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Chapter 8 Solutions Engineering and Chemical Thermodynamics 2e

Milo Koretsky Wyatt Tenhaeff School of Chemical, Biological, and Environmental Engineering Oregon State University [email protected]

8.1 No this will not work. A central idea behind binary phase equilibrium is that when both species are volatile, there will be a mixture of 1 and 2 in both the vapor and the liquid. There will be a higher mole fraction of the lighter component, toluene, in the liquid relative to the mole in the vapor, but there will also be cyclohexane present.

2

8.2 Yes if species b is the lighter component, the phase diagram for the system looks approximately like the one below. We can see that both xa and ya are smaller at the higher pressure, P2:

Pbsat

P2

xa

P1

Pasat

ya

xa y a

3

8.3 The addition of the second species to the liquid will lower the liquid mole fraction of water. Thus, the fugacity of water in the liquid will be lower than its fugacity in the vapor. Since T and P are constant, the fugacity of the steam remains constant. Since the fugacity of water in liquid is lower than vapor, it will all condense.

4

8.4 The addition of the second species to the liquid will lower the liquid mole fraction of water. Thus, the fugacity of water in the liquid will be lower than its fugacity in the vapor and the steam will condense. However, since the container is rigid, the pressure of the vapor will decrease as the steam condenses, lowering its fugacity. Therefore, some of the steam will condense until the pressure in the vapor is low enough that the fugacity in the vapor matches the fugacity in the liquid.

5

8.5 When the third component is added to the liquid, the mole fraction of the water in the liquid decreases so tis fugacity decreases. Thus, the fugacity of water in the vapor will be greater than in the liquid and some water will condense, lowering the number of moles in the vapor.

6

8.6 (a) since i 0. This indicates that like interactions are stronger than unlike interactions. (b)

P  xa a Pasat  xb b Pbsat RT ln  a   A  3B  x2b  4Bx3b  3174.4 J/mol RT ln  b   A  3B  xa2  4Bxa3  334.4 J/mol

a  3.60 b  1.14 From the Antoine equation: 3166.38  Pasat  exp 10.9237   0.0270 bar  298.15  80.15 

3816.44  Pbsat  exp 11.6834   0.0312 bar  298.15  46.13 

Solving for pressure: P  0.048 bar

(c) Solving for mole fraction x  P sat ya  a a a  0.41 P

35

8.30

Determine the temperature and the vapor phase of mole fraction a:   2.74  7600 J mol

  0.6

  74.5kPa

  exp







Low P ideal gas

 P    P

 P    P

P    P    P   exp 

A   A   P   exp    P  

74.5kPa  0.6exp󰇟2.74󰇛0.4󰇜 󰇠  0.4exp󰇟2.74󰇛0.6󰇜 󰇠  → steamtables   → Antoineequation ln P  9.2806 

Guess and check:

2788.51   52.36

TC

 

P 󰇛kPa󰇜 P 󰇛kPa󰇜

P(kPa)

50C

12.35

36

50.3

70C

31.19

73

108

*60C

19.94

52

74.5

  P  0.25 

36

8.31 (a) Given:

x1  0.471 x2  1  x1  0.529 P  0.073 bar From the Antoine Equation at 20 oC: sat

 0.10 bar

sat 2

 0.29 bar

P1 P

Setting fugacities equal: y1 P  x1 1 P1sat y2 P  x2 2 P2

sat

P  x1 1P1sat  x 2 2 P2sat

Applying the two-suffix Margules Equation

 A 2 x2   RT   A 2 2  exp  x1   RT 

1  exp 

So

 A 2  sat  exp  A 2  sat P  x1 exp  x 2  P1 x2  RT x1  P2  RT    Solving for A:  J  A  1220    mol 

(b) We can examine the Margules parameter A and we find that A > 0. This indicates that like interactions are stronger than unlike interactions.

37

(c)

 1   2  exp 

A  1.68  RT 

[ 1   1 P1sat  0.172 bar [ 2   2 P2sat  0.048 bar (d)

x P y1  1 1 1 P

sat

 0.77

(e) Lever rule nl y z 0.77  0.7  1 1  0.23 T n y1  x1 0.77  0.471

nl  2.3 mol

38

8.32 To calculate the pressure, we can use the following:         

To calculate the mole fraction of vapor we use:  

   

Start by calculating  & , using the three-suffix Margules model

RT ln a   A  3 B xb2  4 Bxb3

RT ln  b  A  3B xa2  4 Bxa3

Next we need to use the Antoine equation to calculate Psat for each of the species in our solution.

