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If I'm correct, one and the same EoS is used for both liquid and gas phase. I'm trying the classical Peng-Robinson equation to get pressure from temperature and density: when using values for the gas phase the result is great, but when using values for the liquid phase, the result is very big negative pressure. What am I missing? My alpha function is one-parameter Soave, oxygen at saturation point

Gas: T = 90.062 K, V = 0.22658 m3/kg, result = 1 bar

Liquid: T = 90.062 K, V = 0.00087581 m3/kg, result = -284 bar

May be the alpha function is different for gas and liquid phases?

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    $\begingroup$ At a given temperature and specific volume, the PR equation in the saturation region gives multiple values for the pressure. $\endgroup$
    – Chet Miller
    Commented May 18 at 10:55

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I think i found the reason. Peng-Robinson is very steep and inaccurate at the gas-liquid boundary and even the smallest change in volume leads to enormous changes in pressure. The error could be as much as 16%.

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  • $\begingroup$ I think you are correct, and the reason should make sense conceptually. The density of a liquid phase is not very strongly dependent on pressure, so back calculating from a density to a pressure is not going to be very accurate because the difference in the densities is going to be very small even for a large change in pressure. $\endgroup$
    – cp95
    Commented 2 days ago
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Yes, the P-R EoS ( and cubic equations in general) is very ill-conditioned for the liquid phase.

Your best bet is to first get the pressure corresponding to the gas-phase temperature and volume. With that done, render the equation into its cubic form

$aV^3+bV^2+cV+d=0$

where $a,b,c,d$ are functions of temperature and pressure. Then synthetically divide out the gas phase volume to get a quadratic equation:

$aV^3+bV^2+cV+d=(V-V_g)[aV^2+(aV_g+b)V+(aV_g^2+bV_g+c)]=0$

And solve the quadratic equation indicated by the second factor for its smallest positive root. This root represents the liquid volume at the same pressure according to the EoS (which may be less than fully accurate). The remaining quadraric root is an unstable state.

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