Chapter 8 Solutions Engineering and Chemical Thermodynamics 2e

Transcription

1 Chapter 8 Solutons Engneerng and Chemcal Thermodynamcs e Mlo Koretsky Wyatt Tenhaeff School of Chemcal, Bologcal, and Envronmental Engneerng Oregon State Unversty mlo.koretsky@oregonstate.edu

2 8. No ths wll not work. A central dea behnd bnary phase equlbrum s that when both speces are volatle, there wll be a mxture of and n both the vapor and the lqud. There wll be a hgher mole fracton of the lghter component, toluene, n the lqud relatve to the mole n the vapor, but there wll also be cyclohexane present.

3 8. Yes f speces b s the lghter component, the phase dagram for the system looks approxmately lke the one below. We can see that both x a and y a are smaller at the hgher pressure, : b x a y a a x a y a 3

4 8.3 The addton of the second speces to the lqud wll lower the lqud mole fracton of water. Thus, the fugacty of water n the lqud wll be lower than ts fugacty n the vapor. Snce T and are constant, the fugacty of the steam remans constant. Snce the fugacty of water n lqud s lower than vapor, t wll all condense. 4

5 8.4 The addton of the second speces to the lqud wll lower the lqud mole fracton of water. Thus, the fugacty of water n the lqud wll be lower than ts fugacty n the vapor and the steam wll condense. However, snce the contaner s rgd, the pressure of the vapor wll decrease as the steam condenses, lowerng ts fugacty. Therefore, some of the steam wll condense untl the pressure n the vapor s low enough that the fugacty n the vapor matches the fugacty n the lqud. 5

6 8.5 When the thrd component s added to the lqud, the mole fracton of the water n the lqud decreases so ts fugacty decreases. Thus, the fugacty of water n the vapor wll be greater than n the lqud and some water wll condense, lowerng the number of moles n the vapor. 6

7 8.6 (a) snce < (ln < 0), the unlke nteractons are stronger and the lke nteractons are weaker. (b) [bar] x a, y a Snce the pressure wth a-b nteractons s hgher than the deal soluton case where all the nteractons are the same (blue lne, Raoult s law), the lke nteractons are stronger. (c) Snce t takes more energy to bol the mxture (maxmum bolng azeotrope), the unlke nteractons are stronger and the lke nteractons are weaker. 7

8 8.7 (a) (). For postve devatons from dealty, both actvty coeffcents are greater than unty. Ths mples that the lke nteractons of the speces are stronger than the unlke nteractons, whch s true for a mxture of methane and ethanol. Ethanol con hydrogen bond wth tself, whch leads to strong lke-lke nteractons. The H bond nteractons are much greater than the dsperson and nducton of ethanol methane. (). Both actvty coeffcents are less than one n ths case, so the unlke nteracton are stronger than the lke nteractons. Ths behavor s seen n acetone/chloroform mxtures because the partally postve hydrogen atom n chloroform s attracted to the partally negatve oxygen n acetone leadng to a H bond n the unlke speces. Nether of the lke nteractons are nearly ths strong. (b) Ethane/Methane. Snce methane and ethane are smlar n structure, the unlke nteractons are very smlar to lke nteractons. (c) The cross coeffcent, B, s more negatve than the pure speces B parameters when the unlke nteractons are stronger than ether of the lke nteractons. Ths statement s true for acetone and chloroform (see (a)()). (d) The geometrc average s an accurate estmate of the cross vral coeffcent when dsperson forces govern the nteractons between unlke molecules and the polarzabltes of the two speces are nearly equal. Ths s true for methane/ethane mxtures. 8

9 8.8 In System I at the composton where the two g curves touch, the Gbbs energes are equal and we have vapor-lqud equlbrum. At all other compostons for ths T and, the Gbbs energy of the lqud phase s lower so the lqud more stable. Ths behavor s consstent wth a mnmum bolng azeotrope (see Fgure 8.8(a)), where the temperature of the system s the temperature of the azeotrope. In System II at the composton where the two g curves touch, the Gbbs energes are equal and we have vapor-lqud equlbrum. At all other compostons for ths T and, the Gbbs energy of the vapor phase s lower so the vapor more stable. Ths behavor s consstent wth a maxmum bolng azeotrope (see Fgure 8.8(b)). 9

10 8.9 (a) False the ncrease n entropy upon mxng can allow two speces to mx when the unlke nteractons are weaker. (b) True the energetc effects outwegh the ncrease n entropy upon mxng 0

11 8.0 The purpose of applyng salt to the road s to lower the freezng pont of water. Therefore, we want to fnd the lowest temperature at whch lqud water exsts. From the dagram, the lowest temperature s -. ºC. The smallest composton of salt requred to lower the freezng pont to ths temperature s approxmately 3 wt%.

12 8. For ncreasng x a n System : ure untl frst te lne s reached Mxture of and along the te lne ure untl the second te lne s reached Mxture of and along the second te lne ure for the remanng compostons For ncreasng x a n System : ure untl the frst lne s reached ure and along the te lne ure for the remanng compostons (Along the te lnes, the compostons are determned usng the lever rule.)

13 8. The speces that dssocate nto ons wll lower the fugacty of lqud water the greatest. More ons wth MgCl. Hghest bolng pont 0.05 M MgCl 0.05 M NaCl 0.05 M glucose (C 6 H O 6 ) pure water Lowest bolng pont 3

14 8.3 The speces that dssocate nto ons wll lower the fugacty of water the greatest. More ons wth MgCl. Hghest freezng pont pure water 0.05 M glucose (C 6 H O 6 ) 0.05 M NaCl 0.05 M MgCl Lowest freezng pont 4

15 8.6 A Txy phase dagram dsplays the bubble pont and dew pont over the entre range of lqud and vapor mole fractons. By assumng deal soluton and deal gas behavor of the mxture, we can use the followng relatonshps x x y y We can calculate the uraton pressures usng Antone s Equaton. The temperature that sfes the frst equaton s the bubble pont. The temperature that sfes the second equaton s the dew pont. The followng spreadsheet was created wth these two expressons (the temperatures were found usng a Solver functon): x T bp (K) y T dp (K) The data were plotted to obtan the followng graph: Txy Dagram for Cyclohexane and n-hexane Temperature (K) x, y 5

