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How does evaporation work at the molecular level?

  1. Jan 29, 2010 #1
    The moderators here decided my explanation of evaporation was wrong, but they failed to explain why. They simply said I was an idiot and locked my thread.

    So... I ask the gods of physics here that moderate this board: How does evaporation work at the molecular level?

    Say you have a body of water at the same temperature as the surrounding air. No heat added whatsoever.

    My questions:
    - does evaporation occur?
    - if so, what causes that evaporation and what is happening at the molecular level?
  2. jcsd
  3. Jan 29, 2010 #2


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    Be careful when re-posting a thread that was, for clear reasons, closed !

    But OK, let's consider your question "what is evaporation" under the PoV that you are asking a question and not trying to find a back door to propose a "new theory".

    As this is something pretty elementary, the wiki article on evaporation seems like a good starting point. http://en.wikipedia.org/wiki/Evaporation

    From there:
    looks to me like a good qualitative starting point. Is that sufficient as a start of an answer to your question ?
  4. Jan 29, 2010 #3
    Yeah, welcome to PF.

    Evaporation isn't some kind of interaction between water and air molecules. Apparently that's a common misunderstanding, but evaporation is basically independent of the air (and type thereof) and occurs fine without any air at all. You kept returning the discussion to details you invented for that interaction - some kind of "electric attraction" and a "balloon effect" (which doesn't even make sense.. the buoyancy of a balloon depends on excluding some density whilst such attraction would only raise it).
  5. Jan 29, 2010 #4
    That explanation of evaporation is good, but doesn't describe the detail at the molecular level.

    Water molecules are polar which means they are basically tiny magnets. The two hydrogen atoms sit on one side of the oxygen atom creating a positive charge on one side and negative on the other.

    For a single molecule to evaporate it must overcome the cohesive force of the water (hydrogen bond). Hydrogen bonds are essentially magnetism. (The hydrogen is electromagnetically attracted to the oxygen in another molecule.) While heat excites the molecules and churns them more vigorously, evaporation can happen in two ways:
    1. two water molecules align so their oxygens repel from each other.
    2. a water molecule hydrogen bonds to a molecule or atom in the air.

    Does anyone have an argument to that way of looking at it? We are talking about the same thing, I am merely describing what happens to the individual water molecules. The temperature is a measurement of how vigorously the molecules are moving. If the molecules are moving around more vigorously, their hydrogen bonds are more likely to be broken which allows them to escape into the air.
  6. Jan 29, 2010 #5


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    Yes. Both are wrong.

    Evaporation has nothing to do with the molecular structure of water. (Yes, there is a small factor of the water's polar structure, but that an influencing factor, not a determining factor. Lots of non-polar liquids are perfectly capable of evaporating).

    So you can think of water molecules as merely atoms. They bump into each other, one gets more kinetic energy and another gets less. The one that gets more energy has enough to escape.

    That's it.
  7. Jan 29, 2010 #6
    You are saying the molecules are bumping into each other. I agree. I am describing exactly HOW they are bumping into each other.

    If the H bumps into an O and they don't bump too hard, they will stick together electromagnetically (hydrogen bond).

    If the O bumps into an O they will repel from each other. Partly Newtonian, but the negative charge of the O's will ADD to the repulsion.

    Do you disagree with that?
  8. Jan 29, 2010 #7

    Char. Limit

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    Well, no. The problem is that you're assuming that was the only factor. If that was true, non-polar molecules (such as gasoline, made up of various hydrocarbons of formula C8H18) would never evaporate. We know that gasoline evaporates, so your idea must either simply be a contributing factor (NOT the entirety) or false.
  9. Jan 29, 2010 #8
    My point is that the orientation of the molecule is the key to understanding what is happening as the molecule overcomes the cohesive force of the water. If the water is moving vigorously (high heat), then more molecules escape that cohesion. They are bouncing at odd angles and the electrical charges of the molecules as they collide are an important aspect to evaporation. If you have 4 H's colliding, they will magnetically repel from each other. If you have 2 O's colliding they will magnetically repel from each other. If a single H and a single O collide, then the velocity of the collision will determine if they bounce off or stick electromagnetically (hydrogen bond).
  10. Jan 29, 2010 #9
    Being non-polar does not mean the molecule doesn't have an electrical charge to it. Water just happens to have an electrical charge on each side.
  11. Jan 29, 2010 #10
    At high heat, I do agree the energy level of the water molecule will cause it to spontaneously overcome the hydrogen bond (cohesion) to other water molecules.

