VLE PART 2

CHE553 Chemical Engineering Thermodynamics 3/12/2015

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HENRY’S LAW

Henry’s law states that the partial pressure of the species in the vapor phase is

directly proportional to its liquid-phase mole fraction.

where Hi is Henry’s constant. Values of Hi come from experiment.

Application of Raoult’s law and Henry’s law:

Liquid mixtures that form an ideal solution at low pressure

Use Raoult’s law for all species

Liquid mixtures that do not form an ideal solution and at low pressure

Use Raoult’s law for solvent

Use Henry’s law for solute

Gases that dissolve in a liquid mixture (e.g. air in water)

Use Henry’s law for the dissolved gaseous species

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10.4i i iy P x H

Table 10.1 lists Henry’s constant values at 25oC (298.15K) for a few gases dissolved in water.

Henry’s constant is also an indicator for solubility of a gas in the solvent. The greater the Henry’s constant, the lower is the solubility.

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Least soluble

in water

Very soluble

in water


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Assuming that carbonated water contains only CO2(1) and H2O(2), determine the compositions of the vapor and liquid phases in a sealed can of “soda” and the pressure exerted on the can at 10oC (283.15K). Henry’s constant for CO2 in water at 10oC (283.15K) is about 990 bar.

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Example 10.2 Solution:

Given Henry’s constant for CO2 in water = 990 bar (i.e CO2 dissolved in water)

Henry’s law is applied to species 1 and Raoult’s law is applied to species 2 :

y1P = x1H1 y2P = x2P2sat

F = 2-2+2 = 2, only T is given. Need to specify the mole fraction to solve the problem. Assume the vapor phase is nearly pure CO2 (y11) and the liquid phase is nearly pure H2O (x21). Take x1 = 0.01.

Sum the two equations give

P = x1H1 + x2P2sat

With H1 = 990 bar, x1 = 0.01 and P2sat = 0.01227 bar (Table F.1 at 10oC or

283.15 K),

P = (0.01)(990) + (0.99)(0.01227) = 9.912 bar

Then by Raoult’s law, eq. (10.1) written for species 2:

y1 = 1 - y2 = 1 - 0.0012 = 0.9988, and the vapor phase is nearly pure CO2

as assumed.

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2 2

2

0.99 0.012270.0012

9.912

satx Py

P

In real life, most solutions are non ideal i.e they do not obey Raoult’s law. Non ideal solution deviate from Raoult’s law.

The non ideality exhibit by these solutions due to dissimilarity in the sizes of molecules, shapes of the molecules, and inter-molecular interactions between molecules.

Modified Raoult’s law accounts for non-ideal solutions, or non-idealities in the liquid phase:

where i is activity coefficient.

There are a number of correlations that can be used to predict the value of activity coefficient e.g. Margules, Van Laar, Non-Random Two Liquid (NRTL), Wilson, etc.

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VLE BY MODIFIED RAOULT’S LAW

1,2,..., 10.5sat

i i i iy P x P i N

The simplest correlation to predict activity coefficient for binary mixture is as follow:

ln 1 = Ax22 ln 2 = Ax1

2

where A is a function of temperature only or simply a constant.

This correlation can only be used in some liquid mixtures which exhibit small deviation from ideal solution.

Bubblepoint and dewpoint calculation by modifed Raoult’s law:

Because iyi = 1, eq. (10.5) may be summed over all species to yield

Because ixi = 1, eq. (10.5) may be summed over all species to yield

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10.6sat

i i i

i

P x P

1

10.7/ sat

i i i

i

Py P


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Example 10.3

For the system methanol(1)/methyl acetate(2), the following equations provide a reasonable correlation for the activity coefficients:

ln 1 = Ax22 ln 2 = Ax1

2 where A = 2.771 - 0.00523T

In addition, the following Antoine equations provide vapor pressures:

where T is in kelvins and the vapor pressures are in kPa. Assuming the validity of eq. (10.5), calculate

(a) P and {yi}, for t/T = 45oC/318.15K and x1 = 0.25

(b) P and {xi}, for t/T = 45oC/318.15K and y1 = 0.60

(c) T and {yi}, for P = 101.33 kPa and x1 = 0.85

(d) T and {xi}, for P = 101.33 kPa and y1 = 0.40

(e) The azeotropic pressure, and the azeotropic composition, for t/T =

45oC/318.15K

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1 2

3643.31 2665.54ln 16.59158 ln 14.25326

33.424 53.424

sat satP PT T

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Solution:

(a) BUBL P calculation (given T = 318.15K and x1 = 0.25)

1. Calculate P1sat and P2

sat using Antoine equation for T = 318.15K

2. Calculate activity coefficients from the given equation

ln 1 = Ax22 ln 2 = Ax1

2 where A = 2.771 - 0.00523T

3. Calculate P by eq. (10.6)

4. Calculate yi by eq. (10.5)

1 2

3643.31 2665.54ln 16.59158 ln 14.25326

33.424 53.424

sat satP PT T

1 1 1 2 2 2

sat sat sat

i i i

i

P x P x P x P

sat

i i i iy x P P

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(b) DEW P calculation (given T = 318.15K and y1 = 0.60)

Values of P1sat, P2

sat and A are unchanged from part (a).

