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Chapter 6 Solutions Engineering and Chemical Thermodynamics

Wyatt Tenhaeff Milo Koretsky Department of Chemical Engineering Oregon State University [email protected]

6.1 (a) The Clausius-Clapeyron equation:

2

vap dPisat hi dT  Pisat RT 2

or

3

4

so

 h vap  1 1   Pisat  101  kPaexp  i    R  T 373 [K]  

(b) and (c) Using

we obtain the following table T [K]

Eqn 6.24 [kPa]

273.156 278.15 283.15 288.15 293.15 298.15 303.15 308.15 313.15 318.15 323.15 328.15 333.15 338.15 343.15 348.15 353.15 358.15 363.15 368.15 373.15

0.84 1.16 1.58 2.13 2.84 3.76 4.93 6.40 8.24 10.54 13.36 16.82 21.04 26.13 32.26 39.58 48.28 58.56 70.67 84.84 101.35

Steam Tables % [kPa] Difference 0.6113 0.87 1.23 1.71 2.34 3.17 4.25 5.63 7.38 9.59 12.35 15.76 19.94 25.03 31.19 38.58 47.39 57.84 70.14 84.55 101.35

37.30% 33.02% 28.30% 24.51% 21.51% 18.62% 15.94% 13.68% 11.71% 9.86% 8.19% 6.75% 5.50% 4.40% 3.42% 2.58% 1.87% 1.25% 0.75% 0.34% 0.00%

The logarithmic trend is well-represented. However, at lower temperatures the ClausiusClapeyron equation is up to 37% off. The actual heat of vaporization changes from  kJ    kg 

2501.3 

 kJ 

at 0.01 oC to 2257.0  kg  at 100 oC, a difference of around 10%.  

5

Pressure [kPa]

100.00

10.00

Eqn 6.23 Tables 1.00

0.10 273

283

293

303

313

323

333

343

353

363

373

Temperature [K]

(d) For 100 ºC to 200 ºC, we obtain the following table: T [K]

Eqn 6.24 [kPa]

Steam Tables [kPa]

% Difference

373.15 378.15 383.15 388.15 393.15 398.15 403.15 408.15 413.15 418.15 423.15 428.15 433.15 438.15 443.15 448.15 453.15 458.15 463.15 468.15 473.15

101.35 120.51 142.64 168.11 197.30 230.63 268.55 311.54 360.11 414.82 476.24 545.00 621.74 707.16 801.99 906.98 1022.93 1150.68 1291.10 1445.10 1613.62

101.35 120.82 143.28 169.06 198.53 232.1 270.1 313 361.3 415.5 475.9 543.1 617.8 700.5 791.7 892 1002.2 1122.7 1254.4 1397.8 1553.8

0.00% 0.26% 0.44% 0.56% 0.62% 0.63% 0.57% 0.47% 0.33% 0.16% 0.07% 0.35% 0.64% 0.95% 1.30% 1.68% 2.07% 2.49% 2.93% 3.38% 3.85%

6

Pressure [kPa]

10000.00

Eqn 6.23

1000.00

Tables

100.00 373

383

393

403

413

423

433

443

453

463

473

Temperature [K]

Over this range the Clausius-Clapeyron equation represents the data well and is no more than 4 % off. The actual heat of vaporization changes from 2257.0  kJ/kg  at 100 oC to 1940.7  kJ/kg  at 200 oC, a difference of around 15%. (e) The heat of vaporization can be corrected for temperature as follows hvap  T  

Tb

T

T

Tb

l  c P dT  hvap  Tb  

v

 c P dT

We can acquire heat capacity data from Appendix A.2, but to simplify the analysis, we will use an average heat capacity for the vapor. hvap  T   75.4 373.15  T   40626  34.13 T  373.15 hvap  T   56026  41.27T

Substitute this expression into the Clausius-Clapeyron equation dPisat Pi

sat



 56026  41.27T  dT RT 2

Integrate:  1 1  T   1   56026    41.27 ln    T 373.15   373.15   R

Pisat  101.35 kPa  exp 

Now plot the data as before from 0.01 ºC to 200 ºC. 7

     

T [K]

Eqn 6.24 [kPa]

Steam Tables [kPa]

% Difference

273.16 278.15 283.15 288.15 293.15 298.15 303.15 308.15 313.15 318.15 323.15 328.15 333.15 338.15 343.15 348.15 353.15 358.15 363.15 368.15 373.15 378.15 383.15 388.15 393.15 398.15 403.15 408.15 413.15 418.15 423.15 428.15 433.15 438.15 443.15 448.15 453.15 458.15 463.15 468.15 473.15

0.64 0.91 1.28 1.78 2.43 3.29 4.39 5.81 7.60 9.86 12.66 16.12 20.35 25.49 31.68 39.10 47.91 58.32 70.54 84.80 101.35 120.46 142.40 167.48 196.00 228.29 264.70 305.56 351.26 402.16 458.64 521.10 589.92 665.51 748.27 838.59 936.89 1043.55 1158.98 1283.56 1417.67

0.6113 0.87 1.23 1.71 2.34 3.17 4.25 5.63 7.38 9.59 12.35 15.76 19.94 25.03 31.19 38.58 47.39 57.84 70.14 84.55 101.35 120.82 143.28 169.06 198.53 232.1 270.1 313 361.3 415.5 475.9 543.1 617.8 700.5 791.7 892 1002.2 1122.7 1254.4 1397.8 1553.8

4.95 4.96 4.24 3.88 3.86 3.65 3.35 3.17 3.03 2.78 2.51 2.28 2.06 1.84 1.58 1.34 1.09 0.82 0.57 0.29 0.00 0.30 0.61 0.94 1.28 1.64 2.00 2.38 2.78 3.21 3.63 4.05 4.51 4.99 5.49 5.99 6.52 7.05 7.61 8.17 8.76

8

The agreement between the two values at lower temperatures improves significantly at lower temperatures, but actually worsens at higher temperatures. The agreement could potentially be improved by not averaging the heat capacity.

9

6.2 We can find the required pressure by applying the Clapeyron equation: dP hl  h s  dT vl  v s T





We can find the molar volume of water ice from any number of reference books. At 0 ºC and 1 bar:  kg 

 l  1000  

 s  917  

 m3    mol 

 v l  1.80  10  5 

  m3  kg   v s  1.97  10  5 3 m 

 m3     mol 

Also at 0 ºC and 1 bar, 

J   mol 

h l  h s  6010 









If we assume that h l  h s and v l  v s are independent of temperature and pressure, we can separate variables in the Clapeyron equation and integrate. P2  P1 

hl  h s

v

l

v

s



P2  1  10 5  Pa  

so

 T2  T1

ln

  

1.80  10

5

6010  J/mol  1.97  10

5

P2  66.1  bar 

10

m

3

/ mol



 268.15 K    273.15 K 

ln

6.3 (a) At 1 bar, the gas will act as an ideal gas.  v

J     300 K   m3   mol  K    0 . 0249   1  10 5  Pa   mol  

 8.314 

RT   P

The number of moles of vapor are found as follows (neglect molar volume of liquid) V  v

nv 

 

0.001 m 3  m3  0.0249    mol 

n v  0.0402 mol

(b) At 21 bar, the gas will not behave ideally. Since we are assuming that the molar volume of liquid is negligible and the heat of vaporization is independent of temperature, the Clapeyron equation becomes dP h vap  v dT vT

