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When a cup of water evaporates into air, water molecules collide near the water-air surface in such a way that enables one of the water molecules to escape the water surface. In each such collision, a little kinetic energy $\Delta KE$ is transferred to the energetic molecule that escapes. This process does not require an addition of heat $Q$ from the surroundings, because the kinetic energy $\Delta KE$ is transferred between two water molecules and is not transferred from the surroundings.

This post and this post seem to state that the process of evaporation increases the total entropy of system + the surroundings. The posts/answers in those links offer a variety of ways of explaining this, including that (a) the increase in total entropy is due to a difference in chemical potential µ, and (b) the increase in total entropy is due to the fact that gases have a higher specific entropy. I believe I read elsewhere that gases generally have more microstates than liquids, which would also why the total entropy increases during the transformation of a liquid into a gas.

My question doesn't relate to these answers, but instead is about how the equation $dS=\frac{dQ}{T}$ applies to evaporation. In particular: how does this equation explain why total entropy increases during evaporation? What are the correct values for $dQ$?

One guess that I have is: $dQ$ equals the kinetic energy $\Delta KE$ transferred to the energetic water molecule during the collision at the surface. On the one hand, substituting $\Delta KE$ in for $dQ$ makes some sense to me because I don't know of any other energy exchanges that occur during evaporation except the transfer of $\Delta KE$. On the other hand, my statement $dQ = \Delta KE$ seems wrong because $dQ$ is supposed to be the heat transferred between the surroundings and the system, and $\Delta KE$ is energy transferred within the system. (Only once evaporation occurs does the escaped particle become part of the 'surroundings.')

To clarify: I want to understand how the equation explains the increase in total entropy of the system + surroundings. Let's define the 'system' to be the liquid water in the cup. I think this means the 'surroundings' would be everything else (air + water vapor).

Please also correct me if (a) I am incorrect in stating that evaporation does not require an input of energy from the surroundings or (b) I am incorrect in stating that total entropy of the system + surroundings increases during evaporation. My question is based on those two statements about evaporation.

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  • $\begingroup$ Are you referring to evaporation into the air? In using the equation you refer to, you must identify an initial state and a final state of a system, and determine the change in entropy between these states. Suppose you have liquid water with bone dry air above it in a container as state 1, and, as state 2, you have liquid water with air saturated with water vapor above it. Can you think of an adiabatic reversible process that can take you from state 1 to state 2, say, involving semipermeable membranes? $\endgroup$ – Chet Miller Apr 18 '17 at 12:43
  • $\begingroup$ Thanks for your help and for clarifying! You are right; I mean to ask about evaporation of water into air. Let's consider escape of a single particle: the initial state is when the particle is in the and the final state is when the particle is in the air. I believe this text (people.sca.uqam.ca/~enrico/sca5002/Iribarne/iribarne3.pdf) states that dQ = 0 in the scenario you described involving semipermeable membranes (p. 48-49, under 3.10). Is that the answer--dS = 0 in the process I described? Would this also mean dQ = 0? $\endgroup$ – jdphysics Apr 19 '17 at 1:39
  • $\begingroup$ It doesn't make sense (to me) to talk about single molecules when discussing entropy of macroscopic systems. In the scenario I described in my previous comment, the entropy definitely increases from the initial state of the system to the final state. $\endgroup$ – Chet Miller Apr 19 '17 at 11:40
  • $\begingroup$ Doesn't your final statement contradict the text I cited? If there is no heat transfer then dQ = 0 and hence dS = 0. The statement that dS = 0 also agrees with the following statement about reversible processes: "A reversible process changes the state of a system in such a way that the net change in the combined entropy of the system and its surroundings is zero." en.wikipedia.org/wiki/… $\endgroup$ – jdphysics Apr 20 '17 at 2:40
  • $\begingroup$ Do you mean that we can't discuss entropy unless we consider evaporation of the entire container of water? Or do you mean that there is some threshold of particles that must evaporate in order to invoke the concept of entropy in a sensible way? Can we facilitate the discussion by assuming that a quarter of the water evaporates? $\endgroup$ – jdphysics Apr 20 '17 at 2:43
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how the equation dS=dQT applies to evaporation. In particular: how does this equation explain why total entropy increases during evaporation?

Consider closed container of fixed volume that contains liquid water and less than equilibrium amount of water vapor.

The amount of vapor will increase and the entropy of the system will increase as well. Intuitively, this is because the system gets closer to equilibrium state. Mathematically, this can be derived as increase in the Boltzmann entropy of the system; when an element of liquid evaporates, the resulting macrostate with more molecules in vapor is consistent with more microstates than before.

The equation $dS=dQ/T$ is not really applicable; first, because there is no heat transfer to the container; second, because the process is not thermodynamically reversible.

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  • $\begingroup$ Thank you! One small clarification: when you say the entropy of the system increases, are you considering the system to be everything in the closed container? With Boltzmann entropy, am I correct that it's not as useful to distinguish between (i) the entropy of the liquid water and (ii) the entropy of the vapor + air? $\endgroup$ – jdphysics Apr 20 '17 at 23:37
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    $\begingroup$ I meant entropy of the whole system liquid+vapor in the container. Entropy of the liquid alone probably decreases in such a process, as the liquid loses particles and gets colder. The distinction between entropy of liquid part and entropy of the gaseous part is important, they are two different entropies. $\endgroup$ – Ján Lalinský Apr 21 '17 at 20:45
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I would think $\delta Q$ is heat added or extracted from the system (here vapor / liquid mixture). Introduction of kinetic energy just get thing complicated.

When heat is added to the system, the phase moves toward more vapor (more micro state as free to move and rotate in 3D space) thus more entropy. And if heat is removed, the phase moves towards more liquid (more order or less micro state) and so less entropy.

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  • $\begingroup$ Thanks for the answer and the help! Is my question based on a major misconception about evaporation? Specifically, does evaporation require any input of energy from the air into the water? I'm imaging a scenario where a single water particle escapes by virtue of a collision with another water molecule, not by virtue of any heat exchanged from the air to the water molecules. Does my question make an incorrect assumption about how evaporation occurs? $\endgroup$ – jdphysics Apr 19 '17 at 1:52
  • $\begingroup$ I guess you are trying to understand the mechanism of evaporation, which may not need to use entropy. Entropy is used for a system but not a single or multiple moving particles. $\endgroup$ – user115350 Apr 19 '17 at 5:23
  • $\begingroup$ I have a working understanding of evaporation, but reading your comment made me wonder if my existing understanding was incorrect. I thought you suggested that evaporation requires an input of energy from the surroundings, which conflicts with what I previously understood about evaporation. My question is: can the equation dS = dQ/T explain the entropy increase that occurs during evaporation? While the process of evaporation occurs one particle at a time, I still want to talk about the entropy of the whole system (i.e., the water). $\endgroup$ – jdphysics Apr 20 '17 at 2:48

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