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Deriving Peng-Robinson Entropy Departure Function

  1. Apr 22, 2013 #1
    EDIT: Nevermind. I figured it out. The two expressions are, in fact, equal.


    An excerpt from a book at this link, http://webpages.sdsmt.edu/~ddixon/Departure_Fxns.pdf, states that the entropy departure function for any equation of state is equal to the following (Eqn. 4.4-28):

    [itex]s_{T,P}-s^{IG}_{T,P}=RlnZ+\int^{v_{T,P}}_{v→∞}[(\frac{∂P}{∂T})_{v}-\frac{R}{v}]dv[/itex]

    And that the specific entropy departure function for the Peng-Robinson EOS is (Eqn. 4.4-30):

    [itex]s_{T,P}-s^{IG}_{T,P}=Rln(Z-B)+\frac{\frac{da}{dT}}{2\sqrt{2}b}ln[\frac{Z+(1+\sqrt{2})B}{Z+(1-\sqrt{2})B}][/itex]

    The Peng-Robinson EOS is:

    [itex]P=\frac{RT}{v-b}-\frac{a}{v(v+b)+b(v-b)}[/itex]

    Where:

    [itex]a=0.45724\frac{R^{2}T^{2}_{c}}{P_{c}}[1+κ(1-\sqrt{\frac{T}{T_{c}}})]^{2}[/itex]

    [itex]b=0.07780\frac{RT_{c}}{P_{c}}[/itex]

    κ is a constant that depends on the acentricity of the specific chemical species.

    The other parameters A,B, and Z used in the above equations are:

    [itex]A=\frac{aP}{R^{2}T^{2}}[/itex]

    [itex]B=\frac{Pb}{RT}[/itex]

    [itex]Z=\frac{Pv}{RT}[/itex]

    My problem is that I can't figure out how to derive the Peng-Robinson entropy departure function from the integral definition I gave above. When I evaluate [itex](\frac{∂P}{∂T})_{v}[/itex], I get:

    [itex](\frac{∂P}{∂T})_{v}=\frac{R}{v-b}-\frac{\frac{da}{dT}}{v(v+b)+b(v-b)}[/itex]

    And when I evaluate the integral using this equation, I get:

    [itex]s_{T,P}-s^{IG}_{T,P}=Rln(Z)+Rln(\frac{∞}{v})+Rln(\frac{v-b}{∞-b})+\frac{\frac{da}{dT}}{\sqrt{2}b}ln[(\frac{v+b(1+\sqrt{2})}{\sqrt{(v+b)^{2}-2b^{2}}})(\frac{\sqrt{(∞+b)^{2}-2b^{2}}}{∞+b(1+\sqrt{2})})][/itex]

    After simplifying, I get:

    [itex]s_{T,P}-s^{IG}_{T,P}=Rln[\frac{Z(v-b)}{v}]+\frac{\frac{da}{dT}}{\sqrt{2}b}ln(\frac{v+b(1+\sqrt{2})}{\sqrt{(v+b)^{2}-2b^{2}}})[/itex]

    I've been able to find that [itex]Rln[\frac{Z(v-b)}{v}]=Rln(Z-B)[/itex], thus, the equation becomes:

    [itex]s_{T,P}-s^{IG}_{T,P}=Rln(Z-B)+\frac{\frac{da}{dT}}{\sqrt{2}b}ln(\frac{v+b(1+\sqrt{2})}{\sqrt{(v+b)^{2}-2b^{2}}})[/itex]

    But I can't show whether this is the same as the equation given in the link (the second equation I gave in this post). And if it's not, where did I go wrong in the derivation?

    Thanks.

    EDIT: Nevermind. I figured it out. The two expressions are, in fact, equal.
     
    Last edited: Apr 22, 2013
  2. jcsd
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