16 8.7 The vapor mole fractons of n-pentane () and n-hexane () are y 0.33 y 0.67 We can solve ths problem by fndng the temperature at whch the followng relatonshp s true: y y The uraton pressures can be found usng Antone s Equaton. From Appendx A, Speces A B C n-pentane () n-hexane () Therefore, we must solve the followng equaton for T: We obtan 0.33 bar 0.67 bar exp9.3 exp9.64 T T T 334 K 6

17 8.8 To calculate the total pressure and the vapor composton we can use Equatons 8.8 and 8.9. These equatons requre usng uraton pressures, whch can be calculated wth Antone s Equaton. From Table A.., Speces A B C ropylene () ropane () n-butane (3) Isobutane (4) n-pentane (5) Therefore, the uraton pressures at 50 K are Equaton 8.8: Speces bar ropylene ().74 ropane ().8 n-butane (3) 0.39 Isobutane (4) n-pentane (5) bar Equaton 8.9: x x x x x bar0.5.8 bar bar bar bar y x Usng ths equaton, the followng table can be created Speces y ropylene () 0.4 ropane () 0.48 n-butane (3) Isobutane (4) 0.96 n-pentane (5)

18 8.9 Namng rotocol: Cyclohexane Speces Benzene Speces Toluene Speces 3 n-heptane Speces 4 To determne the temperature and vapor compostons, the uraton pressure of each speces s requred. The vapor pressures can be calculated Antone s equaton and data from Appendx A: exp9.35 T exp9.806 T exp T exp9.535 T 56.5 Expresson for pressure assumng deal gas behavor: x x x x Substtute the uraton pressure expressons and solve for T: T 36.3 K At ths temperature.6 bar.9 bar bar bar To determne the vapor composton, we can use Therefore, y x y y y y

19 8.0 Let the subscrpt a desgnate n-butane and b desgnate sobutane. Frst, perform mole balances: x F x L y V a, feed a a x F x L yv b, feed b b Usng the nformaton provded n the problem statement, V 0.4F L 0.6F x, 0.6x 0.40 y x, 0.6x 0.40y afeed a a bfeed b b Equaton 8.9 states: y a xa a y b x b b Substtutng these expressons nto the mole balances, we obtan x a a 0.4 a xa, feed0.6xa 0.40 xa0.6 x b b 0.4 a xb, feed0.6xb 0.40 xb0.6 We also know that x a x b. Therefore, x x a, feed b, feed 0.4a 0.4 b For the uraton pressures, we can substtute Antone s Equaton. The Antone coeffcents can be found n Appendx A exp exp8.979 T 34.4 T bar bar Solve for T: 9

20 T 66.6 K At ths temperature Therefore, a b 0.8 bar.3 bar y 8 a 0. x a yb. 3xb We can obtan the followng equaton from the mole balance on speces a: x x a, feed 0.6xa 0.5 x a x b 0.9x a x a Now calculate the vapor mole fractons ya 0. 8x a y a y b 434 a 0

21 8. A mass balance on component gves x, feed F y V x L () If we assume deal gas and deal soluton, the equlbrum relaton n the flash drum can be wrtten accordng to Raoult s law: y x () Substtutng Equaton nto gves x, feed Solvng for x : F x V x L x V L x x, feed V F L F (3) An overall balance gves F V L (4) Snce the sum of the mole fractons equals, we have x, feed x (5) V L F F Usng Equaton 4, we get x x, feed V F V F Ths leaves equaton for the unknown (V/F). Once V/F s determned, x s found from Equaton 3. Then y s found from Equaton. The soluton summary s llustrated below:

22 calc Antone coeffcents a b c nc 4 H 0 nc 5 H nc 6 H 4 A B C Intal condtons T [K] calc x calc y x y l/f 0.54 v/f

23 8. Snce the pressure s 0.5 bar, the vapor can also be assumed deal. Therefore, the vapor compostons can be calculated as follows: y a x a a yb xb b To fnd the uraton pressures, we use Antone s Equaton and the approprate values from Appendx A. We obtan, a b.03 bar bar Snce the soluton s deal, we also have the followng relatonshp: x x.03 bar x bar a a a b 0.5 bar xa a a x x Now we can calculate the gas compostons: b y a y b bar 0.5 bar ya

24 8.3 Snce the pressure s bar, we wll assume deal gas behavor. The followng relatonshps hold: y a b x y x a a aa b bb aa x x a Frst, fnd the uraton pressures usng data n Appendx A. a bar bar b b b Now, create expressons for the actvty coeffcents usng the equatons n Table 7.: a exp 86 x a b exp 86x a We have one equaton for one unknown: bar x a exp 86 x a bar x a exp 86x a bar x a We can substtute the lqud mole fracton of n-pentane nto the followng equatons to obtan the vapor compostons y a Therefore, x a exp bar bar 86 x a y a y b y 0.53 a 4

25 8.4 (a) Calculate the actvty coeffcents from A and B: a exp RT A 3B x 3 b 4Bx b b exp RT A 3B x b 4B x b 3 Calculate the uraton pressures wth Antone s equaton: Speces A B C (a) Isobutane (a) Hydrogen Sulfde (b) We can fnd the mole fractons usng the followng equaton: xa aa xb bb x b exp RT A 3B x 3 b 4Bx b a x b exp RT A 3B x b 4B x b Substtutng values, we obtan x b a 0.47 x 0.53 Calculate the actvty coeffcents: 3 b a b.07.4 Now, calculate the vapor mole fractons: y a x a a a 0.53 y b a a y a 0.8 5

26 (b) We can use the followng equlbrum expressons to calculate equlbrum concentratons: y a a x a a a a y b b x b b b b These equatons can be rearranged to yeld x y a a a a a a x y b b b b b b Use the fugacty coeffcent expressons from Table 7.: a expb a a a RT RT b expb b a b RT RT The a and b parameters can be calculated wth the van der Waals EOS: a a a b b a b b Substtute these values nto the expressons for the fugacty coeffcent: a 0.84 a b b Substtute numercal values and actvty coeffcent expressons nto the equatons for pressure: x b exp 8.34 x b exp A 3Bx 3 b 4Bx b y a A 3B x b 4B x b y a We have two equatons for two unknowns. Solve smultaneously: x b 0.47 x a y a 0.4 y b

