# Deriving Peng-Robinson Entropy Departure Function

1. Apr 22, 2013

### cjc0117

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):

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

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

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

The Peng-Robinson EOS is:

$P=\frac{RT}{v-b}-\frac{a}{v(v+b)+b(v-b)}$

Where:

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

$b=0.07780\frac{RT_{c}}{P_{c}}$

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:

$A=\frac{aP}{R^{2}T^{2}}$

$B=\frac{Pb}{RT}$

$Z=\frac{Pv}{RT}$

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 $(\frac{∂P}{∂T})_{v}$, I get:

$(\frac{∂P}{∂T})_{v}=\frac{R}{v-b}-\frac{\frac{da}{dT}}{v(v+b)+b(v-b)}$

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

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

After simplifying, I get:

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

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

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

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