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Chapter 11 - Section B - Non-Numerical Solutions

11.6 Apply Eq. (11.7):

Ti (nT)

ni

P,T,nj

= Tn

ni

T,P,nj

= T Pi (n P)

ni

P,T,nj

= Pn

ni

T,P,nj

= P

11.7 (a) Let m be the mass of the solution, and define the partial molar mass by: mi

m

ni

T,P,nj

LetMk be the molar mass of species k. Then

m = k

nkMk = niMi + j

njMj (j = i)

and mni

T,P,nj

=(niMi)

ni

T,P,nj=Mi Whence, mi =Mi

(b) Define a partial specific property as: Mi

M t

m i

T,P,mj

=

M t

ni

T,P,m j

ni

mi

T,P,m j

IfMi is the molar mass of species i , ni =mi

Miand

ni

mi

T,P,m j

=1

Mi

Because constant m j implies constant nj , the initial equation may be written: Mi =Mi

Mi

11.8 By Eqs. (10.15) and (10.16), V1 = V + x2d V

d x1and V2 = V x1

d V

d x1

Because V = 1 thend V

d x1=

1

2

d

d x1whence

V1 =1

x2

2

d

d x1=

1

1

x2

d

d x1

=

1

2

x2

d

d x1

V2 =

1

+

x1

2

d

d x1 =

1

1 +

x1

d

d x1

=

1

2 + x1

d

d x1

With = a0 + a1x1 + a2x21 and

d

d x1= a1 + 2a2x1 these become:

V1 =1

2[a0 a1 + 2(a1 a2)x1 + 3a2x

21 ] and V2 =

1

2(a0 + 2a1x1 + 3a2x

21)

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11.9 For application of Eq. (11.7) all mole fractions must be eliminated from the given equation by the

relation xi = ni/n:

n M = n1M1 + n2M2 + n3M3 +n1n2n3

n2C

For M1, (n M)n1

T,P,n2,n3

= M1 + n2n3C 1n2

2n1

n3 nn1

T,P,n2,n3

Because n = n1 + n2 + n3,

n

n1

T,P,n2,n3

= 1

Whence, M1 = M1 +n2n3

n2

1 2

n1

n

C and M1 = M1 + x2x3[1 2x1]C

Similarly, M2 = M2 + x1x3[1 2x2]C and M3 = M3 + x1x2[1 2x3]C

One can readily show that application of Eq. (11.11) regenerates the original equation for M. The

infinite dilution values are given by:

Mi = Mi + xjxkC (j, k = i)

Here xj and xk are mole fractions on an i -free basis.

11.10 With the given equation and the Daltons-law requirement that P =

i pi , then:

P =RT

V

i

yiZi

For the mixture, P = Z RT/V. These two equations combine to give Z =

i yiZi .

11.11 The general principle is simple enough:

Given equations that represent partial properties Mi , MRi , or

MEi as functions of com-

position, one may combine them by the summability relation to yield a mixture property.

Application of the defining (or equivalent) equations for partial properties then regenerates

the given equations if and only if the given equations obey the Gibbs/Duhen equation.

11.12 (a) Multiply Eq. (A) of Ex. 11.4 by n (= n1 + n2) and eliminate x1 by x1 = n1/(n1 + n2):

n H = 600(n1 + n2) 180 n1 20n31

(n1 + n2)2

Form the partial derivative ofn H with respect to n1 at constant n2:

H1 = 600 180 20

3n21

(n1 + n2)2

2n31(n1 + n2)3

= 420 60

n21(n1 + n2)2

+ 40n31

(n1 + n2)3

Whence, H1 = 420 60x21 + 40x

31

Form the partial derivative ofn H with respect to n2 at constant n1:

H2 = 600 + 202 n31

(n1 + n2)3or H2 = 600 + 40x

31

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(b) In accord with Eq. (11.11),

H = x1(420 60x21 + 40x

31) + (1 x2)(600 + 40x

31)

