Instructor solution manual to accompany physical chemistry 7th ed by peter atkins chap07

7 Simple mixtures Solutions to exercises Discussion questions E7.1(b) For a component in an ideal solution, Raoult’s law is: p = xp ∗ For real solutions, the activity, a, replaces the mole fraction, x, and Raoult’s law becomes p = ap ∗ E7.2(b) All the colligative properties are a result of the lowering of the chemical potential of the solvent due to the presence of the solute This reduction takes the form µA = µA ∗ + RT ln xA or µA = µA ∗ + RT ln aA , depending on whether or not the solution can be considered ideal The lowering of the chemical potential results in a freezing point depression and a boiling point elevation as illustrated in Fig 7.20 of the text Both of these effects can be explained by the lowering of the vapour pressure of the solvent in solution due to the presence of the solute The solute molecules get in the way of the solvent molecules, reducing their escaping tendency E7.3(b) The activity of a solute is that property which determines how the chemical potential of the solute varies from its value in a specified reference state This is seen from the relation µ = µ−− + RT ln a, where µ−− is the value of the chemical potential in the reference state The reference state is either the hypothetical state where the pure solute obeys Henry’s law (if the solute is volatile) or the hypothetical state where the solute at unit molality obeys Henry’s law (if the solute is involatile) The activity of the solute can then be defined as that physical property which makes the above relation true It can be interpreted as an effective concentration Numerical exercises E7.4(b) Total volume V = nA VA + nB VB = n(xA VA + xB VB ) Total mass m = nA MA + nB MB = n(xA MA + (1 − xA )MB ) m =n xA MA + (1 − xA )MB where n = nA + nB 1.000 kg(103 g/kg) = 4.6701¯ mol (0.3713) × (241.1 g/mol) + (1 − 0.3713) × (198.2 g/mol) V = n(xA VA + xB VB ) = (4.6701¯ mol) × [(0.3713) × (188.2 cm3 mol−1 ) + (1 − 0.3713) × (176.14 cm3 mol−1 )] n= = 843.5 cm3 E7.5(b) Let A denote water and B ethanol The total volume of the solution is V = nA VA + nB VB We know VB ; we need to determine nA and nB in order to solve for VA Assume we have 100 cm3 of solution; then the mass is m = ρV = (0.9687 g cm−3 ) × (100 cm3 ) = 96.87 g of which (0.20) × (96.87 g) = 19.374 g is ethanol and (0.80) × (96.87 g) = 77.496 g is water 77.496 g = 4.30 mol H2 O 18.02 g mol−1 19.374 g nB = = 0.4205 mol ethanol 46.07 g mol−1 nA = INSTRUCTOR’S MANUAL 98 100 cm3 − (0.4205 mol) × (52.2 cm3 mol−1 ) V − n B VB = 18.15 cm3 = VA = nA 4.30¯ mol = 18 cm3 E7.6(b) Check that pB /xB = a constant (KB ) xB (pB /xB )/kPa 0.010 8.2 × 103 0.015 8.1 × 103 0.020 8.3 × 103 KB = p/x, average value is 8.2 × 103 kPa E7.7(b) In exercise 7.6(b), the Henry’s law constant was determined for concentrations