SM_14A - Chapter 14 Section A Mathcad Solutions 14.1 A12:=...

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Φ 2 P T , y 1 , y 2 , ( 29 exp B 22 P Psat 2 - ( 29 P y 1 2 δ 12 + R T = Φ 1 P T , y 1 , y 2 , ( 29 exp B 11 P Psat 1 - ( 29 P y 2 2 δ 12 + R T = δ 12 2 B 12 B 11 - B 22 - = B 12 52 cm 3 mol = B 22 1523 - cm 3 mol = B 11 963 - cm 3 mol = BUBL P calculations with virial coefficients: (b) y 1 x 1 ( 29 0.808 = P bubl x 1 ( 29 85.701kPa = x 1 0.75 = y 1 x 1 ( 29 0.731 = P bubl x 1 ( 29 80.357kPa = x 1 0.50 = y 1 x 1 ( 29 0.562 = P bubl x 1 ( 29 64.533kPa = x 1 0.25 = y 1 x 1 ( 29 x 1 γ 1 x 1 ( 29 Psat 1 P bubl x 1 ( 29 = P bubl x 1 ( 29 x 1 γ 1 x 1 ( 29 Psat 1 1 x 1 - ( 29 γ 2 x 1 ( 29 Psat 2 + = BUBL P calculations based on Eq. (10.5): (a) Psat 2 37.31 kPa = Psat 1 82.37 kPa = γ 2 x 1 ( 29 exp x 1 2 A 21 2 A 12 A 21 - ( 29 1 x 1 - ( 29 + = γ 1 x 1 ( 29 exp 1 x 1 - ( 29 2 A 12 2 A 21 A 12 - ( 29 x 1 + = Margules equations: T 55 273.15 + ( ) K = A 21 1.42 = A 12 0.59 = 14.1 Chapter 14 - Section A - Mathcad Solutions 463
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