The molar volume using pressure expansion of the virial equation is vv 

RT 1  B ' P   1   RT   B'  P  P 

Substituting this expression into the Clapeyron equation yields dP  dT

h vap  1  R  B'  T 2 P  

Separation of variables yields P2  21 10 5 Par



P1 1 10 5 Pa

h vap  1 '  B  dP   R  P 

T2

dT 2 T1 300 K T



and integration results in

11

 P2  P1

ln



  B '  P2  P1   

 h vap R

 1 1     T2 T1 

We can substitute values for given quantities and constants to solve for T2. T2  523.3 K

(c) Using the virial equation,  J   1    B '    8.314   P   mol  K  

v v  RT 



3 1 7  m    1  10   5  J   21  10 Pa



  523.3 K   

 m    mol 

v v  0.00164 

3

We can assume the volume occupied by the liquid is negligible. Therefore, nv 

V  vv

 

0.001 m 3  m3  0.00164    mol 

n v  0.61  mol

12









6.4 We can use the following computational path to solve for pressure at which graphite and diamond are in equilibrium at 25 oC. graphite

diamond

P = 1 [atm]  J   mol 

g  2866 

g1 

P

 v graphdP

g 3 

1

1

 v diam dP

P

P diamond

g 2  0

graphite

Summing together the three steps we get: 

J   g1  g 2  g 3  mol 

g 1[atm]  2866 

To find the change in Gibbs energy with pressure, we apply the fundamental property relation, Equation 5.9. At constant temperature: 0 dgi  v idP  sidT If the solid is assumed incompressible, we can integrate to get

 vi dP  vi P

g i  

Thus the sum of Gibbs energy becomes 

P

1





J    v graph dP  0   v diam dP  v graph  v diam  P  1  mol  1 P

g 1[atm]  2866 

Solving 



J   mol 

 2866  

  1 1           2.26 3.51  

 cm 3     g 

 

 g    12   mol   

or P = 1,514 [MPa] = 15,143 [bar]

13

  

1  m3        10 6  cm 3 

 





 P  1.01  10 5  Pa 



6.5 From the Clausius-Clapeyron equation: (I) T

where h fus is the enthalpy of fusion at temperature T. We can get the molar volumes from the densities:  kg   MW  mol   1.17  10  5 v lAl    kg  ρl 2,300    m3 

 m3     mol 

 kg    mol   1.00  10  5 s v Al    kg  ρs 2,700    m3 

 m3     mol 

0.027 

and

MW

0.027 

so  m3  v lAl  v sAl  1.7  10 6    mol 

We can use the following path to calculate for h Tfus . solid

T

liquid

h Tfus 933

T

T

933

 cPs dT

 cPl dT

T = 933 [K] solid

 J  h fus  10,711   mol

liquid

14

h Tfus 

933.45 T  J  s l cP dT  10,711  cP dT    mol  T 933.45







J





J



s l Using c P  20.608  0.0138T  mol K  and c P  31.748  mol K  , we get:     2 h T fus  5,819.9  11 .68T  0.0069T

Back into Equation (I) gives:  5,819.9  11.68T  0.0069T 2 

1.7  10 6 T

dP   

 dT 

Integrating: 100 [bar]

T

1

933.45 [K] 

 dP 





 5,819.9  11.68T  0.0069T 2 

1.7  10  6 T

 dT 

or

100  1  105 1.7  10  6   5819.9 ln



T  2 2   11.68 T  993.45  0.00345 T  993.45 933.45 

solving for T gives T = 934.91 [K]

15



6.6 We can assume that silver acts as an ideal gas at 1500 K. We can also assume the molar volume of the vapor is much greater than the molar volume of liquid. Therefore, we can use Equation 6.22 dP sat P sat

h vap dT



RT 2

This can be rearranged to show dP sat h vap P sat  dT RT 2

This relation assumes 1. vv>>vl 2. Silver acts as an ideal gas We can differentiate the expression for pressure in the problem statement to obtain dP sat dT dP sat dT

 14260 0.458    14260    0.458 ln  T   12.23   exp 2 T  T  T  



 14260 0.458  sat   P 2 T   T



Therefore, h vap  14260 0.458      2 T  RT 2  T

Therefore, h vap  R 14260  0.458T 

At 1500 K,  J   h vap   8.314   mol  K    kJ  h vap  112 .8   mol 



 14260  0.4581500  K    

16

6.7 For a single component system:   Gi  g i  g

From the fundamental property relation given by Equation 5.9: dg   sdT  vdP

We can identify a phase transition from the vertical line of the g vs. T plot, as indicated below. Since this transition is vertical, i.e., the temperature is constant, the pressure must also be constant. Thus, we can differentiate the Gibbs energy with respect to temperature at constant pressure to get:  g     T 

 s P

Hence the slope of a plot of g (or ) vs. T at any temperature must be the negative of the value of entropy on the plot for s vs. T. The resulting curve is sketched below. 

s

phase transition; vertical line indicates P = const



T 

  Thick line denotes lowest value of 

 Slope = the negative value of s at the same T on the curve above

 T*

T

17

6.8 The ferrite phase has stronger bonds. At room temperature, iron is in the ferrite phase. The heating to 912 ºC has the effect of increasing the entropy contribution to the Gibbs energy. At a high enough temperature, the austenite phase becomes stable, so that its entropy must be greater than the ferrite phase. If the entropy of the austenite phase is greater, the enthalpy of the ferrite phase must be greater or else the austenite phase would be stable over the entire temperature range. Hence, the ferrite phase has stronger bonds.

18

6.9 Since the pressures are low, we can assume ideal gas behavior. We can also assume that the molar volume of the vapor is much greater than the molar volume of liquid and the heat of vaporization is independent of temperature. Therefore, we can rearrange Equation 6.24 to obtain

h vap

h vap

 J    760 torr     8.314   ln   mol  K    400 torr    1 1      353.25 K 333 . 75 K   kJ    32.3   mol 

This value is 7.7% smaller than the reported value.

19

6.10 (a) The freezing point occurs where there is a discontinuity in the g vs. T plot, as indicated below. The liquid is at a temperature higher than the freezing point and the solid at lower temperature. These are demarked below. The melting temperature is 250 K, which occurs at a value g = 3,000 [J/mol] Pure species "a"

Gibb's Energy, g (J/mol)

6000

1 unit  up New Freezing  point

4,000 solid Freezing  point

2,000

liquid

100

200

1.2  units  up

300

Temperature, T (K) (b) At constant pressure, the entropy can be found from Equation 5.14. For the solid we have:  g  s     T 

 P

g 1,000 J     10   T 10  mol K 

And for the liquid, we get:  g  s     T 

 P

g 2,000 J     40  T 50 mol K  

(c) As we change pressure, we can see how the Gibbs energy changes at any given temperature by Equation 5.14:

20

 g     P 

v T

Assuming the molar volumes of the liquid and vapor stay constant over the temperature range around the melting point, we see that the Gibbs energy of the liquid increases by 1.2 times the Gibbs energy of the solid, since the molar volume of the liquid is 20% larger. The Gibbs energy of the new freezing point at higher pressure is schematically drawn on the plot above. For convenience, we choose the solid to increase by 1 unit on the plot. Thus, the liquid increases by 1.2 units. As the sketch shows, the freezing point, where the two lines intersect, will shift to higher temperature.