expressed in mole fractions Thus the concentration in molality must be converted to mole fraction m(A) = 1000 g, corresponding to 1000 g n(A) = = 13.50¯ mol 74.1 g mol−1 n(B) = 0.25 mol Therefore, 0.25 mol = 0.0182¯ 0.25 mol + 13.50¯ mol xB = using KB = 8.2 × 103 kPa [exercise 7.6(b)] p = 0.0182¯ × 8.2 × 103 kPa = 1.5 × 102 kPa E7.8(b) Kf = RT ∗2 M 8.314 J K−1 mol−1 × (354 K)2 × 0.12818 kg mol−1 = 18.80 × 103 J mol−1 fus H = 7.1 K kg mol−1 Kb = RT ∗2 M 8.314 J K−1 mol−1 × (490.9 K)2 × 0.12818 kg mol−1 = 51.51 × 103 J mol−1 vap H = 4.99 K kg mol−1 E7.9(b) We assume that the solvent, 2-propanol, is ideal and obeys Raoult’s law xA (solvent) = p/p ∗ = 49.62 = 0.9924 50.00 MA (C3 H8 O) = 60.096 g mol−1 250 g = 4.1600 mol 60.096 g mol−1 nA nA xA = nA + n B = nA + n B xA nA = SIMPLE MIXTURES 99 nB = nA −1 xA = 4.1600 mol 8.69 g = 273¯ g mol−1 = 270 g mol−1 3.186 × 10−2 mol MB = E7.10(b) − = 3.186 × 10−2 mol 0.9924 Kf = 6.94 for naphthalene mass of B nB MB = nB = mass of naphthalene · bB bB = T Kf so MB = (mass of B) × Kf (mass of naphthalene) × T (5.00 g) × (6.94 K kg mol−1 ) = 178 g mol−1 (0.250 kg) × (0.780 K) nB nB = T = Kf bB and bB = mass of water Vρ MB = E7.11(b) ρ = 103 kg m−3 nB = V RT T = (density of solution ≈ density of water) T = Kf RT ρ Kf = 1.86 K mol−1 kg (1.86 K kg mol−1 ) × (99 × 103 Pa) = 7.7 × 10−2 K (8.314 J K−1 mol−1 ) × (288 K) × (103 kg m−3 ) Tf = −0.077◦ C E7.12(b) mix G = nRT (xA ln xA + xB ln xB ) nAr = nNe , mix G xAr = xNe = 0.5, n = nAr + nNe = = pV 21 ln 21 + 21 ln 21 = −pV ln = −(100 × 103 Pa) × (0.250 L) = −17.3 Pa m3 = −17.3 J E7.13(b) mix G pV RT = nRT xJ ln xJ [7.18] J m3 ln 103 L 17.3 J − mix G = = 6.34 × 10−2 J K−1 T 273 K − mix G xJ ln xJ [7.19] = mix S = −nR T J mix S = n = 1.00 mol + 1.00 mol = 2.00 mol x(Hex) = x(Hep) = 0.500 Therefore, mix G = (2.00 mol) × (8.314 J K −1 mol−1 ) × (298 K) × (0.500 ln 0.500 + 0.500 ln 0.500) = −3.43 kJ INSTRUCTOR’S MANUAL 100 +3.43 kJ = +11.5 J K−1 298 K mix H for an ideal solution is zero as it is for a solution of perfect gases [7.20] It can be demonstrated from mix S mix H E7.14(b) = = mix G + T mix S = (−3.43 × 103 J) + (298 K) × (11.5 J K −1 ) = Benzene and ethylbenzene form nearly ideal solutions, so mix S = −nR(xA ln xA + xB ln xB ) mix S, differentiate with respect to xA To find maximum is zero and find value of xA at which the derivative Note that xB = − xA so mix S use = −nR(xA ln xA + (1 − xA ) ln(1 − xA )) d ln x = x dx xA d ( mix S) = −nR(ln xA + − ln(1 − xA ) − 1) = −nR ln dx − xA =0 when xA = 21 Thus the maximum entropy of mixing is attained by mixing equal molar amounts of two components mB /MB nB = = nE mE /ME mE ME 106.169 = 1.3591 = = 78.115 mB MB mB = 0.7358 mE E7.15(b) Assume Henry’s law [7.26] applies; therefore, with K(N2 ) = 6.51 × 107 Torr and K(O2 ) = 3.30 × 107 Torr, as in Exercise 7.14, the amount of dissolved gas in kg