27 8.5 The followng condtons hold at the azeotrope: We can fnd the partal pressure of ethanol at 60 ºC usng Antone equaton: ln B A T C The A, B, and C parameters can be found n Appendx A. Usng these values, we calculate at 60 ºC: bar 0.56 bar Therefore, 0.64 bar 0.64 bar.37 and bar 0.56 bar To determne A, we can use the approprate equaton from Table 7.. For speces J mol K Kln.37 A J A 4 mol For speces A J A 3 mol J mol K Kln.4 The best value s obtaned by averagng these two: A 366 J mol 7

28 (b) We can use Equaton 8.6 to solve for the vapor mole fractons, but frst fnd the actvty coeffcents at the gven lqud composton: exp exp Equaton 8.6: y x x x y 0.67 y y

29 8.6 (a) The curve that goes through the orgn should be labeled wth and the other curve belongs to. Ths s the case because for a Lews-Randall reference state, As x, ln 0 (b) The lke forces are stronger because the actvty coeffcents are greater than one. The fugacty s greater than the uraton pressure for a mxture; therefore, we can nfer that more molecules are volatlzng due to the weaker unlke nteractons. (c) The total system pressure s calculated as follows: x x From the steam tables,.03 bar The dagram n the problem statement provdes ln ln 0.5 Substtute and calculate: bar bar.58 bar (d) We can assume the vapor phase s deal snce the pressure s relatvely low. For equlbrum to exst, the fugactes of the lqud and vapor phases must be equal. Therefore, y x x bar y.58 bar 9

30 y 0.4 (e) From the graph, we can fnd the actvty coeffcent of speces at nfnte dluton. ln Now calculate the Henry s law constant wth [ 5.bar [ 3.5. bar (f) Ths system does exhbt an azeotrope. Because the pressure calculated n art (c) s greater than ether of the pure speces uraton pressures, the pressure must go through a maxmum, where an azeotrope wll form. 30

31 8.7 An teratve technque wll be used to solve ths problem. We are assumng deal soluton behavor. We calculate the pressure usng y ˆ x y ˆ y ˆ x We can combne the equatons to obtan: x x ˆ ˆ We can calculate the fugacty coeffcents usng the equaton from Table 7.: For the frst round of calculatons, we wll assume the fugacty coeffcents are equal to one. Substtutng values from the problem statement, we get 8.4 bar Now we can calculate the vapor mole fractons: y x ˆ y y We obtan: y 0.7 y 0.78 Usng the mole fractons and pressure obtaned above calculate the fugacty coeffcents from Table 7.: ln ˆ v ln vb mx RT ln ˆ v ln vb mx RT Note: a mx y a y ya y a b mx yb yb b v b mx y a y a a RTv b v b mx y a y a a RTv 3

32 For the mole fractons shown above and crtcal data n Appendx A, a mx.6 b mx Therefore, ˆ v ˆ v 0.79 Now, repeat the process above startng wth the new fugacty coeffcents. We obtan.6bar y 0.70 y 0.79 v v ˆ ˆ Iteratng a thrd tme, we get.3bar y y 0.98 ˆ v ˆ v A fourth teraton provdes.5bar y y ˆ v ˆ v The ffth teraton results n.5bar y y Snce the pressure for the ffth teraton s essentally equal to the pressure obtaned n the fourth repetton, we have solved the problem. The fnal answer to ths problem s.5bar y y

33 8.8 As n Example 8.6, an teratve technque s requred to solve ths problem. Step. Determne all of the constants for the problem. Calculatng the pure speces a and b parameters (a, b, a, and b ) for the van der Waals equaton s descrbed n Chapter 4. Crtcal values are provded n Table A... To fnd the a and b parameters for the mxture, use a mx b mx y y a y ya y a b yb The actvty coeffcents wll also reman constant. Calculate them wth the followng equatons: exp A 3B 4B 3 RT x RT x exp A 3B 4B 3 RT x RT x The uraton pressures can be calculated usng Antone s Equaton. Step. Set the fugacty coeffcents equal to one for the ntal calculaton. Step 3. Calculate the pressure usng x ˆ x ˆ Step 4. Evaluate the vapor mole fractons wth the followng equatons: y x y y ˆ Step 5. Now that the vapor mole fractons are known, calculate the fugacty coeffcents wth the equaton from Table 7.: ln v ˆ ln vb mx RT b y v b mx a y a a RTv 33

34 ln v ˆ ln vb mx RT b y v b mx a y a a RTv Note: You must solve for v wth van der Waals EOS (use the pressure obtaned n Step 3). Step 6. Start a new teraton at Step 3, usng the new values for fugacty coeffcents. Repeat steps 3 through 6 untl the change n pressure from one teraton to another s smaller than your acceptable lmt. 34

35 8.9 (a) We can examne the Margules parameter A and we fnd that A > 0. Ths ndcates that lke nteractons are stronger than unlke nteractons. (b) x x a a a b b b RT A B x Bx 3 ln a 3 b 4 b J/mol 3 ln b 3 a 4 a J/mol RT A B x Bx a b From the Antone equaton: a exp bar b exp bar Solvng for pressure: bar (c) Solvng for mole fracton xa aa ya

36 8.30 Determne the temperature and the vapor phase of mole fracton a: ka J mol exp Low deal gas exp A exp A 74.5 ka 0.6exp exp steam tables Antone equaton ln Guess and check: TC ka ka (ka) 50C C *60C

37 8.3 (a) Gven: x 0.47 x x bar From the Antone Equaton at 0 o C: 0.0 bar 0.9 bar Settng fugactes equal: y x yx x x Applyng the two-suffx Margules Equaton A exp x RT A exp x RT So A A x exp x x exp x RT RT Solvng for A: A 0 J mol (b) We can examne the Margules parameter A and we fnd that A > 0. Ths ndcates that lke nteractons are stronger than unlke nteractons. 37