21

6.11 For a single component system the fundamental property relation, Equation 5.9, gives: dg   sdT  vdP

We can identify a phase transition from the vertical line of the g vs. P plot, as indicated below. Since this transition is vertical, i.e., the pressure is constant, the temperature must also be constant. Thus, we can differentiate the Gibbs energy with respect to pressure at constant temperature to get:  g     P 

v T

Hence the slope of a plot of g vs. P must have a slope that matches the plot for v vs. T. Since the molar volume of phase  is about twice the value of phase , its slope should be twice as big. The resulting curve is sketched below. v

phase transition; vertical line indicates T = const



v v

 P

g

 Slope of top line is

Straight line since v is constant



 Thick line denotes lowest value of g

 P*

P

22

about twice as big as the slope of bottom line

6.12 The saturation pressure can be found using the Clausius-Clapeyron equation with the assumption that the heat of vaporization is independent of temperature. First, we need to use the given data for the 63.5 ºC and 78.4 ºC to find the heat of vaporization.

h vap 

 P sat   R ln 2   P sat   1 

 1 1     T2 T1   kJ  h vap  42.39   mol 

 J    760 torr     8.314   ln   mol  K    400 torr    1 1    351.55 K  336.65 K 

Now we can calculate the vapor pressure at 100 ºC.

  h vap sat sat P3  P2 exp  R  P3sat

  kJ    42.39     1 1 1 1    mol       760 torr  exp       kJ   373.15 K 351.55 K   T3 T2    0.008314   mol  K  

 1760 torr  2.32 atm

In comparison, ThermoSolver gives a value of 2.23 atm, using the Antoine equation.

23













6.13 We can show using the Chain Rule that 

 gi   T    T

      









  Tg i  g i T   T2    T   P

     

  

1  g i  T  T  

 P



gi T2

 P

Using fundamental property relations, Equation 5.14 states  g i      si  T  P

Therefore, 

 gi  T T









 

     

 



 Ts i  g i T

2



 Ts i   hi  Ts i  T

2



 hi T2

 P

24

6.14 Let T1  922 K, T2  1,300 K g 2  h2  T2 s 2 dh  c P dT 1,300 K

h2  h1 

 c P dT

922 K



J   mol 

h2  h1  c P  T2  T1   39,116 

c ds  P dT T  T2   J    85.10   mol K   T1 

s2  s1  c P ln



J   mol 

g 2  h2  T2 s2  71,500 

Alternative solution using the result from Problem 6.13:  dg  T    hT   h1  c P  T  T1    dT  T2 T2    g    T



 g    T

We must leave h as a function of T

 h1  c P  T  T1     dT T2  922 K 

1300 K

2

 g    T



d 1

 1 1  T  J   g  g     c P ln 2   55.01         h1  c PT1    T 2  T1  mol K   T2 T1   T1  

J   mol 

g 2  71,500 

25

6.15 A possible hypothetical solution path is presented below: monoclinic

orthorhombic T = 298 [K]

g1

g3

T = 368.3 [K] monoclinic

g2 = 0

orthorhombic

From the diagram, we see that the Gibbs energy for steps one and three can be calculated as follows:  g m   T 298 K 



368 K

g1 



dT 

P

and  g o   T 368 K 



298 K

g 3 



dT 

P

respectively. We can apply Equation 5.14 from the thermodynamic web  g   s    T  P

At 368 K, sulfur undergoes a phase transition, so o g 3m68 K 0

Using these above relationships, the expression for

mo g 298 K 

368 K m

s

dT  0 

298 K

m o g 298 K

becomes

298 K o

368 K

298 K

368 K

298 K

368 K

 s dT  

 13.8  0.066T  dT   11  0.071T  dT

 mo g 298 K  79.5 

J   mol 

Therefore, the transition from the monoclinic to orthorhombic state occurs spontaneously. The orthorhombic state is more stable.

26

6.16 At the phase transition, the following is true  g  g      T  Sr ( s )  T  Sr (l )

Using the thermodynamic web, the following can be shown (see Problem 6.13) h   g / T      2  T  P T

The enthalpies can be written as follows T

h l  T   49179 

 35.146 dT

 35.146T  3540

 37.656 dT

 37.656T  16305.4

1500 K T

h s  T   20285 

900 K

g can also be calculated at 900 K for solid Sr and 1500 K for liquid Sr. T  gl   T   gs   T 





J   mol  K 



 83.85 



 ref 



J   mol  K 



 68.68 



 ref

 g  at any temperature using the differential equation as follows  T

We can find 

 g   T

 d 

  g / T  T

 

 

 P

dT   

h T2

dT

Substituting our expressions, we get g l /T



 83.85

 g    T

d

T



1500 K

 35.146T  3540 

 

T2

 dT 

 gl     35.146 ln(T )  3540  176.04  T  T   g s /T



 68.68

 g    T

d

T



900 K

 37.656T  16305.4 

 

 dT

T2 

27

 gs  T 



  37.646 ln(T )  16305.4  205.5  T 



Set

 gl   T 



 l   g   T   







and solve for T:

T melt  1059.8 K

The enthalpy of melting is defined as







h fus  h s T melt  h l T melt



Using the expressions developed above  kJ   kJ   kJ   33.71   7.41     mol   mol   mol 

h fus  26.30 

28

6.17 At the phase transition, the temperature and Gibb’s energy of both phases must be equal. Mathematically, this is equivalent to  g  g      T  SiO2 ( s )  T  SiO 2 (l )

Using the thermodynamic web, the following can be shown (see Problem 6.13) h   g / T      2 T  P T 

Using the definition of enthalpy, we can write the following h T 

T

h ref

T ref

 dh 

 cP

dT

 h T   href 

T

 c P dT

Tref

The enthalpies can be written as follows h l  T   738440  h s  T   856840 

T

 85.772 dT

2500 K T

 53.466  0.02706T  1.27  10

5



T 2  2.19  10  9 T 3 dT

1100 K

g can also be calculated at 1100 K for solid SiO2 and 2500 K for liquid SiO2. T  gl     T     g   T 



ref



s

J   mol  K 

 487.3 





  ref

J   mol  K 

 903.5 

We can substitute our expressions for

g and h T  into the above differential equation and T

separate variables to obtain

29

 g l /T





g   T 

d

 487.3

  738440 

T









T

 85.772 dT 

2500 K

 

 

 g s /T



 903 .5



g   T 

d







 53.466  0.02706T



T



1100 K

  856840 

T

 dT

T2 

2500 K

 

T

1100 K







 1.27  10  5 T 2  2.19  10  9 T 3 dT  

 dT

2



  

Integration provides gl  9.5268  10 9  85.772 ln(T )  1052.4T   487.5 T T gs  1.82  10 10 T 4  2.12  10 6 T 3  0.01353T 2  927190.9  53.466T ln(T )  1230T   903.5 T T

If we plot

 gl   g s       T   T     

vs. T, we obtain the following:

There are three solutions, but only the solution between 1100 K and 2500 K is physically meaningful. If we magnify the plot near the middle solution, we find T  1983 K

The enthalpy of fusion is defined as







h fus  h s T melt  h l T melt



Using the expressions developed above

30

 kJ   kJ   kJ      782.78   9.72     mol   mol   mol 

h fus  792.5 

31

6.18 From the Clausius-Clapeyron equation sat dPCS



2

dT

h vap



T vv  vl



Assuming: v v  v l

we get sat dPCS

h vap



2

dT

(I)

Tv v

The saturation pressure is given by: sat ln PCS  62.7839  2

4.7063  10 3  6.7794 ln T  8.0194  10  3 T (II) T

sat 5  At T = 373 K, PCS 2  4.48 10 Pa. Taking the derivative of Equation II sat d ln PCS