of water is n(N2 ) = 103 g 18.02 g mol−1 × p(N2 ) 6.51 × 107 Torr = (8.52 × 10−7 mol) × (p/Torr) For p(N2 ) = xp and p = 760 Torr n(N2 ) = (8.52 × 10−7 mol) × (x) × (760) = x(6.48 × 10−4 mol) and, with x = 0.78 n(N2 ) = (0.78) × (6.48 × 10−4 mol) = 5.1 × 10−4 mol = 0.51 mmol The molality of the solution is therefore approximately 0.51 mmol kg−1 in N2 Similarly, for oxygen, n(O2 ) = 103 g 18.02 g mol−1 × p(O2 ) 3.30 × 107 Torr = (1.68 × 10−6 mol) × (p/Torr) For p(O2 ) = xp and p = 760 Torr n(O2 ) = (1.68 × 10−6 mol) × (x) × (760) = x(1.28 mmol) and when x = 0.21, n(O2 ) ≈ 0.27 mmol Hence the solution will be 0.27 mmol kg−1 in O2 SIMPLE MIXTURES E7.16(b) 101 Use n(CO2 ) = (4.4 × 10−5 mol) × (p/Torr), p = 2.0(760 Torr) = 1520 Torr n(CO2 ) = (4.4 × 10−5 mol) × (1520) = 0.067 mol The molality will be about 0.067 mol kg−1 and, since molalities and molar concentration for dilute aqueous solutions are approximately equal, the molar concentration is about 0.067 mol L−1 E7.17(b) M(glucose) = 180.16 g mol−1 T = Kf bB Kf = 1.86 K kg mol−1 T = (1.86 K kg mol−1 ) × 10 g/180.16 g mol−1 0.200 kg = 0.52 K Freezing point will be 0◦ C − 0.52◦ C = −0.52◦ C E7.18(b) The procedure here is identical to Exercise 7.18(a) ln xB = = fus H R × 5.2 × 103 J mol−1 8.314 J K−1 mol−1 ¯ = −0.0886, xB = 1 − ∗ T T [7.39; B, the solute, is lead] × 1 − 600 K 553 K implying that xB = 0.92 n(Pb) , n(Pb) + n(Bi) implying that n(Pb) = xB n(Bi) − xB 1000 g = 4.785 mol 208.98 g mol−1 Hence, the amount of lead that dissolves in kg of bismuth is For kg of bismuth, n(Bi) = n(Pb) = (0.92) × (4.785 mol) = 55 mol, − 0.92 11 kg or Comment It is highly unlikely that a solution of 11 kg of lead and kg of bismuth could in any sense be considered ideal The assumptions upon which eqn 7.39 is based are not likely to apply The answer above must then be considered an order of magnitude result only E7.19(b) Proceed as in Exercise 7.19(a) The data are plotted in Fig 7.1, and the slope of the line is 1.78 cm/ (mg cm−3 ) = 1.78 cm/(g L−1 ) = 1.78 × 10−2 m4 kg−1 12 10 6 Figure 7.1 INSTRUCTOR’S MANUAL 102 Therefore, (8.314 J K−1 mol−1 ) × (293.15 K) = 14.0 kg mol−1 (1.000 × 103 kg m−3 ) × (9.81 m s−2 ) × (1.78 × 10−2 m4 kg−1 ) M= E7.20(b) Let A = water and B = solute pA 0.02239 atm = 0.9701 [42] = ∗ pA 0.02308 atm nA aA and xA = γA = xA nA + n B 0.122 kg 0.920 kg = 0.506 mol nA = = 51.05¯ mol nB = −1 0.01802 kg mol 0.241 kg mol−1 51.05¯ 0.9701 xA = = 0.990 γA = = 0.980 51.05 + 0.506 0.990 aA = E7.21(b) B = Benzene µB (l) = µ∗B (l) + RT ln xB [7.50, ideal solution] RT ln xB = (8.314 J K−1 mol−1 ) × (353.3 K) × (ln 0.30) = −3536¯ J mol−1 Thus, its chemical potential is lowered by this amount ∗ ∗ [42] = γB xB pB = (0.93) × (0.30) × (760 Torr) = 212 Torr pB = aB pB E7.22(b) Question What is the lowering of the chemical potential in the nonideal solution with γ = 0.93? pA pA = yA = = 