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Homework Help: How to derive Thermodynamic Relations from Volume Data

  1. Oct 5, 2011 #1
    I'm struggling to derive some thermodynamic equations from this http://my.safaribooksonline.com/book/chemical-engineering/9780132441902/thermodynamic-properties-from-volumetric-data/ch03lev1sec1" [Broken]:

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

    I'm trying to derive all the equations from 3.8 to 3.14 for Pressure and Temperature and equations 3.47 to 3.54 for Volume and Temperature.

    2. The attempt at a solution

    I would appreciate some direction in how to proceed. I've been trying to derive eqn. 3.8 for the past hour but I keep on getting an extra PV term. This is how I did it (wrongly!!):

    dH= dU + PdV + VdP

    Dividing the above equation by dP

    dU/dP=dH/dP - PdV/dP - V

    I know (I correctly derived this one ^^) that dH/dP=[V-T(dV/dT)] so substituting back and integrating dP from 0 to P

    U=∫[V-T(dV/dT)|(0 to P) - PV - VP

    also the summation term Ʃni*hi , I'm not so sure where that came from but correct me if I'm wrong but is it ok to assume that since we are considering a gas system of n moles of i components and relating back to u=h-pv we are assuming that the pressure and volumes of the individual molecules are negligible but only take into account their enthalpies h so we get ui=ni*hi then for the whole system we get the summation Ʃni hi ?

    If I'm able to get the correct equation for U then I believe that it would be more or less straight forward to derive the other equations since I can just refer back to the definitions of Gibbs Energy G=U+PV-TS and Helmholtz Energy A=U-TS....I hope so at least
    Last edited by a moderator: May 5, 2017
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  3. Oct 5, 2011 #2


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    [itex]\sum n_ih_i^0[/itex] is the constant from integrating [itex]dH[/itex]. And you're getting multiple [itex]PV[/itex]s because you're not integrating [itex]\int_0^P P(\partial V/\partial P)_T\,dP[/itex] or [itex]\int_0^P V\,dP[/itex] correctly. Just keep it as [itex]d(PV)[/itex] and integrate that to get [itex]PV[/itex].
  4. Oct 6, 2011 #3


    Then substituting for dH=[V-T(∂V/∂P)]dP and integrating

    U=∫[V-T(∂V/∂P)]dP - PV + Ʃni*hi

    I've reached as far as trying to derive the Gibbs Energy equation but I'm stuck again

    Definition: G=H-TS
    dG=dH-d(TS), substituting for dH=[V-T(∂V/∂P)]dP

    Then from the ideal gas law PV=nRT, (∂V/∂P)=nR/P, substituting back in


    This is where I'm stuck how do I proceed with the derivative for d(TS). I expanded it so that d(TS)= TdS + SdT but when I replace it in the above equation and taking derivative P I get the following:

    dG=[V-nRT/P]dP - T(dS/dP) - S(dT/dP)

    From maxwell relations: (dS/dP)= - (dV/dT) = nR/P (from ideal gas law V=nRT/P)
    and (dT/dP)=(dV/dS) what goes on with this term?

    If I keep it as dT/dP and apply it on the ideal gas law, it will give me V/nR....I can't "see" how to progress to get the same form of the equation in the book/eqn 3.12

    It's a similar problem I faced for the Helmholtz Energy equation (3.11)
  5. Oct 6, 2011 #4


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    I got as far as "ideal gas law." These equations don't assume ideality.
  6. Oct 6, 2011 #5
    According to the chapter in the book the equations are for a homogeneous mixture at a fixed composition and temperature and they study effect of pressure and volume, respectively, on enthalpy and entropy.
    Also it is mentioned that the equations are used to calculate the thermodynamic properties U,H,G,etc relative to the ideal gas state. Which is why I referred to the ideal gas law...I don't know if this helps...
  7. Oct 6, 2011 #6


