SM_14A - 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

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 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- ( 29γ 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 y 1 y 2 P kPa 0.812 0.188 85.14 = y 1 y 2 P Find y 1 y 2 , P , ( 29 = y 2 1 y 1- = y 2 Φ 2 P T , y 1 , y 2 , ( 29 ⋅ P ⋅ 1 x 1- ( 29γ 2 x 1 ( 29 ⋅ Psat 2 ⋅ = y 1 Φ 1 P T , y 1 , y 2 , ( 29 ⋅ P ⋅ x 1 γ 1 x 1 ( 29 ⋅ Psat 1 ⋅ = Given x 1 0.75 = y 1 y 2 P kPa 0.733 0.267 79.621 = y 1 y 2 P Find y 1 y 2 , P , ( 29 = y 2 1 y 1- = y 2 Φ 2 P T , y 1 , y 2 , ( 29 ⋅ P ⋅ 1 x 1- ( 29γ 2 x 1 ( 29 ⋅ Psat 2 ⋅ = y 1 Φ 1 P T , y 1 , y 2 , ( 29 ⋅ P ⋅ x 1 γ 1 x 1 ( 29 ⋅ Psat 1 ⋅ = Given x 1 0.50 = y 1 y 2 P kPa 0.558 0.442 63.757 = y 1 y 2 P Find y 1 y 2 , P , ( 29 = y 2 1 y 1- = y 2 Φ 2 P T , y 1 , y 2 , ( 29 ⋅ P ⋅ 1 x 1- ( 29γ 2 x 1 ( 29 ⋅ Psat 2 ⋅ = y 1 Φ 1 P T , y 1 , y 2 , ( 29 ⋅ P ⋅ x 1 γ 1 x 1 ( 29 ⋅ Psat 1 ⋅ = Given x 1 0.25 = y 2 1 y 1- = y 1 0.5 = P Psat 1 Psat 2 + 2 = Guess: 464 Combining this with Eq. (12.10a) yields the required expression ln γ 1 ∞ A 12 = It follows immediately from Eq. (12.10a) that: (a) Psat 2 P 1 = x 2 1 x 1- = i 2 rows P ( ) .. = y 1 0.000 0.2716 0.4565 0.5934 0.6815 0.7440 0.8050 0.8639 = P 12.30 15.51 18.61 21.63 24.01 25.92 27.96 30.12 = x 1 0.000 0.0895 0.1981 0.3193 0.4232 0.5119 0.6096 0.7135 = Data: Pressures in kPa 14.4 Ans....
View Full Document

This note was uploaded on 01/29/2011 for the course CHEM 101 taught by Professor Brown during the Spring '10 term at The University of Akron.

Page1 / 49

SM_14A - 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

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online