38 (c) A exp.68 RT [ 0.7 bar [ bar (d) x y 0.77 (e) Lever rule l n y z T n y x l n.3 mol 38

39 8.3 To calculate the pressure, we can use the followng: To calculate the mole fracton of vapor we use: Start by calculatng &, usng the three-suffx Margules model RT ln RT ln a b 3 A 3Bxb 4Bxb 3 A 3Bx 4Bx a a Next we need to use the Antone equaton to calculate for each of the speces n our soluton. a b exp exp bar bar Ths can be determned by plottng x a and y a vs ressure [bar] xa, ya Where x a s the blue lne and y a s the red lne. We can see that t does form an azeotrope. Next we can determne the pressure at whch that azeotrope occurs va solver or other analyss. = bar 39

40 8.33 At the azeotrope x y x y Therefore, the expressons equatng lqud and vapor fugactes smplfy to (Note: We are also assumng that the vapor behaves deally, whch s reasonable snce the pressure s.3 torr.) We can calculate the uraton pressures at 5 ºC usng Antone s equaton data n Appendx A exp exp bar For the van Laar equaton bar 58.5torr torr exp A RT Bx Ax Bx exp B RT Ax Ax Bx Substtute these expressons nto the above equatons for pressure. We have ln A RT Bx Ax Bx ln B RT Ax Ax Bx Insertng numercal values, we have two equatons for two unknowns: ln ln A B Solve the equatons smultaneously: B0.7 A0.8 B0.7 A0.8 A0.8 B0.7 40

41 A B J mol J mol Now, to calculate the pressure and lqud composton when the vapor mole fracton of ethanol s 0.75, we can use the followng equatons y x y x Therefore, y y y y y y x x x x Substtute the actvty coeffcent expressons for the van Laar equaton: x exp A Bx x exp A B x y RT Ax Bx RT Ax B x y x exp B Ax x RT Ax Bx exp B Ax RT Ax B x Insert numercal values and solve for x : x x Now, we can calculate the pressure torrexp x y 73. torr

42 8.34 (a) The molar volume can be found by manpulatng the expresson gven n the problem statement: RT v Ay y B v m v mol 8.34 We calculate the total volume as follows V n T v 5 mol m 3 /mol / / m 3 The molar volume of speces s calculated by lettng y v v m mol The partal molar volume of speces s calculated by evaluatng the approprate dervatve: V V n n RT n n A Bn n n n n n T,, n V RT Ay B 3 4 m V mol (b) To fnd the pure speces fugacty coeffcent, frst fnd an expresson for the pure speces fugacty: low f d RT ln v v low 4

43 RT B d RT ln f v low low RT ln B low RT ln f low If we let low go to zero, cancel the remanng low s, and smplfy, we obtan B f v exp RT B v exp RT.38 v low To fnd the fugacty coeffcent of speces n the mxture, we use the followng procedure: v fˆ V d RT ln y low low RT Ay v ˆf B d RT ln low y low RT ln v low ˆf low Ay B RT ln y low If we let low go to zero, cancel the remanng low s, and smplfy, we obtan ˆf v y exp RT Ay B ˆ v exp RT Ay B.6 (c) lot the actvty coeffcent of speces versus ts lqud mole fracton: 43

44 lot of Actvty Coeffcent He nry's Henry's x The actvty coeffcent based on Henry s Law s less than one. Thus, the fugacty s less than the fugacty based solely on - nteractons. Consequently, the lke nteractons must be stronger. (d) For phase equlbrum of speces to exst, the followng must be true: y x [ ˆv Henry ' However, we can t calculate the actvty coeffcent untl we know the lqud mole fracton. Therefore, we need to make an expresson for the actvty coeffcent contanng the mole fracton: x y ˆ v Henry's H y ˆ v exp 7 x H We can fnd the mole fracton wth a solver functon: x 0.03 (e) Intally assume the fugacty coeffcent s one at the uraton pressure and the oyntng correcton s neglgble. Use the followng equaton: Henry' s [ 7000barexp bar 44

45 The fugacty coeffcent at uraton can be found accordng the the result of part (b): B exp RT.00 so the assumpton s vald. 45

46 8.35 At the azeotrope x y xa y a b b Therefore, the expressons equatng lqud and vapor fugactes smplfy to a a b b (Note: We are also assumng that the vapor behaves deally.) Snce the lqud-phase nondealty s represented by the two-suffx Margules equaton, the followng expressons are used to calculate the actvty coeffcents: a exp A RT x b exp A RT x a b exp A RT x a Substtute these expressons nto the pressure equatons and equate the pressures. Solve for x a : a exp A RT x a b kaexp 8.34 x a 0.68 exp 38.5 A RT x a x a 46.5 kaexp x a Now, calculate the actvty coeffcent of methanol. a exp Use ths value to calculate the pressure ka 76.4 ka 46

47 Thus, an azeotrope forms at a pressure of 76.4 ka and a temperature of 55 ºC. The composton s 68% methanol and 3% ethyl acetate. Ths s a mnmum-bolng azeotrope (postve devatons from Raoult s Law) because the azeotropc pressure s greater than the two uraton pressures. 47

48 8.36 RTln ln99.7 3,30 J mol RT 4.60 RTln exp RT 0.0, 0.75, The value 0.75 makes physcal sense. 48

49 8.37 System ropertes: 0.57bar 30 o o T C K Toluene Data: Antone A B C T C C o 59.7 K 4.4bar 0.57 fˆ f v v general correlaton R 0.5 T R C From Appendx C and ( ) log o () log 0.03 ( ) () log log o log Vapor s pure, lqud s a bnary mxture f f fˆ x v l f x v To fnd we need to lnearly nterpolate 3.3 R 0.76, TR log R ( o ) () log log

50 0.709 x 0.75 From the graph, and more lnear nterpolaton x 0.7 x x

51 8.38 ln ln suffx Margules ln 0.75 RT ln 953 J mol 0.4 low,,,, 80ka 60ka (a) exp RT exp RT 80.3 ka mol 4 mol mol (b) At azeotrope. 8 6 exp RT RT ln

