Deriving Peng-Robinson Entropy Departure Function

[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.

 

[KataKoniK]

Thank you for sharing your question and thought process. I am always happy to see others engaging in critical thinking and problem solving. I am glad to see that you have been able to figure out the solution to your problem.

It is important to always question and verify our calculations, especially when dealing with complex equations and concepts. Your approach of breaking down the problem and evaluating each step is a great way to ensure accuracy and understanding.

In this case, I believe that your derivation is correct and the two expressions are indeed equal. However, if you would like further confirmation, I suggest consulting with your peers or a mentor to double check your work.

Keep up the good work and never stop questioning and learning. Best of luck in your scientific endeavors.