2

dT



sat dPCS

1

2

sat PCS 2

dT



4.7063  10 3 T

2



6.7794  8.0194  10  3 T

(III)

Plugging Equation III into Equation I,  4.7063  10 3 

T2 

 sat 6.7794 h vap  8.0194  10  3  PCS  2 T Tv v 



 kJ  gives:  mol 

vap Solving for vv using hCS  24.050  2

v

v

 h vap  4.7063  10 3 6.7794    8.0194  10 3   sat T TPCS T2   2

z

Pv B  0.878  1  RT v

or  m3   cm 3    740    mol   mol 

B  7.4  10  4 

32

1

 m3    mol 

 6.08  10 3 

This value is about 50% higher than the reported value. Alternative solution: Following similar development as Problem 6.3: sat dPCS

2

dT

h vap    1 RT 2  sat  B '  PCS 2 

  1 h vap ' sat  sat  B  dPCS  2 dT 2 RT  PCS2 

We must be careful about the limits of integration. We need to pick a value of T close so enthalpy of vaporization is not too different, but far enough away to avoid round off error. If we sat 5  choose T = 378 K, Equation I gives PCS 2  5.04 10 Pa. Integrating: 4.4810 5   Pa

 373 1 h vap ' sat  sat  B  dPCS   2 dT 2 P RT 5     378 5.0410   Pa CS2



 4.48  105  h vap  1 1  ' 5 5 ln  B 4.48 10  5.04  10    5  R  373 378   5.04  10 





Solving for B’ gives: B'  2.55  107   Pa or  m 3  B  B ' RT  7.9 104     mol

33

6.19 Calculate vA, vB, v, VA, VB, and V from the ideal gas law: RT  0.05 m 3 / mol P RT vB   0.05 m 3 / mol P RT v   0.05 m 3 / mol P vA 

V A  n A v A  0.1 m 3 V B  n B v B  0.15 m 3 V

 ntot v  0.25 m 3

We calculate the partial molar volumes as follows  V V A    n A  V V B    n B

 

 T , P, n B 

  T , P, n A



   n A  n B  RT   RT  0.05 m 3 / mol  n A  P  P



   n A  n B  RT   RT  0.05 m 3 / mol n B  P  P

To find the remaining quantities, we can apply Equations 6.44 and 6.46 Vmix  n A V A  v A   n B V B  v B  Vmix  2 0.05  0.05  3 0.05  0.05  0 v mix  x A V A  v A   x B V B  v B  v mix  0

34

6.20 (a) For a pure species property v a  v y a  1

Substitution yields  cm 3    mol 

v a  1001  80 0  2.51 0   100 

(b) From Equation 6.29  V Va    na

 

 n b ,T , P

We can find V by multiplying the given expression for molar volume by the total number of moles.  y y  n n V   na  nb   100 ya  80 yb  2.5 a b   100na  80nb  2.5 a b y a yb  na  nb 

Differentiating with respect to na we get, Va 

 na



 100na  80nb  2.5 

na nb  nb na nb   100  2.5  2.5 na  nb  n na  nb  na  nb  2 b

so Va  100  2.5 yb 1  y a   100  2.5 yb2

To find the molar volume at infinite dilution, we can use the following relation Va  lim Va y a 0

 cm 3    mol 

 Va  102.5 

(c) Since species A contributes more to a mixture than to a pure species,

35

v mix  0

Note: The Gibbs-Duhem equation says that species B also contributes more.

36

6.21 Calculate mole fractions: n1 1  mol   0 .2 ntot 5  mol

y1 

y 2  0.4

y3  0.4

Calculate v. Obtain an expression for v: v

RT P



B   2 A  1  P  RT  y1  y 2   RT      

Substitute values: 

3    82.06  cm  atm    mol  K  v 50  atm 



  500  K    

1   50   9.0  10 2

5



 0.2  0.4  3.0  10  5 

 cm 3    mol 

v  919.0 

Calculate V. V  ntot v   cm 3   V   5  mol   919.0      mol    



V  4595 cm 3



Calculate v1. The value of v1 can be found by substituting y1=1 into the expression for v1. 

3    82.06  cm  atm    mol  K  v1   50  atm 

 



  500  K  

1   50   9.0  10 2

 cm 3    mol 

v1  698 

Calculate v2.

37

5



1  0   3.0  10  5 





3    82.06  cm  atm    mol  K  

v2 

  500  K    

50  atm 

1   50   9.0  10 2

5



 0  1  3.0  10  5 

 cm 3    mol 

v2  1067 

Calculate v3. 



3    82.06  cm  atm    mol  K  

v3 

  500  K   

50  atm 



1   50  3.0  10  2

5

 cm 3  v3  882    mol 

Calculate V1 . From Equation 6.29:  V     nv     V1      n1  n 2 , n3 ,T , P  n1  n 2 , n3 ,T , P

We can substitute the expression for V into this derivative and use the fact that ntot  n1  n2  n3 to obtain V1 

  RT    n1  n2  n3   P 2  A  n1  n2   B  n1  n2  n3     n1  P  RT  RT 

   

  n 2 , n 3 ,T , P

Differentiating we get RT P

V1 



B   2 A  1  P  RT  RT      

Substitute values: 

3    82.06  cm  atm    mol  K  V1   50  atm

 



  500  K  

1   50   9 10 2

 cm 3    mol 

V1  697.5 

38

5

 3  10  5



6.22 (a) By definition:  H H a    na

 

 T , P , nb , n c

n  na  nb  nc H  nh  5,000na  3,000nb  2,200nc  500

na nbnc

na  nb  nc 2

  H  nbn c 2na nbnc     5,000  500  2 3   na  T ,P,n ,n n  n  n n  n  n       a b c a b c  b c  H   Ha    5,000  500xb x c 1 2xa   J/mol  na  T ,P,n ,n b c (b) xa  xb  xc 

1 3

Ha  5,018.5  J/mol (c) xa  1 , xb  xc  0

Ha  5,000  J/mol (d) xb  1 , xa  xc  0

Hb  hb  3,000  J/mol

39

6.23 Let the subscript “1” designate CO2, and “2” designate propane. To calculate the partial molar volumes, the following formulas will be used: V1  v  y 2

dv dy 2

v  y1V1  y 2V2

Expressions can’t be obtained for the molar volume with the van der Waals EOS; therefore, the problem will be solved graphically. First, obtain an expression for the pressure that contains the mole fractions of CO2 and propane: a mix  y12 a1  2 y1 y 2 a1a 2  y 22 a 2

bmix  y1b1  y 2 b2

P

y 2 a  2 y1 y 2 a1a2  y 22 a 2 RT  1 1 v   y1b1  y 2 b2  v2

Solve for a and b using data from the appendices.  J  m3 

a1  0.366 

2

 mol 

b1  4.29  10  5  



 m3 

 mol   J  m3  a 2  0.941    mol 2 

 mol 

b2  9.06  10  5 



 m3 

Now we can create a spreadsheet with the following headings: y1

y2

amix

bmix

v

The last column contains the molar volumes obtained by solving the van der Waals equation with the spreadsheet’s solver function. After the table is completed, we create the following graph.

