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Chapter 11 - Section B - Non-Numerical Solutions

11.6 Apply Eq. (11.7):

Ti ≡

[

∂(nT )

∂ni

]

P,T,n j

= T(

∂n∂ni

)

T,P,n j

= T Pi ≡

[

∂(n P)

∂ni

]

P,T,n j

= P(

∂n∂ni

)

T,P,n j

= P

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

(

∂m∂ni

)

T,P,n j

Let Mk be the molar mass of species k. Then

m = �k

nkMk = niMi + �j

n jM j ( j �= i)

and(

∂m∂ni

)

T,P,n j

=

[

∂(niMi )

∂ni

]

T,P,n j

= Mi Whence, mi = Mi

(b) Define a partial specific property as: Mi ≡

(

∂ M t

∂mi

)

T,P,m j

=

(

∂ M t

∂ni

)

T,P,m j

(

∂ni

∂mi

)

T,P,m j

If Mi is the molar mass of species i , ni =mi

Miand

(

∂ni

∂mi

)

T,P,m j

=1

Mi

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

Mi

11.8 By Eqs. (10.15) and (10.16), V1 = V + x2dVdx1

and V2 = V − x1dVdx1

Because V = ρ−1 thendVdx1

=−1ρ2

dρ

dx1whence

V1 =1ρ

−x2

ρ2

dρ

dx1=

1ρ

(

1 −x2

ρ

dρ

dx1

)

=1ρ2

(

ρ − x2dρ

dx1

)

V2 =1ρ

+x1

ρ2

dρ

dx1=

1ρ

(

1 +x1

ρ

dρ

dx1

)

=1ρ2

(

ρ + x1dρ

dx1

)

With ρ = a0 + a1x1 + a2x21 and

dρ

dx1= a1 + 2a2x1 these become:

V1 =1ρ2

[a0 − a1 + 2(a1 − a2)x1 + 3a2x21 ] and V2 =

1ρ2

(a0 + 2a1x1 + 3a2x21)

679PROPRIETARY MATERIAL. © 2005 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers

and educators for course preparation. If you are a student using this Manual, you are using it without permission.

11.9 For application of Eq. (11.7) all mole fractions must be eliminated from the given equation by therelation xi = ni/n:

nM = n1 M1 + n2 M2 + n3 M3 +n1n2n3

n2C

For M1,[

∂(nM)

∂n1

]

T,P,n2,n3

= M1 + n2n3C

[

1n2

−2n1

n3

(

∂n∂n1

)

T,P,n2,n3

]

Because n = n1 + n2 + n3,(

∂n∂n1

)

T,P,n2,n3

= 1

Whence, M1 = M1 +n2n3

n2

[

1 − 2n1

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 . Theinfinite dilution values are given by:

M∞i = Mi + x j xkC ( j, k �= i)

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

11.10 With the given equation and the Dalton’s-law requirement that P =∑

i pi , then:

P =RTV �

iyi Z i

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

i yi Z i .

11.11 The general principle is simple enough:

Given equations that represent partial properties Mi , M Ri , or M E

i 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 regeneratesthe 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):

nH = 600(n1 + n2) − 180 n1 − 20n3

1

(n1 + n2)2

Form the partial derivative of nH with respect to n1 at constant n2:

H1 = 600 − 180 − 20[

3n21

(n1 + n2)2−

2n31

(n1 + n2)3

]

= 420 − 60n2

1

(n1 + n2)2+ 40

n31

(n1 + n2)3

Whence, H1 = 420 − 60 x21 + 40 x3

1

Form the partial derivative of nH with respect to n2 at constant n1:

H2 = 600 + 202 n3

1

(n1 + n2)3or H2 = 600 + 40 x3

1



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

H = x1(420 − 60 x21 + 40 x3

1) + (1 − x2)(600 + 40 x31)

Whence, H = 600 − 180 x1 − 20 x31

(c) Write Eq. (11.14) for a binary system and divide by dx1: x1d H1

dx1+ x2

d H2

dx1= 0

Differentiate the the boxed equations of part (a):

d H1

dx1= −120 x1 + 120 x2

1 = −120 x1x2 andd H2

dx1= 120 x2

1

Multiply each derivative by the appropriate mole fraction and add:

−120 x21 x2 + 120x2

1 x2 = 0

(d) Substitute x1 = 1 and x2 = 0 in the first derivative expression of part (c) and substitute x1 = 0in the second derivative expression of part (c). The results are:

(

d H1

dx1

)

x1=1=

(

d H2

dx1

)

x1=0= 0

(e)

