APPLICATIONS

• Applications of Raoult’s law

• Qualitative description of phase diagrams for mixtures

Raoult’s law

• Model the vapor phase as a mixture of ideal gases:

• Model the liquid phase as an ideal solution

ivi Pyf ˆ

isati

li xPf ˆ

VLE according to Raoult’s law:

222

111

xPPy

xPPysat

sat

Acetonitrile (1)/nitromethane (2)

• Antoine equations for saturation pressures:

209/

64.972,22043.14/ln

224/

47.945,22724.14/ln

2

1

CTkPaP

CTkPaP

osat

osat

Calculate P vs. x1 and P vs. y1 at 75 oC

66.72

0.75

Bubble line

Dew line

Diagram is at constant T

Calculate the P-x-y diagram

satsatsatsatsat

sat

sat

PxPPxPxPP

:Summing

xPPy

xPPy

21211211

222

111

)()1(

12

111

1 yyP

Pxy

sat

Bubble pressure calculations

Knowing T and x1, calculate P and y1

59.74

0.43

Diagram is at constant T

Knowing T and y1, get P and x1

satsat

satsat

satsat

Py

Py

P

summing

xP

PyxPPy

xP

PyxPPy

2

2

1

1

22

2222

11

1111

1

Dew point calculation

In this diagram, the pressureis constant

78oC

0.51 0.67

Calculate a T-x1-y1 diagram

ii

isati

sat

sat

CPA

BT

xTPPy

xTPPy

ln

)(

)(

222

111

get the two saturation temperaturesThen select a temperature from the range between T1

sat and T2sat

At the selected T,summing (1) and (2) solve for x1

(1)

(2)

Why is this temperaturea reasonable guess?

Given P and y1 solvefor T and x1

Given P and x1, get T and y1

212

1221

2

1

2

2211

222

111

xxPP

PPxx

P

P

P

P

xPxPP

xPPy

xPPy

sat

satsat

sat

sat

sat

satsat

sat

sat

Iterate to find T, then calculate y1

209/

64.972,22043.14/ln

224/

47.945,22724.14/ln

2

1

CTkPaP

CTkPaP

osat

osat

209

64.972,2

224

47.945,20681.0ln

2

1

TTP

Psat

sat

212

12

xxPP

PP

sat

satsat

Estimate P1

sat/P2sat using a guess T

Then calculate P2sat from (III)

Then get T from (I)Compare calculated T with guessed T

(I)

(II)

(III)

Finally, y1 = P1sat x1/P and y2 = 1-y2

In this diagram, the pressureis constant

78oC

0.51

76.4

0.75

Dew pointsBubblepoints

Knowing P and y, get T and x

• Start from point c last slide (70 kPa and y1= 0.6)

sat

satsat

satsat

satsat

satsat

P

PyyPP

P

y

P

yP

P

PyxxPPy

P

PyxxPPy

2

1211

2

2

1

1

2

22222

1

11111

1

Iterate to find T, and then calculate x

sat

satsat

P

PyyPP

2

1211

209/

64.972,22043.14/ln

224/

47.945,22724.14/ln

2

1

CTkPaP

CTkPaP

osat

osat

209

64.972,2

224

47.945,20681.0ln

2

1

TTP

Psat

sat

Estimate P1sat/P2

sat using a guess TThen calculate P1

sat from (III)And then get T from (I)

(I)

(II)

(III)

x1= Py1/P1sat

79.6

0.44

Ki = yi/xi

Ki = Pisat/P

ReadExamples 10.4, 10.5, 10.6

Flash Problem

T and P

1 mol ofL-V mixtureoverallcomposition {zi}

V, {yi}

L, {xi}

mass balance:

L + V =1

mass balance component i

zi = xi L + yi V for i = 1, 2, …n

zi = xi (1-V) + yi V

Using Ki values, Ki = yi/xi

xi= yi /Ki;

yi = zi Ki/[1 + V(Ki -1)]

read and work examples 10.5 and 10.6 1

11

i i

ii

ii KV

Kzy

Flash calculations

F=2-+N

For a binary

F=4-

For one phase:P, T, x (or y)

Subcooled-liquidabove the upper surface

Superheated-vaporbelow the under surface

L is a bubble point

W is a dew point

LV is a tie-line

Line of critical points

Each interior loop represents the PTbehavior of a mixture of fixed composition

In a pure component, the bubble and dewlines coincide

What happens at points A and B?

Critical point of a mixture is the point wherethe nose of a loop is tangent to the envelopecurve

Tc and Pc are functions of composition, and do not necessarilycoincide with the highest T and P

How do we calculate a P-T envelope?

Fraction of theoverall systemthat is liquid

At the left of C, reductionof P leads to vaporization

At F, reduction in P leads tocondensation and then vaporization (retrograde condensation)

Important in the operation ofdeep natural-gas wells

At constant pressure, retrograde vaporization may occur

Class exercise

• From Figure 10.5, take P = 800 psia and generate a table T, x1, y1. We call ethane component 1 and heptane component 2. In the table complete all the T, x1, y1 entries that you can based on Figure 10.5. For example, at T= 150 F, x1 = 0.771, we don’t know y1 (leave it empty for now). Continue for all the points at P = 800 psia. Once the table is complete, graph T vs. x1, y1. Also fill in the empty cells in the table reading the values from the graph.

Minimum and maximum ofthe more volatile speciesobtainable by distillation at this pressure(these are mixture CPs)

This is a mixture of very dissimilarcomponents

azeotrope

The P-x curve in (a) lies belowRaoult’s law; in this case there are strongerintermolecular attractions between unlikethan between like molecular pairs

This behavior may result in a minimumpoint as in (b), where x1=y1 Is called an azeotrope

The P-x curve in (c) lies above Raoult’s law; in this case there are weakerintermolecular attractions between unlikethan between like molecular pairs; it could end as L-L immiscibility

This behavior may result in a maximumpoint as in (d), where x1=y1, it is alsoan azeotrope

Usually distillation is carriedout at constant P

Minimum-P azeotrope is amaximum-T (maximum boiling)Point (case b)

Maximum-P azeotrope is a minimum-T (minimum boiling)Point (case d)

Limitations of Raoult’s lawWhen a component critical temperature is < T, the saturation pressure is not defined.

Example: air + liquid water; what is in the vapor phase? And in the liquid?

Calculate the mole fraction of air in water at 25oC and 1 atm

Tc air << 25oC

Henry’s law

iii HxPy

For a species present at infinite dilution in the liquid phase,

The partial pressure of that species in the vapor phase is directly proportional to the liquid mole fraction

Henry’s constant

Calculate the mole fraction of air in water at 25oC and 1 atm.

First calculate y2 (for water, assuming that air does not dissolve in water)

Then calculate x1 (for air, applying Henry’s law)

See also Example 10.2

Modified Raoult’s law

2222

1111

xPPy

xPPysat

sat

Fugacity vapor

Fugacity liquid

is the activity coefficient, a function of composition andtemperature

It corrects for non-idealities in the Liquid phase