CHAPTER 2:VAPOR/LIQUID EQUILIBRIUM

VLE BY MODIFIED RAOULT’S LAW

LEARNING OUTCOMES• At the end of the lecture, you should be able to apply

modified Raoult’s law by performing VLE calculations.

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VLE BY MODIFIED RAOULT’S LAW• Raoult's law can be modified by multiplying the right side of the equation

by an activity coefficient, γ. This equation accounts for non-ideal solutions, or non-idealities in the liquid phase.

• This equation is a much more realistic representation of vapor-liquid equilibrium behavior when dealing with low pressures.

• Modified Raoult’s law:

where i is activity coefficient• Activity coefficients are functions of temperature and liquid-phase

composition, and ultimately are based on experiment.• For present purposes, the necessary values are assumed known.

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1,2,..., 10.5sati i i iy P x P i N

• 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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10.6sati i i

iP x P

1 10.7/ sat

i i ii

Py P

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Example 10.3For 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 - 0.00523TIn addition, the following Antoine equations provide vapor pressures:

1 23643.31 2665.54ln 16.59158 ln 14.2532633.424 53.424

sat satP PT T

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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Solution:(a)BUBL P calculation

1. Calculate P1sat and P2

sat using Antoine equation for T = 318.15K

P1sat = 44.51 kPa P2

sat = 65.64 kPa 2. Calculate activity coefficients from the given equation

ln 1 = Ax22 ln 2 = Ax1

2 where A = 2.771 - 0.00523TA = 1.107 1 = 1.864 2 = 1.072

3. Calculate P by eq. (10.6)

P = 73.50 kPa 4. Calculate yi by eq. (10.5)

y1 = 0.282 y2 = 0.718

1 23643.31 2665.54ln 16.59158 ln 14.2532633.424 53.424

sat satP PT T

1 1 1 2 2 2sat sat sat

i i ii

P x P x P x P

sati i i iy x P P

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(b) DEW P calculation Values of P1

sat, P2sat 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 for 1 and 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

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4. Return to the first step. Iterate to convergence on a value for P, xi and i .

11 2 1

1 1and x 1sat

y Px xP

1 1 1 2 2 2

1/ /sat satP

y P y P

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Iteration P x1 x2 1 21 51.1034 0.6886 0.3114 1.1133 1.6903

2 63.6626 0.7705 0.2295 1.0600 1.9294

3 63.0141 0.8010 0.1990 1.0448 2.0345

4 62.9267 0.8115 0.1885 1.0401 2.0732

5 62.9163 0.8151 0.1849 1.0386 2.0865

6 62.9152 0.8163 0.1837 1.0381 2.0909

7 62.9150 0.8166 0.1834 1.0379 2.0924

8 62.9150 0.8168 0.1832 1.0379 2.0929

9 62.9150 0.8168 0.1832 1.0378 2.0931

10 62.9150 0.8168 0.1832 1.0378 2.0931

Initial 1 = 2 = 1,

Final values:P = 62.915 kPa x1 = 0.8168 1 = 1.0378 2 = 2.0931

(c) BUBL T calculation Calculate T1

sat and T2sat using Antoine equation for P = 101.33 kPa

T1sat = 337.71 K T2

sat = 330.08 K A mole fraction weighted average of these values then provides an initial T:

T = x1T1sat + x2T2

sat = 336.57 K

An iterative procedure consists of the steps:1. For the current value of T calculate values for A, 1, 2, P1

sat, P2sat, 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

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11 1 2 2

sat PPx x

11

1 1ln satBT C

A P

lni

i isati i

BT CA P

Iterate to convergence on a value for T.Initial T = x1T1

sat + x2T2sat = 336.57 K

Final values:T = 331.194 K or 58.044oCP1

sat = 77.9966 kPa P2sat = 105.3247 kPa

A = 1.0389 1 = 1.0236 2 = 2.1182Calculate vapor phase mole fractions by eq. (10.5)