Whence, H = 600 180x1 20x31

(c) Write Eq. (11.14) for a binary system and divide by d x1: x1d H1

d x1+ x2

d H2

d x1= 0

Differentiate the the boxed equations of part (a):

d H1

d x1= 120x1 + 120x

21 = 120x1x2 and

d H2

d x1= 120x 21

Multiply each derivative by the appropriate mole fraction and add:

120x 21x2 + 120x21x2 = 0

(d) Substitute x1 = 1 and x2 = 0 in the first derivative expression of part (c) and substitute x1 = 0

in the second derivative expression of part (c). The results are:d H1

d x1

x1=1

=

d H2

d x1

x1=0

= 0

(e)

11.13 (a) Substitute x2 = 1 x1 in the given equation for V and reduce:

V = 70 + 58x1 x21 7x

31

Apply Eqs. (11.15) and (11.16) to find expressions for V1 and V2. First,

d V

d x1= 58 2x1 21x

21

Then, V1 = 128 2x1 20x21 + 14x

31 and V2 = 70 + x

21 + 14x

31

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V = x1(128 2x1 20x21 + 14x

31) + (1 x1)(70 + x

21 + 14x

31)

Whence, V = 70 + 58x1 x21 7x

31

which is the first equation developed in part (a).

(c) Write Eq. (11.14) for a binary system and divide by d x1: x1dV1

d x1+ x2

dV2

d x1= 0


dV1

d x1= 2 40x1 + 42x

21 and

dV2

d x1= 2x1 + 42x

21


x1(2 40x1 + 42x21) + (1 x1)(2x1 + 42x

21) = 0

The validity of this equation is readily confirmed.

(d) Substitute x1 = 1 in the first derivative expression of part (c) and substitute x1 = 0 in the second

derivative expression of part (c). The results are:

dV1

d x1

x1=1

=

dV2

d x1

x1=0

= 0

(e)

11.14 By Eqs. (11.15) and (11.16):

H1 = H + x2d H

d x1and H2 = H x1

d H

d x1

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Given that: H = x1(a1 + b1x1) + x2(a2 + b2x2)

Then, after simplification,d H

d x1= a1 + 2b1x1 (a2 + 2b2x2)

Combining these equations gives after reduction:

H1 = a1 + b1x1 + x2(x1b1 x2b2) and H2 = a2 + b2x2 x1(x1b1 x2b2)

These clearly are not the same as the suggested expressions, which are therefore not correct. Note

that application of the summability equation to the derived partial-property expressions reproduces

the original equation for H. Note further that differentiation of these same expressions yields results

that satisfy the Gibbs/Duhem equation, Eq. (11.14), written:

x1d H1

d x1+ x2

d H2

d x1= 0

The suggested expresions do not obey this equation, further evidence that they cannot be valid.

11.15 Apply the following general equation of differential calculus:x

y

z

=

x

y

w

+

x

w

y

w

y

z

(n M)

ni

T,P,nj

=

(n M)

ni

T,V,nj

+

(n M)

V

T,n

V

ni

T,P,nj

Whence,

Mi = Mi + n

M

V

T,n

V

ni

T,P,nj

or Mi = Mi n

M

V

T,n

V

ni

T,P,nj

By definition,

Vi

(nV)

ni

T,P,nj

= n

V

ni

T,P,nj

+ V or n

V

ni

T,P,nj

= Vi V

Therefore, Mi = Mi + (V Vi )

M

V

T,x

11.20 Equation (11.59) demonstrates that ln i is a partial property with respect to GR/RT. Thus ln i =

G i/RT. The partial-property analogs of Eqs. (11.57) and (11.58) are:

ln i

P

T,x

=VRi

RTand

ln i

T

P,x

= HRi

RT2

The summability and Gibbs/Duhem equations take on the following forms:

GR

RT=

i

xi ln i and i

xi dln i = 0 (const T, P)

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11.26 For a pressure low enough that Z and ln are given approximately by Eqs. (3.38) and (11.36):

Z = 1 +B P

RTand ln =

B P

RT

then: ln Z 1

11.28 (a) Because Eq. (11.96) shows that ln i is a partial property with respect to GE/RT, Eqs. (11.15)

and (11.16) may be written for M GE/RT:

ln 1 =GE

RT+ x2

d(GE/RT)

d x1ln 2 =

GE

RT x1

d(GE/RT)

d x1

Substitute x2 = 1 x1 in the given equaiton for GE/RT and reduce:

