds of non-protonated molecules in bulk liquid water will vary but it was encouraging to see that the speeds (say 2000m/s) were of an order of magnitude commensurate with with Boltzmann's distribution for water in Boltzmann distribution for water. On this curve, the experimental speeds would also be out in the flat area of the curve equivalent to my energy curve above.

  • $\begingroup$ chemistry.stackexchange.com/questions/88803/… $\endgroup$ – Alchimista Jan 2 at 16:18
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    $\begingroup$ Does this answer your question? Kinetic energy of molecules in liquid state? $\endgroup$ – Alchimista Jan 2 at 16:19
  • 1
    $\begingroup$ You do know the average kinetic energy of a gaseous water molecule if you know the temperature of the gas ... $\endgroup$ – Karl Jan 2 at 17:08
  • $\begingroup$ This Temperature is the average kinetic energy of a "bulk" liquid is false. $\endgroup$ – Poutnik Jan 2 at 18:18
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    $\begingroup$ At equilibrium, average kinetic energy of evaporating molecules is equal to average kinetic energy of condensing molecules. If not, it would not be possible for vapour and liquid to be in thermal equilibrium. This energy is lower for room temperature, compared e.g. with boiling water temperature. $\endgroup$ – Poutnik Jan 3 at 13:58

[OP] the average kinetic energy of evaporating water molecules

You have to specify whether you are talking about the kinetic energy just before the water molecule breaks the hydrogen bonds to its neighbors or just afterwards. A millisecond before or after the event, of course, the average kinetic energy will be determined by the bulk temperature.

One way to picture this is two water molecules "colliding" at the surface of the liquid. In order for one water molecule to leave the liquid phase, the collision needs to have at least sufficient energy to break the hydrogen bonds (and perhaps some more to overcome an activation energy). When that happens, the total kinetic energy of the molecules in the collision will be lowered (because energy is conserved). What exactly we mean by "collision" in the liquid phase is not so important because the temperature-dependence of the kinetics of the process is still governed by the Arrhenius equation (and the Boltzmann distribution of collision energies) even when we go from simple mono-atomic gases to wicked-complicated liquids.

The energy required to break the water loose (~40 kJ/mol, according to Poutnik's answer) is much higher than the median collision energy at room temperature (~2 kJ/mol). Whether the water molecule has more or less kinetic energy than the average after it enters the gas phase is not so important. The important part is that it had an unusually high energy, and most of that excess energy went into breaking the hydrogen bonds.

[OP] Assuming we could measure the kinetic energy of multiple evaporating water molecules just as they left the surface of liquid water (at standard temp and pressure) over time, would the average kinetic energy of the sample of evaporating water molecules equal the average kinetic energy of boiling water (at 101.3kpa)? https://chemistry.stackexchange.com/a/121656/

The kinetic energy of the water molecules that just evaporated is not exceptionally high or low. As the molecule leaves the liquid phase, the available energy is somehow partitioned between that water molecules and the ones it leaves behind. Your question boils down to whether the evaporative cooling of the liquid is temperature-dependent. I'm not sure if there is an experiment that would address this, and whether it has been done already.

[OP] Diagram for discussion

The diagram has a lot of problems, so I would not use it for discussion. The curve does not change that much with temperature, the shaded area is too big by far (according to Poutnik's answer, a molecule on the surface of the liquid has about a one in a million chance per collision to get into the gas phase at 373 K). Here is a diagram showing how the energy available from collisions changes when the temperature is raised from room temperature to body temperature. You can see that the changes for the bulk of energies are subtle. If you zoom in to the regions of exceptionally high collision energies, however, you can see that the changes are substantial.

enter image description here

  • $\begingroup$ Thanks. Good call. Found and then created similar chart from Boltzmann as shown in edit. Much appreciated. Will take the liquid - just before evaporation option. $\endgroup$ – Jeff Jan 16 at 10:08

At gas/liquid phase equilibrium, average kinetic energy of evaporating molecules, i.e. those just passed to a gas phase, is equal to average kinetic energy of condensing molecules.