y1 = 0.67 y2 = 0.33 10

1 1 11 2 11

satx Py y yP

Iteration A 1 2 P1sat P2sat P1sat new Tnew

1 1.0108 1.0230 2.0757 96.8550 126.3286 0.7667 79.4337 331.6390

2 1.0365 1.0236 2.1147 79.4337 106.9537 0.7427 78.1175 331.2317

3 1.0387 1.0236 2.1179 78.1175 105.4619 0.7407 78.0068 331.1972

4 1.0388 1.0236 2.1182 78.0068 105.3363 0.7406 77.9974 331.1943

5 1.0389 1.0236 2.1182 77.9974 105.3257 0.7405 77.9966 331.1940

6 1.0389 1.0236 2.1182 77.9966 105.3248 0.7405 77.9966 331.1940

7 1.0389 1.0236 2.1182 77.9966 105.3247 0.7405 77.9966 331.1940

8 1.0389 1.0236 2.1182 77.9966 105.3247 0.7405 77.9966 331.1940

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(d) DEW T calculation From part (c),

P = 101.33 kPa T1sat = 337.71 K T2

sat = 330.08 K Initial value of T:

T = (0.4)(337.71) + (0.6)(330.08) = 333.13 K or t = 59.98oC 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 T2. Calculate x1 by eq. (10.5)

3. Calculate 1 and 2 from the correlating equation4. Find a new value for P1

sat from eq. (10.7) written:

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

6. Return to the initial step

1 2

11 2

sat y yP P

11 2 1

1 1and x 1sat

y Px xP

11

1 1ln satBT C

A P

Iterate to convergence on the value of T.Initial T = x1T1

sat + x2T2sat = 333.13 K, 1 = 2 = 1.

Final values:T = 326.69 K or t = 53.54oCP1

sat = 64.64 kPa P2sat = 89.92 kPa

A = 1.0624 1 = 1.3632 2 = 1.2520x1 = 0.4600 x2 = 0.5400

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Iteration A P1sat P2sat x1 x2 1 new 2 new P1sat new Tnew

1 1.0287 84.41 112.55 0.7499 0.4802 0.5198 1.3204 1.2677 66.66 327.42

2 1.0586 66.66 92.29 0.7223 0.4605 0.5395 1.3609 1.2516 64.87 326.78

3 1.0620 64.87 90.19 0.7192 0.4591 0.5409 1.3643 1.2509 64.67 326.70

4 1.0624 64.67 89.95 0.7189 0.4594 0.5406 1.3641 1.2513 64.64 326.69

5 1.0624 64.64 89.93 0.7188 0.4597 0.5403 1.3637 1.2517 64.64 326.69

6 1.0624 64.64 89.92 0.7188 0.4598 0.5402 1.3634 1.2519 64.64 326.69

7 1.0624 64.64 89.92 0.7188 0.4599 0.5401 1.3633 1.2520 64.64 326.69

8 1.0624 64.64 89.92 0.7188 0.4600 0.5400 1.3632 1.2520 64.64 326.69

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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,

1 112

2 210.8y x

y x

sati i i

i

y Px P

1 1

122 2

10.9sat

satPP

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The correlating equations for activity coefficients:ln 1 = Ax2

2 ln 2 = Ax12

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

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

andwhen x1 = 1, ln 2 = Ax1

2 = A(1) = A 2 = exp(A) x2 = 0, ln 1 = Ax2

2 = A(0) = 0 1 = exp(0) = 1

Therefore in these limits,

For T = 318.15 K or 45oC, from part (a) P1

sat = 44.51 kPa P2sat = 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 functionof x1 and must pass through the value of 1.0 at some intermediate composition.

1 1

12 121 0 1 12 2

expand

exp

sat sat

sat satx x

P A PP 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 1

az = Ax22 1

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

With x1az = y1

az, eq. (10.5) becomesPaz = 1

azP1sat = (1.657)(44.51) = 73.76 kPa

x1az = y1

az = 0.325

1

2ln ln1.4747 0.388

1 2

2 1

65.64 1.474744.51

az sat

az satP AP

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

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

REFERENCE• 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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