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Unformatted text preview: 6.2 (a) General: = Ni i where i = Ni a f T , P , N j i and (1) d = i dN i + Ni d i However, we also have that d = FG IJ H TK dT +
V,N FG IJ H V K T, N dV + FG IJ HN K
i dN i
T ,V , N j i (2) Subtracting (2) from (1) yields 0= FG IJ H T K dT  V ,N FG IJ H V K T ,N dV + i  LM F I MN GH N JK
i i T ,V , N j i OP PQdN + N d
i i i At constant T and V 0 = i  LM FG IJ MN H N K
i i T ,V , N OP dN + N d PQ
i i (general equation) For = A , i  i = Ai and FG IJ HN K
i FG IJ HN K =
T ,V , N
ji FG A IJ HN K
i = Gi .
T ,V , N ji Thus, = Ai  Gi =  PVi and
T ,V , N ji NidAi T , V = P Vi dNi T , V specific equation for = A (b) Following the analysis above, we also get Solutions to Chemical and Engineering Thermodynamics, 3e 0= FG IJ H U K dU  V ,N FG IJ H V K U,N dV + i  LM F I MN GH N JK i U ,V , N j i OP PQdN + N d
i i i and, at constant U and V OP dN + N d PQ G F S IJ Now, choosing = S , and using that G = , which is easily HN K T
0=
i i i i i U ,V , N ji LM  F I M GN J N H K i i U ,V , N ji derived, yields T Ni dSi U ,V = Hi dN i U ,V (c) Following a similar analysis to those above, we obtain 0= FG IJ H SK dS 
V ,N FG IJ H V K S, N dV + i  LM F I MN GH N JK
i S ,V , N j i OP PQdN + N d
i i i which, at constant V and S, reduces to LM F I MN GH N JK Finally, using = U , and a U N f
0 = i 
i S ,V , N j i OP PQdN + N d
i i i i S ,V , N ji = Gi yields NidU i S ,V = l PVi + TSiqdNi S ,V
6.3 (a) At constant U and V, S = maximum at equilibrium S = S I + S II = NiI Si I + NiII Si II
i =1 i =1 C C but dS = 0 = FG S IJ H U K F S IJ +G H U K
I I II II dU I + dU II
V ,N V ,N FG S IJ H V K F S IJ +G H V K
I I II II U ,N dV I + dV II U ,N FG S IJ H N K F S IJ + G H N K
I II II i I i U ,V , N dN iI
j i dNiII
U ,V , N j i Since U = U I + U II = constant, dU II = dU I Solutions to Chemical and Engineering Thermodynamics, 3e Chapter 6 Since V = V I +V II = constant, dV II = dV I and since Ni = NiI + NiII = constant, dNiII = dNiI Also, FG S IJ H U K
Thus =
V ,N 1 S ; T V FG IJ H K =
U ,N P S and T Ni FG H IJ K =
U ,V , N ji Gi T (see previous problem) dS = 0 = F 1  1 I dU + FG P HT T K H T
I I II I I  P II G I G II dV I  1I  1II dN iI T II T T i IJ K FG H IJ K T I = T II ; P I = PII ; and Gi I = Gi II for equilibrium in a closed system at constant U and V. (b) For a closed system at constant S and V, U has an extremum. Thus dU = 0 = FG U IJ H S K F U IJ +G H S K
I I II II dS I + dS II
V,N V,N FG U IJ H V K F U IJ +G H V K
I I II II S, N dV I + i dV II
S ,N FG U IJ H N K F U IJ + G H N K
I I i II i dNiI dN iII S ,V , N j i II i U ,V , N j i but S, V and N j , j = 1, L, C are constant. Thus dU = 0 = T I  T II dS I + P I  PII dV I + Gi I  Gi II dN iI
i c h c h c h T I = T II , P I = PII and Gi I = Gi II for equilibrium in a closed system at constant S and V. 6.4 (a) For a closed system at constant T and V, A is a minimum at equilibrium; thus dAV , T = 0 . From Eqn. (6.25) dA =  PdV  SdT + Gi dNi or dA V , T = Gi dNi
But, Ni = Ni , 0 + i X . Thus dN i = i dX and dA V , T = A b G gdX = 0 or FGH X IJK
i i V ,T = i Gi = 0 .
i (b) For a closed system at constant U and V, S = maximum, or dS U ,V = 0 . From Eqn. (6.24) dS = 1 P 1 dU + dV  Gi dN i ; thus T T T Solutions to Chemical and Engineering Thermodynamics, 3e dS U ,V =  and 1 1 Gi dNi or dS U ,V =  T T b G gdX
i i S X 6.5 =
U ,V 1 T iGi = 0
i Let mi = molecular weight of species i. Multiplying Eqn. (6.32a) by mi and summing over all species i yields, for a closed system mi Ni = total mass in system = mi Ni,0 + X i mi
total mass in system initially However, since the total mass is a conserved quantity, mi Ni = mi Ni,0 X i mi = 0 , where X can take on any value.
