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5 Simple mixtures
Answers to discussion questions
D5.1 At equilibrium, the chemical potentials of any component in both the liquid and vapor phases must be
equal. This is justified by the requirement that, for systems at equilibrium under constant temperature and
pressure conditions, with no additional work, G= 0 [see Section 3.5(e) and the answer to Discussion
question 3.3]. HereG = i(v) i(1), for all components, i , of the solution; hence their chemical
potentials must be equal in the liquid and vapor phases.
D5.3 All of the colligative properties are a function of the concentration of the solute, which implies that the
concentration can be determined by a measurement of these properties. See eqns 5.33, 5.34, 5.36, 5.37,
and 5.40. Knowing the mass of the solute in solution then allows for a calculation of its molar mass. For
example, the mole fraction of the solute is related to its mass as follows:
xB = mB/MBmB/MB+ mA/MA
.
The only unknown in this expression is MB which is easily solved for. See Example 5.4 for the details
of how molar mass is determined from osmotic pressure.
D5.5 A regular solution has excess entropy of zero, but an excess enthalpy that is non-zero and dependent on
composition, perhaps in the manner of eqn 5.30. We can think of a regular solution as one in which the
different molecules of the solution are distributed randomly, as in an ideal solution, but have different
energies of interaction with each other.
Solutions to exercises
E5.1(a) Let A denote acetone and C chloroform. The total volume of the solution is
V =nAVA + nCVC.
VA and Vcare given; hence we need to determinenA and nC in 1.000 kg of the solution with the stated
mole fraction. The total mass of the sample is m = nAMA + nCMC (a). We also know that
xA =nA
nA+ nCimplies that (xA 1)nA+ xAnC =0
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SIMPLE MIXTURES 95
and hence that
xCnA+ xAnC =0. (b)
On solving (a) and (b), we find
nA =
xA
xC
nC, nC =
mxC
xAMA+ xCMC.
SincexC =0.4693,xA =1 xC =0.5307,
nC =(0.4693)(1000 g)
[(0.5307)(58.08)+(0.4693)(119.37)] gmol1 =5.404 mol,
nA =
0.5307
0.4693
(5.404)mol = 6.111 mol.
The total volume,V = nAVA+ nBVB, is therefore
V =(6.111 mol)(74.166cm3 mol1)+(5.404 mol)(80.235cm3 mol1)
= 886.8 cm3 .
E5.2(a) Let A denote water and B ethanol. The total volume of the solution isV = nAVA+ nBVB
We are given VA, we need to determine nA and nB in order to solve forVB.
Assume we have 100 cm3 of solution; then the mass of solution is
m = d V = (0.914 g cm3)(100 cm3)= 91.4 g
of which 45.7 g is water and 45.7 g ethanol.
100 cm3 =
45.7 g
18.02 g mol1
(17.4 cm3 mol1)+
45.7g
46.07 g mol1
VB
=44.13 cm3 +0.9920 molVB,
VB =55.87 cm3
0.9920 mol= 56.3 cm3 mol1 .
E5.3(a) Check whetherpB
xBis equal to a constant (KB)
x 0.005 0.012 0.019
p/x 6.4103 6.4103 6.4103 kPa
Hence,KB 6.4103 kPa .
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96 SOLUTIONS MANUAL
E5.4(a) In Exercise 5.3(a), the Henrys law constant was determined for concentrations expressed in mole
fractions. Thus the concentration in molality must be converted to mole fraction.
m(GeCl4) = 1000 g, corresponding to
n(GeCl4)=1000 g
214.39 g mol1 = 4.664 mol, n(HCl) = 0.10 mol.
Therefore,x=0.10 mol
(0.10 mol)+ (4.664 mol)=0.0210.
FromKB =6.4 103 kPa (Exercise 5.3(a)), p = (0.02106.4103 kPa) = 1.3102 kPa .
E5.5(a) We assume that the solvent, benzene, is ideal and obeys Raoults law.
Let B denote benzene and A the solute; then
pB = xBpB and xB =
nB
nA+ nB.
HencepB =nBp
B
nA + nB, which solves to
nA =nB(p
B pB)
pB.
Then, sincenA = mAMA
, wheremA is the mass of A present,
MA =mApB
nB(pB pB)
=mAMBpB
mB(pB pB)
.