GE

RT= 1.8x1 + x

21 + 0.8x

31 whence

d(GE/RT)

d x1= 1.8 + 2x1 + 2.4x

21

Then, ln 1 = 1.8 + 2x1 + 1.4x21 1.6x

31 and ln 2 = x

21 1.6x

31


GE

RT= x1 ln 1 + x2 ln 2 = x1(1.8 + 2x1 + 1.4x

21 1.6x

31) + (1 x1)(x

21 1.6x

31)

Whence,GE

RT= 1.8x1 + x

21 + 0.8x

31

which is the first equation developed in part (a).

(c) Write Eq. (11.14) for a binary system with Mi = ln i and divide by d x1:

x1dln 1

d x1+ x2

dln 2

d x1= 0


dln 1

d x1= 2 + 2.8x1 4.8x

21 and

dln 2

d x1= 2x1 4.8x

21


x1(2 + 2.8x1 4.8x21) + (1 x1)(2x1 4.8x

21) = 0

The validity of this equation is readily confirmed.

(d) Substitute x1 = 1 in the first derivative expression of part (c) and substitute x1 = 0 in the second

derivative expression of part (c). The results are:

dln 1

d x1

x1=1

=

dln 2

d x1

x1=0

= 0

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(e)

11.29 Combine definitions of the activity coefficient and the fugacity coefficients:

i fi/xi P

fi/Por i =

i

i

Note: See Eq. (14.54).

11.30 For CEP = const., the following equations are readily developed from those given in the last column

of Table 11.1 (page 415):

HE = CEP T and SE = GE

T

P,x= CEP

T

T

Working equations are then:

SE1 =HE1 G

E1

T1and SE2 = S

E1 + C

EP

T

T

HE2 = HE1 + C

EP T and G

E2 = H

E2 T2 S

E2

For T1 = 298.15, T2 = 328.15, T = 313.15 and T = 30, results for all parts of the problem are

given in the following table:

I. II. For CE

P = 0

GE1 HE1 S

E1 C

EP S

E2 H

E2 G

E2 S

E2 H

E2 G

E2

(a) 622 1920 4.354 4.2 3.951 1794 497.4 4.354 1920 491.4

(b) 1095 1595 1.677 3.3 1.993 1694 1039.9 1.677 1595 1044.7

(c) 407 984 1.935 2.7 1.677 903 352.8 1.935 984 348.9

(d) 632 208 2.817 23.0 0.614 482 683.5 2.817 208 716.5

(e) 1445 605 2.817 11.0 1.764 935 1513.7 2.817 605 1529.5

(f) 734 416 3.857 11.0 2.803 86 833.9 3.857 416 849.7

(g) 759 1465 2.368 8.0 1.602 1225 699.5 2.368 1465 688.0

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11.31 (a) Multiply the given equation by n (= n1 + n2), and convert remaining mole fractions to ratios of

mole numbers:nGE

RT= A12

n1n2

n+ A13

n1n3

n+ A23

n2n3

n

Differentiation with respect to n1 in accord with Eq. (11.96) yields [(n/n1)n2,n3 = 1]:

ln 1 = A12n2

1

n

n1

n2

+ A13n3

1

n

n1

n2

A23

n2n3

n2

= A12x2(1 x1) + A13x3(1 x1) A23x2x3

Similarly, ln 2 = A12x1(1 x2) A13x1x3 + A23x3(1 x2)

ln 3 = A12x1x2 + A13x1(1 x3) + A23x2(1 x3)

(b) Each lni is multiplied by xi , and the terms are summed. Consider the first terms on the right of

each expression for ln i . Multiplying each of these terms by the appropriate xi and adding gives:

A12(x1x2 x21x2 + x2x1 x

22x1 x1x2x3) = A12x1x2(1 x1 + 1 x2 x3)

= A12

x1x

2[2 (x

1+ x

2+ x

3)] = A

12x

1x

2

An analogous result is obtained for the second and third terms on the right, and adding them

yields the given equation for GE/RT.

