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Why water evaporates below 100 ºC

  1. Dec 27, 2007 #1
    Hello everyone.

    There is a question I have been thinking about for a time. Let's state it the easy way:

    Let's suppose my room is a closed system with constant pressure (1 atm) and temperature (standard 25º C). Under these conditions, if I put a container with water in my room, it will end up empty because water have evaporated. But according to the phase diagram of water, water vapor should not exists as an equilibrium state at those conditions. So, what is wrong?

    I understand evaporation (I know the difference with boiling) as a surface phonemenon, but once the system reachs the equilibrium, the water should be liquid as indicated by the phase diagram.


  2. jcsd
  3. Dec 27, 2007 #2

    D H

    Staff: Mentor

    The phase diagram says water should not exist as a liquid above 1000C at 1 atmosphere of pressure. It does not say that water should not exist as a vapor below 1000C.

    What is happening is that the molecules in the water are "bouncing" around a lot. Some of the molecules have a low velocity while others have a high velocity. The Maxwell-Boltzmann distribution describes the distribution of velocity as a function of temperature. While the average velocity of the molecules in a flask of water at 250C is well below that required to escape the surface of the water, some of the molecules do have sufficient velocity to escape the surface, and that is exactly what they do.
  4. Dec 27, 2007 #3


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    It's due to the uneven distribution of energy among the water molecules at any given time caused by collisons between water moecules due to their random motion. A water moelcule near the surface can acquire enoguh energy due to these random collisions to escape as water vapour.

    Losing the most energetic molecules due to eavporuation will lower the mean energy of the water molecules, meaning temperature of the water will lower which is why evapouration causes the remaining liquid to cool.
  5. Dec 28, 2007 #4
    The phase diagram do says that at 1 atm and 25 ºC the equilibrium state of water is in liquid phase.

    As for both replys: I know evaporation is a surface phenomena different from boling; however, no matter which process it is, the final state is that we get water vapor at 25 ºC and 1 atm, against predeicted by the phase diagram.

    Any more suggestions?
  6. Dec 28, 2007 #5

    D H

    Staff: Mentor

    You are reading the phase transition diagram a bit too naively. The phase transition diagram says that at 1 atm and 25oC a vessel that contains H2O in which the water vapor pressure is at the saturation level will have a liquid phase. It does not say all of the H2O molecules in the vessel will be in the liquid phase.
  7. Dec 28, 2007 #6
    At 25 ºC and 1 atm in that vessel will exists one and only one state of equilibrium, right? Draw that state of equilibrium for each T and P and you have the phase diagram. Ok, now the phase diagram says that a 25ºC and 1 atm what you have in the vessel is liquid water, not vapor, and, hence it makes no sense to talk about water at the saturation level, simply because it is in liquid zone.

    So I insist, how does this reconciliate with the fact that water evaporates at that conditions? Well I think I'm going toward the solution which has to be with the presence of other components in the air. That is, we are dealing with a system with 2 components (H2O and N2) and 2 phases. So chemical potentials will have something to say, but I dont have the correct solution yet.

    When people are asked about how can water evaporates at 25 ºC most people tend to give the surface evaporation phenomena explanation. This is, however, not the question. The question has nothing to do with the explanation of mollecular phenomena (Maxwell distribution of velocities of whatever). We are dealing here only with states of equilibrium, no matter how those states are reached. And what should shock a first glance is that the equilibruim state contains water vapor at ambiental conditions, which does not agree with the water state of equilibrium at those conditions. As i said, I suspect that the phase diagram of water as a PURE SUBSTANCE is not directly applicable when more substances are involved.

    If any has the closed solution, please let me know.
  8. Dec 28, 2007 #7

    D H

    Staff: Mentor

    By the rest of your post, I assume you mean one and only one phase, and that is simply wrong. Partially fill a vessel with water, vent the air at the top of the vessel to vacuum, and close the vessel. The vacuum portion of the vessel will not remain a vacuum for long. Water will evaporate until the pressure in the chamber reaches the partial pressure for water at 25oC. The chamber contains water vapor and water liquid. You're persistence in incorrectly interpreting the meaning of the phase diagram is clouding your understanding (pun intended).

    Wrong again. Take my previous example of a vessel comprising water and a vacuum. Now instead of closing the valve immediately, open another valve that lets dessicated air (or helium, or whatever) in to the upper part of the vessel. Liquid water will still evaporate until the partial pressure of the H2O in the gas side of the vessel reaches the temperature-dependent saturation vapor pressure.
  9. Dec 28, 2007 #8


    Staff: Mentor

    Rather than looking simply at the phase diagram you need to look at the saturated vapor pressure. The boiling point of water is 100ºC because of the fact that at 100ºC the saturated vapor pressure is 1 atm. However, at lower temperatures there is still a vapor pressure. Assuming your room is a closed system then the container with water will evaporate until the partial pressure of water vapor in the room is equal to the saturated vapor pressure (23.76 mmHg at 25ºC).