    However, at lower heat overcoming that bond is more difficult. That is where the electromagnetic repulsion/attraction of the water molecules becomes important.

    So.. assuming low heat: much of the evaporation is caused by the repulsion effect of 2 O's or 4 H's pushing away from each other. In the case of gasoline, and assuming the temperature is the same in both bodies of fluids (air and gasoline), I posit there is an electromagnetic attraction to another molecule in the air that is responsible for the evaporation. If you disagree, and assuming the same temperature, where does the energy come from to evaporate the water/gasoline (separate it from its cohesive bond)?
  12. Jan 29, 2010 #11


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    You are continuing to assume that the polarity plays a critical role in evaporation.

    There is no point in you proceeding until you determine just how much the property of polarity affects evaporation.

    Let's resolve that first.
  13. Jan 29, 2010 #12

    Char. Limit

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    First, I assume that by heat, you mean temperature. In chemistry, heat is an entirely different concept.

    Second, you will have to define "low temperature" and "high temperature". Remember that even ice water has temperature of 273.15 K. It doesn't matter what the temperature of the outside is: that only determines the change
    in the change in temperature, or if you are
    fluent in calculus, the temperature of the surrounding air only determines the second derivative of the water temperature.

    Water at room temperature has close to 3,750 J. That's significant kinetic energy, no matter the surrounding temperature.
  14. Jan 29, 2010 #13
    The electrical charge plays a critical role in evaporation only at low temperatures. At higher temperatures, the kinetic energy is certainly more important.
  15. Jan 29, 2010 #14


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    Not a reliable source but you can do your own research:

  16. Jan 29, 2010 #15
    I don't know what temperature the electrical charge becomes more important. Certainly at 212 F and the electrical charge effect is nearly entirely negated. At room temperature? It becomes a factor.

    At room temperature, both bodies of fluids are the exact same temperature. The cohesion of water is quite strong. What is responsible for a single water molecule to evaporate? It's not moving fast enough to overcome the cohesion, so for it to have the energy to separate from the bond, it must be either repelled or attracted to something. EG. tumbling water molcules. some of which are hydrogen bonded to each other will align in certain ways to attract/repel from each other based on their charges. That's where I think the energy comes from to evaporate.

    And I don't see any reason why the electrical charge of water wouldn't be attracted to the electrical charge of a molecule in the air. That's certainly a factor, given the same temperature and no introduction of another energy source.
  17. Jan 29, 2010 #16
    It's already there. An average speed of water molecules at room temperature is in hundreds of meters per second. Intermolecular bonds keep most molecules inside the main body of water, but some of them occasionally attain enough energy, simply through random interactions with each other, to break the bonds and fly away.

    Air is irrelevant. You'll observe evaporation even in the vacuum.
  18. Jan 29, 2010 #17

    Char. Limit

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    We're not saying that dipole-dipole attraction (what you're describing) isn't a factor. We're saying that at room temperature, it isn't significant.

    Also, please, don't use Fahrenheit. Celsius is ok, but any real scientist in my mind should use Kelvins. It gives you the idea of how much energy room temperature really is. Room temperature is 295 Kelvins. That's a lot, considering that the boiling point of water is 373 Kelvins. 295/373 is highly significant fraction of energy.
  19. Jan 29, 2010 #18
    Describing "phase" or "state change" of water simply because it is in liquid water or floating in the air seems erroneous to me.

    The water molecule is identical in both places. The temperature can be identical. The only difference is its surrounding molecules. To say it's phase is different in that case is not very descriptive.
  20. Jan 29, 2010 #19

    Char. Limit

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    If the temperature (and pressure) are identical, the vast majority of water molecules will be in the same phase. I'm talking 999 out of 1000.
  21. Jan 29, 2010 #20
    andrewbb, on a molecular level evaporation is very simple: some water molecules on the surface happen (after random collisions with their neighbours) to have enough kinetic energy to break away from the attraction of other water molecules (think escape velocity).

    Not only was your air-water attraction theory wrong, but if anything it was closer to the (opposite) process by which some of the air molecules (including H2O from the vapour phase) get dissolved into the water.
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