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

V = 70 + 58 x1 − x21 − 7 x3

1

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

dVdx1

= 58 − 2 x1 − 21 x21

Then, V1 = 128 − 2 x1 − 20 x21 + 14 x3

1 and V2 = 70 + x21 + 14 x3

1




V = x1(128 − 2 x1 − 20 x21 + 14 x3

1) + (1 − x1)(70 + x21 + 14 x3

1)

Whence, V = 70 + 58 x1 − x21 − 7 x3

1

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

(c) Write Eq. (11.14) for a binary system and divide by dx1: x1dV1

dx1+ x2

dV2

dx1= 0


dV1

dx1= −2 − 40 x1 + 42 x2

1 anddV2

dx1= 2 x1 + 42 x2

1


x1(−2 − 40 x1 + 42 x21) + (1 − x1)(2 x1 + 42 x2

1) = 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 secondderivative expression of part (c). The results are:

(

dV1

dx1

)

x1=1=

(

dV2

dx1

)

x1=0= 0

(e)

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

H1 = H + x2d Hdx1

and H2 = H − x1d Hdx1



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

Then, after simplification,d Hdx1

= 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. Notethat application of the summability equation to the derived partial-property expressions reproducesthe original equation for H . Note further that differentiation of these same expressions yields resultsthat satisfy the Gibbs/Duhem equation, Eq. (11.14), written:

x1d H1

dx1+ x2

d H2

dx1= 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

[

∂(nM)

∂ni

]

T,P,n j

=

[

∂(nM)

∂ni

]

T,V,n j

+

[

∂(nM)

∂V

]

T,n

(

∂V∂ni

)

T,P,n j

Whence,

Mi = Mi + n(

∂ M∂V

)

T,n

(

∂V∂ni

)

T,P,n j

or Mi = Mi − n(

∂ M∂V

)

T,n

(

∂V∂ni

)

T,P,n j

By definition,

Vi ≡

[

∂(nV )

∂ni

]

T,P,n j

= n(

∂V∂ni

)

T,P,n j

+ V or n(

∂V∂ni

)

T,P,n j

= Vi − V

Therefore, Mi = Mi + (V − Vi )

(

∂ M∂V

)

T,x

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

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

(

∂ ln φi

∂ P

)

T,x

=V R

i

RTand

(

∂ ln φi

∂T

)

P,x

= −H R

i

RT 2

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

G R

RT= �

ixi ln φi and �

ixi d ln φi = 0 (const T, P)



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

Z = 1 +B PRT

and ln φ =B PRT

then: ln φ ≈ Z − 1

11.28 (a) Because Eq. (11.96) shows that ln γi is a partial property with respect to G E/RT , Eqs. (11.15)and (11.16) may be written for M ≡ G E/RT :

ln γ1 =G E

RT+ x2

d(G E/RT )

dx1ln γ2 =

G E

RT− x1

d(G E/RT )

dx1

Substitute x2 = 1 − x1 in the given equaiton for G E/RT and reduce:

G E

RT= −1.8 x1 + x2

1 + 0.8 x31 whence

d(G E/RT )

dx1= −1.8 + 2 x1 + 2.4 x2

1

Then, ln γ1 = −1.8 + 2 x1 + 1.4 x21 − 1.6 x3

1 and ln γ2 = −x21 − 1.6 x3

1


G E

RT= x1 ln γ1 + x2 ln γ2 = x1(−1.8 + 2 x1 + 1.4 x2

1 − 1.6 x31) + (1 − x1)(−x2

1 − 1.6 x31)

Whence,G E

RT= −1.8 x1 + x2

1 + 0.8 x31

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

(c) Write Eq. (11.14) for a binary system with Mi = ln γi and divide by dx1:

x1d ln γ1

dx1+ x2

d ln γ2

dx1= 0


d ln γ1

dx1= 2 + 2.8 x1 − 4.8 x2

1 andd ln γ2

dx1= −2 x1 − 4.8 x2

1


x1(2 + 2.8 x1 − 4.8 x21) + (1 − x1)(−2 x1 − 4.8 x2

1) = 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 secondderivative expression of part (c). The results are:

(

d ln γ1

dx1

)

x1=1=

(

d ln γ2

dx1

)

x1=0= 0



(e)

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

γi ≡fi/xi Pfi/P

or γi =φi

φi

Note: See Eq. (14.54).