    Last edited: Dec 28, 2007
  10. Dec 28, 2007 #9


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    To the OP: Remember that water can evaporate from solid phase directly to vapor without going through the liquid phase, under the right conditions. Look up "sublimation".
  11. Dec 28, 2007 #10
    I should think the answer is quite simple. Any liquid will have a vapor pressure which depends on only the temperature. Whenever the ambient pressure is equal to the vapor pressure at the given tempeature, then the boiling phenomenon occurs. Water has vapor pressure of 23.7 mmHg at 25oC.
    A water tank in a room at 25oC and 1 atm will finally evaporate because the water vapor will condense at any surfaces inside the room. The amount of water vapor in the air is constant (of course if the room is gas tight).
    If the water is replaced by motor oil, the same phenomenon happens, but it would take a very long time, which I guess tens of years or even hundreds.
  12. Dec 28, 2007 #11
    You should clarify your understanding of thermodynamic concepts instead of telling my insistence is wrong. It seems to me that you talk more like an engineer than like a physics.

    No. What a phase diagram represents is the existing phase(s) at THERMODYNAMICAL EQUILIBRIUM STATES of any substance. It doesn't matter if there exist one, two or n phases, the important is that you take two independent variables (T and P), fix them to a value and the diagram says you that AT THE EQUILIBRIUM STATE corresponding that T and P, the phase(s) corresponding to that substance is this or that (liquid, vapor, liquid-vapor. triple point, etc).

    So, take T=300K, P=1 atm and by no means will you get a vessel which contains ONLY water at these conditions to contain vapor. You can get it if you vacuum the vessel, but not at 300K and 1 atm!!!.

    Take a look the images I attach. The first one is the phase diagram for water without concrete values on the axis. The second one tells that at 1 atm and 300K WATER CAN EXIST only as liquid once the equilibrium has been reached.

    Attached Files:

  13. Dec 28, 2007 #12

    D H

    Staff: Mentor

    Yes, you will. Lets put the empty vessel in a room with completely dessicated air (zero percent humidity). Now partially fill the vessel with liquid water and close the vessel off. The pressure in the vessel will rise to 1.03 atmospheres as some of the water in the vessel evaporates. If you only put a small amount of liquid water in the vessel (i.e., not enough to bring the vapor pressure up to 0.03 atmospheres), all of the water will evaporate, leaving no water in the liquid phase even though the temperature is well below the boiling point.
  14. Dec 28, 2007 #13
    I think the answer goes more or less like this:

    For equilibrium to exists, not only must be temperature and pressure definited (uniform trough the system) but also the chemical potential of a component in each phase must be equal (Gibbs rule).

    Take a closed room at 300K and 1 atm inside of which there is a open vessel of water. We have two components H2O and N2 (supposing dry air is mostly N2) and two phases: gas and liquid.

    Ok, now if there weren't N2 the water would reach the equilibrium state which, accorrding to water phase diagram, would be liquid water. This is what "people" would expect because they don't account for the existence of N2 and, as said in elementary school, water is liquid at 300K.

    Now let's account for the existence of N2. As many times in thermodynamics, lets think of a constrained equilibrium state, that is, we imagine vapor water and liquid water separated chemically by mean of a diathermal and mechanical wall which allow mechanical and thermal equilibrium. With this in mind let's answer some questions:

    First of all, why can water vapor exist at that pressure and temperature? Well, the key point is that water vapor is surrounded by a pressure of 1 atm, but its parcial pressure is much lower than that; and, according to the phase diagram, water vapor at 300K can exist when the pressure is small enough. So, water vapor exists at that (parcial) pressure.

    Now, if we remove the constrain the system will evolve to its global equilibrium state (Zero Law of equilibrium principle of thermodynamics). Note that we come from a constrained equilibrium state, not a global one. That is, the whole system was not at global equilibrium (i.e., not at its maximum possible entropy among constrained equilibrium states) but each part was at equilibrium. Ok, now we know that the chemical potential of vapor water overheated over its saturated level is lower than that of liquid water at 1 atm, so liquid water evaporates in order for chemical equilibrium to exists.

    Well, more or less the solution wold have to follow this way, but there is some dark points already, like how to know which chemical potential is actually lower and which higher.

    Someone can complete the explanation?
    Last edited: Dec 28, 2007
  15. Dec 28, 2007 #14

    D H

    Staff: Mentor

    What do you think the term relative humidity means? Vapor pressure? Partial pressure?