3166.38   Pasat  exp 10.9237   0.0270 bar  298.15  80.15  3816.44   Pbsat  exp 11.6834   0.0312 bar  298.15  46.13 

 This can be determined by plotting xa and ya vs P 0.06

0.05

Pressure[bar]

0.04

0.03

0.02

0.01

0.00 0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

xa,ya

Where xa is the blue line and ya is the red line. We can see that it does form an azeotrope. Next we can determine the pressure at which that azeotrope occurs via solver or other analysis. P = 0.0477 bar

39

8.33 At the azeotrope

x1  y1

x2  y2

Therefore, the expressions equating liquid and vapor fugacities simplify to

P   1P1sat

P   2 P2sat

(Note: We are also assuming that the vapor behaves ideally, which is reasonable since the pressure is 122.3 torr.) We can calculate the saturation pressures at 25 ºC using Antoine’s equation data in Appendix A.

3803.98    0.078 bar 58.5 torr P1sat  exp12.2917  298.15  41.68   2788.51    0.127 bar 95.2 torr  P2sat  exp 9.2806  298.15  52.36   For the van Laar equation

 A  Bx 2  1  exp  RT  Ax1  Bx2 

  

2

  

 B  2  exp   RT 

 Ax1   Ax1  Bx 2

2

  

  

Substitute these expressions into the above equations for pressure. We have

 P ln  P sat  1

2  Bx 2    A    RT  Ax 1  Bx 2  

 P ln  P sat  2

 Ax1   B   RT  Ax1  Bx2 

Inserting numerical values, we have two equations for two unknowns:   B  0.72  122.3 A   ln   58.5  8.314  298.15   A 0.28   B  0.72 

2

  A  0.28  122.3 B   ln   95.2  8.314  298.15  A 0.28   B  0.72 

2

Solve the equations simultaneously:

40

  

2

 J  A  6416.0   mol  J  B  2856.6   mol Now, to calculate the pressure and liquid composition when the vapor mole fraction of ethanol is 0.75, we can use the following equations

y1P  x1 1P1sat

y 2 P  x 2 2 P2sat

Therefore,

x1 1P1sat y1 P y1   y2 P y 2 x2  2 P sat 2 y1 P2sat y2 P1sat

x  1 1 x2  2

Substitute the activity coefficient expressions for the van Laar equation:

y1P2sat y2 P1sat

2 2   A  Bx   A  B 1 x1    2  x1 exp   x1 exp      RT  Ax1  B1 x1     RT  Ax1  Bx2       2 2      B  Ax1   Ax1 B   x2 exp     1 x1  exp  RT  Ax1  B 1 x1     RT  Ax1  Bx2    

Insert numerical values and solve for x1 :

x1  0.935  x 2  0.065 Now, we can calculate the pressure

P

x11 P1sat y1

2    2856.6 0.065 6416      0.935 58.5 torr  exp   8.314 298.15  6416 0.935  2856.6 0.065    0.75

P  73.1  torr 

41

8.34 (a) The molar volume can be found by manipulating the expression given in the problem statement:

RT  P Ay 1y 2  B  P  8.314 298.15  90 10 5 8.314 298.15  2 10 14 1 / 3 2 / 3  8 1015 v        90  105 3 4  m  v  3.55  10    mol

v









We calculate the total volume as follows

V  n T v  15 mol 3.55 104 m3 / mol  5.33103  m3  The molar volume of species 2 is calculated by letting y 2  1 .

v2 

 8.314  298.15  90  10

5

 90 10  8.314 298.152 10   01  8 10 14

5

15

 

 m3  v2  4.54 10  4    mol The partial molar volume of species 2 is calculated by evaluating the appropriate derivative:

 nn   V    n1  n 2 RT  V2     P  A 1 2  Bn1  n2    P  n1  n2   n 2 T , P , n1 n 2  RT V2   P  Ay12  B P 3 4  m  V2  4 .04  10    mol (b) To find the pure species fugacity coefficient, first find an expression for the pure species fugacity:

 f 2v   v dP RT ln    2 Plow  Plow P

42

 f2v    RT  PB dP  RT ln      Plow  Plow  P P

 P  B 2  f2v  2 RT ln  RT ln P  P      low   Plow   Plow  2 If we let Plow go to zero, cancel the remaining Plow’s, and simplify, we obtain

