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The physics behind the Dew Point phase shift

  1. Oct 25, 2014 #1
    Many websites gives the results of Dew Point but not the reason.

    What is going on at the dew point? Not the result (at a certain VSP vapour becomes liquid if it can find nuclei to coat) but the reason for the result.

    I have been reading up on this – just out of curiosity – and would like someone to tell me if I’m on the right lines or way off.

    As far I understand it when Water vapour molecule slow down (due to a heat loss) they can no longer be a gas simply as they are moving too slowly to maintain their distance apart. When they move closer they become a liquid.

    Is that right?

    Or can anyone give a simple to understand reason behind phase shift.

    Thank you

    K
     
  2. jcsd
  3. Oct 25, 2014 #2

    russ_watters

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    Staff: Mentor

    That's relates, but I like to look at it a different way: The air's moisture holding capacity varies with temperature. Dew point is the lowest temperature a parcel of air can be and still hold on to the moisture it has.
     
  4. Oct 25, 2014 #3
    I know that at a certain saturation vapour pressure the gas will shift to liquid.

    What I'm trying to work out is why?

    There must be something going on with the molecules - I think it may be down to their speed (maybe they slow down and at a certain speed they can't rebound so much - but that doesn't sound right :)
    But also why (with no nuclei) will the H2O stay as gas all the way down to around -37 or less

    I know the results but what I'm trying to get an insight into is what is really going on

    The "air" doesn't really have a moisture holding capacity as the H2O gas can exist in a vacuum (apart from itself)
     
  5. Oct 25, 2014 #4
    Perhaps if you look at it from a liquid point of view.
    Well, what does the vapour pressure of a liquid depend upon. Surely temperature of the liquid has some say in the matter.
    And if you have heard of the chemical forces that hold the molecules of a liquid together, such as hydrogen bonds and the like, you realize that as the molecules of the liquid jostle about, some may obtain enough kinetic energy to overcome these bonds, and escape the liquid surface and become vapour. Increase the liquid temperature, and a greater percentage of molecules would have enough kinetic energy to escape.
    If a quantity of the liquid is sealed in container, the vapour pressure above the liquid will represent the percentage of molecules that have enough energy to escape at a certain temperature.

    Why does the vapour pressure only reach a certain value at a particular temperature. Surely if the vapour is saturated, one can promote more evaporation by just increasing the volume above the liquid.
    Good point. But that reduces the number of collisions of the molecules in the vapour state with the liquid surface. Any vapour molecule of any kinetic energy can become part of the liquid. a molecule of high kinetic energy can barge right in, and one of low kinetic energy can be attracted by the other liquid molecules.

    Karen lorr, can you connect the dots now? on saturation pressure and dew point?
     
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