40

From the line of best fit, we find v  8  10  5 y 22  8  10  5 y 2  0.00147

Therefore, dv  1.6  10  4 y 2  8  10  5 dy 2

and



V1  v  y 2  1.6  10 4 y 2  8  10 5



We can find the partial molar volume of propane from the following relationship v  y1V1  y 2V2

V2 

v  y1V1 y2

Tabulate the values of the partial molar volumes in the spreadsheet and create the following graph

41

42

6.24 (a) ga  40 

kJ mol

G a  g  x b

dg dx b

dg  40  60  RT  ln xa  1  ln x b  1  5x a  5xb   dx b G a  40xa  60x b  RT xa ln x a  x b ln xb   5xa x b  

          x b40  60  RT  ln xa  1  ln x b  1  5x a  5xb 

G a  40xa  xb   RT xa  xb  ln x a  5xb2 G a  40  RT ln x a  5xb2 G a  40 

8.314(300) kJ kJ ln 0.2  5(0.64)   40.8  1000 mol mol

G a   Gmix  ng  x a ga  xb gb  g  xa ga  x bgb   RT x a ln xa  xb ln x b   5x a xb Gmix  nT RT x a ln xa  xb ln x b   5xa x b   2.2 kJ (b) gmix  hmix  Tsmix Assume the entropy of mixing is ideal: hmix  5x a xb  0 so

so

h  hmix  hsensibleheat  0

hsensibleheat  0 and T goes down

43

6.25 To find V1 and V2 , we can read values directly from the graphs. Calculate mole fractions x1 

1  0.2 5

At x1  0.2 ,  cm 3    mol 

V1  46.5 

 cm 3    mol 

V2  69.8 

The following relationships are employed to calculate the molar volumes of pure species v1  lim V1 x1 1

v 2  lim V2  lim V2 x 2 1

x1  0

From the graph  cm 3    mol 

v1  50 

 cm 3    mol 

v2  70 

Therefore,



V1  50 cm 3





V2  280 cm 3



To calculate the total volume, we can use V  n1V1  n 2V2

Substituting the values, we find



V  1 46.5  4 69.8  325.7 cm 3



Therefore

44

v

V 325.7  ntot 1 4  cm 3    mol 

v  65.14 

Using Equation 6.43 we can calculate the change in volume. Vmix  V  V1  V2   325.7   50  280 



Vmix  4.3 cm 3



45

6.26 (a) Expression for hmix : hmix   X i H i  X i hi  X Cd H Cd  X Sn H Sn    X Cd hCd  X Sn hSn 

Multiply both sides by the total number of moles H mix   nCd H Cd  n Sn H Sn    nCd hCd  n Sn hSn 

Therefore,

 H mix  Cd

 H mix    H Cd  hCd  nCd  n Sn ,T , P

 

(b) We can show by repeating Part (a) for Sn that

 H mix  Sn  H Sn  hSn Equation 6.65:

 H mix  Cd  hmix  X Sn dhmix dX Sn  H mix  Sn  hmix  X Cd dhmix dX Cd

Since,

hmix  13000 X Cd X Sn

We get, dhmix d 13000 X Cd X Sn   13000 X Cd  X Sn   dX Sn dX Sn dhmix d 13000 X Cd X Sn   13000 X Sn  X Cd   dX Cd dX Cd

Therefore, for 3 moles of cadmium and 2 moles of tin at 500 ºC:

46

2 H Cd  hCd   H mix  Cd  13000 X Cd X Sn  X Sn13000 X Cd  X Sn   13000 X Sn



J   mol 

H Cd  hCd   H mix  Cd  2080 

and 

J   mol 

2 H Sn  hSn   H mix  Sn  13000 X Cd  4680 

(c) Equation 6.37: nCd d  H mix  Cd  n Sn d  H mix Sn  0

Differentiate with respect to XCd: d  H mix  Cd

nCd

dxCd

 n Sn

d  H mix Sn dxCd

0

where d  H mix  Cd dX Cd

d  H mix Sn dX Sn

 26000 X Sn  26000  26000 X Cd  26000 X Cd

Therefore,

nCd

d  H mix  Cd dxCd

 nSn

d  H mix Sn dxCd

 xCd  ntot   26000  26000 xCd   1  xCd  ntot  26000 xCd

Inspection of the above expression reveals that nCd

d  H mix  Cd dxCd

 n Sn

d  H mix Sn dxCd

0

(d) A graphical solution can be found using the tangent-slope method discussed on pages 285-287: A plot of a line tangent to the enthalpy of mixing curve at XCd = 0.6, is given below:

47

Heat of mixin g in cadmium (Cd )­Tin  (Sn) s ystem 6000 5400

    J   hmix      mol

4800 4200 3600

data

3000 2400 1800 1200 600 0 0

0 .2

fit to:   J   hmix  13, 000 xCd x Sn     mol  

0.4

0.6

0.8

1

xCd

The intercepts give the respective partial molar quantities as follows: 

J   mol   J   4800   mol 

H Cd  hCd  2050  H Sn  hSn

The values using the graphical method are reasonably close to the analytical method.

48

6.27 The following can be shown with the Gibbs-Duhem equation 0  x1V1  x 2V2

Differentiation with respect to x1: 0  x1

dV1 dV 2  x2 dx1 dx1

If the partial molar volume of species 1 is constant, the Gibbs-Duhem equation simplifies to 0

dV 2 dx1

Therefore, the partial molar volume of species 2 is also constant. Note that in this case, since the partial molar volume of species 1 is constant: V1  v1

and similarly for species 2: `

V2  v2

Hence, the molar volume can be written: v  x1V1  x2V2  x1v1  x2v2

This is known as Amagat’s law.

49

6.28 (a) Let species 1 represent HCl and species 2 represent H2O. An expression for the enthalpy of the solution is h  x1h1  x 2 h2  hmix

which can be written ~ h  x1h1  x2 h2  x1hs

Therefore, ~ H  n1h1  n2 h2  n1hs

To use the heat of solution data in Table 6.1, we need to determine the values of n1 and n2 consistent with the convention used in the table. As seen in Example 6.6,

x1 

n1 1   n1  n2  1  n

For this problem x1  0.2

Therefore, n1  1 n2  n  4

Now we can find expressions for the partial molar enthalpies.  H H H 2 O  H 2    n2

 

 n1 ,T , P

~ dhs H H 2 O  H 2  h2  n1 dn2

 H H 2 O  hH 2 O  H 2  h2  n1

~ dhs dn2

Using the data in Table 6.1 for n  4 ,

50

~  hs  61,204 

J   mol solute 

J J      64049      56852  ~ ~ ~  dhs hs  n  5  hs  n  3  mol solute   mol solute    dn2  5  mol  3  mol   5  mol  3  mol 

~ dhs J    3598.5  dn2  mol  mol solute 

Therefore,  J   H H 2 O  hH 2 O  1  mol solute    3598.5   mol  mol solute  







J   mol 





  3599 

We can calculate H HCl  hHCl using Equation 6.46





~ hmix  x H 2 O H H 2 O  hH 2 O x HCl hs  x H 2 O H H 2 O  hH 2 O H HCl  hHCl   x HCl x HCl

  H HCl  hHCl 

  J   0.2  61204   mol   



  J    0.8  3599   mol    0 .2

(b) For n1  2 and n2  80 ,  n n  82 n   1 2   1   1  40 n1  2 

The new values for the number of moles consistent with Table 6.1 n1  1  mol

n2  40  mol

Using the data in the table for n  40 , ~ ~ ~ dhs hs  n  50  hs  n  30    50  mol  30  mol  dn2

Interpolating the data in Table 6.1

51

  