0.314 pA + p B 760 Torr pA = (760 Torr) × (0.314) = 238.64 Torr pB = 760 Torr − 238.64 Torr = 521.36 Torr pA 238.64 Torr aA = ∗ = atm pA (73.0 × 103 Pa) × × 760 Torr 101 325 Pa aB = = 0.436 atm pB 521.36 Torr ∗ = atm 760 Torr pB (92.1 × 10 Pa) × 1011 325 atm Pa × = 0.755 aA 0.436 = = 1.98 xA 0.220 aB 0.755 γB = = 0.968 = xB 0.780 γA = Solutions to problems Solutions to numerical problems P7.3 Vsalt = ∂V mol−1 [Problem 7.2] ∂b H2 O = 69.38(b − 0.070) cm3 mol−1 with b ≡ b/(mol kg−1 ) Therefore, at b = 0.050 mol kg−1 , Vsalt = −1.4 cm3 mol−1 SIMPLE MIXTURES 103 The total volume at this molality is V = (1001.21) + (34.69) × (0.02)2 cm3 = 1001.22 cm3 Hence, as in Problem 7.2, V (H2 O) = (1001.22 cm3 ) − (0.050 mol) × (−1.4 cm3 mol−1 ) = 18.04 cm3 mol−1 55.49 mol Question What meaning can be ascribed to a negative partial molar volume? P7.5 Let E denote ethanol and W denote water; then V = nE VE + nW VW [7.3] For a 50 per cent mixture by mass, mE = mW , implying that nE ME = nW MW , Hence, V = nE VE + which solves to nE = Furthermore, xE = or nW = n E M E VW MW V VE + ME VW MW nE ME MW , nW = ME V VE M W + M E V W nE = ME nE + n W 1+ M W Since ME = 46.07 g mol−1 and MW = 18.02 g mol−1 , xE = 0.2811, ME = 2.557 Therefore MW xW = − xE = 0.7189 At this composition VE = 56.0 cm3 mol−1 Therefore, nE = VW = 17.5 cm3 mol−1 [Fig.7.1 of the text] 100 cm3 (56.0 cm3 mol−1 ) + (2.557) × (17.5 cm3 mol−1 ) = 0.993 mol nW = (2.557) × (0.993 mol) = 2.54 mol The fact that these amounts correspond to a mixture containing 50 per cent by mass of both components is easily checked as follows mE = nE ME = (0.993 mol) × (46.07 g mol−1 ) = 45.7 g ethanol mW = nW MW = (2.54 mol) × (18.02 g mol−1 ) = 45.7 g water At 20◦ C the densities of ethanol and water are, ρE = 0.789 g cm−3 , ρW = 0.997 g cm−3 Hence, VE = mE 45.7 g = = 57.9 cm3 of ethanol ρE 0.789 g cm−3 VW = mW 45.7 g = = 45.8 cm3 of water ρW 0.997 g cm−3 INSTRUCTOR’S MANUAL 104 The change in volume upon adding a small amount of ethanol can be approximated by V = dV ≈ VE dnE ≈ VE nE where we have assumed that both VE and VW are constant over this small range of nE Hence (1.00 cm3 ) × (0.789 g cm−3 ) (46.07 g mol−1 ) V ≈ (56.0 cm3 mol−1 ) × P7.7 T 0.0703 K = 0.0378 mol kg−1 = Kf 1.86 K/(mol kg−1 ) Since the solution molality is nominally 0.0096 mol kg−1 in Th(NO3 )4 , each formula unit supplies 0.0378 ≈ ions (More careful data, as described in the original reference gives ν ≈ to 6.) 