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Clapton equation for water ice coexistence

  1. Jul 21, 2012 #1
    I meant to put Clapeyron equation, auto correct had other ideas...

    1. The problem statement, all variables and given/known data

    The densities of ice and water are 0.917 g cmE-1 and 1.00 g cmE-1 respectively. The latent heat of fusion of ice is 333 J gE-1. Use this information to derive the slope dp/dT of the water-ice coexistence line in atm/K at 1 atm pressure at 273K

    2. Relevant equations

    dp/dT= ΔH/TΔv = (Δv + pΔv)/ TΔv

    And of course ρ = m/v

    3. The attempt at a solution

    V ice = vI, v water = vW
    VI=1/0.917 vW=1 ∴ Δv= -0.0905125
    ∴ dp/dT=2*(-0.0905125)/-273*0.0905125 = 0.007326 atm/K

    However I think this is wrong because I've not used the latent heat of fusion. A suspicion is that I may be using the wrong formula, my lecturer didn't have very distinguishable u's and v's... Is the equation for change in enthalpy Δu+pΔv or Δv+pΔv. If the former is true (seems more likely) how do I use the lhf?

    Thanks :)
  2. jcsd
  3. Jul 21, 2012 #2


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  4. Jul 21, 2012 #3
    :smile: Got it! I wish I had been more of a pain in the backside to the lecturer when he was writing down notes. He would never reference what the meaning of the symbols, just write them down.

    1st year undergrads, be annoying to lecturers. I sure as hell know I'm going to be annoying next year!
  5. Jul 22, 2012 #4
    Just a quick check of my answer, hope you don't mind:

    = 333/ 273*(-0.0905125) = -13.48 atm/ K

    I have assumed that this is for 1g of water.
  6. Jul 22, 2012 #5


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    Looks ok except you haven't kept track of your units. So you still have some work to do there.
  7. Jul 22, 2012 #6
    Is it that I've used cm-3 instead of m-3 which I would have to use because of the standard units to calculate pressure?
  8. Jul 22, 2012 #7


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    Yes, that's part of the problem. Also, you need to think about your pressure unit.
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