From the data
MA =(19.0 g)(78.11 g mol1)(51.5 kPa)
(500g)(53.351.5) kPa= 82gmol1
E5.6(a) MB =mass of B
hB[B = compound].
nB =mass of CCl4 bB [bB =molality of B].
bB =T
Kf[5.37]; thus
MB =mass of BKf
mass of CCl4 TKf =30 K/(mol kg
1)[Table 5.2],
MB =(100g)(30Kkgmol1)
(0.750 kg)(10.5K)= 381g mol1 .
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SIMPLE MIXTURES 97
E5.7(a) T =KfbB[5.37], bB =nB
mass of water
nB
V
[dilute solution].
103 kg m3[density of solution density of water].
nB V
RT[5.40], T Kf
RT
withKf =1.86 K/(mol kg1)[Table 5.2].
T (1.86K kgmol1)(120103 Pa)
(8.314JK1 mol1)(300K)(103 kg m3)=0.089 K.
Therefore, the solution will freeze at about 0.09C .
COMMENT. Osmotic pressures are inherently large. Even dilute solutions with small freezing point
depressions have large osmotic pressures.
E5.8(a) mixG = nRT{xAlnxA+ xBlnxB}[5.18], xA =xB =0.5, n=pV
RT.
Therefore,
mixG = (pV)
1
2ln
1
2+
1
2ln
1
2
= pVln 2
= (1.0)(1.013105 Pa)(5.0103 m3)(ln 2)
= 3.510
2
J = 0.35 kJ .
mixS= nR{xAlnxA + xBlnxB} =mixG
T[5.19] =
0.35 kJ
298K= +1.2 J K1 .
E5.9(a) mixS= nR
J
xJ lnxJ[5.19].
Therefore, for molar amounts,
mixS= R
J
xJlnxJ
= R[(0.782 ln 0.782)+(0.209 ln 0.209)+(0.009 ln 0.009)+(0.0003 ln 0.0003)]
=0.564R = +4.7 J K
1
mol
1
.
E5.10(a) Hexane and heptane form nearly ideal solutions; therefore eqn 5.19 applies.
mixS= nR(xAlnxA + xBlnxB)[5.19].
We need to differentiate eqn 5.19 with respect toxAand look for the value ofxAat which the derivative
is zero. SincexB =1 xA, we need to differentiate
mixS= nR{xAlnxA+ (1 xA) ln(1 xA)}.
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This gives using d lnxdx
=1
x
dmixS
dxA= nR{lnxA+ 1ln(1xA)1} = nR ln
xA
1 xA
which is zero when xA =1
2. Hence, the maximum entropy of mixing occurs for the preparation of a
mixture that contains equal mole fractions of the two components.
(a)n(Hex)
n(Hep)=1 =
m(Hex)
M(Hex)
m(Hep)
M(Hep)
.
(b)
m(Hex)
m(Hep) =
M(Hex)
M(Hep) =
86.17 g mol1
100.20 g mol1 = 0.8600 .
E5.11(a) With concentrations expressed in molalities, Henrys law [5.26] becomespB = bBK.
Solving forb, the solubility, we have bB =pB
K.
(a) pB =0.10 atm = 10.1 kPa.
b =10.1 kPa
3.01103 kPa kg mol1 = 3.4 mmol kg1 .
(b) pB =1.00 atm = 101.3 kPa.
b =
101.3 kPa
3.01103 kPa kg mol1 = 34 mmol kg1
.
E5.12(a) As in Exercise 5.11(a), we have
bB =pB
K=
5.0101.3 kPa
3.01103 kPa kg mol1 = 0.17 mol kg1 .
Hence, the molality of the solution is about 0.17 mol kg1 and, since molalities and molar con-
centrations for dilute aqueous solutions are approximately equal, the molar concentration is about
0.17 mol dm3.
E5.13(a) The solubility in grams of anthracene per kg of benzene can be obtained from its mole fraction with use
of the equation.
lnxB =fusH
R
1
T
1
T
[5.39; B, the solute, is anthracene]
=
28.8103 Jmol1
8.314 J K1 mol1
1
490.15 K
1
298.15K
= 4.55.
Therefore,xB =e4.55 =0.0106.
SincexB 1, x(anthracene)n(anthracene)
n(benzene).