52 exp RT 85.9ka (c) Snce A>0, the lke nteractons are stronger 5

53 8.39 (a) If ths mxture s n equlbrum wth vapor under these condtons, what s the vapor phase mole fracton of a? mol A 400 J mol 3 mol Table 7. B 50 J mol 4 mol RTln A 703 T bar 50 ka ln (b) Determne uraton pressure of b ka 50 A RTln B A B ln ka 53

54 8.40 (a) lot shows xy phase dagram for a bnary mxture of speces and at 300K. Vapor, y =0.67 (b) ~ 50 ka y A ~0.77 x A ~

55 8.4 (a) 7850 J mol J mol 0.4 A RTln B A B 050 J mol ln From steam tables: 70. ka ka 55.3 ka (b) (c) 55 ka [ [ 0 RTln 340 J mol ln [ ka 7 ka 55

56 8.4 (a) bar (b) (c) () lqud () x = 0. () mol z = 0.6 () phases, vapor and lqud () x = 0.4, y = 0.74 () apply the lever rule l n v n l n 0.8 mol v n.8 mol (d) y x x 0.74 y 56

57 A RT ln x J,040 mol (e) (f) y, A 0.74 x, A 0.4 FF VA LA F y V x L V L, A A, A A A A mol.8 s mol 0.8 s y, B x, B VA VB LB y V y V x L mol VB 0.5 s mol LB 0.67 s, A A, B B, B B 57

58 8.43 exp B RT RT ln,,, RT V RT B B B RT V B B B RT RT V V,, B B RT ln RT V V B B ln ln V RT B B V RT ln ln RT B B ln ln ln (I) (II) guess solve II for solve RHS for repeat y v RHS (I)

59 From Table B.3 60 a a LHS I: ln 5.64 y =

60 8.44 (b) (d) 0.8 At azeotrope:.06 RT ln.65 RT ln A 3630 J mol A 3798 J mol 370 J mol (e) F mol s L.46 mol s V. F0.54mol..,

61 , 0.3 V V 0.3 mol s 6

62 8.45 (a) In order to be consstent wth Henry s law, the actvty coeffcent of speces should approach one as the mole fracton of speces goes to zero. For the gven equaton lm Henry's 30.5 x 0 0 Henry's Indeed, the actvty coeffcent expresson s consstent wth Henry s law. It s also clear from the expresson that the actvty coeffcent wll always be greater than unty. Therefore, the fugacty of the mxture s always greater than the deal fugacty based on Henry s law reference state. Ths suggests that the tendency for the molecules to escape nto the vapor phase (vaporze) n the real system s greater than the tendency when only unlke nteractons exst. The unlke nteractons are stronger. (b) Frst, start wth the Gbbs-Duhem as follows ln x x ln x Henry' s LR x For the gven actvty coeffcent expresson 0 ln Henry' s x 6.0x Hence, LR ln x x LR ln 6x 0 x 6.0x x 0 Separate varables and ntegrate: LR ln 6x 30. 5x C For the Lews/Randall reference state when x. Therefore, LR 6

63 and C C 30.5 LR ln 6x 30.5x 30.5 LR ln 30. 5x (c) Frst, solve the problem for the solute (speces ). For vapor-lqud equlbrum Therefore, ˆ ˆ v l f f ' y x Henry s [ (Note: The fugacty coeffcent does not appear n the equaton because we are assumng the vapor behaves deally.) Henry ' s y [ [ exp 30.5x x Now, consder speces. Therefore, ˆ ˆ v l f f y x LR y LR x exp 30.5x exp 30.5 x (d) The followng must be true for an azeotrope to form y x x y Substtute the expressons from art (c) nto the above relatonshp: [ exp 30.5 x exp 30.5x 63

64 [ exp 30.5 x exp 30.5 x Insert numercal values and solve for x : x 0.86 x 0.4 Now, we can calculate the pressure. LR 0.0 bar exp ka 0.0 bar 64

65 8.46 Snce the pressure s bar, we can assume the gas s deal. We wll also assume the lqud s deal. We can fnd the solublty of oxygen (speces ) n methanol (speces ) by solvng for x n the followng equaton. x[ x x x [ Usng Antone s Equaton, we fnd 0.68 From Table 8. [ bar bar Now, we can solve for the mole fracton bar [ bar x At 00 bar, the Henry s constant wll change. We calculate the new value as follows. V d ln [ d RT [ ln bar 8.34 [ 00 bar 57.4bar Now calculate the mole fracton as before a 0.68 bar [ bar x

66 8.47 Snce the pressure s bar, we can assume the vapor behaves deally. Also, snce the lqud mole fracton of carbon doxde s so low, we can assume the soluton s deal (Henry s lmt). Therefore, y x [ CO CO CO To calculate the left hand sde of the equaton, we do the followng: y y CO H O y y CO H O By equatng fugactes, we also have y Therefore, H O x H O H O y CO x H O H O 035 a a 0.704bar Now we can calculate Henry s constant: [ CO y CO x CO bar bar 66

67 8.48 The followng s true: V RT ln [ T lot the data gven n the problem statement: lnh From the slope we can fnd the desred quantty: slope R 8.34 T 9.55 V 8.4 J/mol bar m 3 /mol 67

68 8.49 Use the followng relatonshp ln[ v H h O O O / T R Therefore, the followng plot should be created wth a lnear ft to the data. 4.5 lot of Henry's Constant Data ln [ y = x /T (K ) From the trendlne, we have H O h R H v O h O v O K K 8.34 J mol K 904 J mol 68

