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Homework Help: Why is the Gibbs Free energy equal to this:

  1. Apr 3, 2017 #1
    1. The problem statement, all variables and given/known data
    the isothermal compressibility of graphite is about ##3*10^{-6} bar^{-1}##, while that of diamond is more than ten times less and hence negligible in comparison. (isothermal compressibility is the fractional reduction in volume per unit increase in pressure, as defined in problem 1.46.) Use this information to make a revised estimate of the pressure at which diamond becomes more stable than graphite (at room temperature).

    2. Relevant equations
    ##G_g=V_g*P_g## graphite
    ##G_d = V_d*P_d## diamond

    I am not sure what the relevant equation is for this: that is my question. Chegg says:
    ##(\frac{∂G}{∂P})_T = V_o * (1-κ_TP)##

    3. The attempt at a solution

    I set the formulas equal to each other because when G is equal, then P will be equal. Using this knowledge, I solved them for P and found that it was equal to ##15.344 kbar##
    This is the pressure when the compressibility factor is neglected.

    if we consider the compressibility factor, the formula is supposed to be ##(\frac{∂G}{∂P})_T = V_o * (1-κ_TP)##

    the formula for G, that I know, is ##G = U + PV - TS##

    if I take the derivative with respect to P and constant T then ##(\frac{∂G}{∂P})_T = V## ???

    I don't know... Please show me how they got that formula
  2. jcsd
  3. Apr 3, 2017 #2
    You need to integrate the equation with respect to P to get the correct effect of P on G. Did you do that?
  4. Apr 3, 2017 #3
    so ##UP + \frac{1}{2}P^2V - TSP## ?
  5. Apr 3, 2017 #4
    I don't think so. What is the initial state or states of the materials?
  6. Apr 3, 2017 #5
    They are solids. I am trying to figure out how they got that equation above. When I do the partial derivative of G, I don't get that. Are they using a different formula?

    this is what I need to get. ##(\frac{∂G}{∂P})_T = V_o * (1-κ_TP)##

    after I get that, I integrate it from 0 to P etc.
  7. Apr 3, 2017 #6
    That is all correct. My question about STATE was not about the state of aggregation. It was about the reference temperature and pressure of each solid at which the free energy is taken to be zero. Tell me about the starting states of the two solids, and how they are related in terms of G.
  8. Apr 3, 2017 #7
    sorry... :(

    that is all of the information that the problem gives me. there is a chart but it gives the free energy when the pressure is zero instead. that is that diamond is 2.9 kJ greater than graphite at 0 kbar.

    graphite = PV
    diamond = PV + 2.9
  9. Apr 3, 2017 #8
    OK. I understand what's going on here now, and how to do it. You should not be using PV. You should be using ##\int{PdV}##


    Find the value of P for which G for graphite is equal to G for diamond
  10. Apr 3, 2017 #9
    but the first step was to find the pressure when the compressibility factor was neglected. that's the PV and PV+2.9
    that value of P is then used in in the formula given after integrating it. why did they do it that way. is that wrong?
  11. Apr 3, 2017 #10
    Yes. That's wrong. In the first step, the compressibility of the material was neglected, and a certain value of the pressure was obtained. Now, we are solving the problem over again, this time with the compressibility of the material not neglected; and we will be comparing the value of the pressure we get with this more accurate calculation with the cruder approximation we got in the first step.
  12. Apr 3, 2017 #11
    Got it. How did you get that formula? I'm flipping through the textbook and I can't find anything that remotely resembles it :( (not saying it is wrong, just saying that I am putting forth effort)
  13. Apr 3, 2017 #12
    Well, we know that, in general, $$dG=-SdT+VdP$$
    We know that the temperature is being held constant and the pressure is being increased. So, for such a change, ##dG=VdP##. Our initial condition for the integrations is G=G(300,0) at T= 300K and P=0. So, for each material,
    $$G(300,P)=G(300,0)+\int_0^P{VdP}$$Then you just do the integration for each of the materials.
  14. Apr 3, 2017 #13
    I don't understand where these formulas came from. why is it the integral of PdV = to this?

    I'm trying to solve it... I'm stuck trying to get that P on its own. that one of them is squared is making it difficult for me :(
    I need to try a different approach than what I just did lol.
  15. Apr 3, 2017 #14
    What is this integral equal to: $$\int_0^P{V_0(1-\kappa P')dP'}$$where P' is a dummy variable of integration.
  16. Apr 3, 2017 #15
    is that just a formula to memorize? I am trying to figure out how you got to that formula. That's the one they give me in the solution as the derivative of G with respect to P with constant T.

    by the way. for P I got ##P = \frac{2(V_{od}-V_{og})}{-V_{og}k_g + V_{od}k_d}##

    I'm double checking it. But so far...

    edit: YIKES! forgot the 2.9kJ :(
  17. Apr 3, 2017 #16
    Are you not familiar with the equation ##dG=-SdT+VdP##? You are aware that ##V=V_0(1-\kappa P)##, correct? So, $$\int{VdP}=\int_0^P{V_0(1-\kappa P')dP'}$$

    I'm having trouble understanding where your disconnect is?
  18. Apr 3, 2017 #17
    No I wasn't. My professor doesn't speak very good English. I'm doing my best to teach myself from the textbook. :( I'm passing, but I occasionally miss things.
    Thanks for your help : ).

    I'm still working on solving for P. That 2.9 is making it difficult.
  19. Apr 3, 2017 #18
    You need to use the quadratic formula.
  20. Apr 3, 2017 #19
    I was afraid of that.
  21. Apr 3, 2017 #20


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    You might try using the above statement to arrive at Chestermiller's equation ##V=V_0(1-\kappa P)##
  22. Apr 3, 2017 #21
    We can tackle that problem (1.46) next... It's pretty long...
  23. Apr 3, 2017 #22
    ok I got a result for P. It is too long to type...
    but I will plug in ##.189 \frac{kJ}{kbar}## (in the chapter the book gives us that the slope for graphite is ##V = 5.31*10^{-6} m^3## and the slope for diamond is ##V = 3.42*10^{-6} m^3##) for ##V_{og}-V_{od}##
    and those values for V

    ##K_g = 3*10^{-6} bar^{-1}##

    ##K_d## is negligible in comparison. so do I will just replace it with 0?
    Last edited: Apr 3, 2017
  24. Apr 3, 2017 #23
    I filled out my quadratic equation and I am getting

    ##\frac{kJ^2}{kbar^2}## for b^2
    4ac = ##\frac{kJ}{kbar^2}##

    I am certain I am supposed to have kJ^2 in 4ac, then everything would work out. but I can't find where it would come from? only the ##V_{og}## term has kJ for units... right?

    something isn't right because I am getting a crazy number. if I pretend I just forgot the kJ somewhere.
    Last edited: Apr 3, 2017
  25. Apr 3, 2017 #24
    Wait! I got it. the B minus version worked of the quadratic formula. at first I made a I-D-10-T error and got the wrong answer, but when I tried again, I got it. 16.53 which is the right number. as for the units...

    It is the same thing except rather than using the quadratic formula, they subtracted G_od-G_og and got 16.33
    Last edited: Apr 3, 2017
  26. Apr 3, 2017 #25
    double post
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