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SIMPLE MIXTURES 99
Therefore, in 1 kg of benzene,
n(anthr.) x(anthr.)
1000 g
78.11 g mol1
(0.0106)(12.80 mol)= 0.136 mol.
Themolality of thesolutionis therefore0.136 molkg1. SinceM=178 g mol1, 0.136 molcorresponds
to 24 g anthracene in 1 kg of benzene.
E5.14(a) The best value of the molar mass is obtained from values of the data extrapolated to zero concentration,
since it is under this condition that eqn 5.40 applies.
V =nBRT[5.40], so =mRT
MV=
cRT
M, c =
m
V.
= gh [hydrostatic pressure], so h = RTgM
c.
dm3 Figure 5.1
Hence, plothagainstcand identify the slope asRT
gM. Figure 5.1 shows the plot of the data.
The slope of the line is 0.29 cm/(g dm3), so
RT
gM=
0.29 cm
g dm3 =0.29 cmdm3 g1 =0.29102 m4 kg1.
Therefore,
M =RT
(g)(0.29102 m4 kg1)
=(8.314 J K1 mol1)(298.15K)
(1.004103 kg m3)(9.81m s2)(0.29102 m4 kg1)= 87 kg mol1 .
E5.15(a) ForA (Raoults law basis; concentration in mole fraction)
aA =pA
pA[5.43] =
250 Torr
300 Torr= 0.833 ; A =
aA
xA=
0.833
0.90= 0.93 .
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100 SOLUTIONS MANUAL
ForB (Henrys law basis; concentration in mole fraction)
aB =pB
KB[5.50] =
25 Torr
200 Torr= 0.125 ; B =
aB
xB=
0.125
0.10= 1.25 .
ForB (Henrys law basis; concentration in molality)
An equation analogous to eqn 5.50 is used, aB =pB
KBwith a modified Henrys law constant KBthat
corresponds to the pressure of B in the limit of very low molalities,
pB =bB
b KB,
is analogous to pB = xBKB. Since xBand bBare related as bB =xB
MAxA, KBand KBare related as
KB = xAMAbKB.
We also needMA,
MA =xB
xAbB=
0.10
(0.90)(2.22 mol kg1)=0.050 kg mol1.
Then,KB =(0.90)(0.050 kgmol1)(1molkg1)(200 Torr) =9.0 Torr
andaB =25 Torr
9.0 Torr= 2.8 , B =
aBbBb
= 2.82.22
= 1.25 .
COMMENT. The two methods for the solute B give different values for the activities. This is reasonable
since the chemical potentials in the reference states and are different.
Question. What are the activity and activity coefficient of B in the Raoults law basis?
E5.16(a) In an ideal dilute solution the solvent (CCl4) obeys Raoults law and the solute (Br2) obeys Henrys law;
hence
p(CCl4)= x(CCl4)p(CCl4)[5.24] =(0.950)(33.85Torr)= 32.2 Torr ,
p(Br2)= x(Br2)K(Br2)[5.26] =(0.050)(122.36Torr) = 6.1 Torr ,
p(Total)= (32.2+6.1)Torr = 38.3 Torr .
The composition of the vapor in equilibrium with the liquid is
y(CCl4)=p(CCl4)
p(Total)=
32.2 Torr
38.3 Torr= 0.841 ,
y(Br2)=p(Br2)
p(Total)=
6.1 Torr
38.3 Torr= 0.16 .
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SIMPLE MIXTURES 101
E5.17(a) Let A =acetone and M =methanol.
yA =pA
pA+ pM[Daltons law] =
pA
101.3 kPa=0.516.
pA =52.3 kPa, pM =49.0 kPa.
pA =pA
pA[5.43] =
52.3 kPa
105 kPa= 0.499 , aM =
pM
pM=
49.0 kPa
73.5 kPa= 0.668 .
A =aA
xA=
0.499
0.400= 1.25 , M =
aM
xM=
0.668
0.600= 1.11 .
E5.18(a) I =1
2
(bi/b
)z2i[5.70]
and for an MpXqsalt,(b+/b) =p(b/b), (b/b
) = q(b/b), so
I =1
2(pz2++ qz
2)
b
b
.
I(KCl) =1
2(11+11)
b
b
=
b
b
.
I(CuSO4) =1
2(122 +122)
b
b
= 4
b
b
.