Consequently, if this equation is to be satisfied for all values of X, then imi = 0 ! Similarly, in the mu ltireaction case, starting from Ni = Ni ,0 + ij X j , we get
j =1 M mi Ni = mi Ni,o + mi ij X j mi ij X j = 0 = X j ijmi
i =1 i= 1 i =1 j =1 i =1 j =1 j =1 i =1 C C C M C M M C Since the X j 's are not, in general, equal to zero, we have ijmi = 0
i =1 C In particular, for the reaction H 2 O = H 2 + 1 2 O 2 , or H 2 + 1 2 O 2  H 2 O = 0 , we have a f
1 a f ijmi = (+ 1)(2) + F 2I (32) + ( 1)(18) = 0 . HK
i 6.6 From Eqns. (6.64) we have V1 = V 1 + V mix + x2 V mix x1 a f
T ,P (1) and Solutions to Chemical and Engineering Thermodynamics, 3e Chapter 6 V2 = V 2 + V mix + x1 V mix x1 a f
T ,P (2) Now since T, P and X, are the independent variables, we have that
0 since pure component volume is a function of T and dV1 T, P = = = dV 1 T, P + d V a P only
mix f T, P +d V mix x 1
x 2 a f
T , P + 2 V mix 2 x1 a f V mix x 1
dx 1 since a f LMN a f OPQ x 2 V mix x1 T , P T , x 2 x 1 P dx 1 + x2 2 V mix 2 x 1 a f
T, P dx 1 T , P x 2 = 1 x 1 Similarly dV2 T , P =  x1 2 V mix
2 x1 a f
T ,P dx1 Thus xiVi T , P = x1x2 2 V mix
2 x1 a f
T, P dx1  x2 x1 2 V mix
2 x1 a f
T, P dx1 0 Thus, V1 and V2 given by equations (1) and (2) identically satisfy the GibbsDuhem equation xid i T , P = 0 . A similar argument applies for the partial molar enthalpies of Eqn. (6.69). 6.7 (also available as a Mathcad worksheet) The students can solve this problem by drawing tangent lines to the V mix curves. Polak and Lu smoothed their data using the RedhichKister equation (see Eqn. (6.65a)). That is, they fitted their data to V mix = x1 x2 C j x2  x1
j =1 n a f j 1 = x1 1  x1 a f C (1  2 x)
j j 1 Now V mix = 1  x1 x1
1 Thus f C a1  2x f  x C a1  2 x f  2 x a1  x f C ( j  1)a1  2 x f a V f V  V = a V f  x = a1  x f k A  2 x Bp x
j j 1 1 j j 1 1 1 1 j 1 1 1 mix 2 mix 1 2 1 1 a f a j 2 (1) and Solutions to Chemical and Engineering Thermodynamics, 3e V2  V 2 = V mix  x1 where A= a f V mix = x12 A + 2 x2 B x1 a f k p
j 2 (2) Cj a1  2 x1f j 1
n j =1 and B = C j ( j  1) 1  2 x1
j =1 n a f Taking species 1 to be methyl formate, Polak and Lu found C1 methyl formate  Methanol methyl formate  Ethanol 0.81374 C2 0.00786
3 C3 0.0846
3 C4 0.0448  0.33259  010154  0.0516 0.0264 . to get m kmol ] [units are cc/mol; multiply by 10 I have used the equations above and the constants given to find V1  V 1 and V2  V2 , since this leads to more accurate results than the graphical method. The results are tabulated and plotted below. Methyl formate  Methanol V mix acc molf
xMF xMF 0 0 0.459 0 0.6 0.075 0.035 0.136 0.1 0.039 0.329 0.007 0.7 0.063 0.021 0.162 0.2 0.065 0.225 0.025 0.8 0.047 0.011 0.192 0.3 0.080 0.148 0.051 0.9 0.027 0.004 0.236 0.4 0.085 0.093 0.080 1.0 0 0 0.309 0.5 0.083 0.058 0.109 V1  V 1 V2 V 2 V mix acc molf b V1  V 1 V2 V 2 Thus VMF = 6278 + V1 V 1 cc/mol or 103 m3 kmol . . VM = 4073 + V2 V 2 . .
