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Homework Help: Determing triple point temperature

  1. Dec 4, 2011 #1
    1. The problem statement, all variables and given/known data

    The metal Eborium (Eb) has three solid phases: a, b, c. At a pressure of 0.45 atm, the a and b phase coexist at 70 K (temperature). The molar volume of Eb(a) is 1.23 liter/mole and that of Eb(b) is 1.47 liter/mole. The heats of transformation are as follows:

    Eb(a) ---> Eb(c) delta H (a --> c) = 375.8 J/mole

    Eb(b) ---> Eb(c) delta H (b --> c)= 24.3 J/mole

    At a pressure of 0.59 atm, three phases may be present. Determine the temperature at which the system must be brought to insure that three phases are present (Assume this is close to 77 K (temperature)).

    2. Relevant equations

    Clausius-Clapeyron: 1)dp/dT= deltaH/(T deltaV) or 2)ln(p2/p1)= - (deltaH/R)(1/T2 - 1/T1)

    3. The attempt at a solution

    I integrated eqn 1) on both sides and got p= (deltaH/deltaV) lnT2/T1 and got 2 equations after that by plugging eveything in :

    p= (375.8/1.23) ln T2/77 and p= (24.3/1.47) ln T2/77. When I equate these two equations I get T2=77 K which doesn't seem right. What am I doing wrong? What's confusing me is that the question says "three solid phases a, b and c" so am I using the right equations?
  2. jcsd
  3. Dec 4, 2011 #2

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    Welcome to PF, ebunny91! :smile:

    I'm afraid your integration of equation 1 and your subsequent substitution are both not quite right.
    Do you have a reason to assume that deltaV is independent of temperature?
    Furthermore your LHS should become p2-p1 instead of just p (there is an integration constant).
    Then you substituted the volumes instead of the delta-volume.
    And you substituted a T1 of 77 K, while your problem states that the measurement is at 70 K (or is there a typo involved?)

    I believe your second equation should not be used, since it assumes a transition to a gas phase that behaves like an ideal gas.

    However, there is third way to approach this problem.
    For small changes, you can apply ΔT≈(dT/dp) Δp...
  4. Dec 4, 2011 #3
    It was a typo yea. It's supposed to be 77 K.

    I know that at the triple point all three solid phases should be in equilibrium at the same temperature and pressure. The problem already gives us the pressure (0.59 atm) so now we need to find the temperature at the triple point. (I'm just typing this out to help me understand).

    We can't use the 2nd equation because there aren't any gases involved, right?

    The amount of given info is confusing me. I think I am overlooking a really simple concept, but I'm not sure what it is.
  5. Dec 4, 2011 #4

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    There are 3 p-T phase lines involved that intersect.

    You have a point (p1,T1) on one of the three lines, and you can calculate the slope dp/dT of that specific line.
    Furthermore you have the pressure p2 of the triple point...
  6. Dec 4, 2011 #5
    Oh ok, so dP/dT= ΔP/ΔT = P2-P1/T2-T1 = 0.59-0.45 atm/T2 - 77K

    and then maybe I can use dP/dT= ΔH/TΔV to get the slope of the same line? But I don't have ΔH (a-->b)...
  7. Dec 4, 2011 #6

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    Can you perhaps think up a process that starts with a and ends in b?
  8. Dec 4, 2011 #7
    Well maybe I can use the two equations to get the ΔH (a-->b)

    First eqn:
    Eb(a) ---> Eb(c) ΔH (a --> c) = 375.8 J/mole

    and flip/negate the second eqn:

    Eb(c) ---> Eb(b) ΔH (b --> c) = - 24.3 J/mole

    So Eb(a) ---> Eb(b) ΔH = ΔH (a --> c) + ΔH (b --> c)
    = 375.8 - 24.3 = 351.5 J/mol

    so dP/dT = ΔH/TΔV
    = ΔH/T2 (V(b) - V(a))
    = 351.5/T2 (1.47 - 1.23)

    So by equating the two equations:

    351.5/T2 (1.47 - 1.23) = 0.59-0.45 atm/T2 - 77K

    Then solving for T2, I get 77.00736116 K. Which seems about right since the question says to assume the temperature at the triple point is close to 77 K (temperature)). Is this the right method and is my answer correct?
  9. Dec 4, 2011 #8

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    Looks good!

    However, you're not supposed to "solve" T2.
    You're just supposed to calculate dp/dT at the measured temperature T1.
    The formula dp/dT is for one specific point (p1,T1) where the volumes were measured.
    At a different T, the volumes will be different, so the formula wouldn't hold.
  10. Dec 4, 2011 #9
    But then I would only have the slope of the line. I have to determine the temperature at which the system must be brought to insure that the three phases are present. Do I have to do something more to get the temperature of the triple point?
  11. Dec 4, 2011 #10

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    Huh? :confused:

    You have a point on the line, the slope of the line at that point, and you have the final pressure.

    Just don't fill in a variable T to find the slope at that point.
  12. Dec 4, 2011 #11
    I have two equations for the slope of a line:

    351.5/T2 (1.47 - 1.23) and 0.59-0.45 atm/T2 - 77K

    Since it is the same line, the two equations (slopes) should be equal. So I equated them to solve for final temperature.

    The question tells me to assume that the final temperature is close to 77K so can't we assume that the molar volumes are the same?
  13. Dec 4, 2011 #12

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    Your first equation for the slope of the line is wrong.
    It's not at T2.

    And no, you can't assume the molar volumes are the same at another temperature.
    Molar volumes typically change with temperature.
  14. Dec 4, 2011 #13
    So in equation 1, it's at T1 = 77K. Then by equating the equations I only have one unknown which is T2 and that turns out to be 77.00736044 K
  15. Dec 4, 2011 #14

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    Yes, that's what I meant.

    One more thing though.
    I think you have forgotten to convert your units to SI units.
    Did you calculate with "atmosphere" and "liter"?
  16. Dec 4, 2011 #15
    Oh I forgot to convert the atm to joules/liter. Thanks!!
  17. Dec 4, 2011 #16

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    Cheers! :smile:

    (I hope you meant atm to Pascal, and liter to m3.)
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