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H20 Molecular shift - what is the initial state just below the Dew Point

  1. Oct 14, 2011 #1
    Hi all

    Another question (sorry)

    Water vapour in the air will phase shift back to liquid when it reaches its Dew Point temp (Partial Pressure = the Saturated Vapour Pressure), OK I have this.

    For this phase shift I assume that there are Condensation Nuclei and these nuclei are at a lower temperature than the vapour – so any vapour entering the local area will enter an area of reduced humidity and condense back to liquid and stick to the nuclei.

    But (there is always a but ;-)

    Assuming that the air /vapour/water temp (I’ll treat them all as about the same) is below 0c then why do some of the vapour molecules shift to water on contact with the nuclei and why do some shift directly to ice.

    I am asking about the initial shift of a single molecule. I understand that “after” shifting the Bergeron Process will mean vapour dissipates whilst ice increases, but this requires some ice to be present near the vapour.

    But, again, when initially shifting, in the vicinity of condensation nuclei do some shift to water and other shift to ice.

    Am I missing something here ?

    *Do all vapour molecules (when there is no ice present) shift to water first, then freeze?
    *Do all molecules shift to ice first, which then evaporates ?
    *Is it due the vapour shifting then releasing the latent energy it picked up in its evaporation –the energy reduction causing less agitation so a rigid hexagonal lattice forms?
    * etc
    * etc

    Sorry am a little confused by this.

    Can any explain the physics behind this – the technical reason for the shift to water first (if that is what happens)
    Last edited: Oct 14, 2011
  2. jcsd
  3. Oct 17, 2011 #2
    It's OK, I worked it out.

    Just in case anyone else was wondering - the answer is that due to high surface tension (pure) water needs a supersaturation of at least 120% down to -20c, whereas vapour will form ice around a nuclei at higher temps. The crystals then act nuclei for vapour so the Bergeron Process will ensure there are more crystals than droplets.
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