Methyl formate  Ethanol xMF 0 V mix b g g acc molf
xMF 0.1 0.080 0.682 0.013 0.7 0.174 0.081 0.390 0.2 0.136 0.507 0.043 0.8 0.134 0.037 0.522 0.3 0.174 0.381 0.085 0.9 0.077 0.010 0.680 0.4 0.196 0.285 0.137 1.0 0 0 0.861 0.5 0.203 0.205 0.201 0 0.935 0 0.6 0.196 0.138 0.284 V1  V 1 V2 V 2 V mix acc molf V1  V 1 V2 V 2 Solutions to Chemical and Engineering Thermodynamics, 3e Chapter 6 Thus VMF = 6278 + V1 V 1 cc/mol. Multiply by 103 for m3 kmol . . VE = 5868 + V2  V 2 . b b g g 6.8 This problem is similar to the last one, and will be treated in a similar fashion. Fenby and Ruenkrairergasa give their data in the form H mix J mol = x2 1  x2 a f a f C a1  2 x f
n j =1 j 2 j 1 (1) where component 2 is the fluorobenzene. The constants given in the aforementioned reference and Fenby and Scott J. Phys. Chem 71, 4103 (1967) are given below System C1 C2 C3 C4 C6 H 6  C6 F5Cl C6 H 6  C6 F5Br C6 H 6  C6 F5I C6 H 6  C6 F6 C6 H 6  C6 F5H 2683 3087 4322 1984 230 929 356 161 +1483 +578 970 696 324 +1169 +409 0 0 0 0 +168 Solutions to Chemical and Engineering Thermodynamics, 3e If we replace x2 with 1  x1 in Eqn. (1), we regain the equation of the previous illustration, except for a factor of (1) j 1 in the sum and the corresponding places in the other equations. xC 6 H 6
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 [Note: J/mol] H mix 0 252 463 609 679 671 590 453 284 119 0 bH  Hg C 6H6 bH  Hg C 6 F5Cl xC 6 F5Cl
1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 2642 2171 1790 1466 1175 903 646 409 205 57.8 0 0 39.2 130 242 349 439 506 555 601 666 784 C6 H 6  C6 F5Br xC 6 H 6
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 H mix 0 263 488 654 751 772 717 595 420 212 0 bH  H g bH  H g
C6 H 6 2747 2248 1829 1469 1149 861 600 370 181 50.0 0 C6 F5 Br 0 42.9 153 306 486 683 893 1120 1374 1671 2035 C6 H 6  C6 F5I H mix 0 359 657 883 1026 1081 1042 910 688 382 0 C6 H 6 3837 3119 2489 1937 1456 1040 689 402 187 48.9 0 bH  H g bH  H g
C6 F5 I 0 52.1 200 431 740 1121 1572 2095 2695 3379 4159 xC 6 F5 x
1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 Solutions to Chemical and Engineering Thermodynamics, 3e Chapter 6 C6 H 6  C6 F6 xC 6 H 6
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 H mix 0 218 392 502 536 496 394 253 108 4.5 0 bH  H g bH  H g
C6 H 6 2298 1899 1590 1332 1097 867 637 413 212 60.9 0 C6 F6 C6 H 6  C6 F5H H mix C6 H 6 bH  H g bH  H g
C6 F5 H xC 6 F5 x
1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0 0 61.0 0 31.2 2.2 36.2 1.1 93.0 3.9 2.8 +6.8 146 13.5 42.3 +37.4 162 31.4 72.3 +100 125 57.5 87.0 +202 28.9 86.9 84.5 344 +121 110 66.7 524 +308 116 39.4 +737 +503 85.9 12.6 +973 +688 0 0 1217 Note: Changes in sign in column Solutions to Chemical and Engineering Thermodynamics, 3e Relations among the unknowns T S = TV , PS = P V , no phase equilibrium relations, but 3 chemical equilibrium relations of the form ij Gi = 0 . 8 unknowns 5 eqns. = 3 unspecified unknowns or 3 degrees of freedom 6.10 (a) In general, for a binary, twophase mixture C = 2, M = 0, P = 2 F = C  M  P + 2 = 2  0  2 + 2 = 2 degrees of freedom. However, for an azeotrope there is the additional restriction x1 = y1 , which
