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The Peng Robinson Equation of State gives a better approximation of the critical compressibility factor $Z_c$, where experimental data ranges between $0.24$ and $0.28$ (Source). Since the calculated value should be $0.307$ for any substance (Source), I'd expect to obtain a rather close number by re-applying numerically the Equation of State, given a point $(T_c,P_{c})$ on the Clapeyron's Diagram.

However, when I try to solve the Equation of State for $Z_c$ it gives $Z_c=0.321379$ (Mathematica), a solution with discrepancy of $4.6\%$ from the theoretical value.

Also when I use sightly different parameters for this EoS from Wikipedia, the numerical result with the same code is $Z_c=0.311155$.

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  • $\begingroup$ @hazzey in PR EOS, Zc is an independent parameter. please check my solution $\endgroup$ Aug 22 '15 at 14:55
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The original Peng Robinson (1976) paper has the equation rearranged in cubic form with respect to Z.

$$Z^3-(1-B)Z^2+(A-3B^2-2B)Z-(AB-B^2-B^3)=0$$ $$A=\frac{aP}{R^2T^2}$$ $$B=\frac{bP}{RT}$$ $$Z=\frac{Pv}{RT}$$ $$a(T_c)=0.45724\frac{R^2T_c^2}{P_c}$$ $$b(T_c)=0.07780\frac{RT_c}{P_c}$$

At the critical point, A and B are 0.45724 and 0.07780, respectively.

When the cubic equation is solved for its three roots at the critical point, it yields one real root and two imaginary roots. $$Z_c=0.321379$$ $$Z_c=0.30041-0.01199i$$ $$Z_c=0.30041+0.01199i$$

The average of these three roots is: $$Z_c=0.30739967$$

At the critical point, there is only one phase in the system. However, the cubic equation is judged based on how good it is at finding the three roots corresponding to each phase in the system. I assume the standard practice in comparing cubic equations of state is to take the average of all three roots whether they are real or not.

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    $\begingroup$ Out of curiosity, I tried this with the Redlich-Kwong EOS. The roots are 0.30492, 0.34754-0.029245i, and 0.34754+0.029245i. The average of these three is 0.333, same as the published critical compressibility factor. $\endgroup$
    – morristtu
    Nov 4 '15 at 16:17
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When a cubic EOS is solved at the critical point, it MUST have 3 equal roots, because the parameter equations were derived in order to satisfy the critical constraints (inflection point with zero slope at the critical point). E.g. for the R-K Eos Zc=1/3 (Giorgio Soave)

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The critical compressibility factor Zc for the Peng Robinson equation of state is reported as 0.307 in Walas, Phase Equilibria in Chemical Engineering. The coefficients were 0.45724 and 0.07780. Using Excel, I found Zc = 0.3213. Using 0.457235 and 0.077796 as reported in Stryjek and Vera's paper, Zc = 0.3111. You can see that the solution is very sensitive to the values used. Still, I can't get 0.307 either.

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  • $\begingroup$ did you check @morristtu answer? $\endgroup$ Jan 19 '17 at 21:49
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It is important to understand that the critical point of a pure substance is characterized by certain properties. For example, the first and second partial derivative of the pressure with respect to the molar volume at constant temperature vanish. Secondly, it is important to understand that the critical point given by an equation of state (EoS) is not necessarily the same point as determined by measurements.

To find the critical point of the Peng-Robinson (PR) EoS one can calculate the first and second partial derivative with respect to the molar volume, and look for a given volume/temperature pair where both partial derivatives vanish. Doing this with a solver for non-linear equation systems (e.g. a Levenberg-Marquardt method) I found the critical point of CO2 to be close to $1.0T_c$ and $0.89\rho_c$, giving a critical pressure of $1.0p_c$, which yields a critical compressibility of $Z_c=0.3074$.

To get good results I recommend the following:

  • If you use iterative solvers be sure to check residuals and convergence of the method.
  • Use the same constants and critical state variables everywhere, as the solution is quite sensitive.
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    $\begingroup$ It just so happened that I stumbled upon morristtu's answer. Yesterday, I didn't take the time to furmulate a "complete" answer - sorry. However, I'd like to hear some comments on why it is justified to leave the real domain, and which three distinct phases should be present at the critical point - Two aspects I found rather confusing and maybe misleading in the accepted answer. $\endgroup$ Apr 5 '17 at 18:53
  • $\begingroup$ Thanks for taking the time to edit your answer. I have cleaned up the now obsolete comments. $\endgroup$
    – user16
    Apr 5 '17 at 23:55

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