0.0096 The data are plotted in Figure 7.2 The regions where the vapor pressure curves show approximate straight lines are denoted R for Raoult and H for Henry A and B denote acetic acid and benzene respectively 300 Extrapolate for KB R Henry 200 B p / Torr P7.9 mB = = +0.96 cm3 Raoult 100 Henry A H Raoult 0 0.2 0.4 0.6 xA R H 0.8 1.0 Figure 7.2 pA pB ∗ and γB = x p ∗ for the Raoult’s law activity xA p A B B pB coefficients and γB = for the activity coefficient of benzene on a Henry’s law basis, with K xB K ∗ ∗ determined by extrapolation We use pA = 55 Torr, pB = 264 Torr and KB∗ = 600 Torr to draw up As in Problem 7.8, we need to form γA = SIMPLE MIXTURES 105 the following table: xA pA /Torr pB /Torr aA (R) aB (R) γA (R) γB (R) aB (H) γB (H) 0 264 1.00 — 1.00 0.44 0.44 0.2 20 228 0.36 0.86 1.82 1.08 0.38 0.48 0.4 30 190 0.55 0.72 1.36 1.20 0.32 0.53 0.6 38 150 0.69 0.57 1.15 1.42 0.25 0.63 0.8 50 93 0.91 0.35 1.14 1.76 0.16 0.78 1.0 55 1.00[pA /pA∗ ] 0[pB /pB∗ ] 1.00[pA /xA pA∗ ] —[pB /xB pB∗ ] 0[pB /KB ] 1.00[pB /xB KB ] GE is defined as [Section 7.4]: GE = mix G(actual) − mix G(ideal) = nRT (xA ln aA + xB ln aB ) − nRT (xA ln xA + xB ln xB ) and with a = γ x GE = nRT (xA ln γA + xA ln γB ) For n = 1, we can draw up the following table from the information above and RT = 2.69 kJ mol−1 : xA xA ln γA xB ln γB GE /(kJ mol−1 ) P7.11 0 0 0.2 0.12 0.06 0.48 0.4 0.12 0.11 0.62 0.6 0.08 0.14 0.59 0.8 0.10 0.11 0.56 1.0 0 (a) The volume of an ideal mixture is Videal = n1 Vm,1 + n2 Vm,2 so the volume of a real mixture is V = Videal + V E We have an expression for excess molar volume in terms of mole fractions To compute partial molar volumes, we need an expression for the actual excess volume as a function of moles V E = (n1 + n2 )VmE = n n2 n1 + n a0 + a1 (n1 − n2 ) n1 + n n1 n2 a1 (n1 − n2 ) a0 + n1 + n n1 + n The partial molar volume of propionic acid is so V = n1 Vm,1 + n2 Vm,2 + V1 = a0 n22 a1 (3n1 − n2 )n22 ∂V = Vm,1 + + ∂n1 p,T ,n2 (n1 + n2 )2 (n1 + n2 )3 V1 = Vm,1 + a0 x22 + a1 (3x1 − x2 )x22 That of oxane is V2 = Vm,2 + a0 x12 + a1 (x1 − 3x2 )x12 INSTRUCTOR’S MANUAL 106 (b) We need the molar volumes of the pure liquids Vm,1 = M1 74.08 g mol−1 = = 76.23 cm3 mol−1 ρ1 0.97174 g cm−3 86.13 g mol−1 = 99.69 cm3 mol−1 0.86398 g cm−3 In an equimolar mixture, the partial molar volume of propionic acid is and Vm,2 = V1 = 76.23 + (−2.4697) × (0.500)2 + (0.0608) × [3(0.5) − 0.5] × (0.5)2 cm3 mol−1 = 75.63 cm3 mol−1 and that of oxane is V2 = 99.69 + (−2.4697) × (0.500)2 + (0.0608) × [0.5 − 3(0.5)] × (0.5)2 cm3 mol−1 = 99.06 cm3 mol−1 P7.13 Henry’s law constant is the slope of a plot of pB versus xB in the limit of zero xB (Fig 7.3) The partial pressures of CO2 are almost but not quite equal to the total pressures reported above pCO2 = pyCO2 = p(1 − ycyc ) Linear regression of the low-pressure points gives KH = 371 bar 80 60 40 20 0.0 0.1 0.2 0.3 Figure 7.3 The activity of a solute is aB = pB = xB γB KH so the activity coefficient is γB = pB yB p = xB K H xB K H SIMPLE MIXTURES 107 where the last equality applies Dalton’s law of partial pressures to the vapour phase A spreadsheet applied this equation to the above data to yield p/bar 10.0 20.0 30.0 40.0 60.0 80.0 P7.16 ycyc 0.0267 0.0149 0.0112 0.009 47 0.008 35 0.009 21 xcyc 0.9741 