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I How can water evaporate?

  1. Apr 22, 2016 #1
    Some facts: Water does evaporate.
    The average kinetic energy of a water molecule at 373 K is ~0.03 eV
    The average strength of the H bond is ~0.24 eV
    In the liquid phase the average H2O molecule is bound by 2 to 4 H bonds.
    Therefore - it is bound in the liquid phase by bonds which are stronger then
    the agitational (kinetic) energy of the molecules.
    In order to evaporate these bonds must be broken. How is this possible?
    An eV is an energy unit equal to 1.6E-19 J
     
  2. jcsd
  3. Apr 22, 2016 #2

    DrClaude

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    Averages are meaningless here. What is important is that there is a distribution of kinetic energy, and some water molecules have sufficient energy to escape from the bulk of the liquid.

    Note: this is not an A level question. I have lowered it to I.
     
  4. Apr 22, 2016 #3

    CWatters

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    Kazys.. Have you wondered why evaporation cools liquids?
     
  5. Apr 22, 2016 #4
    It is an A level question because it has not been answered. Not just by you and me, by
    anybody. Your answer just restates the question in other words. Yes, indeed, some
    molecules do gain sufficient energy to escape, that is clear from the first stated
    condition -water does evaporate. But how does this happen?? What is the distribution
    of agitational energy of the water molecules? Strictly speaking not just the molecules
    the "clusters" because the H-bond is sufficiently strong that water molecules do
    act individually. A common answer is that the distribution is Maxwellian. That
    is not just an unproved assumption, it is by definition false.
     
  6. Apr 22, 2016 #5

    PeterDonis

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    An "A" level question requires graduate level knowledge of the subject matter. Your question does not indicate that level of knowledge.

    The distribution does not have to be Maxwellian to explain why some water molecules have sufficient energy to escape, i.e., to explain how it is possible for water to evaporate. All that has to be true is that the distribution is such that there are some water molecules with kinetic energy larger than the energy binding them to adjacent water molecules. There are many, many possible distributions that have that property.

    Also, even the requirement as I have stated it above is actually too strong. It is also possible for a water molecule to gain sufficient kinetic energy to escape as a result of a collision with an incoming molecule of N2 or O2 or some other gas contained in the air. So even if the distribution happened to be such that no water molecule, at some instant of time, had sufficient kinetic energy to escape due to the distribution, water molecules could still gain such energy as a result of collisions, and could therefore escape.

    I don't understand what you mean by this.
     
  7. Apr 22, 2016 #6

    Nugatory

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    Was that supposed to be "do NOT act individually"?

    In any case, do you have an accepted source or a calculation to back up the claim about the strength of the hydrogen bond between molecules?
     
  8. Apr 22, 2016 #7

    CWatters

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    How hard did you look?

    http://journals.aps.org/prl/abstract/10.1103/PhysRevLett.115.236102

     
  9. Apr 23, 2016 #8
    To PeterDonis
    It seems that I started this post improperly, for this I apologize. My intention was to illustrate that some common and very familiar
    phenomena are still not completely understood. The evaporation of water is an example. Water in the liquid phase is a very complex
    medium. The molecule is deceptively simple, but it has an exceptionally high dielectric constant, because it is so small and has
    a high H to O ratio it forms strong cohesive H bonds. As a result, in the liquid phase it does not exist as separate molecules,
    but as constantly reforming clusters of molecules. The distribution of agitational energy for molecules within these clusters
    and/or between the clusters is not known. You reflect this lack of knowledge when you state: "there are many possible distributions",
    Indeed so, but until the actual distribution is known the precise mechanism used by an escaping molecule cannot be determined.
    Re. the Maxwellian distribution - it provides a good approximation for the distribution of kinetic energy of water molecules in the
    vapor phase, by assumption it has often been applied also to the liquid phase, that is not correct. The "by definition" comment refers
    to the conditions that Maxwell used to derive a distribution of kinetic energy for a statistically large number of interacting entities.
    These are: interaction occurs between 2 entities at a time, these collisions are elastic, scattering is isotropic. None of
    these conditions apply for liquid water.
     
