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Suppose I leave a glass of water on a table at STP. The water molecules at the surface of the water can be assumed to obey the Maxwell-Boltzmann distribution (at the least they obey a distribution of speeds comparable to the MB distribution). Only the most energetic molecules, near the surface of the liquid, have enough kinetic energy to overcome the attractive inter-molecular forces present in the liquid. As these highly energetic molecules escape from the liquid (i.e., evaporate), the average kinetic energy of the molecules in the liquid decreases. The temperature of the water is nothing but a measure of the average kinetic energy of the molecules in that water and hence the waters temperature decreases. This is effectively the mechanism behind sweating.

But now, if only the most energetic molecules were able to escape the inter-molecular forces and end up as water vapour, does this not mean that the water vapour above the surface now has a higher temperature than that of the water beneath it since the vapour is composed only of the most energetic molecules and the remaining liquid contains only the least energetic molecules? If this is the case, then surely it is an example of energy flowing from a cold sink to a hot sink? I realise this is impossible and so my thinking is definitely incorrect although I can't seem to build an intuition for why I am incorrect. All I can do is simply state the second law and tell my self I'm wrong. But that's no way to understand so if anyone can help me out on this issue it would be most appreciated!

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  • $\begingroup$ how come you say that energy flowed from lower temperature to a higher one ? Temperature of the surrounding increased after the molecules evaporated. Had evaporation not occurred temperature would not have increased. And after evaporation, temperature of glass decreased while of surrounding increased. $\endgroup$
    – Ankit
    Feb 17, 2021 at 9:05

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But now, if only the most energetic molecules were able to escape the inter-molecular forces and end up as water vapour, does this not mean that the water vapour above the surface now has a higher temperature than that of the water beneath it since the vapour is composed only of the most energetic molecules and the remaining liquid contains only the least energetic molecules? [emph added]

No. Some energy went into breaking the intermolecular bonds of the liquid to obtain the gaseous state. At equilibrium, the temperatures, pressures, and chemical potentials (meaning the partial pressures) of the liquid and vapor are identical. This is actually a great example of the Second Law at work: gradients in the intensive variables (temperature, pressure, chemical potential) are eliminated through shifting and exchange of the corresponding extensive variables (entropy, volume, matter).

Note that this equilibrium temperature (of the liquid and gas) will be lower than the original temperature of the liquid. The reason is that the molecules in the liquid were bonded to each other to some degree, meaning that they were in a low-energy state relative to a gas. Upon evaporation, energy was required to break these bonds, and—assuming a simple isolated system—this energy could come only from the thermal energy of the substance. So yes, you might that the liquid originally at 25°C is now a liquid–gas mixture at 24.8°C.

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  • $\begingroup$ Thanks for the response! Okay that makes reasonable sense to me. Is it fair to view those molecules that are energetic enough to escape as particles at the bottom of a potential well (at the equilib position of the well) which posses kinetic energy greater than the limit of the potential energy as $r$->$\infty$ and hence when they arrive at infinity they still have some kinetic energy left but significantly less than before they escaped the well? Also, its clear to me that the water must decrease in temp when evaporation occurs. does this mean that the vapor is also at that reduced temp? $\endgroup$
    – SalahTheGoat
    Feb 16, 2021 at 18:57
  • $\begingroup$ (1/n) Is my thinking correct that the evaporated molecules should have the same temperature as the liquid they are evaporating from? Say we have water in a vacuum at 25C. The most energetic molecules will boil off expending some of their kinetic energy in favour of greater potential energy. Thus the liquid cools to say 24.9C(as its lost energetic molecules) but the vapor is also at the same temp as it used up kinetic energy to gain potential energy and separate into a gas. More molecules continue to boil off but these have to overcome greater potential energy and hence have... $\endgroup$
    – SalahTheGoat
    Feb 17, 2021 at 9:14
  • $\begingroup$ (2/n) a lower kinetic energy when they successfully escape into gas form. Thus the water will now have a temp of say 24.8C (since its cooled due to more energetic molecules escaping) but the vapor will now also have a temp of 24.8C because the recently boiled off molecules had to overcome greater potential and hence have lower kinetic energy than those that boiled off first when the water was still at 25C? Hope this all makes sense $\endgroup$
    – SalahTheGoat
    Feb 17, 2021 at 9:17
  • $\begingroup$ I edited my answer to discuss the temperature change. $\endgroup$
    – Chemomechanics
    Feb 17, 2021 at 19:19
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There is no violation of the second law here -- you're neglecting that a glass of water is an open system.

In other words, it interacts with the environment not only via "heat transfer" but also via mass transfer as well.

So, while heat is transferred from a "colder/liquid" phase to a "hotter/vapor" phase, the entropy lost by the liquid water is more than offset by the entropy gained by the environment from the contribution of higher-energy water molecules that got vaporized away into it (away from the liquid phase). So ultimately there is a net increase in the "total entropy" of universe, fully consistent with the Second Law.

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    $\begingroup$ @SalahTheGoat Yes -- the system (i.e., just the water molecules, to be precise) does zero work ($W=0$) expanding against a vacuum. And since the container is adiabatic as well ($Q=0$), then by the First Law it follows that $\Delta U=Q−W=0$ so there is no net change in the internal energy of the system (as a whole). Now, if it were an ideal gas, this would simply be an irreversible isothermal expansion generating $\Delta S=nR\ln\left(V_f/V_i\right)$ of entropy. $\endgroup$
    – ManRow
    Feb 17, 2021 at 9:02
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    $\begingroup$ But for a real gas like water vapor, this expansion will reduce the temperature of the system if done below its inversion temperature, indicating that some of the kinetic energy must be getting converted to potential energy instead. Either way, your water vapor molecules will have greater kinetic energy than those in the liquid phase, but the total internal energy of your system will still stay the same (per energy conservation). $\endgroup$
    – ManRow
    Feb 17, 2021 at 9:02
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    $\begingroup$ @SalahTheGoat First of all ideal gases (i.e., gas-phase molecules!) do not expend any energy expanding against a vacuum. It is really simply a free expansion (i.e., an irreversible process which generates entropy). The energy of a gas phase molecule does not change unless it transfers energy to its surroundings (doesn't apply in your case since the container is adiabatic, so no heat transfer with surroundings/environment), or it does some sort of work against an external pressure (but a vacuum has no pressure so no work is done either). $\endgroup$
    – ManRow
    Feb 17, 2021 at 9:33
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    $\begingroup$ Second of all, at the "phase transition" between liquid water and vapor, remember that at equilibrium (i.e., when saturation vapor pressure is attained), having a "vapor molecule" transfer some amount of heat $\delta Q$ to a "liquid molecule" will simply convert the vapor molecule (energy donor) to a liquid one and liquid molecule (energy recipient) to a vapor one -- so whatever "heat flow" you could have at equilibrium really doesn't change anything at all. $\endgroup$
    – ManRow
    Feb 17, 2021 at 9:33
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    $\begingroup$ So yes you can perfectly say "the vapor molecules heat up the liquid ones" or equivalently there is "heat flow" from the vapor to the liquid -- like a "dynamic equilbrium" -- except that, of course, in the end, we'll ultimately have no "net" change in the system occurring at all (so, "nothing really changes")! $\endgroup$
    – ManRow
    Feb 17, 2021 at 9:33

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