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Edit: This was not a homework problem. I simply pulled an example from a textbook to easily illustrate and convey the problem at hand.

I am seeking some assistance in evaluating the entropy change of a real gas using BWRS-EOS. I have successfully accomplished this using Peng-Robinsons which will be used as a point for comparison. Any feedback or suggestions would be highly appreciated. Consider Example 8.8. Enthalpy and entropy from the Peng-Robinson equation - Introductory Chemical Engineering Thermodynamics by J. Richard Elliott.

As evaluated via BWRS-EOS for Propane - Python script: https://pastebin.com/80w9npaT

T1 = 378.15 K, P1 = 500 kPaA, Enthalpy Departure = -407.06 J/mol, Entropy Departure = -52.43 J/mol, Cp Ideal = 89.82 J/mol.K

T2 = 463.15 K, P2 = 2500 kPaA, Enthalpy Departure = -1324.7 J/mol, Entropy Departure = -67.12 J/mol, Cp Ideal = 105.04 J/mol.K

$\Delta H = \Delta H_{ideal} + \Delta H_{real} = C_p * (T2 - T1) + (\Delta H_{departure2} - \Delta H_{departure1})$

enthalpy_delta = 0.5*(89.82+105.04)*(463.15-378.15) + (-1324.7-(-407.06)) = 7363.91 J/mol CORRECT

$\Delta S = \Delta S_{ideal} + \Delta S_{real} = C_p * ln(\frac{T2}{T1})-R*ln(\frac{P2}{P1}) + (\Delta S_{departure2} - \Delta S_{departure1})$

entropy_delta = 0.5*(89.82+105.04)*ln(463.15/378.15) - 8.31447 * ln(2500/500) + (-67.12-(-52.43)) = -8.32 J/mol.K. INCORRECT

The results for enthalpy are correct, however, entropy is horribly incorrect. I strongly believe this is due to the entropy departure values which differ from what I calculate via Peng-Robinson by over 1000%. As summarised in the comparison between PR-EOS and BWRS-EOS below, the entropy departure as calculated by BWRS is an order of magnitude larger than PR-EOS.

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Any advice would be highly appreciated, I've been struggling for a while and cannot get this to work.

Entropy departure per K.E Starling:

enter image description here

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  • $\begingroup$ Consider to spell out acronyms. $\endgroup$
    – Qmechanic ♦
    Commented Feb 22, 2023 at 7:34
  • $\begingroup$ What do you get if you omit the first term on the right hand side of the departure equation? $\endgroup$
    – Chet Miller
    Commented Feb 22, 2023 at 12:36
  • $\begingroup$ Significantly closer: Entropy Delta = 5.595 J/mol.K. state1 Entropy Departure = -0.438, state2 Departure = -1.22. Is this technically correct? I'll check this against some other examples and compare the results $\endgroup$
    – Harry Ferrier
    Commented Feb 22, 2023 at 12:47
  • $\begingroup$ does not seem to produce correct results for other cases I have tried :( $\endgroup$
    – Harry Ferrier
    Commented Feb 22, 2023 at 13:16

2 Answers 2

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The BWRS approach is based on molar volume = 1/rho. As such, to apply the Starling entropy departure equation, you start with $C_v^{IG}\ln{(T2/T1)}$ and apply the Starling entropy departure corrections to this.

Alternately, you can start with the full ideal gas relationship $C_v^{IG}\ln{(T_2/T_1)}-R\ln{(\rho_2/\rho_1)}$ and then apply the Starling entropy departure equation, with the first term omitted because it represents the ideal gas contribution of molar volume.

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    $\begingroup$ STATE 1: Z: 0.9617, Molar Volume: 6047.7cm3/mol, Enthalpy Departure: -407.1J/mol, Entropy Departure: -52.4J/mol, CpID: 89.8J/mol.K, CvID: 81.5 J/mol.K. $\endgroup$
    – Harry Ferrier
    Commented Feb 24, 2023 at 9:31
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    $\begingroup$ STATE2: Z: 0.9028, Molar Volume: 1390.6cm3/mol, Enthalpy Departure: -1234.7J/mol, Entropy Departure: -67.1J/mol, CpID: 105.0J/mol.K, CvID: 97.7J/mol.K. $\endgroup$
    – Harry Ferrier
    Commented Feb 24, 2023 at 9:31
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    $\begingroup$ CvID * ln(T2/T1) = (97.7+81.5)/2*ln(463.15/378.15) = 18.16 J/mol.K Very close to your 18.0 K/mol.K $\endgroup$
    – Harry Ferrier
    Commented Feb 24, 2023 at 9:31
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    $\begingroup$ Calculating dS with CpID: dS = (89.8+105.0)/2*ln(463.15/378.15)+(-67.1-(-52.4)) = 5.05 J/mol.K $\endgroup$
    – Harry Ferrier
    Commented Feb 24, 2023 at 9:32
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    $\begingroup$ Calculating dS with CvID: (81.5+97.7)/2*ln(463.15/378.15)+(-67.1-(-52.4)) = 3.47 J/mol.K $\endgroup$
    – Harry Ferrier
    Commented Feb 24, 2023 at 9:32
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Found another resource suggesting the same derivation for entropy departure. Again produces unsatisfactory results. Could there be a units issue?

enter image description here

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