(c) For infinite dilution of species 1, x1 = 0: ln 1(x1 = 0) = A12x2 + A13x3 A23x2x3

For pure species 1, x1 = 1: ln 1(x1 = 1) = 0

For infinite dilution of species 2, x2 = 0: ln 1(x2 = 0) = A13x23

For infinite dilution of species 3, x3 = 0: ln 1(x3 = 0) = A12x22

11.35 By Eq. (11.87), written with M G and with x replaced by y: GE = GR i

yi GRi

Equations (11.33) and (11.36) together give GR

i = Bii P. Then for a binary mixture:GE = B P y1B11 P y2B22 P or G

E = P(B y1B11 y2B22)

Combine this equation with the last equation on Pg. 402: GE = 12 P y1y2

From the last column of Table 11.1 (page 415): SE =

GE

T

P,x

Because 12 is a function of T only: SE =

d12

d TP y1y2

By the definition of GE, HE = GE + T SE; whence, HE =

12 T

d12

d T

P y1y2

Again from the last column of Table 11.1: CEP =

HE

T

P,x

This equation and the preceding one lead directly to: CEP = Td212

d T2P y1y2

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11.41 From Eq. (11.95):

(GE/RT)

T

P

=HE

RT2or

(GE/T)

T

P

=HE

T2

To an excellent approximation, write:

(GE/T)

T

P

(GE/T)

T

HE

T2mean

From the given data:(GE/T)

T=

785/323 805/298

323 298=

0.271

25= 0.01084

andHE

T2mean=

1060

3132= 0.01082

The data are evidently thermodynamically consistent.

11.42 By Eq. (11.14), the Gibbs/Duhem equation, x1d M1

d x1+ x2

d M2

d x1= 0

Given that M1 = M1 + Ax 2 and M2 = M2 + Ax 1 thend M1

d x1

= A andd M2

d x1

= A

Then x1d M1

d x1+ x2

d M2

d x1= x1A + x2A = A(x2 x1) = 0

The given expressions cannot be correct.

11.45 (a) For ME = Ax 21x22 find

ME1 = Ax 1x22(2 3x1) and

ME2 = Ax21x2(2 3x2)

Note that at both x1 = 0 (x2 = 1) and x1 = 1 (x2 = 0), ME1 =

ME2 = 0

In particular, ( ME1 ) = ( ME2 )

= 0

Although ME has the same sign over the whole composition range, both ME1 andME2 change

sign, which is unusual behavior. Find also that

d ME1d x1

= 2Ax 2(1 6x1x2) andd ME2d x1

= 2Ax 1(1 6x1x2)

The two slopes are thus of opposite sign, as required; they also change sign, which is unusual.

For x1 = 0d ME1d x1

= 2A andd ME2d x1

= 0

For x1 = 1d ME1d x1

= 0 andd ME2d x1

= 2A

(b) For ME = A sin(x1) find:

ME1 = A sin(x1) + Ax2 cos(x1) and ME2 = A sin(x1) Ax1 cos(x1)

d ME1d x1

= A2x2 sin(x1) andd ME2d x1

= A 2x1 sin(x1)

The two slopes are thus of opposite sign, as required. But note the following, which is unusual:

For x1 = 0 and x1 = 1d ME1d x1

= 0 andd ME2d x1

= 0

PLOTS OF THE FUNCTIONS ARE SHOWN ON THE FOLLOWING PAGE.

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Pb. 11.45 (a) A 10 i ..0 100 xi .00001..01 i

MEi..A xi

21 xi

2MEbar1i

..A xi 1 xi2

2 .3 xi

MEbar2i....A xi xi 1 xi 2

.3 1 xi

0 0.2 0.4 0.6 0.8 10.5

0

0.5

1

1.5

2

MEi

MEbar1i

MEbar2i

xi

Pb. 11.45 (b) MEi.A sin .p xi (pi prints as bf p)

