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Verifying That Diamond Is More Stable At Certain Conditions

  1. Mar 26, 2014 #1
    1. The problem statement, all variables and given/known data
    Verify that diamond becomes more stable than graphite at approximately 15 kbars



    2. Relevant equations



    3. The attempt at a solution

    So, I understand that to show one substance is more stable than another at particular environmental conditions, I just have to show that the one with a lower Gibbs free energy is more stable.

    In the section, the formula [itex]\left( \frac{\partial G}{\partial P} \right)_{T,N} = V[/itex]. Since the partial derivative is given approximately by [itex]\frac{\partial G}{\partial P} = \frac{\Delta G}{\Delta P}[/itex], I can write [itex]\frac{\Delta G}{\Delta P} = V[/itex].

    In the back of the textbook I am using is a table containing the Gibbs free energy of formation of substances at at temperature of T=293 K and a pressure of P = 1 bar, call it [itex]G_1[/itex] Using this information, I figured I could obtain the Gibbs free energy at P = 15 kbar, call it [itex]G_{15}[/itex]:

    [itex]\frac{\Delta G}{\Delta P} = V \implies[/itex]

    [itex]\Delta G = \Delta P V \implies[/itex]

    [itex]G_{15} = \Delta P V + G_1[/itex].

    Now, if we assume that the change in volume of the substance is negligible as the pressure changes from P = 1 bar to P = 15 kbar, which is an assumption the author of the textbook uses, then the formula above is applicable. The volume of diamond at T=293 and P=1 bar is 5.31 x 10^-6. Calculating the Gibbs free energy at 15 kbars of the diamond:

    [itex]G_{15,D} = (15 \times 10^{3}~bar -1~bar)(5.31 \times 10^{-6} ~m^3)+ 2.900 \times 10^{3} ~J = 2900 ~J[/itex]

    And for graphite, where the volume is 3.42 x 10^-6

    [itex]G_{15,G} = (15 \times 10^{3}~bar -1~bar)(3.42 \times 10^{-6}) + 0 = 0.0513 ~J[/itex].

    Clearly the Gibbs free energy is less for graphite. What did I do incorrectly?
     
  2. jcsd
  3. Mar 26, 2014 #2

    haruspex

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    Where does the 3.42 come from? You seem to have a smaller volume for the graphite than for the diamond, which is clearly wrong.
     
  4. Mar 26, 2014 #3
    Whoops...I read it incorrectly. The diamond should have the volume of 3.42. But this still does not change the fact that the Gibbs free energy of graphite is much smaller than that of diamond's.
     
  5. Mar 26, 2014 #4

    haruspex

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    ok, so next I question the G1 values. What exactly does it say in the back of the book, and how did you use that to calculate the G1s?
     
  6. Mar 27, 2014 #5
    I uploaded a screen shot of the reference data that I am using, and the paragraph that precedes it. Does this answer your question regarding the G1's, or do you require more information?
     

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  7. Mar 27, 2014 #6
    Okay, I think I understand now. Number that is closer to being negative is the substance that is less stable. Therefore, the graphite would be less stable at these particular environmental conditions.
     
  8. Mar 27, 2014 #7

    haruspex

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    How many Pascals to the Bar?
     
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