    Google these, please. I give up.
  16. Dec 28, 2007 #15
    "Dessicated air" -> so you are taking N2 to the experiment also. I said only water, only H2O. Carry out that experiment in a room with contain nothing else than H2O.

    As the book of Kestin says, "no matter how small amount of an inert gas you push into a vessel containing vapor-liquid equilibrium water, the water will evaporte." So, what you say is just the experiment of Kelvin-Helmholtz, i.e., perturbation of vapor-liquid equilibrium with a gas. But, if there is no added gas, there will be no vapor water at 300K and 1 atm. These last statement is pretty clear anyway, I don't know if you are trying to find water vapor in any single condition, but at equilibrium state, water alone does not exists as vapor at 300K and 1 atm.

    Nevertheless, you are getting closer to the real core of the problem: the existence of another substance (N2) in the room.
  17. Dec 28, 2007 #16

    D H

    Staff: Mentor

    Where in the heck do you think the 1 atmosphere comes from? The ether? Of course water alone does not exist as vapor at 1 atmosphere at 300K. I never said it did. Water does exist as vapor at 0.03 atmospheres at 300K, and liquid water will evaporate to achieve that 0.03 atmospheres if the partial pressure of water is less than 0.03 atmospheres.

    Liquid water will continue to evaporate after the partial pressure reaches 0.03 atmospheres, but gaseous water will condense into liquid at the same rate. That is your equilibrium point at 25oC.

    I have given and others have given several examples of and several reasons why your reasoning is wrong.
  18. Dec 28, 2007 #17
    My reasoning is right, I dont have doubts about what I have argued. That is not what worry me.

    What I would to like to find is an explanation of WHY the state of equilibrium at 300K and 1 atm is what we all know it is.

    You have just given me descriptions of what everyone now that happens, but not the reasons on the ground of rigurous thermodynamics. I want some who say "well, the reason why the system reaches this equilibrium state and water evaporate is because the Gibbs potential of vapor in presence of N2... bla bla bla"

    There is a big difference between actual last-causes or theoretical explanations and just enumeration of facts that is what you have provided so far. One have only understood the underlying physics when one is capable to answer "this happens because" and not just "this happens and this other also happen" and so on.

    I know what happens when you put water in a vacuum vessel; I know where pressure comes from, I know those thing that anyone not ver intelligent would know. But, up to now, no one has been capable of explain WHY water MUST be in vapor state.

    When I read a thermodynamic textbook, a good one such as Kestin, Callen, Atkins or Zemansky one problem could be "Given N2 and H2O in a room, proove that the equilibrium state at 300k will be N2 and H2O vapor" Then, on the ground of the theoretical lessons I would proove the assertion. However, in no place of the book is said "vapor pressure of water in pressence of N2 this or that". You just can deduce it from the very basic principles and mathematics.

    Barely could you write a book with the enumeration of facts you have provided. It is nothing against you, I just want a theoretical explanation and not the dumb-engineer recipe. And, I believe strongly that such an explanation is impossibe if you nor even touch the concept of Gibbs chemical potential.

    So, I wait for what seem to be a unknown explanation.

    By the way, I bet you are an engineer.
  19. Dec 28, 2007 #18

    D H

    Staff: Mentor

    Your first two posts (the OP and #4) illustrated to me and to others an incredible lack of understanding. Freshman level at best, and a poor understanding of freshman-level concepts. We don't know you; we can only infer your level of understanding based on what you post. Given the low caliber of those first few posts, we explained what was going on in simple terms.

    If you have a serious question don't come in looking like a fool.

    It now appears that you are into psychoceramics and have a hidden agenda. We don't tolerate that here.
  20. Dec 28, 2007 #19
    Ok you are right at to some extent, but not totally.

    I admitt that my first posts belied my level of knowledge. Im sorry if that leaded you to think I was at freshman level. Anyway, I didnt mean this to get to you so strongly. In any case Im sorry for that.

    However I think it became clear that I am looking for something deeper that just a recipe. So, there is a piece of guilty in you also.

    Really, I dont want to bother you and I'm sorry if I have. I hope to keep talking to you sharing our interest in physics.

    And, once again, if anyonce could shed light on my question I would be grateful.
  21. Dec 28, 2007 #20


    Staff: Mentor

    What is wrong with being an engineer? This is the second time you have said that as though "engineer" were an insult.

    You have had essentially the same answer to your question from 4 or 5 different people now, and your responses to DH have become progressively more deliberately ignorant, impolite, and immature.

    Do some studying and answer your own questions from now on. It is clear that you need to learn the hard way.
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