eliminates one degree of freedom. Thus, there is only 1 degree of freedom for a binary, azeotropic system. (b) In osmotic equilibrium P I PII , since the membrane is capable of supporting a pressure difference, and G2I G2II , where 2 is the species which does not pass through the membrane. Therefore, the independent unknowns are T I , I II PI , x1I , T II , P II and x1II . [Note, x2 and x2 are not independent unknowns
I I II II since x2 = 1  x1 and x2 = 1  x1 ]. a f There are two equilibrium relations
I between these six unknowns: viz. T = TII and G1I = G1II . Consequently, there are four degrees of freedom ... that is, as we shall see in Sec. 8.7, if T, PI , P II and x1I are specified, x1II will be fixed. Case II: M = 0, C = 2, P = 3 F = 2  0  3 + 2 = 1 6.11 (a) Gibbs Phase Rule: F = C  M  P + 2 C = 2 , M = 0 F = 2  0  P + 2 = 4  P degrees of freedom. Therefore, a maximum of 4 phases can exist at equilibrium (for example a solid, two liquids and a vapor, or two solids, a liquid and a vapor, etc.) (b) Gibbs Phase Rule: F = C  M  P + 2 C = 2 , M = 1 F = 2  1  P + 2 = 3  P degrees of freedom. Therefore, a maximum of e phases can exist at equilibrium (for example a two liquids and a vapor, or a solid, a liquid and a vapor, etc.) (c) Case I: M = 0, C = 2, P = 2 F = 2  0  2 + 2 = 2 6.12 (a) dNi & & = Ni + Ni ,rxn dt dU 0 dV & & = Ni Hi + Q  W s P dt dt & dS Q & & = Ni Si + + Sgen dt T dS & & & T  T Ni Si  TSgen = Q dt Solutions to Chemical and Engineering Thermodynamics, 3e Chapter 6 dU dS dV & & & = N i Hi + T  T Ni S i  TS gen  P dt dt dt dU dV dS & & +P T = Ni Hi  TSi  TS gen dt dt dt dU dV dS dNi dX & & & +P T = Ni i  TSgen =  i i  TSgen dt dt dt dt dt General expression Now System is only permeable to species 1 dU dV dS dN 1 dX & +P T   1 1 =  TS gen 0 dt dt dt dt dt When T and P constant d d (U + PV  TS )  N1  1 X 1 0 dt dt d G  N1  1 X 1 0 dt G  N1  1 X 1 = minimum at equilibrium (b) When T and V are constant d d (U  TS )  N1  1 X 1 0 dt dt A  N1  1 X 1 = minimum at equilibrium b g F H I K F H I K a f a a a f f a f f 6.13 (a) 2N N 2 2O O 2 2N + O N 2 O 2N + 2O 2NO 2N + 4O N 2 O 4 2N + 4O 2NO 2 1 N2 + O2 N2O 2 N 2 + O 2 2NO N 2 + 2O2 N 2 O 4 (b) F = C  M  P + 2 = 7  5  1+ 2 = 9  6 = 3 F = 3 degrees of freedom (c) 1 degree of freedom used in O 2 : N 2 ratio 2 degres of freedom N 2 + 2O2 2NO 2 5 2N + 5O N 2 O 5 N2 + O2 N2O5 2 5 independent reactions 6.14 Mass balance: M1 + M2 = M f $ $ $ Energy balance: M1U 1 + M2U 2 = M f U f Molecular weight H 2O = 18.02 g mol In each case the system is M1 kg of solution 1 + M2 kg of solution 2. Since Q = 0 , Ws = 0 (adiabatic mixing) $ $ For liquids U H . Thus we have
$ $ M H + M2 H 2 $ Hf = 1 1 M1 + M2 Solutions to Chemical and Engineering Thermodynamics, 3e 1 $ $ $ when M1 = M2 ; H f = H1 + H 2 . 2 (a) Read from Figure 6.11 $ H1 = 6.9 103 J kg $ H = 6.1 103 J kg
2 c h 1 $ Thus H f = 5.410 104 = 2 .705 10 4 J kg 2 To find the composition, so a sulfuric acid balance 1 1 M1 + 2 M 2 = f M f f = 1 + 2 2 where i = weight percent of ith flow stream. c h a f since M1 = M2 Thus f = 1 (10 + 90) = 50 wt % sulfuric acid. From Figure 6.11 2 50 wt % H2SO 4 Tf ~ 110 C $ $ H = U = 2 .705 10 4 J kg $ (b) Here H1 = 69 103 J kg , .