69 8.50 Frst, format the top of the spreadsheet as follows T (ºC) 0 R (J mol - K - ) 8.34 A (J/mol) Intal Guess T (K) 83.5 Measured Data Caculated usng data gven n the problem statement Calculated usng "A" x x y y (a) (J/mol) (J/mol) g E,exp (J/molg E,calc (J/mol) (ge -ge,calc) Measured Data secton: Smply copy the gven data nto the cells below the approprate column. Note that x x y y Calculatons wth Data Gven n the roblem Statement secton: The next secton contans values of the actvty coeffcents and excess Gbbs energy calculated wthout usng A. The actvty coeffcents are calculated usng Equatons 8. and 8.: y where s equal to the pressure when x x Calculate the excess Gbbs energy: gexp E RT x ln x ln Calculated usng A secton: We are fttng the two-suffx Margules equaton; therefore, g E calc Ax x The value of A s equal to an ntal guess that you enter as shown n the spreadsheet above. Calculatng g E E exp g calc should be straghtforward. After enterng all of the avalable data and equatons n the spreadsheet, the value of A s found usng a solver functon whch mnmzes the sum of g E E exp g calc by varyng A. 69

70 T (ºC) 0 R (J mol - K - ) 8.34 A (J/mol) T (K) 83.5 Measured Data Caculated usng data gven n the problem statement Calculated usng "A" x x y y (a) (J/mol) (J/mol) g E (J/mol) g E,calc (J/mol) (ge -ge,calc) SUM E g A 398 J/mol OF s equal to the sum of the entres n the g E exp E g calc column. OF g E J /mol 70

71 8.5 Frst, format the top of the spreadsheet as follows Measured Data Caculated usng data gven n the problem statement Calculated usng "A" x x y y (a) (J/mol) (J/mol),calc (J/mol),calc (J/mol) -,calc)/ ] -,calc)/ ],calc)/ ] -,calc)/ ] + [( Measured Data secton: Smply copy the gven data nto the cells below the approprate column. Note that x x y y The next secton contans values of the actvty coeffcents and excess Gbbs energy calculated wthout usng A. The actvty coeffcents are calculated usng Equatons 8.5 and 8.6: y where s equal to the pressure when x x Calculated Usng A secton: To calculate ã and calc ã we use the followng expressons calc calc exp Ax RT calc exp Ax RT The calculatons for the remanng columns should be obvous. The value of A s found usng a solver functon whch mnmzes the sum of the of the entres n the followng column calc calc by changng A. We obtan A 44 J mol OF s equal to the sum of the entres n the OF calc calc column. 7

72 8.5 Ths problem can be solved by creatng a spreadsheet. Frst, format the top of the spreadsheet as follows T (ºC) 60 R (J mol - K - ) 8.34 A (J/mol) Intal Guess T (K) Measured Data Caculated usng data gven n the problem statement Calculated usng "A" x x y y (a) (J/mol) (J/mol) g E,exp (J/molg E,calc (J/mol) (ge -ge,calc) Measured Data secton: Smply copy the gven data nto the cells below the approprate column. Note that x x y y The next secton contans values of the actvty coeffcents and excess Gbbs energy calculated wthout usng A. The actvty coeffcents are calculated usng Equatons 8. and 8.: y where x s equal to the pressure when x Calculate the excess Gbbs energy: gexp E RT x ln x ln Calculated usng A secton: We are fttng the two-suffx Margules equaton; therefore, g calc E Ax x The A value s equal to an ntal guess that you enter as shown n the spreadsheet above. Calculatng g E E exp g calc should be straghtforward. After enterng all of the avalable data and equatons n the spreadsheet, the value of A s found usng a solver functon whch mnmzes the sum of the g E E exp g calc column by varyng A. 7

73 T (ºC) 60 R (J mol A (J/mol) T (K) Measured Data Caculated usng data gven n the problem statement Calculated usng "A" x x y y (a) g E (J/mol) g E,calc (J/mol) (g E -g E,calc) SUM Therefore, A 37. J mol In the example n the text, A was calculated to be 43 J/mol, whch s 0.5% larger than the value obtaned usng the objectve functon. The value obtaned usng the objectve functon s more accurate because t utlzes more emprcal data. 73

74 8.53 (a) We wll fnd the 3-suffx Margules parameters usng a spreadsheet. Frst, format the top of the spreadsheet as follows: T 40 ºC R 8.34 J mol - K - A Guess J/mol T 33.5 K B Guess J/mol Measured Data Calculated usng 3-Suffx Margules Equaton x x y y (a) calc (a) (- calc ) Measured Data secton: Copy the gven data nto the cells below the approprate column. Note that x x y y Calculated usng 3-Suffx Margules Equaton secton: Reference the cells contanng the A and B parameters to calculate the actvty coeffcents usng 3 3 RT ln A3B x 4Bx RT ln A3B x 4Bx The pressure s then calculated as follows: where x x when x when x 0 The last column s self-explanatory. Use the solver functon to smultaneously fnd the values of A and B that mnmze the sum of (- calc ). Dong so, we obtan A 335 J/mol B 44 J/mol (b) Format the spreadsheet as follows 74

75 T 40 ºC R 8.34 J mol - K - A Guess J/mol T 33.5 K B Guess J/mol Caculated usng data gven Measured Data n the problem statement g E x x y y (a) (J/mol) Calculated usng "A" and "B" g E calc (g E - g E (J/mol) calc) Measured Data secton: Copy the gven data nto the cells below the approprate column. Note that x x y y Calculatons wth Data Gven n the roblem Statement secton: Ths secton contans values of the actvty coeffcents and excess Gbbs energy calculated wthout usng A and B. The actvty coeffcents are calculated usng Equatons 8. and 8.: y where x s equal to the pressure when x Calculate the excess Gbbs energy usng gexp E RT x ln x ln Calculated usng A and B secton: We are fttng the two-suffx Margules equaton; therefore, g E calc x x A B x x Calculatng g E E g calc should be straghtforward. The values of A and B are found usng a solver functon whch mnmzes the sum of the entres n the g E E g calc column. We obtan A 364 J/mol B 47.5 J/mol (c) Format the spreadsheet as follows: 75

