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Enthalpies of Vaporization

  1. May 19, 2012 #1
    Hey guys,

    I was just hoping to clear something up regarding enthalpies of vaporization.

    The idea of the heat of vaporization for a substance being "temperature dependent" is confusing me, as I had been under the impression that these quantities were only considered at the boiling point of a substance. That is, you impart a certain amount of energy to raise the temperature of liquid water at some pressure to 100°C, and then the ΔH[itex]_{vap}[/itex] represents the extra energy which needed to be imparted to then convert that liquid water to water vapour (without any raise in temperature).

    What, then, would a ΔH[itex]_{vap}[/itex] at some other temperature represent, given that you are not at a sufficiently high temperature for vaporization to occur...?

    Thanks!
     
  2. jcsd
  3. May 19, 2012 #2

    Borek

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    So you are suggesting that water at room temperature doesn't vaporise and will stay in the glass forever?
     
  4. May 19, 2012 #3
    Hmm, no, but I've always understood that phenomenon in a probabilistic context. That is, at any given time there is some probability that any given molecule of water possesses sufficient energy to escape into a gaseous state.

    Is it this "sufficient energy", then, which is represented by the heat of vaporization at this lower temperature...?
     
  5. May 20, 2012 #4

    Borek

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    Is this different at boiling point?
     
  6. May 20, 2012 #5
    No, but the probability is much higher, because the average kinetic energy of the water molecules is higher? Hmm.
     
  7. May 20, 2012 #6

    Borek

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    I feel like you are starting to see it is exactly the same process, don't you?
     
  8. May 20, 2012 #7
    I think so... And thus, that enthalpies of vaporization tend to decline with temperature represents the greater probability of molecule-escape?

    I suppose the intuition clash came from considering ΔH[itex]_{vap}[/itex] as "the energy which needs to be added to the system" as opposed "the energy required for molecule escape". The former notion lends confusion to a system below the boiling point - any energy in such a scenario provided, say, by something like a flame would raise the temperature, rather than being latent heat, whereas at the boiling point it would be latent heat... I think. Hopefully that makes sense! Haha.
     
  9. May 20, 2012 #8

    Borek

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    Imagine an open pot of water. Water vaporises and temperature goes down, so the heat is not latent - that is, we observe temperature change. Now imagine the same pot but the content is thermostatted now, so that the temperature is constant. Water vaporises and takes heat away, but the temperature is constant, so the heat is - in a way - a latent one. Now imagine water is boiling. Temperature is kept constant because boiling occurs at a well defined temperature, but apart from the reason why the temperature is constant, is there any substantial difference between both scenarios? Temperature is constant, you add heat, water vaporises taking this heat away. Same thing.
     
  10. May 21, 2012 #9
    Ahh yes, this makes sense. Thank you!
     
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