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Chapter 12 - Section B - Non-Numerical Solutions

12.2 Equation (12.1) may be written: yi P = xi i P sati .

Summing for i = 1, 2 gives: P = x1

1P sat

1+ x

2

2P sat

2.

Differentiate at constant T :d Pdx1

= P sat1 x1d 1d x1

+ 1 + P sat2 x2d 2dx1

2

Apply this equation to the limiting conditions:

For x1 = 0 : x2 = 1 1 = 1 2 = 1d 2dx1

= 0

For x1 = 1 : x2 = 0 1 = 1 2 = 2d 1dx1

= 0

Then,d Pdx1 x1 = 0

= P sat1 1 P sat2 ord Pdx1 x1 = 0

+ P sat2 = P sat1 1

d Pdx1 x1= 1

= P sat1 Psat

2

2 ord Pdx1 x1= 1

P sat1 = Psat

2

2

Since both P sati and

i are always positive denite, it follows that:

d Pdx1 x1 = 0

P sat2 andd Pdx1 x1 = 1

P sat1

12.4 By Eqs. (12.15), ln 1 = Ax22 and ln 2 = Ax21

Therefore, ln 1 2

= A( x22 x21 ) = A( x2 x1) = A(1 2 x1)

By Eq. (12.1), 1 2

= y1 x2 P sat2 y2 x1 P sat1

= y1/ x1 y2/ x2

P sat2P sat1

= 12 r

Whence, ln ( 12 r ) = A(1 2 x1)

If an azeotrope exists, 12 = 1 at 0 x az1 1. At this value of x1 , ln r = A(1 2 xaz1 )

The quantity A(1 2 x1) is linear in x1 , and there are two possible relationships, depending on thesign of A. An azeotrope exhists whenever | A| | ln r | . NO azeotrope can exist when | A| < | ln r | .

12.5 Perhaps the easiest way to proceed here is to note that an extremum in ln 1 is accompanied by theopposite extremum in ln 2 . Thus the difference ln 1 ln 2 is also an extremum, and Eq. (12.8)becomes useful:

ln 1 ln 2 = ln 1 2

=d (G E / RT

dx1

Thus, given an expression for G E / RT = g( x1) , we locate an extremum through:

d 2(G E / RT )

dx 21=

d ln( 1 / 2)dx1

= 0

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For the van Laar equation , write Eq. (12.16), omitting the primes ( ):

G E

RT = A12 A21

x1 x2 A

where A A12 x1 + A21 x2

Moreover,d Adx1 = A12 A21 and

d 2 A

d x 21 = 0

Then,d (G E / RT )

dx1= A12 A21

x2 x1 A

x1 x2 A2

d Adx1

d 2(G E / RT )

dx 21= A12 A21

2 A

x2 x1

A2d Adx1

x1 x2 A2

d 2 A

dx 21

d Adx1

2 x1 x2

A3d Adx1

+ x2 x1

A2

= A12 A21 2 A

2( x2 x1)

A2d Adx1

+2 x1 x2

A3d Adx1

2

=2 A12 A21

A3 A2 ( x2 x1) A

d Adx1

+ x1 x2d Adx1

2

=2 A12 A21

A3 A + x2

d Adx1

x1d Ad x1

A

This equation has a zero value if either A12 or A21 is zero. However, this makes G E / RT everywherezero, and no extremum is possible. If either quantity in parentheses is zero, substitution for A andd A/ dx1 reduces the expression to A12 = 0 or A21 = 0, again making G E / RT everywhere zero. Weconclude that no values of the parameters exist that provide for an extremum in ln ( 1 / 2) .

The Margules equation is given by Eq. (12.9b), here written:

G E

RT = Ax1 x2 where A = A21 x1 + A12 x2

d Adx1

= A21 A12d 2 A

dx 21= 0

Then,d (G E / RT )

d x1= A( x2 x1) + x1 x2

d Adx1

d 2(G E / RT )

dx 21= 2 A + ( x2 x1)

d Adx1

+ ( x2 x1)d Adx1

+ x1 x2d 2 A

dx 21

= 2 A + 2( x2 x1)d A

dx1= 2 ( x1 x2)

d A

dx1 A

This equation has a zero value when the quantity in square brackets is zero. Then:

( x2 x1)d Adx1

A = ( x2 x1)( A21 A12 ) A21 x1 A12 x2 = A21 x2 + A12 x1 2( A21 x1 + A12 x2) = 0

Substituting x2 = 1 x1 and solving for x1 yields:

x1 = A21 2 A12

3( A21 A12 )or x1 =

(r 2)3(r 1)

r A21 A12

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When r = 2, x1 = 0, and the extrema in ln 1 and ln 2 occur at the left edge of a diagram suchas those of Fig. 12.9. For values of r > 2, the extrema shift to the right, reaching a limiting value forr = at x1 = 1/ 3. For positive values of the parameters, in all of these cases A21 > A12 , and theintercepts of the ln 2 curves at x1 = 1 are larger than the intercepts of the ln 1 curves at x1 = 0.

When r = 1/ 2, x1 = 1, and the extrema in ln 1 and ln 2 occur at the right edge of a diagramsuch as those of Fig. 12.9. For values of r < 1/ 2, the extrema shift to the left, reaching a limitingvalue for r = 0 at x1 = 2/ 3. For positive values of the parameters, in all of these cases A21 < A12 ,and the intercepts of the ln 1 curves at x1 = 0 are larger than the intercepts of the ln 2 curves at x1 = 1.

No extrema exist for values of r between 1/2 and 2.