0.9464 0.9204 0.892 0.836 0.773 γCO2 1.01 0.99 1.00 0.99 0.98 0.94 GE = RT x(1 − x){0.4857 − 0.1077(2x − 1) + 0.0191(2x − 1)2 } with x = 0.25 gives GE = 0.1021RT Therefore, since mix G(actual) mix G = mix G(ideal) + nG E = nRT (xA ln xA + xB ln xB ) + nGE = nRT (0.25 ln 0.25 + 0.75 ln 0.75) + nGE = −0.562nRT + 0.1021nRT = −0.460nRT Since n = mol and RT = (8.314 J K −1 mol−1 ) × (303.15 K) = 2.52 kJ mol−1 , mix G = (−0.460) × (4 mol) × (2.52 kJ mol−1 ) = −4.6 kJ Solutions to theoretical problems P7.18 xA dµA + xB dµB = [7.11, Gibbs–Duhem equation] Therefore, after dividing through by dxA xA ∂µA + xB ∂xA p,T ∂µB =0 ∂xA p,T or, since dxB = −dxA , as xA + xB = xA ∂µA − xB ∂xA p,T ∂µB =0 ∂xB p,T ∂µA = ∂ ln xA p,T dx ∂µB d ln x = ∂ ln xB p,T x f ∂ ln fA Then, since µ = µ−− + RT ln −− , = p ∂ ln xA p,T ∂ ln pA ∂ ln pB On replacing f by p, = ∂ ln xA p,T ∂ ln xB p,T or, ∂ ln fB ∂ ln xB p,T ∗ If A satisfies Raoult’s law, we can write pA = xA pA , which implies that ∗ ∂ ln pA ∂ ln pA ∂ ln xA = + =1+0 ∂ ln xA p,T ∂ ln xA ∂ ln xA ∂ ln pB =1 ∂ ln xB p,T ∗ which is satisfied if pB = xB pB (by integration, or inspection) Hence, if A satisfies Raoult’s law, so does B Therefore, INSTRUCTOR’S MANUAL 108 P7.20 ln xA = − fus G (Section 7.5 analogous to equation for ln xB used in derivation of eqn 7.39) RT d ln xA d =− × dT R dT xA d ln xA = ln xA = T fus G T fus H dT RT T∗ = − fus H × R ≈ fus H RT fus H R [Gibbs–Helmholtz equation] T dT T∗ T 1 − ∗ T T The approximations ln xA ≈ −xB and T ≈ T ∗ then lead to eqns 33 and 36, as in the text P7.22 Retrace the argument leading to eqn 7.40 of the text Exactly the same process applies with aA in place of xA At equilibrium µ∗A (p) = µA (xA , p + ) which implies that, with µ = µ∗ + RT ln a for a real solution, p+ and hence that p p+ ) + RT ln aA = µ∗A (p) + µ∗A (p) = µ∗A (p + p Vm dp + RT ln aA Vm dp = −RT ln aA For an incompressible solution, the integral evaluates to Vm , so Vm = −RT ln aA In terms of the osmotic coefficient φ (Problem 7.21) Vm = rφRT r= xB nB = xA nA φ=− xA ln aA = − ln aA xB r For a dilute solution, nA Vm ≈ V Hence, V = nB φRT and therefore, with [B] = nB V = φ[B]RT Solutions to applications P7.24 By the van’t Hoff equation [7.40] cRT Π = [B]RT = M Division by the standard acceleration of free fall, g, gives Π c(R/g)T = M (a) This expression may be written in the form cR T Π = M which has the same form as the van’t Hoff equation, but the unit of osmotic pressure (Π ) is now force/area (mass length)/(area time2 ) mass = = area length/time length/time2 SIMPLE MIXTURES 109 This ratio can be specified in g cm−2 Likewise, the constant of proportionality (R ) would have the units of R/g (mass length2 /time2 ) K−1 mol−1 energy K −1 mol−1 = = mass length K −1 mol−1 2 length/time length/time This result may be specified in g cm K−1 mol−1 R = 8.314 51 J K−1 mol−1 R = g 9.806 65m s−2 = 0.847 844 kg m K−1 mol−1 103 g kg × 102 cm m R = 84 784.4 g cm K−1 mol−1 In the following we will drop the primes giving cRT Π= M and use the Π units of g cm−2 and the R units g cm K−1 mol−1 (b) By extrapolating the low concentration plot of /c versus c (Fig 7.4 (a)) to c = we find the intercept 230 g cm−2 /g cm−3 In this limit van’t Hoff equation is valid so RT RT = intercept or M n = intercept Mn (84 784.4 g cm K−1 mol−1 ) × (298.15 K) Mn = (230 g cm−2 )/(g cm−3 ) M n = 1.1 × 105 g mol −1 500 450 400 350 300 250 200 0.000 0.010 0.020 0.030 0.040 Figure 7.4(a) INSTRUCTOR’S MANUAL 110 (c) The plot of Π/c versus c for the full concentration range (Fig 7.4(b)) is very nonlinear We may conclude that the solvent is good This may be due to the nonpolar nature of both solvent and solute 7000 6000 5000 4000 3000 2000 1000 0.00 0.050 0.100 0.150 0.200 0.250 0.300 Figure 7.4(b) (d) Π/c = (RT /M n )(1 + B c + C c2 ) Since RT /M n has been determined in part (b) by extrapolation to c = 0, it is best to determine the second and third virial coefficients with the linear regression fit (Π/c)/(RT /M n ) − =B +C c c R = 0.9791 B = 21.4 cm3 g−1 , C = 211 cm6 g−2 , standard deviation = 2.4 cm3 g−1 standard deviation = 15 cm6 g−2 (e) Using 1/4 for g and neglecting terms beyond the second power, we may write Π 1/2 = c RT 1/2 Mn (1 + 21 B c) SIMPLE MIXTURES 111 We can solve for B , then g(B )2 = C   1/2   Π c RT Mn  1/2  − = 21 B c RT /M n has been determined above as 230 g cm−2 /g cm−3 We may analytically solve for B from one of the data points, say, /c = 430 g cm−2 /g cm−3 at c = 0.033 g cm−3 430 g cm−2 /g cm−3 1/2 − = 21 B × (0.033 g cm−3 ) 230 g cm−2 /g cm−3 × (1.367 − 1) B = = 22.2¯ cm3 g−1 0.033 g cm−3 C = g(B )2 = 0.25 × (22.2¯ cm3 g−1 )2 = 123¯ cm6 g−2 Better values of B and C can be obtained by plotting Π 1/2 RT 1/2 c Mn against c This plot is shown in Fig 7.4(c) The slope is 14.03 cm3 g−1 B = × slope = 28.0¯ cm3 g−1 C is then 196¯ cm6 g−2 The intercept of this plot should thereotically be 1.00, but it is in fact 0.916 with a standard deviation of 0.066 The overall consistency of the values of the parameters confirms that g is roughly 1/4 as assumed 6.0 5.0 n 4.0 3.0 2.0 1.0 0.0 0.00 0.05 0.10 0.15 0.20 0.25 0.30 Figure 7.4(c) ... (0.314) = 238.64 Torr pB = 760 Torr − 238.64 Torr = 521.36 Torr pA 238.64 Torr aA = ∗ = atm pA (73.0 × 103 Pa) × × 760 Torr 101 325 Pa aB = = 0.436 atm pB 521.36 Torr ∗ = atm 760 Torr pB (92.1 ×... determined by extrapolation We use pA = 55 Torr, pB = 264 Torr and KB∗ = 600 Torr to draw up As in Problem 7.8, we need to form γA = SIMPLE MIXTURES 105 the following table: xA pA /Torr pB /Torr aA... (0.500 ln 0.500 + 0.500 ln 0.500) = −3.43 kJ INSTRUCTOR S MANUAL 100 +3.43 kJ = +11.5 J K−1 298 K mix H for an ideal solution is zero as it is for a solution of perfect gases [7.20] It can be demonstrated