   46808 

J   mol 

~  hs  n  30   72428 

J   mol solute 

Therefore,

J J      73729      72428  ~  dhs J  and  mol solute   mol solute   65.05     50  mol H 2O  30  mol H 2 O  dn2  mol  mol solute   J   H H 2 O  hH 2 O  1  mol solute    65.05   mol  mol solute  

52





 J   mol H 2 O 



  65.05 

6.29 First perform an energy balance on the mixing process. hmix  q

We can calculate hmix using data from Table 6.1. Referring to Equation E6.7A, we find ~ hmix  x HCl hs

Calculate x HCl :

x HCl

wHCl 0.30  MW  HCl 36.46    0.175 wH 2 O 0.30 0.70 wHCl    MW  HCl  MW  H 2O 36.46 18.0148

Heats of data are tabulated for a solution containing one mole of the solute for various amounts of water. Thus, we need to calculate how many moles of water must be added to HCl to obtain the above mole fraction. x HCl 

1  mol HCl ; where n is the number of moles of H2O 1  mol HCl  n

n  4.71  mol H 2 O 

By interpolation of data from Table 6.1, we get ~  hs  63224 

J   mol 

(for n  4.71 )

Therefore,   J  hmix   0.175   63224   mol  





J   mol 

  11064  

and 

J   mol 

q  hmix  11064 

53

6.30 To calculate the enthalpy of mixing from Table 6.1, we must use the following expression ~ hmix  x H 2 SO 4 hs

The mole fraction of sulfuric acid is x H 2 SO 4 

1 1 n

where n is the number of moles of water. Equation 6.47 states



hmix  74.4 x H 2 SO 4 x H 2 O 1  0.561x H 2 SO

For n  1 , x H

2

4



SO 4  0.5 and x H 2 O  0.5

Table 6.1:

~  hs  31087 

J   mol 

  J    J   hmix  0.5  31087    15543.5    mol    mol  

Equation 6.47: hmix  74.4 0.5 0.5 1  0.561 0.5  

J   mol 

hmix  13383 

The following table was made n [mol H2O]

x H 2 SO 4

1 2 3 4 5 10 20 50 100

0.5 0.333333 0.25 0.2 0.166667 0.090909 0.047619 0.019608 0.009901

hmix [ kJ/mol]

hmix [kJ/mol]

(Table 6.1)

(Eq. 6.47)

-15543.5 -14978.7 -13001.8 -11414 -10174.2 -6367.27 -3548.43 -1497.22 -762.238

-13382.7 -13441.6 -11993.5 -10568.4 -9367.17 -5835.17 -3284.01 -1414.49 -725.289

% Difference 14.94 10.82 8.07 7.69 8.26 8.72 7.74 5.68 4.97

As you can see, the percent difference between the two methods decreases as the mole fraction of sulfuric acid decreases. Although Equation 6.47 fit data at 21 ºC, while the Table 6.1 tabulates

54

data taken at 25 ºC, we do not expect the temperature dependence to account for all the observed difference. The table and equation come from different experimental data sets, and also represent measurement uncertainty. Nevertheless, the agreement is reasonable.

55

6.31 To calculate the enthalpy of mixing from Table 6.1, we must use the following expression ~ hmix  x HCl hs

The mole fraction of HCl is x HCl 

1 1 n

where n is the number of moles of water. The following table was made using these two equations. n [mol H2O]

x1

~  J  hs   mol HCl 

 J  hmix   mol 

1 2 3 4 5 10 20 50 100

0.5 0.333 0.25 0.2 0.167 0.091 0.048 0.020 0.0099

-26225 -48819 -56852 -61204 -64049 -69488 -71777 -73729 -73848

-13112.5 -16273 -14213 -12240.8 -10674.8 -6317.09 -3417.95 -1445.67 -731.168

56

6.32 A schematic for the process is given below. The inlet streams are labeled “1” and “2” and the exit stream “3”. q

50 wt% NaOH

10 wt% NaOH

50 wt% H2O

90 wt% H2O

Stream 1

Stream 3

 H2O

Stream 2

The energy balance for this process reduces to Q  H 3  H 2  H1

We first convert from weight percentage to mole fraction. For stream 1,

x NaOH ,1

wNaOH 0.50  MW  NaOH 40    0.311 wH 2 O 0.50 0.50 wNaOH    MW  NaOH  MW  H 2O 40 18.0148

and for stream 3,

x NaOH ,3

wNaOH 0.10  MW  NaOH 40    0.048 wH 2 O 0.10 0.90 wNaOH    MW  NaOH  MW  H 2O 40 18.0148

We now calculate the moles of water per mole of NaOH so that we can use Table 6.1: x NaOH 

1 1  nH 2O

Therefore, for every mole of NaOH n H 2 O,1  2.21 57

n H 2 O ,3  19.8

Since enthalpy is a state function, we can choose any hypothetical path to calculate the change in enthalpy. One such path is shown below. The box in our original schematic is depicted with dashed lines below. We pick a basis of 1 mole NaOH. In step A, the inlet stream is separated into its pure components. In step B, 17.6 additional moles of water are added to the pure water stream. Finally the H2O and NaOH streams are remixed

Step A

1 mol NaOH

Step C

1 mol NaOH

2.2 mol H2O

19.8 mol H2O

Step B

2.2 mol H2O

hmix =0

1 mol NaOH

19.8 mol H2O

17.6 mol H2O

The enthalpy change is found by adding each step H 3  H 2  H 1  H A  H B  H C

Since H B represents the mixing of water with water, H B  0 . The enthalpies of mixing for steps A and C can be related to enthalpy of solution data from Table 6.1: ~  hs ,1  23906  

~  hs ,3  42858  

J  mol NaOH  J  mol NaOH 

Note: The enthalpy of solution for Stream 1 is calculated by extrapolation. Generally, extrapolation should be avoided, but it is necessary to complete this problem, and we are not extrapolating very far. For step A, we need the negative value of the heat of solution of stream 1. Thus for a basis of 1 mole NaOH: H A    23906   J 

while for step C:

58

H C  42858  J 

Now adding the enthalpies of each step per 1 mole of NaOH: H  23906  0  42858  18952  J 

To get the total heat that must be removed per mole of product solution, we divide by the number of moles of product per mol of NaOH: q

H  J   910  n NaOH  mol 

59

6.33 The partial molar property can be written as follows:  nT K  n1

K1  

  

T , P , n2 , n3

Applying the chain rule to the above relationship:  nT K1  k   n1

  

T , P , n2 , n3

 k  n1

K 1  k  nT 

 k    n1 

 nT 

  

(1) T , P , n2 , n3

 k 

Now focus on nT  n   1 

T , P , n2 , n3

. At constant T and P, we can write, T , P , n2 , n3

 k   k   k     dk   dx1   dx2   dx3  x1  T , P , x2 , x3  x2  T , P, x1 , x3  x3  T , P , x1 , x2

Therefore,  k   n1 



 T , P , n2 , n3

 k    x1 

 

T , P , x2 , x3

 k    x3

 x1     n1 





 T , P , x1 , x2

T , P , n2 , n3

 x3   n1

 k  x 2

 





 T , P , x1 , x3

 x 2   n1 



 T , P , n2 , n3





 T , P , n2 , n3

but, x1 

n1 n1  n2  n3

so  x1     n1 

 T , P , n2 , n3

n1 1 1 1  x1    2 n1  n2  n3  n1  n2  n3  nT

(3)

