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Calculating the boiling point of water

  1. Apr 2, 2009 #1

    fluidistic

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    Hi,
    I'm looking through wikipedia for a formula to calculate the boiling point of liquids in function of the atmospheric pressure but I didn't find any.
    In fact I'm curious what it would be for water on the Moon, Jupiter and so on.
    By the way, is the fusion point pressure-dependent? I guess no.
    So if a planet has a high pressure at ground level, I could heat up water and put a piece of iron and watching it being liquefied.
     
  2. jcsd
  3. Apr 2, 2009 #2

    mgb_phys

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  4. Apr 2, 2009 #3

    Mapes

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    The melting temperature is indeed pressure-dependent, but the dependence is small compared to that of the boiling temperature. It's a function of the volume difference between the two phases, which is small and doesn't change much for solids and liquids.
     
  5. Apr 2, 2009 #4

    fluidistic

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    Thank you very much to both.
    Do you have the formula?
     
  6. Apr 2, 2009 #5

    Mapes

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    It's the same that mgb_phys mentioned, the Clausius-Clapeyron relation. For a phase change to occur (at constant temperature and pressure), the Gibbs free energies must be equal; i.e., [itex]\Delta G=0[/itex] when comparing the two phases. By definition, [itex]G=U+PV-TS[/itex], so [itex]\Delta G=\Delta U+P\Delta V-T\Delta S=0[/itex] and the change in phase change temperature for a given change in pressure is

    [tex]\frac{\partial T}{\partial P}=\frac{\Delta V}{\Delta S}[/tex]

    For small changes, we can assume that [itex]\Delta V[/itex] and [itex]\Delta s[/itex] are constant and use the fact that [itex]\Delta S_M=\Delta H_M/T_M[/itex] at the melting temperature ([itex]G=H-TS[/itex] by definition). [itex]\Delta H_M[/itex] is the easily found enthalpy of fusion. So we end up with

    [tex]\Delta T_M\approx\frac{T_M\Delta V}{\Delta H_M}\Delta P[/tex]
     
  7. Apr 2, 2009 #6

    fluidistic

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    Thanks mapes.
     
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