Iteration procedure is required to find the liquid phase composition

and to calculate the activity coefficients.

Initial values are provided by Raoult’s law, for which 1 = 2 = 1.0.

The required steps, with the current value of 1 and 2, are:

1. Calculate P by eq. (10.7), written

2. Calculate x1 by eq. (10.5)

3. Evaluate activity coefficients;

ln 1 = Ax22 ln 2 = Ax1

2

4. Return to the first step.

Iterate to convergence on a value for P.

11 2 1

1 1

and x 1sat

y Px x

P

1 1 1 2 2 2

1

/ /sat satPy P y P

(c) BUBL T calculation (given P = 101.33 kPa and x1 = 0.85)

Calculate T1sat and T2

sat using Antoine equation for P = 101.33 kPa

A mole fraction weighted average of these values then provides an initial T:

T = x1T1sat + x2T2

sat

An iterative procedure consists of the steps:

1. For the current value of T calculate values for A, 1, 2, P1sat, P2

sat, and = P1

sat/P2sat

2. Find a new value for P1sat from eq. (10.6) written:

3. Find a new value for T from the Antoine equation for species 1

4. Return to the initial step

Iterate to convergence on a value for T.

12

1

1 1 2 2

sat PP

x x

11

1 1lnsat

BT CA P

ln

sat ii i

i

BT C

A P


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(d) DEW T calculation (given P = 101.33 kPa and y1 = 0.40)

Calculate T1sat and T2

sat using Antoine equation for P = 101.33 kPa

Initial value of T:

T = y1T1sat + y2T2

sat

Because the liquid phase composition is unknown, the activity coefficients

are initialized as 1 = 2 = 1.

Iterative procedure:

1. Evaluate A, P1sat, P2

sat, and = P1sat/P2

sat for the current value of T

2. Calculate x1 by eq. (10.5)

3. Calculate 1 and 2 from the correlating equation

4. Find a new value for P1sat from eq. (10.7) written:

5. Find a new value of T from Antoine equation for species 1

6. Return to the initial step

Iterate to convergence on the value of T.

1 21

1 2

sat y yP P

11 2 1

1 1

and x 1sat

y Px x

P

11

1 1lnsat

BT CA P

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(e) First determine whether or not an azeotrope exists at the given

temperature.

This calculation is facilitated by the definition of a quantity called the

relative volatility:

At an azeotrope y1 = x1, y2 = x2, and 12 = 1. By eq. (10.5),

Therefore,

1 1

12

2 2

10.8y x

y x

sat

i i i

i

y P

x P

1 112

2 2

10.9sat

sat

P

P

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The correlating equations for activity coefficients:

ln 1 = Ax22 ln 2 = Ax1

2

when x1 = 0, ln 2 = Ax12 = A(0) = 0 2 = exp(0) = 1

x2 = 1, ln 1 = Ax22 = A(1) = A 1 = exp(A)

and

when x1 = 1, ln 2 = Ax12 = A(1) = A 2 = exp(A)

x2 = 0, ln 1 = Ax22 = A(0) = 0 1 = exp(0) = 1

Therefore in these limits, eqn. (10.9) becomes

For T = 318.15 K or 45oC, from part (a)

P1sat = 44.51 kPa P2

sat = 65.64 kPa A = 1.107

The limiting values of 12 are therefore

(12)x1=0 = 2.052 (12)x1=1 = 0.224

Because the value at one limit is greater than 1, whereas the value at the

other limit is less than 1, an azeotrope does exist. Because 12 is a

continuous function of x1 and must pass through the value of 1 at some

intermediate composition.

1 112 121 0 1 1

2 2

expand

exp

sat sat

sat satx x

P A P

P P A

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For the azeotrope, 12 = 1, and eq. (10.9) becomes

The difference between the correlating equations for ln 1 and ln 2 provides

the general relation:

Thus the azeotropic occurs at the value of x1 for which this equation is

satisfied when the activity coefficient ratio has its azeotrope value of 1.4747

(from eq. (A)).

When ln 1/2 = 0.388, substitute into eq. (B), gives x1az = 0.325. For this

value of x1,

ln 1az = Ax2

2 1az = exp[(1.107)(1-0.325)2] = 1.657

With x1az = y1

az, eq. (10.5) becomes

Paz = 1azP1

sat = (1.657)(44.51) = 73.76 kPa

x1az = y1

az = 0.325

1

2

ln ln1.4747 0.388

1 2

2 1

65.641.4747

44.51

az sat

az sat

PA

P

2 212 1 2 1 2 1 2 1 1 1 1

2

ln 1 1 2Ax Ax A x x x x A x x A x x A x B


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Smith, J.M., Van Ness, H.C., and Abbott, M.M. 2005. Introduction to Chemical Engineering Thermodynamics. Seventh Edition. Mc Graw-Hill.

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REFERENCE

PREPARED BY: MDM. NORASMAH MOHAMMED MANSHOR FACULTY OF CHEMICAL ENGINEERING, UiTM SHAH ALAM. [email protected] 03-55436333/019-2368303

VLE PART 2

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