The latter is then approximately proportional to $T$.

If these average values were not equal, the system would not be in thermal equilibrium.

Feedback to comments:

@theorist In case the gas phase consists exclusively ( or almost ) from water molecules, then majority of gas molecules collisions with liquid water leads for them to becoming a part of the liquid.

But still, if the mean kinetic energy of condensing molecules differs from the mean kinetic energy of molecules that during the same time interval evaporated liquid, Houston has a problem.

Those collisions that are not part of mass exchange would have to cause nonzero thermal transfer, what is not possible when temperatures of phases is equal.

BTW, I do not think, imply nor say that liquid leaving probability is KE independent. I suppose there is a KE threshold to leave liquid, to overcame the mean bonding energy. OR, perhaps, rather a threshold of the normal component of the velocity vector ( or equivalent energy for 1 degree of freedom ), as the molecule motion has to be properly oriented.

I have created a quick and dirty model to estimate the percentage of vapour molecules that condense after hitting the liquid phase.

I assumed the Boltzman energy distribution for 1 degree of freedom - the normal direction wrt the liquid surface. Then I recalculated the molar water enthalpy of evaporation to 1 water molecule as the energy to overcome when evaporating and calculated the fraction of molecules with enough energy in this degree of freedom.

The water molar enthalpy of evaporation: $\Delta H_{w,mol,evap}=\text{40.7 kJ/mol}$

The Boltzmann statistic: $\exp{\left(-\frac{\Delta H_{w,mol,evap}}{N_\mathrm{A} \cdot kT}\right)} \approx \exp{\left(-\frac{4895}{T}\right)}$

Than I have calculated saturated vapour pressure and related vapour density from the Clausius-Clapeyron equation.

Then I estimated the condensing fraction as the Boltzmann fraction multiplied by water/vapour density ratio.

But still, it seems to me the system would be just much closer to equilibrium than the prior assuming of all condensing vapour molecules, that is not in equilibrium.

t$\pu{[^{\circ}C]}$ T$\pu{[K]}$ x(Boltzmann) p$\pu{[Pa]}$ Vap. Density$\pu{[kg/m^3]}$ Rel. density x condensing
0 273.15 1.65E-08 831 0.0066 6.60E-06 0.0025
10 283.15 3.11E-08 1566 0.0120 1.20E-05 0.0026
20 293.15 5.60E-08 2824 0.0209 2.09E-05 0.0027
30 303.15 9.71E-08 4899 0.0350 3.50E-05 0.0028
40 313.15 1.63E-07 8205 0.0568 5.68E-05 0.0029
50 323.15 2.64E-07 13309 0.0892 8.92E-05 0.0030
60 333.15 4.16E-07 20972 0.1364 1.36E-04 0.0030
70 343.15 6.38E-07 32182 0.2032 2.03E-04 0.0031
80 353.15 9.56E-07 48200 0.2957 2.96E-04 0.0032
90 363.15 1.40E-06 70603 0.4213 4.21E-04 0.0033
100 373.15 2.01E-06 101325 0.5884 5.88E-04 0.0034