1 $ $ H 2 = 3186 105 J kg H f = ( 6.9  318.6) 103 = 156 105 J kg . . 2 1 = 10 wt % , 2 = 60 wt % f = 35 wt % . Using Figure 6.11, Tf ~ 22 C . and $ Notice that there is a balance between the energy released in mixing, Hmix , $ and the energy absorbed in heating the mixture, CP T . In case (a), Hmix is $ very large, and Tf > T or T2 , while in case (b) Hmix is smaller, so that 1 Tf ~ T1 .
6.15 (a) MW H 2 O = 18.02 g mol ; MW H 2SO 4 = 98.08 g mol 100 g H 2 O = 555 mol . 100 g H 2 SO 4 = 1.02 mol Note: When these are mixed, a solution containing 5.44 mol H 2 O /mol acid is formed. H s for such a solution is 58,390 J/mol acid. Thus, total heat released = 1.02 mol acid 58,390 J mol acid = 59,558 J (Negative sign means that heat is released!) (b) Adding another 100 grams of water produces a solution which contains 10.88 mol H 2 O /mol acid. From the graph H s = 64,850 J mol acid . However, 58,390 J/mol of acid were released in preparing the first solution, so that only 6,460 J/mol acid, or 6,590 J, are released on this further dilution. a f Solutions to Chemical and Engineering Thermodynamics, 3e Chapter 6 (c) 60 wt % H 2 SO 4 40 18.02 = 3.629 moles H 2 O moles acid 60 98.08 for which H s = 52,300 J mol acid , and 60 mol aci d = 31,990 J 98.08 Note: Enthalpy of 60 WT% solution is 31,990 J relative to pure components at the same temperature. Similarly 25 wt % H 2SO 4 16.27 mol H 2 O mol acid , H s ~ 68,830 J mol acid Hs = 52,300 J mol acid and 0.25 75 = 13,160 J 98.08 Final solution =175 grams ; 78.75 grams acid = 0803 mol, . Hs = 68,830 J mol acid 96.25 grams water = 5347 mol 6.66 mol H 2 O mol acid . So that . H s = 60,670 J mol acid H s = 48,720 J Thus, enthalpy change on mixing, H mix is Thus, 3570 J = 357 kJ must be removed to keep solution isothermal! N (d) For 1 mole of solute: 1 + N 2 H mix = H 1 + N2 H 2 + 1 H s 2 (argument of N1 Hmix = 48,720  ( 31,990  13160 ) = 3570 J , a f FG IJ H K H s ) and for N1 moles of solute and N2 moles of solvent. a N + N fH
1 2 mix = N1 H 1 + N 2 H 2 + N1 H s Now H1 = or FG H IJ H N K
mix 1 = H1
T ,P FG N IJ = H HN K F N IJ + N a H f aN N f + H G H N K a N N f N
2 1 mix 2 1 s 1 s 2 1 2 1 T, P 1 2 s 2 1 T ,P FG N IJ  N LM H aN N f OP H N K N N a N N f Q F H IJ we obtain Similarly, starting from H = G H N K H a N N f H H = a N N f
H1  H 1 = H s
2 1 1 2 1 T, P 2 mix 2 T ,P 2 s 2 1 2 2 1 since N 2 N1 N = 2 2 N1 N1 a f T ,P (e) 50 wt % acid 50 18.02 = 5.443 mol H 2 O mol acid 50 98.08 H s (5.443) =  58,370 J mol and, from the accompanying graph H s N2 N1 N2 N1 a a f f
at N 2 N1 =5.443 = ( 91,630 )  ( 46,030) = 2,280 J mol 20 so that H2  H 2 = 2,280 J mol and Solutions to Chemical and Engineering Thermodynamics, 3e H1  H1 = (58,370)  544(2,280) = 45,967 J mol . . PL[ 1RWH 6RUU\ DERXW VHW RI GDWD EHLQJ JLYHQ LQ DOFRKRO ZW DQG RWKHU LQ ZDWHU PROHEXWWKLVLVWKHZD\WKHGDWDDSSHDUHGLQWKH,QWHUQDWLRQDO&ULWLFDO7DEOHV D )LUVWZLOOFRQYHUWWKHGDWDWRPROHIUDFWLRQV ZW$ NJ$ u [ NJ$ NJ:
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This homework help was uploaded on 02/06/2008 for the course CBE 103 taught by Professor Rethwisch during the Spring '07 term at University of Iowa.
 Spring '07
 Rethwisch

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