I =I(KCL)+ I(CuSO4) =
b
b
(KCL)+4
b
b
(CuSO4)
=(0.10)+ (4)(0.20) = 0.90 .
COMMENT. Note that the strength of a solution of more than one electrolyte may be calculated by summing
the ionic strengths of each electrolyte considered as a separate solution, as in the solution to this exercise,
by summing the product 1
2
bi
b
z
2i
for each individual ion as in the definition ofI [5.71].
E5.19(a) I =I(KNO3)=
b
b
(KNO3)=0.150.
Therefore, the ionic strengths of the added salts must be 0.100.
(a) I(Ca(NO3)2) =1
2
(22 +2) b
b= 3
b
b .
Therefore, the solution should be made1
3 0.100 mol kg1 =0.0333 mol kg1 in Ca(NO3)2. The
mass that should be added to 500 g of the solution is therefore (0.500 kg) (0.0333 molkg1)
(164 g mol1) = 2.73 g .
(b) I(NaC1)=
b
b
; therefore, withb= 0.100 mol kg1,
(0.500 kg)(0.100 mol kg1)(58.4 g mol1)= 2.92 g .
(We are neglecting the fact that the mass of solution is slightly different from the mass of solvent.)
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102 SOLUTIONS MANUAL
E5.20(a) Theseconcentrationsaresufficiently dilutefor theDebyeHckel limiting lawto givea good approximate
value for the mean ionic activity coefficient. Hence
log = |z+z|AI1/2[5.69].
I =1
2
i
z2i
bi
b
[5.71] =
1
2[(40.010)+ (10.020)+(10.030)+ (10.30)
= 0.060 .
log = 210.509(0.060)1/2 = 0.2494; =0.563 = 0.56 .
E5.21(a) log = A|z+z|I
1/2
1+BI1/2 [5.72].
Solving forB,
B =
1
I1/2 +
A|z+z|
log
.
For HBr,I =
b
b
and |z+z+| =1; so
B =
b
(b/b)1/2 +
0.509
log
.
Hence, draw up the following table.
(b/b) 5.0103 10.0103 120.0103
0.930 0.907 0.879
B 2.01 2.01 2.02
The constancy ofBindicates that the mean ionic activity coefficient of HBr obeys the extended Debye
Hckel law very well.
Solutions to problems
Solutions to numerical problems
P5.1
pA = yApandpB =yBp(Daltons law). Hence, draw up the following table.
pA/kPa 0 1.399 3.566 5.044 6.996 7.940 9.211 10.105 11.287 12.295
xA 0 0.0898 0.2476 0.3577 0.5194 0.6036 0.7188 0.8019 0.9105 1
yA 0 0.0410 0.1154 0.1762 0.2772 0.3393 0.4450 0.5435 0.7284 1
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SIMPLE MIXTURES 103
pB/kPa 0 4.209 8.487 11.487 15.462 18.243 23.582 27.334 32.722 36.066
xB 0 0.0895 0.1981 0.2812 0.3964 0.4806 0.6423 0.7524 0.9102 1
yB 0 0.2716 0.4565 0.5550 0.6607 0.7228 0.8238 0.8846 0.9590 1
The data are plotted in Fig. 5.2.
Figure 5.2
We can assume, at the lowest concentrations of both A and B, that Henrys law will hold. The Henrys
law constants are then given by
KA =pA
xA= 15.58 kPa from the point atxA =0.0898.
KB =pB
xB= 47.03 kPa from the point atxB =0.0895.
P5.3 Vsalt =
V
b
H2O
mol1 [Problem 5.2]
=69.38(b0.070)cm3 mol1 withb b/(mol kg1).
Therefore, atb = 0.050 mol kg1, Vsalt = 1.4 cm3 mol1 .
The total volume at this molality is
V =(1001.21)+ (34.69)(0.02)2 cm3 =1001.22 cm3.
Hence, as in Problem 5.2,
V(H2O) =(1001.22 cm3) (0.050 mol)(1.4 cm3mol1)
55.49 mol= 18.04 cm2 mol1 .
Question. What meaning can be ascribed to a negative partial molar volume?
P5.5 Let E denote ethanol and W denote water; then
V =nEVE + nWVW[5.3].