  10. Apr 23, 2016 #9
    To: CWaters

    Thank you very much for this reference. Knowledge exists at several levels of certitude. Knowledge obtained
    by using simulations of virtual models, as is the case in the noted study, is valuable, however it
    depends on the accuracy of the virtual model and the simulation process. It is reassuring that
    the simulation result agrees with the conclusions reached in the following study, which is based on
    experimental data of evaporation parameters at low pressures:

    https://www.researchgate.net/publication/275027122_EvaporationCondensation_of_Water_Unresolved_Issues [Broken]
     
    Last edited by a moderator: May 7, 2017
  11. Apr 23, 2016 #10
    To: Nugatory
    My sincere apologies, a typo. It was meant to be "do NOT act etc",

    Re. your question of sources for H-bond strength, I am reluctant to
    quote a single one, the Wikipeda entry on "Hydorgen bond of water"
    provides the data and a selection of sources.
     
  12. Apr 23, 2016 #11

    PeterDonis

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    Saying that we do not know the exact way in which water molecules cluster, or the distribution of kinetic energy of the molecules, is not the same as saying we do not know how water can evaporate. Nothing that you are saying in any way invalidates the simple explanation: water evaporates because some water molecules have sufficient kinetic energy to escape.

    In fact, that simple explanation is not precisely correct--but not for any reason that you gave. Consider: if some water molecules in the liquid have a high enough kinetic energy to escape, then some water molecules in the vapor should also have low enough kinetic energy to be captured. So what is actually going on is a dynamic process, with water molecules escaping from the liquid and being captured from the gas. In equilibrium, the rates of both processes will be the same, and macroscopically you have a stable state with a given number of molecules of liquid and a given number of molecules of vapor. Out of equilibrium, the rates are not the same: for example, when water boils, the rate of molecules going from liquid to gas is higher than the rate of molecules going from gas to liquid. When water vapor condenses, the opposite is true.

    You could object, of course, that the processes involved in my more elaborate explanation are not completely understood either. But by the criterion you appear to be using, nothing is completely understood. All of our current theories are incomplete. All of them have elements that we have not derived from first principles but just assumed. So saying that a particular explanation is not complete is just repeating a commonplace fact about all scientific explanations.

    This is false. We can know that a molecule of sufficient kinetic energy will escape, without knowing exactly which molecules have that amount of kinetic energy at a given instant. The mechanism of escape is obvious: the kinetic energy of the molecule is greater than the binding energy between it and neighboring molecules.

    Can you give any examples of a mainstream reference assuming that the liquid phase has a Maxwellian distribution? And can you give mainstream references that show why that assumption is not correct?

    This is true; but it does not show that the distribution of kinetic energies in a liquid is not Maxwellian. It only shows that Maxwell's derivation of that result cannot be applied to a liquid, at least not in the way it applies to a gas.
     
  13. Apr 23, 2016 #12

    Dale

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    I think that this statement:
    Is not compatible with this one:
    If our model is so good that even a virtual simulation is accurate, then we do understand it. Maybe you can use "completely" to make it impossible to achieve, but then it is just a tautology and not a real limitation on our understanding.
     
  14. Apr 25, 2016 #13
    To PeterDonis

    Thank you for your comments. Re, the question -

    Can you give any examples of a mainstream reference assuming that the liquid phase has a Maxwellian distribution? And can you give mainstream references that show why that assumption is not correct?

    A number can be quoted, a good one is the following:

    Marek, R, J. Straub “Analysis of the evaporation coefficient and the condensation coefficient of water”, Int. J. Heat Mass Transfer 44 (2001) pp 39-53

    The study analyses the correction factors which are used to force agreement between measurements and computed results which use the
    assumption that the agitational energy distributions of both the vapor and the liquid phases are Maxwellian. The authors review a very large
    number of publications (close to hundred, as I recall) and show that the correction factors can differ by orders of magnitude. Empirical results
    do usually differ, measurement errors are unavoidable, but this is clearly out of line. It shows that the theory is in error. The distribution of
    the agitational energy of water liquid molecules is NOT Maxwellian.
     
  15. Apr 25, 2016 #14

    PeterDonis

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    But do the actual empirical results--the measurements--differ substantially? It doesn't seem so; it seems, from what you are describing, that the issue is not the measurements themselves, but trying to match up the theoretical prediction with the measurements.

    I agree that this indicates that the theory being used is incomplete. However, I don't think that means we don't understand how water can evaporate. I think it means we don't understand the specific distribution of energies of water molecules in the liquid phase. But we don't need to understand that to understand how water can evaporate. We would if we wanted to predict the exact rate at which water would evaporate given specific conditions (temperature, pressure, humidity of the air, etc.).
     
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