76 T 40 ºC R 8.34 J mol - K - A Guess J/mol T 33.5 K B Guess J/mol Measured Data Caculated usng data gven n the problem statement Calculated usng "A" and "B" x x y y (a),calc,calc -,calc)/ ] -,calc)/ ] -,calc)/ ] + [(,calc)/ ] Measured Data secton: Smply copy the gven data nto the cells below the approprate column. Note that x x y y The next secton contans values of the actvty coeffcents and excess Gbbs energy calculated wthout usng A and B. The actvty coeffcents are calculated usng Equatons 8.5 and 8.6: y where s equal to the pressure when x x Calculated Usng A and B secton: To calculate and we use the followng expressons RT ln calc A 3Bx 3 4Bx RT ln calc A 3Bx 3 4Bx The calculatons for the remanng columns should be obvous. The values of A and B are found usng a solver functon whch mnmzes the sum of the entres n the followng column: calc calc Dong so, we obtan A 369 J/mol B 58 J/mol 76

77 (d) Usng the excess Gbbs energy calculated n art (b), create the followng spreadsheet: x -x g E /x *x If we plot of g E x versus x x, the slope of the resultng lne s B and the ntercept s A. x ge/xx (J/mol) g E /x x vs. x -x y = 55.7x R = x -x 77

78 As you can see from the graph, A 37 J/mol B 55 J/mol The answers obtaned n parts A D agree well wth each other. The average value of A s 360 J/mol, and the standard devaton of the answers s 7 J/mol. The average value of B s 5 J/mol, and the standard devaton s 6.5 J/mol. From ThermoSolver, we obtan the followng values: Objectve Functon A (J/mol) B (J/mol) ressure Gbb s Energy Actvty Coeffcent These values are equal to those obtaned n arts (a) (c). 78

79 8.54 To test for thermodynamc consstency, we use: ln dx 0 0 However, the value of the above ntegral s not useful unless t s referenced to ts absolute value. We can calculate the followng 0 0 ln dx ln dx and confrm that t s approxmately equal to zero. We can obtan the values of the actvty coeffcents from expermental data usng the followng equatons, whch assume deal behavor of the gas phase: y x y x y x y x To fnd the uraton pressures from the provded data, we must realze x x 0 Wth the above expressons, the followng table was made n created n a spreadsheet x ln( / )

80 Use the spreadsheet to create the followng graph: ln( / ) vs. x ln( / ) ln(/) = x x R = x Integrate the trendlne provded n the plot: x x dx x x dx 0.07 Ths ntegral s approxmately zero, so the data are thermodynamcally consstent. 80

81 8.55 To test for thermodynamc consstency, we use: ln dx 0 0 However, the value of the above ntegral should be referenced to ts absolute value. We can obtan the values of the actvty coeffcents from expermental data usng the followng equatons, whch assume deal behavor of the vapor phase: y x x y y x y x To fnd the uraton pressures from the provded data, we must realze x x 0 Wth the above expressons, the followng table was made n created n a spreadsheet x ln( / ) Use the spreadsheet to create the followng graph: 8

82 0.6 ln( / ) vs. x ln(/) x Clearly, the area above the lne y=0 s less than the area below. The value of the followng ntegral can be estmated graphcally. ln dx 0 ln dx 0 There are a number of ways to do ths. One possblty s to prnt the graph. The area above the lne y=0 s cut out, as s the area below. The two peces of paper are weghed. Assumng a constant densty of the prnter paper, the weghts should be equal. If not, the areas are dfferent; thus, the data are not thermodynamcally consstent. The followng data are obtaned for the above graph Mass (mllgrams) Area above y= Area below y= Total area were extended to 0 It should be noted that the trends n ln / x and x to estmate the entre areas above and below the x-axs. Usng these data, we estmate 0 0 ã ln ã ã ln ã dx dx Therefore, the data are not thermodynamcally consstent. 8

83 8.56 The data are thermodynamcally consstent f: ln dx 0 0 To fnd the actvty coeffcents, we can use the followng equatons: x y y x The pressure s constant at atm (.03 bar), but the uraton pressures are dependent upon temperature. We can fnd water s uraton pressures from the steam tables. Use Antone s equaton data n Appendx A to calculate the. pressure of acetone. We can create the followng table. x y T (ºC) (bar) (bar) ln(/) Note: The last row of data (x =) was obtaned by fndng the temperature that results n a uraton pressure of atm n Antone s equaton. 83

84 Now, create the followng plot. ln( / ) vs. x.5.5 ln( / ) x Clearly, the area above the x-axs s greater than the area below. The value of the followng ntegral can be estmated graphcally. ln dx 0 ln dx 0 There are a number of ways to do ths. One possblty s to prnt the graph. The area above lne y=0 s cut out, as s the area below. The two peces of paper are weghed. Assumng a constant densty and thckness of the prnter paper, the weghts should be equal. If not, the areas are dfferent; thus, the data are not thermodynamcally consstent. The followng data are obtaned for the above graph Mass (mllgrams) Area above y=0 0.0 Area below y= Total area It should be noted that the trends n ln were extended to x 0 and x to estmate the entre areas above and below the x-axs. Usng these data, we estmate 84

85 0 0 ã ln ã ã ln ã dx dx Therefore, the data are not thermodynamcally consstent. 85

86 8.57 Let desgnate benzene and desgnate sooctane. To solve ths problem, we wll need to know molar volumes and uraton pressures. Frst, calculate these quanttes from gven data. v MW v 3 5 m mol v 3 4 m.66 0 mol B exp A 4. bar 8.86 bar C T In order to fnd the constants A and B, we need values for both actvty coeffcents. The soluton method for fndng the actvty coeffcent of benzene wll be shown, and then the actvty coeffcent of sooctane can be found analogously. Frst, equate fugactes: o y ˆ x f The reference fugacty s defned as follows o l f f exp l v RT d Fnd the uraton fugacty coeffcent from the followng relatonshp: where ln low z d B' d B'd low B' B a RT If we ntegrate the expresson and let low go to zero, we obtan Therefore, a f o 4. bar exp d a We also need the fugacty coeffcent of speces n the mxture. low.8 bar 86