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104 SOLUTIONS MANUAL
For a 50 per cent mixture by mass, mE = mW, implying that
nEME =nWMW, ornW =nEME
MW.
Hence,V =nEVE+nEMEVW
MW
which solves tonE =V
VE +MEVW
MW
, nW =MEV
VEMW + MEVW.
Furthermore,xE =nE
nE + nW=
1
1+ME
MW
.
SinceME =46.07 g mol1andMW =18.02 g mol1, MEMW
=2.557. Therefore
xE =0.2811, xW =1xE =0.7189.
At this composition
VE =56.0 cm3mol1, VW =17.5 cm
3 mol1[Fig.5.1 of the text].
Therefore,nE =100 cm3
(56.0 cm3 mol1)+ (2.557)(17.5 cm3 mol1)=0.993 mol,
nW = (2.557)(0.993 mol)=2.54mol.
The fact that these amounts correspond to a mixture containing 50 per cent by mass of both components
is easily checked as follows:
mE = nEME =(0.993 mol)(46.07 g mol1)=45.7g ethanol,
mW = nWMW = (2.54 mol)(18.02 g mol1)=45.7g water.
At 20C the densities of ethanol and water are,
E =0.789 g cm3,W =0.997 g cm
3. Hence,
VE =mE
E=
45.7 g
0.789 g cm3 = 57.9 cm3 of ethanol,
VW = mWW
= 45.7 g0.997 g cm3
= 45.8 cm3 of water.
The change in volume upon adding a small amount of ethanol can be approximated by
V =
dV
VEdnE VEnE
where we have assumed that both VEand VWare constant over this small range ofnE. Hence
V (56.0 cm3mol1)
(1.00 cm3)(0.789 g cm3)
(46.07 g mol1)
= +0.96 cm3 .
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SIMPLE MIXTURES 105
P5.7bB =
T
Kf
=0.0703K
1.86 K/(mol kg1
)
=0.0378 molkg1.
Since the solution molality is nominally 0.0096 mol kg1 in Th(NO3)4, each formula unit supplies0.0378
0.0096 4 ions . (More careful data, as described in the original reference gives 5 to 6.)
P5.9 Thedata areplotted in Figure5.3. Theregions where thevaporpressurecurves show approximate straight
lines are denoted R for Raoult and H for Henry. A and B denote acetic acid and benzene respectively.
H
A
R H
R
B
40.0
26.7
13.3
kPa
00 0.2 0.4 0.6
xA
0.8 1.0
Henry
Henry
Raoult
Raoult
Extrapolate
for KB
Figure 5.3
As in Problem 5.8, we need to formA =pA
xApA
and B =pB
xBpB
for theRaoults lawactivitycoefficients
and B =pB
xBKfor the activity coefficient of benzene on a Henrys law basis, with Kdetermined by
extrapolation. We use pA =7.3 kPa, pB =35.2 kPa, and K
B =80.0 kPa to draw up the following
table.
xA 0 0.2 0.4 0.6 0.8 1.0
pA/kPa 0 2.7 4.0 5.1 6.7 7.3
pB/kPa 35.2 30.4 25.3 20.0 12.4 0
aA(R) 0 0.36 0.55 0.69 0.91 1.00[pA/pA]
aB(R) 1.00 0.86 0.72 0.57 0.35 0[pB/pB]
A(R) 1.82 1.36 1.15 1.14 1.00[pA/xApA]
B(R) 1.00 1.08 1.20 1.42 1.76 [pB/xBpB]
aB(H) 0.44 0.38 0.32 0.25 0.16 0[pB/KB]
B(H) 0.44 0.48 0.53 0.63 0.78 1.00[pB/xBKB]
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106 SOLUTIONS MANUAL
GEis defined as [Section 5.4]
GE = mixG(actual) mixG(ideal)= nRT(xAln aA + xBln aB) nRT(xAlnxA+ xBlnxB)
and, witha = x,
GE = nRT(xAln A+ xAln B).
Forn =1, we can draw up the following table from the information above and RT =2.69 kJ mol1.
xA 0 0.2 0.4 0.6 0.8 1.0
xAln A 0 0.12 0.12 0.08 0.10 0
xBlnB 0 0.06 0.11 0.14 0.11 0GE/(kJ mol1) 0 0.48 0.62 0.59 0.56 0
P5.11 (a) The volume of an ideal mixture is
Videal =n1Vm,1 + n2Vm,2
so the volume of a real mixture is
V = Videal+ VE.