Interesting would be the result, if

  • The Maxwell-Boltzmann statistics would be taken as the energy distribution model for liquid phase molecules
  • It would be taken its subset, fitting the Boltzmann 1-degree-of-freedom energy threshold criterium
  • It would be calculated the mean kinetic energy of this subset, then subtracted the heat of evaporation.
  • Finally, it would be calculated the equivalent temperature.
  • 1
    $\begingroup$ I don't believe that is the case. Thermal equilibrium can be maintained by the enormous number of ongoing collisions between non-condensing gas molecules (both air and water vapor water) and the surface of the liquid water. Besides, it seems non-physical that the likelihood of a molecule evaporating is independent of its KE. We know the likelihood of molecules reacting (getting over an activation barrier) depends on KE, because their KE needs to be converted to PE to break bonds. So why wouldn't this also apply to an evaporating molecule, which needs to break the water-water bonds to escape? $\endgroup$ – theorist Jan 4 at 5:08
  • $\begingroup$ "In case the gas phase consists exclusively ( or almost ) from water molecules, then majority of gas molecules collisions with liquid water leads for them to becoming a part of the liquid." What is your basis for that assertion? Yes, at equil., the rate of evaporation and the rate of condensation are equal. And if the vapor pressure is, say, 0.1 atm, and there are no other gases, the concentration of gas molecules at the liquid-vapor boundary will only be ~1/10,000 that of the liquid molecules. But that still doesn't mean that most of the gas-liquid collisions have to lead to condensation..... $\endgroup$ – theorist Jan 7 at 5:59
  • $\begingroup$ ....For instance, even with the enormous difference in density, it could still be that the rate of evaporation from the liquid phase is only 1/100 the rate at which gas molecules strike the liquid phase, in which case only 1% of the collisions of gas molecules with the liquid phase would lead to condensation. The remaining 99% of collisions could maintain the thermal equilibrium. $\endgroup$ – theorist Jan 7 at 6:03
  • $\begingroup$ @theorist About this 1%, you seem to be actually very close to real thing, see the incoming answer update after a while. But it would IMHO just mean the state would be 100 times closer to equilibrium, not that it would be in equilibrium. (BTW, I hope you have not taken the prior Vdp + pdV thing personally.) $\endgroup$ – Poutnik Jan 7 at 7:47
  • $\begingroup$ Your "interesting results" comments @Poutnik, was where I was headed too....see edit $\endgroup$ – Jeff Jan 9 at 5:52

This is a very interesting question!

Also, I don't believe it is a duplicate of the other cited questions, since those seem to concern the average kinetic energy of liquid vs. gaseous water, while the OP is asking whether those that evaporate from the liquid are at the tail of the distribution, and whether there's a relationship between the KE of this subpopulation and the average KE of those in boiling water.

I'll give a somewhat speculative answer, which might serve as a springboard for others to contribute more.

We know that, for chemical reactions, the subpopulation that reacts comes from those with sufficiently high energy (which can be both kinetic and potential) to get over the activation barrier.

Intuitively, one would expect that to be the same for evaporation—it's those molecules that are at the surface, and have sufficiently high KE, that evaporate. And the reason the evaporation rate increases with T is that a greater proportion of molecules meet that criterion.

So what you're asking is whether there is some relationship between the cutoff for sufficient KE to evaporate, and the KE needed for boiling.

I suspect the answer is no, since the boiling temp changes with ambient pressure, yet I don't believe the ambient pressure significantly affects the KE needed to evaporate (it has little effect on the chemical potential of the liquid water).

For instance, the boiling point of water at 1 atm is 100 C, while at 0.1 atm it is only 41 C. If the boiling point determined the KE needed to escape from the liquid to to the gas, then the KE needed at 100 C would be much greater than that needed at 41 C.

However, I don't believe the KE needed to escape the liquid would be significantly affected by ambient pressure because, as I mentioned above, there is little difference between the chemical potential of liquid water under 1 atm pressure and 0.1 atm pressure.

Rather, it appears that, to the extent that the ambient pressure does affect the evaporation rate, it's because reduced ambient pressure reduces the density of the ambient gas and thus increases the diffusion rate, and hence the evaporation rate. See:

Sefiane, K., et al. "On the effect of the atmosphere on the evaporation of sessile droplets of water." Physics of Fluids 21.6 (2009): 062101.