87 ln ˆ low Z d where and Z ' n B mx n ' B mx y B ' ' y y B y B Therefore, and Z ' y y B ln ˆ ' y y B low ' y y y B ' y y y B y ' B y ' B d ˆ (Note: When you calculate ' B usng a geometrc average, ts value s negatve, even though the product of two negatves s postve.) Now we can calculate the actvty coeffcent. y ˆ o x f bar bar From a smlar method.03.6 We have two equatons wth two unknowns: A and B. RT ln A 3Bx 3 4Bx RT ln A 3B x 3 4Bx Solve smultaneously: A 66 J mol B J mol 87

88 (b) A mxture of benzene and sooctane wll splt nto two partally mscble phases f x g 0 The expresson for Gbbs energy s g x g xg RT ln x ln x x x x x A Bx x Therefore, g x RT A Bx 6B x x Assume that A and B are ndependent of temperature. For the mxture to splt nto two phases T A Bx 6B R x x T 8.K The freezng ponts of benzene and sooctane are 78 K and 65 K, respectvely. It s unlkely ths mxture forms a eutectc at ths low a temperature, so t wll soldfy before t reaches the lmt of lqud nstablty. 88

89 8.58 For a pure speces: v l For the eng-robnson equaton of state: RT a v b v bv b where RTc, a, T r,, and c, RTc, b c, The values of the crtcal propertes from Appendx A.and the resultng calculated parameters, a,, and b are: T c [K] c [bar] a b 3 m 3 Speces Jm mol mol n-c 5 H x 0-5 Knowng the eng-robnson parameters, we can calculate the molar volume, v. When the soluton has three real roots, we assgn the smallest root to the lqud phase and the largest root to the vapor phase. These values can then be used to fnd, the fugacty coeffcent of the lqud and vapor, respectvely, usng the expresson: b ln z ln ln RT brt v b v b a v To fnd the uraton pressure at a gven temperature, we solve for lqud and vapor volumes and then use those values for the fugacty coeffcent. Ths process requres teratve soluton and the uraton pressure s found as the pressure where the fugacty of the vapor and lqud are equal. The values that sfy ths crtera l v v, v Speces when 3 3 m m Antone error v l, [bar] mol mol [bar] n-c 5 H.46 x x % 89

90 8.59 For a pure speces: v l For the eng-robnson equaton of state: RT a v b v bv b where RTc, a, T r,, and c, RTc, b c, The values of the crtcal propertes from Appendx A.and the resultng calculated parameters, a,, and b are: T c [K] c [bar] a b 3 m 3 Speces Jm mol mol C 3 H x 0-5 Knowng the eng-robnson parameters, we can calculate the molar volume, v. When the soluton has three real roots, we assgn the smallest root to the lqud phase and the largest root to the vapor phase. These values can then be used to fnd, the fugacty coeffcent of the lqud and vapor, respectvely, usng the expresson: b ln z ln ln RT brt v b v b a v To fnd the uraton pressure at a gven temperature, we solve for lqud and vapor volumes and then use those values for the fugacty coeffcent. Ths process requres teratve soluton and the uraton pressure s found as the pressure where the fugacty of the vapor and lqud are equal. The values that sfy ths crtera l v v, v Speces when 3 3 m m Antone error v l, [bar] mol mol [bar] C 3 H x x % 90

91 8.60 For a pure speces: v l For the eng-robnson equaton of state: RT a v b v bv b where RTc, a, T r,, and c, RTc, b c, The values of the crtcal propertes from Appendx A.and the resultng calculated parameters, a,, and b are: T c [K] c [bar] a b 3 m 3 Speces Jm mol mol C 6 H x 0-5 Knowng the eng-robnson parameters, we can calculate the molar volume, v. When the soluton has three real roots, we assgn the smallest root to the lqud phase and the largest root to the vapor phase. These values can then be used to fnd, the fugacty coeffcent of the lqud and vapor, respectvely, usng the expresson: b ln z ln ln RT brt v b v b a v To fnd the uraton pressure at a gven temperature, we solve for lqud and vapor volumes and then use those values for the fugacty coeffcent. Ths process requres teratve soluton and the uraton pressure s found as the pressure where the fugacty of the vapor and lqud are equal. The values that sfy ths crtera l v v, v Speces when 3 3 m m Antone error v l, [bar] mol mol [bar] C 6 H x x % 9

92 8.6 Followng Example 8.5, we get: T, x Antone Coeffcents A B C expa T C x x y B Intal Estmate Raoult s Law () (), y Speces x A B C Antone [bar] T c, [K] c, [bar] a 3 Jm mol b 3 m mol methane () x 0-5 n-pentane () x 0-4 EOS arameters T c, c, a b a y a y y aa y a b yb y b RT a v v, v v b v l ya k k b v b k v l ln ˆ ln ˆ, ˆ v b RT RTv l x ˆ y v ˆ m y? No j j j y Yes, y 9

93 n a v Jm 3 mol b v m 3 mol v v m 3 mol v l m 3 mol ˆv ˆv ˆ l ˆ l y y y (k+) [bar]

94 8.6 Followng Example 8.6, we get: T, x Antone Coeffcents A B C expa T C x x y B Intal Estmate Raoult s Law () (), y Speces x A B C Antone [bar] T c, [K] c, [bar] a 3 Jm mol b 3 m mol methane () x 0-5 n-pentane () x 0-5 EOS arameters T c, c,i a a y a yy a a k ya b b yb y b RT a v l v, v vb v bvb b v b ln ˆ z ln b RT a b v b brt v b l x ˆ y v ˆ m v l y ln ˆ, ˆ k a k b a k y? Yes No j j j y, y 94

95 n a v Jm 3 mol b v m 3 mol v v m 3 mol v l m 3 mol ˆv ˆv ˆ l ˆ l y y y (k+) [bar]

96 8.63 Followng Example 8.5, we get: T, x Antone Coeffcents A B C expa T C x x y B Intal Estmate Raoult s Law () (), y Speces x A B C Antone [bar] T c, [K] c, [bar] a 3 Jm mol b 3 m mol Carbon Doxde () x 0-5 Benzene () x 0-4 EOS arameters T c, c, a b a y a y y aa y a b yb y b RT a v v, v v b v l ya k k b v b k v l ln ˆ ln ˆ, ˆ v b RT RTv l x ˆ y v ˆ m y? No j j j y Yes, y 96