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.
VE = (n1+ n2)VEm =
n1n2
n1+ n2
a0+
a1(n1 n2)
n1+ n2
soV = n1Vm,1+ n2Vm,2 +n1n2
n1+ n2
a0+
a1(n1 n2)
n1+ n2
.
The partial molar volume of propionic acid is
V1 =
V
n1
p,T,n2
= Vm,1+a0n
22
(n1+ n2)2
+a1(3n1 n2)n
22
(n1+ n2)3
,
V1 = Vm,1 + a0x22 + a1(3x1 x2)x
22 .
That of oxane is
V2 =Vm,2 + a0x21 +a1(x13x2)x
21 .
(b) We need the molar volumes of the pure liquids,
Vm,1 =M1
1=
74.08 g mol1
0.97174 g cm3 =76.23 cm3mol1
andVm,2 =86.13 g mol1
0.86398 g cm3 =99.69 cm3mol1.
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SIMPLE MIXTURES 107
In an equimolar mixture, the partial molar volume of propionic acid is
V1 =76.23+(2.4697)(0.500)2 +(0.0608) [3(0.5)0.5] (0.5)2cm3mol1
= 75.63 cm3 mol1
and that of oxane is
V2 =99.69+ (2.4697)(0.500)2 +(0.0608) [0.53(0.5)] (0.5)2 cm3 mol1
= 99.06 cm3 mol1 .
P5.13 Henrys law constant is the slope of a plot ofpBversusxBin the limit of zero xB(Fig. 5.4). The partial
pressures of CO2are almost but not quite equal to the total pressures reported above.
pCO2 =pyCO2 =p(1ycyc).
Linear regression of the low-pressure points gives KH = 371 bar .
0.0 0.1 0.2 0.30
20
40
60
80
Figure 5.4
The activity of a solute is
aB =pB
KH=xBB
so the activity coefficient is
B =pB
xBKH=
yBp
xBKH
where the last equality applies Daltons lawof partial pressures to thevapor phase. A spreadsheet applied
this equation to the above data to yield
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108 SOLUTIONS MANUAL
p/bar ycyc xcyc CO2
10.0 0.0267 0.9741 1.01
20.0 0.0149 0.9464 0.99
30.0 0.0112 0.9204 1.00
40.0 0.00947 0.892 0.99
60.0 0.00835 0.836 0.98
80.0 0.00921 0.773 0.94
P5.15 GE = RTx(1 x){0.48570.1077(2x1)+0.0191(2x1)2}
withx= 0.25 gives GE =0.1021RT. Therefore, since
mixG(actual)= mixG(ideal)+ nGE
,
mixG= nRT(xAlnxA+ xBlnxB)+nGE = nRT(0.25 ln 0.25+0.75 ln 0.75)+ nGE
= 0.562nRT+0.1021nRT = 0.460nRT.
Sincen =4 mol andRT = (8.314 JK1 mol1)(303.15 K) =2.52 kJ mol1,
mixG= (0.460)(4mol)(2.52 kJ mol1)= 4.6 kJ .
Solutions to theoretical problems
P5.17 A =GnA
nB
[5.4] =oA+ nA
(nGE)
nB
[oAis ideal value = A+ RTlnxA],
nGE
nA
nB
= GE +n
GE
nA
nB
= GE +n
xA
nA
B
GE
xA
B
= GE +nxB
n
GE
xA
B
[xA/nA =xB/n]
= gRTxA(1xA)+ (1 xA)gRT(12xA)
= gRT(1xA)2 = gRTx2B.
Therefore, A =A+ RTlnxA+ gRTx2B .
P5.19 nAdVA + nBdVB =0 [Example 5.1].
HencenA
nBdVA = dVB.
Therefore, by integration,
VB(xA) VB(0) =
VA(xA)VA(0)
nA
nBdVA =
VA(xA)VA(0)
xAdVA
1 xA[nA =xAn, nB =xBn].
Therefore,VB(xA,xB) = VB(0, 1)
VA(xA)
VA(0)
xAdVA
1 xA.
