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Lead's melting point as a function of pressure (Thermo')

  1. Jan 14, 2015 #1
    1. The problem statement, all variables and given/known data
    When lead is melted at atmospheric pressure, the melting point is 600K, the density decreases from 1.101x104 to 1.065x104 kgm-3. The latent heat is 24.5 kJ kg-1

    Estimate the melting point of lead at a pressure of 100 atm.

    2. Relevant equations

    I haven't been able to find any equations for how the melting of a solid varies with pressure. Intuitively I'd say the melting point will go down, but that's really just a guess.

    3. The attempt at a solution
    If the density decreases, the material expands. This means it is doing work on the surroundings, so some energy that goes in does that. I looked at some data from a paper (not what the question wants) that tells me that the melting point of lead goes up as pressure goes up in the range of 800 to 1200 MPa.
     
  2. jcsd
  3. Jan 14, 2015 #2

    DrClaude

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    Staff: Mentor

    Look up the Clausius-Clapeyron relation.
     
  4. Jan 15, 2015 #3
    Okay, yup, I've looked it up

    dp/dT = L/T(Vf-Vi)

    So by subbing in the values I get the gradient of the coexistance curve, which turns out to be 0.113 at 1 atm.
    I'm not sure how I would find the gradient of the curve for 600 atm?
     
    Last edited: Jan 15, 2015
  5. Jan 15, 2015 #4

    DrClaude

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    I guess you mean 100 atm. You can simply make the assumption that the gradient is constant. If you look at a phase diagram, you will see that the coexistance curve can be reasonably approximated as a straight line.
     
  6. Jan 15, 2015 #5
    Okay well in that case the melting point would be the same? If the gradient of the line is independent of pressure then so is the melting point given fixed values of L and deltaV.
     
  7. Jan 15, 2015 #6

    DrClaude

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    That is not correct. The only assumption you are making is that the coexistance curve is a straight line.
     
    Last edited: Jan 15, 2015
  8. Jan 15, 2015 #7
    Okay, so if the coexistance curve is a straight line, then dP/dT is constant. This means L/T(Vf-Vi) is constant. So if L and (Vf -Vi) are constant then doesn't that make T constant?
     
  9. Jan 15, 2015 #8

    DrClaude

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    [Edit: I made a mistake, forget about the coexistance curve being a straight line, which it is not.] You have ##\Delta P / \Delta T = \mathrm{const.}##, therefore what happens for ##\Delta P \neq 0##?
     
    Last edited: Jan 15, 2015
  10. Jan 15, 2015 #9
    Well if delta p doesn't equal zero then delta T can't equal zero either.


    Is the trick to integrate both sides? If I do that I get P = ln(L/T(Vf-Vi))+C.
    Subbing in the values for 1 atm I get C=3.18 but does this value only stand for 1atm? I don't think I can use it for 100atm...

    I'm tempted to just guess and say that if you multiply the pressure by 100 the melting point also gets multiplied by 100.
     
  11. Jan 15, 2015 #10
    Right, got it. If dp/dT is constant then P=kT + C where k is he constant of proportionality 0.113. Using the initial values in the question, C = -66.8.
    The equation now stands as P = 0.113T -66.8
    Letting P = 100 (as opposed to the previous 1), we can find T to be 1476K

    Thanks a lot for all your help DrClaude!
     
  12. Jan 15, 2015 #11

    DrClaude

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    I'm sorry, but I have been leading you in a the wrong direction. I made a mistake with that idea that the coexistance curve was linear. Let me think a bit to get you back in the right direction.
     
  13. Jan 15, 2015 #12

    DrClaude

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    Ok, I got my brain back in gear. What is constant is ##L/\Delta V##, where ##\Delta V = V_\mathrm{l} - V_\mathrm{s}##, the difference in volume between the two phases, which we can approximate as constant for a solid-liquid transition.

    You have to be careful here: it is a definite integral you have to do.

    Wouldn't that give you the same result whatever the value of ##L/\Delta V##?
     
  14. Jan 15, 2015 #13
    So if dP/dT = k/T where k is L/ΔV.
    Integrating both sides, we get P2 - P1 = kln(T2/T1) This is the definite integral?
     
  15. Jan 15, 2015 #14

    DrClaude

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    Yes, that looks correct. You now need to rewrite this as an equation for ##T_2##.
     
  16. Jan 15, 2015 #15
    Okay now I'm getting absolutely collossal numbers.
    T2=T1exp(P2-P1/k)
    T2=600exp(107/68.06)
    That's with the pressure in pascals to match the rest of the units. Surely the melting point shouldn't depend on units like that?
     
    Last edited: Jan 15, 2015
  17. Jan 15, 2015 #16

    DrClaude

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    Your value for k is incorrect.
     
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