Chap08 solutions

Physical Chemistry

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8 Phase diagrams Solutions to exercises Discussion questions E8.1(b) What factors determine the number of theoretical plates required to achieve a desired degree of separation in fractional distillation? The principal factor is the shape of the two-phase liquid-vapor region in the phase diagram (usually a temperature-composition diagram). The closer the liquid and vapour lines are to each other, the more theoretical plates needed. See Fig. 8.15 of the text. But the presence of an azeotrope could prevent the desired degree of separation from being achieved. Incomplete miscibility of the components at speci±c concentrations could also affect the number of plates required. E8.2(b) See Figs 8.1(a) and 8.1(b). Liquid A and B Liquid A & B Solid B Solid B and Solid AB 2 Liquid A & B Solid AB 2 Liquid A & B Solid A Eutectic Solid AB 2 and Solid A 0.33 Figure 8.1(a) Liquid Vapor Vapor and liquid 0.67 B Figure 8.1(b)
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PHASE DIAGRAMS 113 E8.3(b) See Fig. 8.2. Liquid A&B Liquid (A & B) Solid B Liquid (A&B) Liquid Liquid Liquid Solid A 2 B Solid A 2 B Solid B 2 A Solid B 2 A Solid B 2 A Solid B Two solid phases Two solid phases Solid A 2 B Solid A 2 B Solid A Two solid phases Solid A 0.333 0.666 Figure 8.2 Numerical exercises E8.4(b) p = p A + p B = x A p A + ( 1 x A )p B x A = p p B p A p B x A = 19 kPa 18 kPa 20 kPa 18 kPa = 0.5 A is 1,2-dimethylbenzene y A = x A p A p B + (p A p B )x A = ( 0 . 5 ) × ( 20 kPa ) 18 kPa + ( 20 kPa 18 kPa ) 0 . 5 = 0 . 5 26 0.5 y B = 1 0 . 5 26 = 0 . 4 74 0 . 5 E8.5(b) p A = y A p = 0 . 612 p = x A p A = x A ( 68 . 8kPa ) p B = y B p = ( 1 y A )p = 0 . 388 p = x B p B = ( 1 x A ) × 82 . 1kPa y A p y B p = x A p A x B p B and 0 . 612 0 . 388 = 68 . 8 x A 82 . 1 ( 1 x A ) ( 0 . 388 ) × ( 68 . 8 )x A = ( 0 . 612 ) × ( 82 . 1 ) ( 0 . 612 ) × ( 82 . 1 )x A 26 . 6 94 x A = 50 . 2 45 50 . 2 45 x A x A = 50 . 2 45 26 . 6 94 + 50 . 2 45 = 0.653 x B = 1 0 . 653 = 0.347 p = x A p A + x B p B = ( 0 . 653 ) × ( 68 . ) + ( 0 . 347 ) × ( 82 . ) = 73 . 4kPa
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114 INSTRUCTOR S MANUAL E8.6(b) (a) If Raoult’s law holds, the solution is ideal. p A = x A p A = ( 0 . 4217 ) × ( 110 . 1kPa ) = 46 . 43 kPa p B = x B p B = ( 1 0 . 4217 ) × ( 94 . 93 kPa ) = 54 . 90 kPa p = p A + p B = ( 46 . 43 + 54 . 90 ) kPa = 101 . 33 kPa = 1 . 000 atm Therefore, Raoult’s law correctly predicts the pressure of the boiling liquid and the solution is ideal . (b) y A = p A p = 46 . 43 kPa 101 . 33 kPa = 0.4582 y B = 1 y A = 1 . 000 0 . 4582 = 0.5418 E8.7(b) Let B = benzene and T = toluene. Since the solution is equimolar z B = z T = 0 . 500 (a) Initially x B = z B and x T = z T ; thus p = x B p B + x T p T [8 . 3] = ( 0 . 500 ) × ( 74 Torr ) + ( 0 . 500 ) × ( 22 Torr ) = 37 Torr + 11 Torr = 48 Torr (b) y B = p B p [4] = 37 Torr 48 Torr = 0.77 y T = 1 0 . 77 = 0.23 (c) Near the end of the distillation y B = z B = 0 . 500 and y T = z T = 0 . 500 Equation 5 may be solved for x A [A = benzene = B here] x B = y B p T p B + (p T p B )y B = ( 0 . 500 ) × ( 22 Torr ) ( 75 Torr ) + ( 22 74 ) Torr × ( 0 . 500 ) = 0 . 23 x T = 1 0 . 23 = 0 . 77 This result for the special case of z B = z T = 0 . 500 could have been obtained directly by realizing that y B ( initial ) = x T ( Fnal )y T ( initial ) = x B ( Fnal ) p( Fnal ) = x B p B + x T p T = ( 0 . 23 ) × ( 74 Torr ) + ( 0 . 77 ) × ( 22 Torr ) = 34 Torr Thus in the course of the distillation the vapour pressure fell from 48 Torr to 34 Torr.
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Chap08 solutions - 8 Phase diagrams Solutions to exercises...

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