 B 2 P f 2v  P exp   2RT   B 2 P 1.38   v2  exp  2RT  To find the fugacity coefficient of species 2 in the mixture, we use the following procedure:  fˆ v   V2 dP  RT ln  y 2 P2low  Plow   P  fˆv    RT 2  B dP  RT ln  P Ay  2    P  1    y2 Plow  Plow P

 fˆ v   P  1 2 2 2  P  B  RT ln  2  RT ln    P  low  Ay1  Plow  2  y 2 Plow  If we let Plow go to zero, cancel the remaining Plow’s, and simplify, we obtain

 P2  fˆ2v  y2 P exp  Ay12  B    2RT  2   P  ˆv2  exp Ay12  B  1.26   2RT  (c) Plot the activity coefficient of species 2 versus its liquid mole fraction:

43

Henry's

Plot of Activity Coefficient  2 1.2

2

Henry's

1 0.8 0.6 0.4 0.2 0 0

0.2

0.4

0.6

0.8

1

1.2

x2

The activity coefficient based on Henry’s Law is less than one. Thus, the fugacity is less than the fugacity based solely on 1-2 interactions. Consequently, the like interactions must be stronger. (d) For phase equilibrium of species 2 to exist, the following must be true:

y2 ˆ2v P  x2 2Henry 's[ 2 However, we can’t calculate the activity coefficient until we know the liquid mole fraction. Therefore, we need to make an expression for the activity coefficient containing the mole fraction:  x2 

y2ˆ2v P y2 ˆ2v P  's  Henry H2 exp 7 1 1 x2 2  H 2 2  





We can find the mole fraction with a solver function:

x2  0.013 (e) Initially assume the fugacity coefficient is one at the saturation pressure and the Poynting correction is negligible. Use the following equation: Henry 's 2



P2sat [2

 P2sat   7000 bar exp 7 1 0 2   6.38  bar

44

The fugacity coefficient at saturation can be found according the the result of part (b):

 B 2 P 1.00  2RT 

 2sat  exp 

so the assumption is valid.

45

8.35 At the azeotrope

xa  y a

xb  y b

Therefore, the expressions equating liquid and vapor fugacities simplify to

P   a Pasat P   b Pbsat (Note: We are also assuming that the vapor behaves ideally.) Since the liquid-phase nonideality is represented by the two-suffix Margules equation, the following expressions are used to calculate the activity coefficients:

 A 2  A 2 xb   exp 1 x a     RT   RT   A 2  b  exp xa  RT 

 a  exp

Substitute these expressions into the pressure equations and equate the pressures. Solve for xa:

 A 2 A 2 x Pasat exp  1 xa    Pbsat exp  RT a  RT      2900 2900 2 2 xa  68.8 kPa exp 1 xa     46.5 kPa exp   8.314  328.15   8.314   328.15  

xa  0.68 Now, calculate the activity coefficient of methanol.

 2900 0.32 2    1.11  8.314 328.15 

 a  exp

Use this value to calculate the pressure.

P  1.11 68.8  kPa   76.4 kPa

46

Thus, an azeotrope forms at a pressure of 76.4 kPa and a temperature of 55 ºC. The composition is 68% methanol and 32% ethyl acetate. This is a minimum-boiling azeotrope (positive deviations from Raoult’s Law) because the azeotropic pressure is greater than the two saturation pressures.

47

8.36

  RTln  8.314󰇛70  273.15󰇜ln99.7  13,130 J mol

  4.60 RT

    

RTln    󰇛1   󰇜

   1 →     exp 

 

 󰇛1   󰇜   1 RT

  0.011, 0.75, 1

The value 0.75 makes physical sense.

