Engineering & Science
Pitzer Acentric Factor (ω) from Reduced Vapor Pressure at Tr = 0.7
Calculate Kenneth Pitzer’s acentric factor ω = -1.000 - LOG10(P_sat(0.7 * T_c) / P_c) and invert it to estimate saturation vapor pressure at T_r = 0.7 for cubic equations of state.
Why Pitzer anchored ω = 0 to spherical noble gases at T_r = 0.70!
Two-parameter corresponding-states theory claimed every fluid should share the exact same reduced vapor pressure P_r^sat = P_sat / P_c at the same reduced temperature T_r = T / T_c. Spherical noble gases (Argon, Krypton, Xenon) obey that rule beautifully, all hitting P_r^sat ≈ 0.1000 (LOG10 = -1.000) right at T_r = 0.70! Non-spherical alkane chains and polar molecules exhibit steeper vapor-pressure slopes so P_r^sat(0.7) drops well below 0.10—prompting Kenneth S. Pitzer (JACS, 1955) to define the acentric factor: ω = -1.000 - LOG10(P_sat(0.7 * T_c) / P_c)!
Watch how cleanly ω classifies molecular geometry: spherical Argon (P_sat(0.7 T_c) = 4.864 bar, P_c = 48.98 bar) gives ω = +0.0030 (~0); chain-like Propane (P_sat(0.7 T_c) = 2.992 bar, P_c = 42.48 bar) gives ω = 0.1522; hydrogen-bonded Water (P_sat(0.7 T_c) = 10.02 bar, P_c = 220.64 bar) climbs to ω = 0.3428!
| Pure Fluid Class | Psat at T = 0.7 Tc / Pc | Pitzer Acentric Factor ω | Molecular Force Symmetry |
|---|---|---|---|
| Argon Noble Gas (4.864 / 48.98 bar) | Pr = 0.0993 | 0.0030 (~0.00) | Spherical Lennard-Jones Core |
| Propane Alkane (2.992 / 42.48 bar) | Pr = 0.0704 | 0.1522 | Elongated Non-Spherical Chain |
| Water Polar Fluid (10.02 / 220.64 bar) | Pr = 0.0454 | 0.3428 | Strong Hydrogen-Bond Dipoles |
Pitzer Acentric Factor ω = -1 - LOG10(P_sat,0.7 / P_c)
Computes -1 - LOG10(B2 / B3) from saturation pressure B2 measured at T = 0.70 * T_c and critical pressure B3 (in matching pressure units).
=IF(OR(B2 <= 0, B3 <= B2), "Invalid Vapor Pressure Inputs", ROUND(-1 - LOG10(B2 / B3), 4))Reduced Saturation Pressure P_r^sat at T_r = 0.70
Returns dimensionless P_sat(0.7 T_c) / P_c (= 0.1000 for ideal simple spherical fluids).
=IF(OR(B2 <= 0, B3 <= B2), "Invalid Vapor Pressure Inputs", ROUND(B2 / B3, 4))Try the example
Use the original values or change them to test your understanding.