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SIMPLE MIXTURES 109
We should now plot xA/(1 xA)against VA and estimate the integral. For the present purpose we
integrate up toVA(0.5, 0.5) = 74.06 cm3
mol1
[Fig. 5.5], and use the data to construct the followingtable.
VA(cm3 mol1) 74.11 73.96 73.50 72.74
xA 0.60 0.40 0.20 0
xA/(1 xA) 1.50 0.67 0.25 0
The points are plotted in Fig. 5.5, and the area required is 0.30. Hence,
1.5
1.0
0.5
072 73 74 75
xA
/(1xA
)
VA,m/cm3 mol1
( )
Figure 5.5
V(CHCl3; 0.5, 0.5) =80.66 cm3 mol1 0.30 cm3 mol1
= 80.36 cm3 mol1 .
P5.21 = ln aA
r.
Therefore, d = 1
rd ln aA+
1
r2ln aAdr,
d ln aA =1
rln aAdr rd.
From the GibbsDuhem equation, xAdA = xBdB = 0, which implies that (since = +
RTln a, dA = RTd ln aA, dB =RTd ln aB)
d ln aB = xA
xBd ln aA =
d ln aA
r
= 1
r2ln aAdr=d[from(b)] =
1
rdr=dr=d[from(a)]
= d ln r+d.
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110 SOLUTIONS MANUAL
Subtract d ln rfrom both sides, to obtain
d lnaB
r=( 1) d ln r+d =
(1)
rdr+d.
Then, by integration and noting that lnaB
r
r=0
=lnBxB
r
r=0
=ln (B)r=0 =ln 1= 0,
lnaB
r= (0)=
r0
1
r
dr .
P5.23 A(s) A(l).
A(s) = A(l)+RTln aA
andfusG = A(l)
A(s)= RT ln aA.
Hence, ln aA =fusG
RT.
d ln aA
dT=
1
R
d
dT
fusG
T
=
fusH
RT2 [GibbsHelmholtz eqn].
ForT = Tf
T, dT = dTand
d ln aA
dT=
fusH
RT2
fusH
RT2f
.
ButKf =RT2
fMA
fusH.
Therefore,
d ln aA
dT=
MA
Kfand d lnaA =
MAdT
Kf.
According to the GibbsDuhem equation
nAdA+ nBdB =0
which implies that
nAd ln aA + nBd ln aB =0[= +RTln a]
and hence that d ln aA = nB
nAd ln aB.
Hence,d ln aB
dT=
nAMA
nBKf=
1
bBKf[fornAMA =1kg]
We know from the GibbsDuhem equation that
xAd ln aA+ xBd ln aB =0
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SIMPLE MIXTURES 111
and hence that d ln aA = xB
xA
d ln aB.
Therefore ln aA = xB
xAd ln aB
The osmotic coefficient was defined in Problem 5.21 as
= 1
rln aA =
xA
xBln aA.
Therefore,
=xA
xB
xB
xAd ln aB =
1
b
b0
b d ln aB =1
b
b0
b d ln b =1
b
b0
b d ln b+1
b
b0
b d ln
=1+1
b
b
0
b d ln .
From the DebyeHckel limiting law,
ln = Ab1/2 [A =2.303A].
Hence, d ln = 1
2Ab1/2dband so
=1+1
b
1
2A b
0
b1/2 db= 11
2
A
b
2
3b3/2 = 1
1
3A 1/2 .
COMMENT. For the depression of the freezing point in a 1,1-electrolyte
ln aA =fusG
RT+
fusG
RT
and hencer =fusH
R
1
T
1
T
.
Therefore, =fusHxA
RxB
1
T
1
T
=
fusHxA
RxB
T T
TT
fusHxAT
RxBT2
fusHT
RbBT2MA
where =2. Therefore, sinceKf =MRT2
fusH
,
=T
2bBKf.
Solutions to applications
P5.25 In this case it is convenient to rewrite the Henrys law expression as
mass of N2 = pN2 mass of H2OKN2 .
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112 SOLUTIONS MANUAL
(1) AtpN2 =0.784.0 atm=3.1 atm,
mass of N2 =3.1atm100 g H2O0.18g N2/(g H2O atm) = 56g N2 .
(2) AtpN2 =0.78 atm, mass of N2 = 14 g N2 .
(3) In fatty tissue the increase in N2concentration from 1 atm to 4 atm is
4(5614)g N2 = 1.7102g N2 .