Similarly,  x 2     n1 

 T , P ,n2 , n3

n2 x  2 2 nT  n1  n2  n3 

(4)

and 60

(2)

 x3     n1 

 T , P ,n2 , n3

n3 x  3 2 nT  n1  n2  n3 

(5)

Substituting Equations 2, 3, 4, and 5 into the expression 1 for K1 and simplifying:    k     K1  k   1  x1    k    x2    k    x3   x1  T , P, x 2 , x 3  x2  T , P , x1 , x 3  x3  T , P , x1 , x 2

Utilize the fact that x1  x 2  x3  1   k   k  K1  k  x 2       x1  T , P, x , x  x 2 2 3 

   k     x3   k      x    1  T , P, x 2 , x 3  T , P , x1 , x 3    x3  T , P, x1 , x 2





 



(6) 



When we hold species 3 constant:  k   k  dk   dx1    x 1  T , P , x2 , x3   x 2

 k   x 2

  

 k    x1 

T , P , x3

 

T , P , x2 , x3

 

 T , P, x1 , x3

 dx1   dx 2

dx 2



 k      x2

  

T , P , x1 , x3

 dx 2   dx 2

  

Thus,  k   x2

  

T , P , x3

 k      x1 

T , P , x2 , x3

 k  x2



 

 

(7) T , P , x1 , x3

When we hold species 2 constant, a similar analysis shows:  k   x3 



 T , P , x2

 k      x1 

T , P , x2 , x3

 k  x3

 







(8) T , P , x1 , x2

Substituting Equations 7 and 8 into Equation 6 gives  k  x 2

K1  k  x2 





 T , P , x3

 k  x3

 x3 





 T , P , x2

Furthermore, the above analysis can be extended to m components. In general, 

k  xm

K i  k   xm  mi





 T , P , x j i ,m

61

62

6.34 The expression can be found by employing the Gibbs-Duhem equation: 0  n1V1  n 2V2

Differentiate with respect to x1 and then divide by the total number of moles: 0  x1

dV1 dV2 dV dV  x2  x1 1  1  x1  2 dx1 dx1 dx1 dx1

Differentiate the expression given in the problem statement. dV1  5.28  5.28 x1 dx1

Substitution of this result into the Gibbs-Duhem equation and rearrangement yields dV2 5.28 x12  5.28 x1   5.28 x1 dx1 x1  1

Integrate:  cm 3    mol 

V2  2.64 x12  C 

To determine C, we can use the density information given in the problem statement. V2  x 2  1  V2  x1  0  C  v 2

where C  v2  

 

1 2 MW2 







1





 0.768 g/cm 3     84.16  g/mol   

 cm 3    mol 

 109.58 

Therefore,  cm 3    mol 

V2  2.64 x12  109.58 

63

6.35 An expression for the enthalpy of the solution is h  x1h1  x 2 h2  hmix

which is equivalent to ~ h  x1h1  x 2 h2  x1hs

Multiplication by the total number of moles yields ~ H  n1h1  n2 h2  n1hs

To use the heat of solution data in Table 6.1, we need to determine the values of n1 and n2 consistent with the convention used in the table. As seen in Example 6.6,

x1 

n1 1   n1  n2  1  n

For this problem x1  0.33

Therefore, n1  1 n2  n  2

Now we can find expressions for the partial molar enthalpies.  H  n2

H H 2 O  H 2  

 

 n1 ,T , P

~ dhs H H 2 O  H 2  h2  n1 dn2

~ dhs  H H 2 O  hH 2 O  H 2  h2  n1 dn2

Using the data in Table 6.1 for n  2 ,

64

J J      2,787      812  ~ ~ ~  dhs hs  n  3  hs  n  1  mol solute   mol solute     3  mol  1  mol   3  mol H 2O  1  mol H 2O  dn2

~ dhs J    987.5  dn2  mol  mol solute 

Therefore,  J   H H 2 O  hH 2 O  1  mol solute    987.5   mol  mol solute  







Calculate the partial molar enthalpy: 

J   mol 

H H 2 O  hH 2 O  987.5 

From the saturated steam tables at 25 ºC:  kJ  hˆH 2 O  104.87    kg   kJ   hH 2 O  1.89   mol 

Now we can find the partial molar enthalpy 

     kJ J J   0.988    0.90    mol H 2 O   mol H 2 O   mol H 2 O 

H H 2 O  1.89 

65

J   mol 

  987.5 

6.36 (a) Calculate the mole fraction of sulfuric acid

x1 

w1  MW  1 w1 w2   MW  1  MW  2

0.20 98.078   0.044 0.20 0.80  98.078 18.0148

Calculate n to use in Table 6.1: x1 

1  0.044 1 n

n  21.7  mol H 2 O Interpolating from Table 6.1 ~  hs  74621 

J   mol 

Now calculate the heat transfer ~   J  q  hmix  x1hs   0.044    74621    mol  

  



J   mol 

q  3283 

(b) Calculate the mole fraction of pure sulfuric acid. Consider a mixture of 20 kg of 18 M sulfuric acid and 80 kg of water. Find the mass of sulfuric acid present. VH 2 SO4 

mH 2 SO4



 H 2 SO4

20 kg  10.9 L 1.84  kg/L

nH 2 SO4  VM  10.9 L 18 mol/L  196.2 mol





m H 2 SO4  n H 2 SO 4 MW H 2 SO4  196.2 mol 0.098 kg/mol   19.2 kg

Since both the initial (i) and final (f) states contain mixtures, to get the enthalpy of mixing, we need to calculate the relative differences follows: ~ ~ q  hmix  x1, f hs , f  x1, i hs ,i

66

calculate the mole fraction in the final state

w1  MW 1

0.192 98.078 x1, f    0.042 w1 w2 0.192 0.808    MW 1  MW  2 98.078 18.0148 Calculate n to use in Table 6.1: x1, f 

1  0.042 1 n

n  22.8  mol H 2 O Interpolating from Table 6.1 ~  hs, f  74689 

J   mol 

For the initial 18 M sulfuric acid:

w1  MW 1

0.192 98.078 x1,i    0.81 w1 w2 0.192 0.008    MW 1  MW  2 98.078 18.0148 Calculate n to use in Table 6.1: x1,i 

1  0.81 1 n

n  0.23  mol H 2 O We must extrapolate from Table 6.1. To do this we wish to extend the trend at low water concentration. A plot of the data in Table 6.1 is useful. A semi-log plot follows:

67

0

3200 [J /mol]

en th alp o f so lu tio n [J /m o l so lu te]

-10000 -20000 -30000 -40000 -50000 -60000 -70000 -80000 -90000 0.1

1

10

100

n

~  hs,i  3200 

J   mol 

Now calculate the heat transfer ~ ~    J    J   q  hmix  x1, f hs, f  x1, i hs, i   0.042   74689     0.81   3200     mol    mol     

J   mol 

q  710 

(c) Calculate the mole fraction of sodium hydroxide

x1 

w1  MW  1 w1 w2   MW  1  MW  2

0.20 40   0.101 0.20 0.80  40 18.0148

Calculate the n value to use in Table 6.1: x1 

1  0.101 1 n

n  8.9  mol H 2 O Interpolating from Table 6.1 ~  hs  41458 

J   mol 

68

Now calculate the heat transfer ~   J  q  hmix  x1hs   0.101   41458   mol  

  