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

of Table 11.1 (page 415):

�H E = C EP �T and �SE = −�

(

∂G E

∂T

)

P,x= C E

P�T〈T 〉

Working equations are then:

SE1 =

H E1 − G E

1

T1and SE

2 = SE1 + C E

P�T〈T 〉

H E2 = H E

1 + C EP �T and G E

2 = H E2 − T2SE

2

For T1 = 298.15, T2 = 328.15, 〈T 〉 = 313.15 and �T = 30, results for all parts of the problem aregiven in the following table:

I. II. For C EP = 0

G E1 H E

1 SE1 C E

P SE2 H E

2 G E2 SE

2 H E2 G E

2

(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



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

nG E

RT= A12

n1n2

n+ A13

n1n3

n+ A23

n2n3

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

ln γ1 = A12n2

(

1n

−n1

n2

)

+ A13n3

(

1n

−n1

n2

)

− A23n2n3

n2

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

Similarly, ln γ2 = A12x1(1 − x2) − A13x1x3 + A23x3(1 − x2)

ln γ3 = −A12x1x2 + A13x1(1 − x3) + A23x2(1 − x3)

(b) Each ln γi is multiplied by xi , and the terms are summed. Consider the first terms on the right ofeach expression for ln γi . Multiplying each of these terms by the appropriate xi and adding gives:

A12(x1x2 − x21 x2 + x2x1 − x2

2 x1 − x1x2x3) = A12x1x2(1 − x1 + 1 − x2 − x3)

= A12x1x2[2 − (x1 + x2 + x3)] = A12x1x2

An analogous result is obtained for the second and third terms on the right, and adding themyields the given equation for G E/RT .

(c) For infinite dilution of species 1, x1 = 0: ln γ1(x1 = 0) = A12x2 + A13x3 − A23x2x3

For pure species 1, x1 = 1: ln γ1(x1 = 1) = 0

For infinite dilution of species 2, x2 = 0: ln γ1(x2 = 0) = A13x23

For infinite dilution of species 3, x3 = 0: ln γ1(x3 = 0) = A12x22

11.35 By Eq. (11.87), written with M ≡ G and with x replaced by y: G E = G R − �i

yi G Ri

Equations (11.33) and (11.36) together give G Ri = Bi i P . Then for a binary mixture:

G E = B P − y1 B11 P − y2 B22 P or G E = P(B − y1 B11 − y2 B22)

Combine this equation with the last equation on Pg. 402: G E = δ12 Py1 y2

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

(

∂G E

∂T

)

P,x

Because δ12 is a function of T only: SE = −dδ12

dTPy1 y2

By the definition of G E , H E = G E + T SE ; whence, H E =

(

δ12 − Tdδ12

dT

)

Py1 y2

Again from the last column of Table 11.1: C EP =

(

∂ H E

∂T

)

P,x

This equation and the preceding one lead directly to: C EP = −T

d2δ12

dT 2Py1 y2



11.41 From Eq. (11.95):(

∂(G E/RT )

∂T

)

P=

−H E

RT 2or

(

∂(G E/T )

∂T

)

P=

−H E

T 2

To an excellent approximation, write:(

∂(G E/T )

∂T

)

P≈

�(G E/T )

�T≈

−H E

T 2mean

From the given data:�(G E/T )

�T=

785/323 − 805/298323 − 298

=−0.271

25= −0.01084

and−H E

T 2mean

=−10603132

= −0.01082

The data are evidently thermodynamically consistent.

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

dx1+ x2

d M2

dx1= 0

Given that M1 = M1 + Ax2 and M2 = M2 + Ax1 thend M1

dx1= −A and

d M2

dx1= A

Then x1d M1

dx1+ x2

d M2

dx1= −x1 A + x2 A = A(x2 − x1) �= 0

The given expressions cannot be correct.

11.45 (a) For M E = Ax21 x2

2 find M E1 = Ax1x2

2(2 − 3x1) and M E2 = Ax2

1 x2(2 − 3x2)

Note that at both x1 = 0 (x2 = 1) and x1 = 1 (x2 = 0), M E1 = M E

2 = 0

In particular, (M E1 )∞ = (M E

2 )∞ = 0

Although M E has the same sign over the whole composition range, both M E1 and M E

2 changesign, which is unusual behavior. Find also that

d M E1

dx1= 2Ax2(1 − 6x1x2) and

d M E2

dx1= −2Ax1(1 − 6x1x2)

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

For x1 = 0d M E

1

dx1= 2A and

d M E2

dx1= 0

For x1 = 1d M E

1

dx1= 0 and

d M E2

dx1= −2A

(b) For M E = A sin(πx1) find:

M E1 = A sin(πx1) + Aπx2 cos(πx1) and M E

2 = A sin(πx1) − Aπx1 cos(πx1)

d M E1

dx1= −Aπ2x2 sin(πx1) and

d M E2

dx1= Aπ2x1 sin(πx1)

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

For x1 = 0 and x1 = 1d M E

1

dx1= 0 and

d M E2

dx1= 0

PLOTS OF THE FUNCTIONS ARE SHOWN ON THE FOLLOWING PAGE.



