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Chapter 7 Solutions Engineering and Chemical Thermodynamics 2eMilo Koretsky Wyatt TenhaeffSchool of Chemical, Biological, and Environmental Engineering Oregon State Universitymilo@oregonstateThe fugacity in the liquid is greater. At equilibrium, the fugacities are equal. For this to be the case, the...

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Chapter 7 Solutions Engineering and Chemical Thermodynamics 2e

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

7.1 The fugacity in the liquid is greater. At equilibrium, the fugacities are equal. For this to be the case, the air in the room would have to be saturated (100% relative humidity). Since the air contains less water than saturation, water will spontaneously evaporate, and the fugacity of the vapor is smaller.

7.2 hmix = 0. The molecular basis for the Lewis fugacity rule is that all the intermolecular interactions are the same. Therefore, the energy of the mixture is equal to that of the sum of the pure species.

7.3 Mixture A. The molecular basis for the Lewis fugacity rule is that all the intermolecular interactions are the same. n-pentane and n-hexane both have dispersion interactions and they are approximately the same size (polarizability is similar)

7.4 The van der Waals parameter b approximates repulsive interactions with a hard sphere model that is determined by the size of the molecules. Hence it is the weighted average of the size of each of the species in the mixture. That can be seen in the form for a binary mixture:

b  y1b1  y 2b2 The parameter a represents van der Waals attractive interactions and is a “two-body” interactions. Thus, you must sum together all the possible pair-wise interactions. That can be seen in the form for a binary mixture: a  y1y1a1  y1y 2a 12  y 2y 1a 21  y 2y 2a 2  y12 a1  2 y1 y 2a12  y 22a 2

7.5 There are many ways to approach this problem. One approach is shown below. If we assume an ideal solution, for water (species 1), we get: y1P  x1 P1

sat

Solving for x1 at equilibrium x eq  1

y1P  0.9 P1 sat

Since we are at 90% RH. If we calculate the mole fraction of water for 96.55 mass % water, we , water will have a tendency to evaporate so the fugacity of get x1 = 0.99 in sweat. Since x 1  xeq 1 water in the liquid is greater than the fugacity of water in the vapor.

7.6 (a) The magnitude of the Henry’s Law constant is governed by the unlike (1-2) interactions. In the case of acetone and water, strong intermolecular attraction exists due to dipole-dipole interactions and hydrogen bonding. With methane and water, only dispersion is present. Therefore, the interactions between methane and water are weaker, and the fugacity is greater. There is a greater partial pressure for the methane-water system. The Henry’s Law constant is greater for this system. (b) The Henry’s Law constant describes the unlike interactions. Since the unlike interactions result in a fugacity that is equal to the pure fugacity of a, the unlike and like interactions are equal in magnitude. Therefore, the solution is ideal for the entire composition range. Activity Coefficient vs. Composition 1.2 1

ga

0.8 0.6 0.4 0.2 0 0

0.2

0.4

0.6 xa

0.8

1

7.7 (a) First, we must realize that

h E  hmix Since  gE  T   T  

   hmix   0  T2  P , ni

E we see that g T is independent of temperature when the pressure and number of moles is held constant. Therefore,

 gE   T 

  Ax x   a b   P , ni  T

   P, ni

is independent of temperature, which implies A    T P, ni

is independent of temperature at constant pressure and moles. A ~ T  (b) Equation 7.24 provides

 g E   P 

   v E  v mix  T , n i

Substituting the two suffix Margules equation in for the excess Gibbs energy, we find  Axa xb  0   P  T , ni

Therefore, we can see that the A parameter is independent of pressure at constant temperature and moles.

7.9 (a) See graph below. Since the Henry’s law constant which represents a-b interactions is less than the pure species fugacity, the “tendency to escape” of a-b is lower and the a-b interactions are stronger. 200

fˆa  xa fa 150

[bar] 100

ˆf  x  a a a 50

0 0

0.5

xa

1

(b) Both lnga and lngb go to 1 and xi goes to 0. Therefore the Henry’s law reference state is being used. Since gi> 0 the tendency to escape for some a-a or b-b interactions is greater than all a-a interactions. Therefore the a-b interactions are stronger.