J    mol 

q  4187 

(d) Calculate the mole fraction of ammonia

x1 

w1  MW  1 w1 w2   MW  1  MW  2

0.20 17.03   0.209 0.20 0.80  17.03 18.0148

Calculate the n value to use in Table 6.1: x1 

1  0.209 1 n

n  3.78  mol H 2 O Interpolating from Table 6.1 ~  hs  33153 

J   mol 

Now calculate the heat transfer ~   J  q  hmix  x1hs   0.209    33153   mol   

J   mol 

q  6929 

69







6.37 Let species 1 designate ethanol and species 2 designate water. We need to obtain an expression for the molar volume, so first, convert the given the mass fractions and densities to mole fractions and molar volumes.

w1  MW  1

Mole fractions:

x1 

Specific molar volumes:

v  vˆ MW  mixture

w1 w2   MW  1  MW  2

where  MW  mixture  x1  MW  1  x 2  MW  2

Using this set of equations, the following table was made. Mole Frac. EtOH 0.000 0.042 0.089 0.144 0.207 0.281 0.370 0.477 0.610 0.779 1.000

Mole Frac. H2O v (ml/mol) 1.000 18.05 0.958 19.54 0.911 21.18 0.856 23.11 0.793 25.47 0.719 28.34 0.630 31.85 0.523 36.19 0.390 41.65 0.221 48.73 0.000 58.36

The following graph plots the data. The trendline relates v to x1. 60.00

v (ml/mol)

50.00 40.00 v = 4.5491x12 + 35.918x1 + 17.957

30.00

R2 = 1

20.00 10.00 0.00 0.000

0.200

0.400

0.600

Mole Fraction EtOH (x 1)

70

0.800

1.000

Now, we can calculate V1 .  V     nv     V1      n1  n 2 , n3 ,T , P  n1  n 2 , n3 ,T , P

We can substitute the trendline for V into this derivative and use the fact that ntot  n1  n2 to obtain V1 

 n12   4.5491  35.918n1  17.957 n1  n2     n1  n2  n1  

n 2 ,T , P

Differentiating we get  2n  n  n   n 2  2 1  53.875  ml  V1  4.5491 1 1   mol   n1  n2  2  





 ml   mol 

V1  4.5491 2 x1  x12  53.875 

Calculate V2 : V2 

 n2



 4.5491  

 n12  35.918n1  17.957 n1  n2     n1  n2  

Differentiating we get   n12   ml  V2  4.5491   17.957  2  mol    n1  n2    ml   mol 

V2  4.5491x12  17.957 

Plotting V1 and V2 vs. x1 we obtain

71

n1 , T , P

60 59 58 57 56 55 54 53 52

20 19 18 17 16 15 14 13 12 0

0.2

0.4

0.6

0.8

H2O Partial Molar Volume (ml/mol)

EtOH Partial Molar Volume (ml/mol)

Partial Molar Volumes vs. EtOH Mole Fractions

EtOH H2O

1

Mole Fraction EtOH (x1)

(b)

v mix can be calculated using Equation 6.46





v mix  x H 2 O V H 2 O  v H 2 O  x EtOH V EtOH  v EtOH 

From the data table in Part (a),  ml   mol   ml  v H 2 O  v x1  0  18.05   mol  v EtOH  v x1 1  58.36 

Using the expressions for partial molar volumes





 ml   mol 

V1  4.5491 2 0.5   0.5 2  53.875  57.29   ml    mol 

V 2  4.5491 0.5 2  17.957  16.82 

Therefore,    ml    ml   v mix   0.5  16.82  18.05    0.5 57.29  58.36     mol    mol      ml  vmix  1.15    mol 

6.38 72

We can use the density data given in the problem statement to determine the pure species properties. For pure ethanol  x1  1 : MW1

v1 

1



46  g/mol



0.7857 g/cm 3

 cm 3    mol 



 58.55 



  cm 3   3 V1  n1v1   3  mol   58.55     175.7 cm   mol      



For pure formamide  x1  0  :  cm 3  MW2 45  g/mol   39.77   2 1.1314 g/cm 3  mol    cm 3   3 V2  n 2 v 2  1  mol   39.77     39.77 cm   mol    



v2 







To calculate V and v, interpolate in the data table to obtain the density of the mixture when x1  0.75 :



  0.8550 g/cm 3



Therefore, v

where

MW



MW  0.75 46  g/mol   0.25 45  g/mol   45.75  g/mol

Substitute numerical values: v

45.75  g/mol



0.8550 g/cm

3





 53.51 cm 3 /mol

 

  cm 3   3 V  nv   4  mol   53.51     214.0 cm   mol    



Now, we can calculate the volume change of mixing by using Equation 6.43:











Vmix  V  V1  V2  214 cm 3  175.7 cm 3  39.77 cm 3



Vmix  1.47 cm

3



The intensive volume change of mixing:

73



v mix 

 



Vmix  1.47 cm 3   0.368 cm 3 / mol n 4



We can use Equation 6.65 to determine the partial molar volume of formamide: V2  v  x1

dv v  v  x1 dx1 x1

From the provided data table 45.8 45.7  v  0.8401 0.8701  19.50 cm 3 / mol x1 0.8009  0.6986





Therefore,





 







V2  53.51 cm 3 / mol  0.75 19.5 cm 3 / mol  38.89 cm 3 / mol

Now calculate the partial molar volume of ethanol: V  n1V1  n 2V2  V1 

 









214.0 cm 3  1  mol  38.89 cm 3 /mol  58.37 cm 3 / mol 3  mol

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

6.39 Using the definition of G, the Gibbs energy of mixing of an ideal gas can be rewritten in terms of the enthalpy of mixing and the entropy of mixing: ideal gas ideal gas ideal gas g mix  hmix  Tsmix

Since an ideal gas exerts no intermolecular interactions, ideal gas hmix 0

From Equation 6.48: ideal gas smix   R xa ln xa  xb ln xb 

so ideal gas g mix  RT  xa ln xa  xb ln xb 

To find the partial molar Gibbs energy of mixing of species a, we apply Equation 6.29:

 Gmix  a     nng mix   

a

 T , P , nb

Applying the expression above  na nb    RT  na ln na  nb ln nb   na  nb  ln  na  nb   ng mix  RT  na ln  nb ln n  n n  n a b a b  where the mathematical relation of logarithms was used. Thus,

Gmix  a     nng mix   

a

 n n  n    RT  ln na  a  0  ln  na  nb   a b   RT ln xa na  na  nb    T , P , nb 

at infinite dilution xa goes to zero, and the ln term blows up,

 Gmix  a   a  g a   As chemical engineers, we are often interested in the limiting case of infinite dilution. We see that even for ideal gas mixtures the chemical potential in this limit is not mathematically well-.behaved, In Chapter 7, we will develop a different function, the fugacity, which behaves better.

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6.40 At equilibrium l v H  H O O 2 2

or l v GH  GH O O 2 2

Since the water in the liquid phase is pure

GHl 2O  g Hl 2O  hHl 2O  Ts Hl 2O The enthalpy and entropy of liquids are not sensitive to pressure changes. We can use data from the saturated steam tables at 25 ºC to determine the Gibbs energy. l gˆ H  104.87  kJ/kg    298.15 K   0.3673  kJ/kg  K   2O l gˆ H  4.640  kJ/kg  2O







l l gH  MWH 2 O gˆ H   0.0180148  kg/mol    4.640  kJ/kg    0.0836  kJ/mol 2O 2O

Therefore, v H

2O

 g lH O  83.6  J/mol 2

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