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    Yes, relative to the ideal gas state. The Gibbs free energy of an ideal gas is just [itex]G=\sum_i n_i(h_i^0-Ts_i^0)[/itex].
  8. Oct 6, 2011 #7
    I see, yes...but maybe I don't follow you that well, how would I get the same form of the equation as in the book, equation 3.12

    I've been checking another book I have (Introduction to Chemical Engineering Thermodynamics by van Ness, Smith, Abbott) and I THINK that it's related to the ideal gas mixture model which uses partial properties to estimate the properties of a mixture...I'm really sorry but thermodynamics really isn't that easy for me to follow.
  9. Oct 6, 2011 #8
    Gibb's theorem: "a partial molar property (other than volume) of a constituent species in an ideal-gas mixture is equal to the corresponding molar property of the species as a pure ideal gas at the mixture temperature but at a pressure equal to its partial pressure in the mixture"

    Now just go along with me, I think I'm getting the concept but not the math...baby steps, baby steps!

    Considering T and P first. So for Gibbs energy G(ideal mixture)=Gi*pi+G(ideal gas)
    with pi being the individual species i partial pressure and Gi is the gibbs energy of an individual species i.

    And as you pointed out Gibbs energy of an ideal gas is G=∑ni(hi−Tsi).

    So in eqn 3.12, the first 2 terms represent the gibbs free energy of the ideal gas mixture and the last one is purely that of an ideal gas.

    Am I close? :(
  10. Oct 6, 2011 #9


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    No. The last term is the Gibbs free energy of an ideal gas mixture; the first two terms are the deviations due to nonideality.

    This derivation is harder than the ones for U and H. First try to get S by integrating [itex]dS=dU/T+(P/T)dV[/itex], using what you've already calculated for U. You're also going to need


    [itex]PV=znRT[/itex], where z is the compressibility factor, and [itex]f_i=y_iP[/itex]. Once you have S, it's relatively simple to get A and G.
  11. Oct 6, 2011 #10
    This is what I've done earlier for S but after the discussion I think it's wrong

    TdS=dH-VdP, substituting for dH and dividing by T
    dS=[V/T - (dV/dT)]dP - (V/T)dP

    Using PV=nRT replace V/T by nR/P

    dS=[nR/P - (dV/dT)]dP - (nR/p)dP , integrating wrt to P

    S=[nR/P - (dV/dT)]dP - R∑ni*ln pi + ∑ni*si

    pi (partial pressure)=yi*P
    ∑ni*si is from the summability relation: nS=∑ni*si

    Finally: S=∫[nR/P - (dV/dT)]dP - R∑ni*ln yi*P + ∑ni*si
  12. Oct 6, 2011 #11


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    Yeah, that's playing a little too fast and loose. If you're going to assume PV=nRT then the integrand is identically zero, for example (as is the next term).
  13. Oct 6, 2011 #12
    I figured it was too good to be true...this is a nightmare!

    I just got my hands on one of the papers which the book referred to to derive the equations and it looks much more complicated...I could send it to you if you would like to have a look.
  14. Oct 6, 2011 #13


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    See if you can rewrite it using the results you've already gotten and the identities given in the book!
  15. Oct 6, 2011 #14
    What I've got so far:


    dU=[V-T(dV/dT)]dP - d(PV)

    For Entropy: dS=dU/T + (P/T)dV

    substituting for dU I get: dS=[V/T - (dV/dT)]dP - (1/T)d(PV) + (P/T)dV

    dS=[V/T - (dV/dT)]dP - (V/T)dP - (P/T)dV + (P/T)dV cancelling the last 2 terms

    dS=[V/T-(dV/dT)]dP - (V/T)dP

    By looking at the final equation I suppose that Ʃni*si is the constant of integrating (V/T)dP but where does the term RƩni*ln yi*P come from? It seems I keep stopping at the same point no matter the equation...
  16. Oct 6, 2011 #15


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    This is when you want to apply PV = znRT to V/T in the integrand, then manipulate it to get nRT/P plus another term. That term should look familiar to Eq. 3-17.
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