P5.27 (a) i = 1 only, N1 =4, K1 =1.0107 dm3 mol1,
[A]=
410dm3 mol1
1+10dm3mol1 [A].
The plot is shown in Fig. 5.6(a).
Figure 5.6(a)
(b) i = 1; N1 =4, N2 =2; K1 =1.0105 dm3 mol1 =0.10 dm3 mol1,
K2 =2.0106 dm3 mol1 =2.0 dm3 mol1.
[A]=
40.10 dm3mol1
1+0.10 dm3 mol1 [A]+
22.0 dm3 mol1
1+2.0 dm3 mol1 [A].
The plot is shown in Fig. 5.6(b).
P5.28 By the vant Hoff equation [5.40],
= [B]RT =cRT
M.
Division by the standard acceleration of free fall,g, gives
8=
c(R/g)T
M.
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SIMPLE MIXTURES 113
Figure 5.6(b)
(a) This expression may be written in the form
=cRT
M,
which has the same form as the vant Hoff equation, but the unit of osmotic pressure () is now
force/area
length/time2 =
(mass length)/(area time2)
length/time2 =
mass
area.
This ratio can be specified in g cm2. Likewise, the constant of proportionality (R) would have the
units ofR/g,
energyK 1 mol1
length/time2 =
(mass length2/time2)K1 mol1
length/time2 =mass lengthK1mol1.
This result may be specified in g cm K1 mol1 .
R =R
g=
8.31447J K1 mol1
9.80665m s2
=0.847840kg m K1 mol1
103
gkg
102
cmm
R =84784.0 g cm K1 mol1 .
In the following we will drop the primes giving
=cRT
M
and use theunits of g cm2 and theRunits g cm K1mol1.
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114 SOLUTIONS MANUAL
(b) By extrapolating the low concentration plot of/cversus c (Fig. 5.7 (a)) to c = 0 we find the
intercept 230g cm2/g cm
3. In this limit vant Hoff equation is valid so
RT
M=intercept orM =
RT
intercept,
M =(84784.0gcmK1 mol1) (298.15K)
(230gcm2)/(g cm3),
M=1.1105 g mol1 .
500
450
400
350
300
250
200
0.000 0.010 0.020 0.030 0.040
Figure 5.7(a)
(c) The plot of/cversus c for the full concentration range (Fig. 5.7(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.
(d) /c = (RT/M)(1+Bc+ Cc2).
SinceRT/Mhas been determined in part (b)by extrapolation to c =0, it is best to determine thesecond and third virial coefficients with the linear regressionfit
(/c)/(RT/M)1
c= B +Cc,
R =0.9791.
B =21.4 cm3 g1,
C =211cm6 g2,
standard deviation =2.4 cm3 g1.
standard deviation =15 cm6 g2.
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SIMPLE MIXTURES 115
7000
6000
5000
4000
3000
2000
1000
0
0.3000.2500.2000.1500.1000.0500.00
Figure 5.7(b)
(e) Using 1/4 forgand neglecting terms beyond the second power, we may write
c
1/2=
RT
M
1/2(1+
1
2Bc).
We can solve forB; theng(B)2 =C,
c
1/2RT
M
1/21=
1
2Bc.
RT/Mhas been determined above as 230 g cm 2/g cm3. We may analytically solve for B from
one of the data points, say,/c = 430 g cm2/g cm3 atc =0.033g cm3.
430gcm2/g cm3
230gcm2/g cm3
1/21=
1
2B (0.033 g cm3).
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116 SOLUTIONS MANUAL
B =2 (1.3671)
0.033 g cm3
=22.2 cm3 g1.
C = g(B)2 =0.25 (22.2 cm3 g1)2 =123 cm6 g2.
Better values ofB andC can be obtained by plotting
c
1/2/
RT
M
1/2againstc. This plot is
shown in Fig. 5.7(c). The slope is 14.03 cm3 g1. B =2 slope = 28.0 cm3 g1 . C is then
196 cm6 g2 . The intercept of this plot should theoretically 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 thatg
is roughly 1/4 as assumed.
6.0
5.0
4.0
3.0
2.0
1.0
0.0
0.00 0.05 0.10 0.15 0.20 0.25 0.30
n
Figure 5.7(c)