MEbar1i.A sin .p xi

...A p 1 xi cos.p xi

MEbar2i.A sin .p xi

...A p xi cos.p xi

0 0.2 0.4 0.6 0.8 110

0

10

20

30

40

MEi

MEbar1i

MEbar2i

xi

sin

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11.46 By Eq. (11.7), Mi =

(n M)

ni

T,P,nj

= M + n

M

ni

T,P,nj

At constant T and P, d M =

k

M

xk

T,P,xj

d xk

Divide by dni with restriction to constant nj (j = i):M

ni

T,P,nj

=

k

M

xk

T,P,xj

xk

ni

nj

With xk =nk

n

xk

ni

nj

=

nk

n2(k = i)

1

n

ni

n2(k = i)

Mni

T,P,nj

= 1n

k=i

xkMxk

T,P,xj

+ 1n(1 xi)

Mxi

T,P,xj

=1

n

M

xi

T,P,xj

1

n

k

xk

M

xk

T,P,xj

Mi = M +

M

xi

T,P,xj

k

xk

M

xk

T,P,xj

For species 1 of a binary mixture (all derivatives at constant T and P):

M1 = M +

M

x1

x2

x1

M

x1

x2

x2

M

x2

x1

= M + x2

M

x1

x2

M

x2

x1

Because x1 +x2 = 1, the partial derivatives in this equation are physically unrealistic; however, they

do have mathematical significance. Because M =M(x1,x2), we can quite properly write:

d M =

M

x1

x2

d x1 +

M

x2

x1

d x2

Division by d x1 yields:

d Md x1

=Mx1

x2

+Mx2

x1

d x2d x1

=Mx1

x2

Mx2

x1

wherein the physical constraint on the mole fractions is recognized. Therefore

M1 = M + x2d M

d x1

The expression for M2 is found similarly.

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11.47 (a) Apply Eq. (11.7) to species 1: ME1 =

(n ME)

n1

n2

Multiply the given equation by n and eliminate the mole fractions in favor of mole numbers:

n M

E

= An1n2 1

n1 + Bn 2 +

1

n2 + Bn 1

ME1 = An2

1

n1 + Bn 2+

1

n2 + Bn 1

+ n1

1

(n1 + Bn2)2

B

(n2 + Bn1)2

Conversion back to mole fractions yields:

ME1 = Ax 2

1

x1 + Bx2+

1

x2 + Bx1

x1

1

(x1 + Bx2)2+

B

(x2 + Bx1)2

The first term in the first parentheses is combined with the first term in the second parentheses

and the second terms are similarly combined:

ME1 = Ax 2

1

x1 + Bx2

1

x1

x1 + B x2

+

1

x2 + Bx1

1

Bx1

x2 + Bx1

Reduction yields:

ME1 = Ax22

B

(x1 + B x2)2+

1

(x2 + B x1)2

Similarly,

ME2 = Ax21

1

(x1 + B x2)2+

B

(x2 + B x1)2

(b) The excess partial properties should obey the Gibbs/Duhem equation, Eq. (11.14), when written

for excess properties in a binary system at constant T and P:

x1d ME1d x1

+ x2d ME2d x1

= 0

If the answers to part (a) are mathematically correct, this is inevitable, because they were derived

from a proper expression for ME. Furthermore, for each partial property MEi , its value and

derivative with respect to xi become zero at xi = 1.

(c) ( ME1 ) = A

1

B+ 1 (

ME2 ) = A1 +

1

B11.48 By Eqs. (11.15) and (11.16), written for excess properties, find:

d ME1d x1

= x2d2ME

d x 21

d ME2d x1

= x1d2ME

d x 21

At x1 = 1, d ME1 /d x1 = 0, and by continuity can only increase or decrease for x1 < 1. Therefore the

sign ofd ME1 /d x1 is the same as the sign ofd2ME/d x 21 . Similarly, at x1 = 0, d

ME2 /d x1 = 0, and by

the same argument the sign ofd ME2 /d x1 is of opposite sign as the sign ofd2ME/d x 21 .

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11.49 The claim is not in general valid.

1

V

V

T

P

Vid =

i

xi Vi

id =1

i

xi Vi

i

xiV

iT

P

=1

i

xi Vi

i

xi Vii

The claim is valid only if all the Vi are equal.

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Chapter11 B