2

1.6

1.2

lng b 0.8

0.4

0 0

0.5

xa

1

7.10

7.11 200 oC and 1.56 MPa (Pisat = 1.55 MPa) ( isat  1 ) so

70 oC and 1 bar (Pisat = 31.2 kPa) ideal

fi 1 Pi sat

fi 1 Pi sat

70 oC and 1.56 MPa (Pisat = 31.2 kPa) Poynting correction

fi 1 Pi sat

7.12 Initially the system contains water and nitrogen in vapor liquid equilibrium at 1 atm. The liquid is mostly water and the vapor contains a mole fraction of water where the fugacity of the vapor equals the fugacity of the liquid. When the third component is added to the liquid, the mole fraction of water in the liquid decreases, so its fugacity also decreases. To maintain equilibrium, the fugacity of water in the vapor must also decrease to the point where it equals the fugacity of water in the liquid. Therefore, some water will condense and the number of moles of water in the vapor will decrease.

7.13

fˆi l  ˆfi v  yi P  Pi sat  2.34 kPa

7.14 (a) Use Equation 7.7:  fv  g i  g i¼  RT ln i   Plow   

Assume Plow  10 kPa . From the steam tables at 500 ºC   kJ   kJ    kJ  Plow  10 kPa : gˆºi  3489.0     773.15K  9.8977    4163.4     kg   kg  K    kg     kJ   kJ    kJ  P  2 MPa : gˆi  3467.6     773.15K  7.4316     2278.1     kg   kg  K    kg  

Therefore,

   kJ   kJ     2278.1     4163.4    1000 J/kJ 0.0180148 kg/mol    kg   kg    fi v  10 kPa  exp    J    8.314  773.15 K           mol  K     

fiv  1970 kPa  1.97 MPa

 i  0.985 (b) For 500 ºC and 50 MPa, we obtain the following from the steam tables   kJ   kJ   kJ  gˆi  2720.1     773.15K  5.1725     1279.0     kg   kg  K    kg   Using data from Part (a), we can calculate the fugacity and fugacity coefficient:

   kJ   kJ     1279.0       4163.4    1000 J/kJ 0.0180148 kg/mol    kg   kg    fi v  10 kPa  exp    J      8.314  mol  K   773.15 K       v fi  32.4 MPa  and

 i  0.648

7.15 (a) Equation 7.8 states

 fv  P RT ln  i    vi dP  Plow   Plow  However, the Berthelot equation is not explicit in molar volume, so the integral must be transformed.   RT 2a   dP   dv 2 3 Tvi   vi  b 

Substituting this result into the integral, we get

 fiv  vi  RT ln  Plow    RT 

  RTv i 2a    dv  vi  b2 Tvi2 

Plo w

To determine the integral of the first term above, we use decomposition by partial fractions:

b v 1   2 2 v  b  v  b v  b so vi



 RT Plo w

 b  vi dv  ln v i  b     2 v i  b  v i  b

vi

RT Plo w

and v

i  fv   b 2a  i   ln  ln(v i  b )   2  Plow   v i  b RT v  RT i   

Plow

        1  f iv  vi  b  2 a 1      ln  b   ln  RT 2  RT  Plow   v i  b RT  b   b      Plow   Plow

 1 Plow     RT   vi

Since

RT  b , the expression simplifies to Plow

 fv   1 P   P v  b 2 a  low   ln low i ln  i   b    Plow   v b RT RT   RT 2 i    

 1 Plow      vi RT 

If we add ln Plow to both sides and let Plow  0 , we obtain

ln f iv 

v  b  2a b  ln  i  2 vi  b  RT  RT v i

Therefore,

fiv 

 b 2a  RT  exp  vi  b   v i b RT 2vi 

To obtain an expression for  vi , we divide our expression for fugacity by total pressure:

 vi 

 b RT 2a  fi v exp     Pvi  b  v i  b RT 2v i  P

(b) From Problem 4.29, we got v c  3b 

3RTc 8Pc

and a

9 v c RTc2 8

If we substitute into the definition for fugacity coefficient above, we get

 vi 

 1 9  8Tr fi v exp   2   Pr3v i,r 1  3v i,r 1 4Tr v i,r  P

7.16 Equation 7.8:

 f iv    ln  v dP RT i  Plow    Plow P

For the Redlich-Kwong EOS P

RT a  1/ 2 v  b T v v  b

so

 RT  a a  dP   dv  2 2 1/ 2 2  v  b T v v  b T1/ 2v v  b   Therefore, v

 RT Plo w

 vRT   f iv  a a    RT ln dv     2 1/ 2 2 1/ 2  v  b  T v v  b T v  b   Plow 

To determine the integral of the first two terms term above, we use decomposition by partial fractions. For the first term,

1 b v   2 v  b  v  b v  b2 so v





RT Plo w

 b  v dv  ln v  b     2 v  b v  b 

For the second term:

1 1  1 1    vv  b b v v  b so

v

RT Plo w

v



RT Plo w

1 1  v  dv  ln  b  v  b vv  b

v

RT Plo w

Thus, we get  fv  b   v  a a RT ln i   RT lnv  b  RT    1/ 2 ln   1/ 2 v  b  T b v  b  T v  b  Plow 

If we note that

v

RT Plo w

RT  b and let Plow  0, we obtain Plow

a a  RT   b   v  RT ln f iv  RT ln    RT    1 / 2 ln    1/ 2 v  b  v  b  T b v  b  T v  b  Therefore,

fi v 

RT exp vb

  b  a a  v  ln       v  b  RT 3 / 2b  v  b  RT 3 / 2 v  b 

and

 vi 

 b    v  a a fi v RT  exp  ln     Pv  b  v  b  RT 3 / 2b  v  b  RT 3 / 2 v  b P

7.17 Equation 7.8:

 f iv    ln  v dP RT i  Plow    Plow P

The Peng-Robinson EOS can be written P

RT a(T )  2 v  b  v  2vb b 2

Thus,

  2a(T) v  b   RT   dv dP   2  v  b v 2  2vb b 2 2    Therefore, v



RT Plo w

   fv  2a  (T)v  b  RT dv  v   RT ln  i   v  b2 v 2  2vb b 2 2  Plo w   

Simplifying, we get v  fv  v 2a (T ) ln  i      2 dv RT RT  v  b   Plow  Plow

v v  b 

v



RT Plow

v

2

 2vb  b 2 

2

dv

(1)

To determine each of the integrals in Equation 1, we use decomposition by partial fractions. For the first integral:

1 b v   2 v  b  v  b v  b2 so v





RT Plo w

 b  v dv  ln v  b     2 v  b v  b 

v

(2) RT Plo w

For the second integral, decomposition leads to:

v v  b 

v

 2vb b2 

2

2

so



1 v  b  b 2 2 v  2vb b v 2  2vb b2  2

v v  b

v

 RT Plo w

v

2

v

 2vb b 2 

2

dv 

 RT Plo w

v v  b 1 dv dv b  2 2 2 2 2 v  2vb b RT v  2vb b 

(3)

Plo w

We again use partial fractions. For the first term in Equation (3):  1  1 1     2 2 2 2b  v  1  2 b v  1  2 b  v  2vb  b 1









so v

 

 

v

1 v 1 2 b  v 2  2vb  b 2 dv  2 2b ln v  1  2 b RT RT 1

(4)

Plow

Plow

For the second term in Equation (3):

v  b

v

2

 2vb b

2



 v 2

v  b 2

 2vb b

2

2b



 v 2

2

(5)

 2vb b2 

2

Equation (5) can be substituted into Equation (3) to give two terms. The first term gives: v

b

 RT Plo w

v

v  b 

b dv  2 2 2 v 2  2vb b 2 2v  2vb b  RT

(6)

Plo w

The second term is somewhat more problematic. Again decomposition gives:   1  1 1         2 4 b  v  1  2 b   v  1  2 v 2  2vb  b 2



 2b









  b 



2

7.13

ˆf l  ˆf v  y P  Psat  2.34 kPa i i i i

7.14 (a) Use Equation 7.7: g i  g i¼  RT

 fv  ln i   Plow   

Assume Plow  10 kPa . From the steam tables at 500 ºC   kJ   kJ    kJ  Plow  10 kPa : gˆºi  3489.0     773.15K  9.8977    4163.4     kg   kg  K    kg     kJ   kJ    kJ  P  2 MPa : gˆi  3467.6     773.15K  7.4316     2278.1     kg   kg  K    kg  

Therefore,

   kJ   kJ     2278.1     4163.4    1000 J/kJ 0.0180148 kg/mol    kg   kg    fi v  10 kPa  exp      J  8.314 773.15 K       mol  K        

fiv  1970 kPa  1.97 MPa

 i  0.985 (b) For 500 ºC and 50 MPa, we obtain the following from the steam tables   kJ   kJ   kJ  gˆi  2720.1     773.15K  5.1725     1279.0     kg   kg  K    kg   Using data from Part (a), we can calculate the fugacity and fugacity coefficient:

   kJ   kJ     1279.0       4163.4    1000 J/kJ 0.0180148 kg/mol    kg   kg    fi v  10 kPa  exp    J     773.15 K 8.314       mol  K     

fi  32.4 MPa  v

and

 i  0.648

7.15





(a) Equation 7.8 states

 fi v  P   v dP RT ln   Plow   i   Plow However, the Berthelot equation is not explicit in molar volume, so the integral must be transformed.   RT 2a   dP   dv  vi  b 2 Tvi3 