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

MEi..A xi

2 1 xi2 MEbar1i

..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

687A

PROPRIETARY MATERIAL. © 2005 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachersand educators for course preparation. If you are a student using this Manual, you are using it without permission.

11.46 By Eq. (11.7), Mi =

[

∂(nM)

∂ni

]

T,P,n j

= M + n(

∂ M∂ni

)

T,P,n j

At constant T and P , d M =∑

k

(

∂ M∂xk

)

T,P,x j

dxk

Divide by dni with restriction to constant n j ( j �= i):

(

∂ M∂ni

)

T,P,n j

=∑

k

(

∂ M∂xk

)

T,P,x j

(

∂xk

∂ni

)

n j

With xk =nk

n

(

∂xk

∂ni

)

n j

=

−nk

n2(k �= i)

1n

−ni

n2(k = i)

(

∂ M∂ni

)

T,P,n j

= −1n

∑

k �=i

xk

(

∂ M∂xk

)

T,P,x j

+1n(1 − xi )

(

∂ M∂xi

)

T,P,x j

=1n

(

∂ M∂xi

)

T,P,x j

−1n

∑

k

xk

(

∂ M∂xk

)

T,P,x j

Mi = M +

(

∂ M∂xi

)

T,P,x j

−∑

k

xk

(

∂ M∂xk

)

T,P,x j

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

d M =

(

∂ M∂x1

)

x2

dx1 +

(

∂ M∂x2

)

x1

dx2

Division by dx1 yields:

d Mdx1

=

(

∂ M∂x1

)

x2

+

(

∂ M∂x2

)

x1

dx2

dx1=

(

∂ M∂x1

)

x2

−

(

∂ M∂x2

)

x1

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

M1 = M + x2d Mdx1

The expression for M2 is found similarly.



11.47 (a) Apply Eq. (11.7) to species 1: M E1 =

[

∂(nM E)

∂n1

]

n2

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

nM E= An1n2

(

1n1 + Bn2

+1

n2 + Bn1

)

M E1 = An2

{(

1n1 + Bn2

+1

n2 + Bn1

)

+ n1

(

−1(n1 + Bn2)2

−B

(n2 + Bn1)2

)}

Conversion back to mole fractions yields:

M E1 = Ax2

{(

1x1 + 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 parenthesesand the second terms are similarly combined:

M E1 = Ax2

{

1x1 + Bx2

(

1 −x1

x1 + Bx2

)

+1

x2 + Bx1

(

1 −Bx1

x2 + Bx1

)}

Reduction yields:

M E1 = Ax2

2

[

B(x1 + Bx2)2

+1

(x2 + Bx1)2

]

Similarly,

M E2 = Ax2

1

[

1(x1 + Bx2)2

+B

(x2 + Bx1)2

]

(b) The excess partial properties should obey the Gibbs/Duhem equation, Eq. (11.14), when writtenfor excess properties in a binary system at constant T and P:

x1d M E

1

dx1+ x2

d M E2

dx1= 0

If the answers to part (a) are mathematically correct, this is inevitable, because they were derivedfrom a proper expression for M E . Furthermore, for each partial property M E

i , its value andderivative with respect to xi become zero at xi = 1.

(c) (M E1 )∞ = A

(

1B

+ 1)

(M E2 )∞

= A(

1 +1B

)

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

d M E1

dx1= x2

d2 M E

dx21

d M E2

dx1= −x1

d2 M E

dx21

At x1 = 1, d M E1 /dx1 = 0, and by continuity can only increase or decrease for x1 < 1. Therefore the

sign of d M E1 /dx1 is the same as the sign of d2 M E/dx2

1 . Similarly, at x1 = 0, d M E2 /dx1 = 0, and by

the same argument the sign of d M E2 /dx1 is of opposite sign as the sign of d2 M E/dx2

1 .



11.49 The claim is not in general valid.

β ≡1V

(

∂V∂T

)

PV id =

∑

i

xi Vi

β id =1

�i

xi Vi

∑

i

xi

(

∂Vi

∂T

)

P=

1

�i

xi Vi

∑

i

xi Viβi

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



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