  • $\begingroup$ Thanks @theorist. Precisely; is a "relationship between the cutoff for sufficient KE to evaporate, and the KE needed for boiling." I will need to consider your in depth response further. Off the cuff I would have thought a lower air pressure would reduce the energy required for particles to evaporate at room temperature. (KE = 3/2kT, PV = nRT, so less P = less T = reduced kinetic energy). Or, less weight of air pushing on the liquid surface means less energy is required for particles to escape. Not sure, will need to cogitate on your thoughts further. $\endgroup$ – Jeff Jan 3 at 13:16
  • $\begingroup$ @Jeff Less p does not mean less T here. T can be whatever you designate it to be. There are other variables in pV = nRT. Think about what you're comparing when you want to understand the effect of changing ambient pressure: You're comparing the rate of water evaportating at some T (say 25 C) under 1 atm pressure, with the rate of water evaporating at the same T at a lower p. So you already know that, in this comparison, when you are changing p, you are not changing T. $\endgroup$ – theorist Jan 4 at 5:14
  • $\begingroup$ Pressure Evaporation happens faster if there is less exertion on the surface keeping the molecules from launching themselves. (Wiki). I was referring to your earlier comment that pressure may not affect the KE required to evaporate. I'm assuming that this Wiki comment infers that less pressure on the surface means less energy is required to escape. $\endgroup$ – Jeff Jan 9 at 5:47
  • $\begingroup$ @Jeff (a) I don't know what "pressure evaporation" is supposed to mean. (b) "Wiki" is not a reference. You need to give the actual link. Wikipedia could be talkng about pressure and evaporation in any of numerous diff. articles, so there's no way to tell which article you saw it in without a specific link. In the article I reference, the authors explain that ambient pressure from inert gases reduces the evaporation rate by reducing the diffusion rate, not by changing the KE needed to evaporate. It is true that high p can reduce the energy needed to evap., but it's a tiny effect. $\endgroup$ – theorist Jan 9 at 6:21
  • $\begingroup$ Thanks @Theorist. It was Pressure (heading). Evaporation happens......Link : en.wikipedia.org/wiki/Evaporation. Apologies, just getting used to this site. $\endgroup$ – Jeff Jan 10 at 9:06

How about looking at the partial pressures of two gases in a perfectly insulated pair of insulated bottles connected by a Maxwell's demon opening that can tell us which kind of gas molecule went through the opening between the bottles and which way the molecule was moving. Let us say the bottles are exactly the same size and there are in total 10 molecules of gas C and 10 molecules of gas H. In bottle A let us say there are x molecules of gas C and y molecules of gas H. In bottle B let us say there are 10-x molecules of gas C and 10-y molecules of gas H. Let us start with x=6, bottle A has 6x bottle B has 4x. For the flow through the Maxwell gate to be 0, the count of molecules on each side must be equal. Bottle A must have x=6 and y=4 while bottle B must have x=4 and y=6. Bottle A has 10 and Bottle B has 10 molecules each. In this very simple system, the temperature is the same in each bottle. If you increase the temperature of one bottle(a very small amount), warmer gas molecule will flow to the other bottle until the temperature on each side is again the same. Referring back to " do my drying undies effectively boil?" I think this model indicates whatever temperature they started at, they will converge on the same partial pressures as the surrounding environment.

  • $\begingroup$ Thanks Lee, a useful conceptualisation. I think they will too - as you've illustrated. Before that time, just prior to launching from the undies, what are the molecules doing? Have made some summaries in the edited question. Thanks for your help. $\endgroup$ – Jeff Jan 16 at 10:11

As the water body is colder, evaporating water obviously has a lower speed of gas particles as they leave the surface. If that were otherwise, evaporating water would cool tremendously, and at the same time heat the air above.

While it certainly does the former (minus the "tremendously"), the latter for sure does not happen.

The quantitative answer is that any water molecule that leaves the surface has just lost the evaporation enthalpy (strictly speaking its binding energy to the molecules around and below it). If it has any kinetic energy left, it can fly off.

  • $\begingroup$ Thanks @Karl. Help me understand how it's obvious that the molecules have lower speeds. According to Maxwell-Boltzmann, water, even at room temperature can have molecules travelling at very high speeds - with very high kinetic energies, even though the average (bulk liquid) kinetic energy is relatively low. You've nailed exactly the problem at hand I think - how to determine the speed and therefore kinetic energy of evaporating particles so it can be compared to the average kinetic energy of boiling water. The quantitative answer requires maths and I don't know what math to use. $\endgroup$ – Jeff Jan 3 at 13:28
  • $\begingroup$ @Jeff Note that the M-B distribution is applicable to gases only. For liquids, it is just a very rough approximation. $\endgroup$ – Poutnik Jan 3 at 15:08

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