Substituting this result into the integral, we get

 fiv  vi   RTv i 2a     dv RT ln  Plow    v  b2 Tv2  i    RT  i Plo w

To determine the integral of the first term above, we use decomposition by partial fractions:

b v 1   2 2 v  b  v  b v  b so vi



 RT Plo w

 b  vi  ln v  b dv      i 2 v i  b  v i  b

vi

RT Plo w

and v

 fv   b  i a 2 i   ln  ln(v i  b )  2  Plow   v i  b RT v  RT i    Plow

       1   f iv  vi  b  2 a 1     b    ln  ln 2    Plow  RT RT v b    i   b  b  RT    Plow  Plow 

 1 Plow     RT   vi

Since

RT  b , the expression simplifies to Plow

 fv   1 P   P v  b 2 a  low   ln low i ln  i   b    Plow   v b RT RT   RT 2 i     If we add ln Plow to both sides and let Plow  0 , we obtain

ln f iv 

v  b  2a b  ln  i  2 vi  b  RT  RT v i

Therefore,

fiv 

 b RT  exp  v  b   v b

2a  2 

 1 Plow      vi RT 

vi

b

 vi

b

RT vi 

To obtain an expression for  vi , we divide our expression for fugacity by total pressure:

 vi 

 b RT 2a  fi v exp     Pvi  b  v i  b RT 2v i  P

(b) From Problem 4.29, we got v c  3b 

3RTc 8Pc

and a

9 v c RTc2 8

If we substitute into the definition for fugacity coefficient above, we get

 1 9  8Tr fi v   exp   2   Pr3v i,r 1  3v i,r 1 4Tr v i,r  P v i

7.16 Equation 7.8:

 f iv    ln  v dP RT i  Plow    Plow P

For the Redlich-Kwong EOS P

RT a  1/ 2 v  b T v v  b

so

 RT  a a  dP   dv  2 2 1/ 2 2  v  b T v v  b T1/ 2v v  b   Therefore, v

 RT Plo w

 vRT   f iv  a a    RT ln dv  1/ 2  1/ 2   2 2  Plow   v  b  T v v  b T v  b 

To determine the integral of the first two terms term above, we use decomposition by partial fractions. For the first term,

1 b v   2 v  b  v  b v  b2 so

v





RT Plo w

 b  v  ln v  b dv      2 v  b v  b 

For the second term:

1 1  1 1    vv  b b v v  b so

v

RT Plo w

v

 RT Plo w

1 1  v  dv  ln  b  v  b vv  b

v

RT Plo w

Thus, we get  fv  v   b  a a RT ln i   RT lnv  b  RT    1/ 2   1/ 2 ln v  b  T b v  b  T v  b  Plow 

If we note that

v

RT Plo w

RT  b and let Plow  0, we obtain Plow

a a  RT   b   v  RT ln f iv  RT ln    RT    1 / 2 ln    1/ 2 v  b  v  b  T b v  b  T v  b  Therefore,

fi v 

RT exp vb

 b   a a  v  ln      3 / 2b  v  b  RT 3 / 2 v  b v b    RT   

and  b    v  a a fi v RT  exp  ln       Pv  b  v  b  RT 3 / 2b  v  b  RT 3 / 2 v  b P v i

7.17 Equation 7.8:

 f iv    ln  v dP RT i  Plow    Plow P

The Peng-Robinson EOS can be written RT

a(T )

P

v  b 



v

2

 2vb b 2

Thus,

  2a  (T) v  b RT   dv  dP   2  v  b v 2  2vb b 2 2    Therefore, v



RT Plo w

   fv  2a  (T)v  b  RT  dv  v  RT ln  i   v  b2 v 2  2vb b 2 2  Plo w   

Simplifying, we get v  f iv  v 2a (T ) ln     2 dv RT RT  v  b   Plow  Plow

v v  b 

v



RT Plow

v

2

 2vb  b 2 

2

dv

(1)

To determine each of the integrals in Equation 1, we use decomposition by partial fractions. For the first integral:

1 b v   2 v  b  v  b v  b2 so v





RT Plo w

 b  v dv  ln v  b     2 v  b v  b 

v

(2) RT Plo w

For the second integral, decomposition leads to:

v v  b 

v

 2vb ...