48

8.37

System Properties:

Toluene Data:

P  20.57bar T  301o C  574.15 o K

Antoine A  9.3935 B  3096.52

TC 591.7 o K PC  41.14bar

C   53.67

  0.257

fˆ1v  f1v  1 P

1  general correlation

PR 

P  0.5 PC

TR 

574.15  0.77 591.7

From Appendix C log  ( o)   0.086 and log  (1)   0.023

log 1  log ( o)   log (1)   0.092

1  0.810 Vapor is pure, liquid is a binary mixture

f1  1P  f1v  fˆ1l  x1 11sat P1sat x1 1 

v

f1 16.6   1sat P1sat  1sat 31.3

To find 1s at we need to linearly interpolate PRsat 

574.15 31.3  0.77  0.761 , TR  591.7 41.14 0.7

0.8

0.76

-0.126

-0.148

-0.139

log  (1) -0.034

-0.042

-0.039

PR

log 

( o)

sat log1   0.149

49

1sat  0.709 x1 1  0.75 From the graph, and more linear interpolation

x1

1

x11

0.7 0.75

1.06 1.04

0.742 0.78

x1  0.71

50

ln   ln  → 2suffixMargules → ln   0.75

8.38

 





 0.4

J RT ln       1953 mol

low,  , 

           

  80kPa   60kPa

        

(a)

  80.3kPa

     exp  󰇛1   󰇜    󰇛1   󰇜exp     RT RT  

   

 0.45

        10mol

          4mol     󰇛10   󰇜  4

 10  4    3.90mol    (b) At azeotrope.

  ,  ,  → 1

      

  8  exp 󰇩 󰇟  󰇛1   󰇜 󰇠󰇪   6 RT  RT ln󰇛1.33󰇜  1  2  0.38  1  2

2 

1  0.38  0.69 2

51

        exp RT 󰇛1   󰇜    85.9kPa

(c) Since A>0, the like interactions are stronger

52

8.39 (a) If this mixture is in equilibrium with vapor under these conditions, what is the vapor phase mole fraction of a?   1mol

      󰇡

  3mol



 

Table 7.1  󰇤  

   4mol

RTln  A 󰇣

P  0.50bar  50kPa

ln  0.687

T  25

   0.25  





 1703

  1.99



   0.75

A  4010 Jmol

󰇢

 

  P 0.25󰇛1.99󰇜75kPa  0.745  50 P

(b) Determine saturation pressure of b.

P    P    P

 A RTln  B    303 A  B

P 

ln  0.122   1.13

50  0.25󰇛1.99󰇜75  15kPa 0.75󰇛1.13󰇜

53

B  2501 Jmol

8.40

(a) Plot shows Pxy phase diagram for a binary mixture of species 1 and 2 at 300K.

Vapor, y1=0.67 (b)

P ~ 50 kPa yA~0.77 xA~0.47

54

8.41 (a)

  7850 Jmol

  0.6

  3410 Jmol

  0.4

RTln  B 

 A   2050 Jmol A  B

ln 

2050  0.679 8.314363

From steam tables:   70.1kPa   1.97

(b)

󰆹      0.4󰇛1.97󰇜󰇛70.1kPa󰇜  55.3kPa 󰆹     55kPa

(c)

 

[







 → 0

[





RTln    3410 Jmol ln  1.13

[



  3.09

 3.09󰇛70.1kPa󰇜  217kPa

55

8.42

(a) 1 bar (b) (i) liquid (ii) x1 = 0.2 (iii) 1 mol (c) z1 = 0.6 (i) 2 phases, vapor and liquid (ii) x1 = 0.4, y 1 = 0.74 (iii) apply the lever rule n 0.74  0.6   0.7 nv 0.6  0.4 n l  0.82 mol l

n  1.18 mol v

(d)

y1 P  x11 P1sat sat

1 

x1P1  0.74 y1P

56

A

RT ln  1 J  2, 040 2 mol x2

(e) y1, A  0.74 x1, A  0.4 F1  F2  V A  L A F1  y1, AVA  x1, A LA mol s mol LA  0.82 s

VA  1.18

(f) y1, B  0.9 x1, B  0.62 VA  VB  LB y1, AVA  y1, BVB  x1, B LB mol s mol LB  0.67 s

VB  0.51

57

8.43  󰇟    󰇠 󰆹             exp    1 RT ln







RT RT  B ⟹   󰇟 B  2  B   B󰇠   

 󰆹          1 ,,  



󰇛   󰇜RT RT    󰇛 B    B  B󰇜 V V



1 ,, 

RT ln ln

 ,



RT RT  󰇛 B  2 B󰇜 V V

 󰆹  RT      



1 1 󰇧   󰇛2 B  2 B 󰇜󰇨  V V

󰆹 1 P VP  󰇛2 B  2 B 󰇜       ln  RT   V  RT

ln 󰆹   ln

  2 B  2 B  RT 

󰇢 ln 󰆹  ln P   P  P    ln 󰇡   

...