12.7 Equations (11.15) and (11.16) here become:

ln 1 =G E

RT + x2

d (G E / RT )d x1

and ln 2 =G E

RT x1

d (G E / RT )d x1

(a ) For simplicity of notation, omit the primes that appear on the parameters in Eqs. (12.16) and(12.17), and write Eq. (12.16) as:

G E

RT = A12 A21

x1 x2 D

where D A12 x1 + A21 x2

Then,d (G E / RT )

dx1= A12 A21

x2 x1 D

x1 x2 D2

( A12 A21 )

and ln 1 = A12 A21 x1 x2

D+ x2

x2 x1 D

x1 x2 D2

( A12 A21 )

= A12 A21 D

x1 x2 + x22 x1 x2 x1 x22

D( A12 A21 )

= A12 A21 x22

D2( D A12 x1 + A21 x1) =

A12 A21 x22 D2

( A21 x2 + A21 x1)

= A12 A221 x

22

D2= A12

A21 x2 D

2

= A12D

A21 x2

2

= A12 A12 x1 + A21 x2

A21 x2

2

ln 1 = A12 1 + A12 x1 A21 x2

2

The equation for ln 2 is derived in analogous fashion.

(b) With the understanding that T and P are constant, ln 1 =(nG E / RT )

n1 n2and Eq. (12.16) may be written:

nG E

RT =

A12 A21 n 1n 2n D

where n D = A12 n1 + A21 n2

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Differentiation in accord with the rst equation gives:

ln 1 = A12 A21 n 21

n D

n 1(n D )2

(n D )n 1 n 2

ln 1 = A12 A21 n2n D1 n 1

n D A12 = A12 A21 x2 D

1 A12 x1 D

= A12 A21 x2

D2( D A12 x1) =

A12 A21 x2 D2

A21 x2 = A12 A221 x

22

D2

The remainder of the derivation is the same as in Part ( a ).

12.10 This behavior requires positive deviations from Raoults law over part of the composition range andnegative deviations over the remainder. Thus a plot of G E vs. x1 starts and ends with G E = 0 at x1 = 0 and x1 = 1 and shows positive values over part of the composition range and negative valuesover the remainder, with an intermediate crossing of the x1 axis. Because these deviations are usuallyquite small, the vapor pressures P sat

1and P sat

2must not be too different, otherwise the dewpoint and

bubblepoint curves cannot exhibit extrema.

12.11 Assume the Margules equation, Eq. (12.9b), applies:

G E

RT = x1 x2( A21 x1 + A12 x2) and

G E

RT (equimolar) =

18

( A12 + A21 )

But [see page 438, just below Eq. (12.10b)]: A12 = ln 1 A21 = ln

2

G E

RT (equimolar) =

18

(ln 1 + ln

2 ) orG E

RT (equimolar) =

18

ln( 1

2 )

12.24 (a ) By Eq. (12.6):G E

RT = x1 ln 1 + x2 ln 2

= x1 x22 (0.273 + 0.096 x1) + x2 x21 (0.273 0.096 x2)

= x1 x2(0.273 x2 + 0.096 x1 x2 + 0.273 x1 0.096 x1 x2)

= x1 x2(0.273 )( x1 + x2)

G E

RT = 0.273 x1 x2

(b) The preceding equation is of the form from which Eqs. (12.15) are derived. From these,

ln 1 = 0.273 x22 and ln 2 = 0.273 x21

(c) The equations of part ( b) are not the reported expressions, which therefore cannot be correct. SeeProblem 11.11.

12.25 Write Eq. (11.100) for a binary system, and divide through by d x1:

x1d ln 1

dx1+ x2

d ln 2dx1

= 0 whenced ln 2

dx1=

x1 x2

d ln 1dx1

= x1 x2

d ln 1d x2

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Integrate, recalling that ln 2 = 1 for x1 = 0:

ln 2 = ln(1) + x10 x1 x2 d ln 1dx2 dx1 = x10 x1 x2 d ln 1dx2 dx1(a ) For ln 1 = Ax

22 ,

d ln 1d x2 = 2 Ax2

Whence ln 2 = 2 A x10 x1 d x1 or ln 2 = Ax 21By Eq. (12.6),

G E

RT = Ax1 x2

(b) For ln 1 = x22 ( A + Bx2) ,

d ln 1dx2

= 2 x2( A + Bx2) + x22 B = 2 Ax2 + 3 Bx22 = 2 Ax2 + 3 Bx2(1 x1)

Whence ln 2 = 2 A x10 x1 dx1 + 3 B x10 x1 dx1 3 B x10 x21 dx1ln 2 = Ax 21 +

3 B2

x21 Bx31 or ln 2 = x

21 A +

3 B2

Bx1 = x21 A + B2

(1 + 2 x2)

Apply Eq. (12.6):G E

RT = x1 x22 ( A + Bx2) + x2 x

21 ( A +

3 B2

Bx1)

Algebraic reduction can lead to various forms of this equation; e.g.,

G E

RT = x1 x2 A +

B2

(1 + x2)

(c) For ln 1 = x22 ( A + Bx2 + Cx22 ) ,

d ln 1dx2

= 2 x2( A + Bx2 + Cx 22 ) + x22 ( B + 2Cx 2) = 2 Ax2 + 3 Bx

22 + 4Cx

32

= 2 Ax2 + 3 Bx2(1 x1) + 4Cx 2(1 x1)2

Whence ln 2 = 2 A

x1

0

x1 d x1 + 3 B

x1

0

x1(1 x1)d x1 + 4C

x1

0

x1(1 x1)2dx1

or ln 2 = (2 A + 3 B + 4C ) x10 x1 dx1 (3 B + 8C ) x10 x21 dx1 + 4C x10 x31 dx1ln 2 =

2 A + 3 B + 4C 2

x21 3 B + 8C

3 x31 + Cx

41

ln 2 = x21 A +3 B2

+ 2C B +8C 3

x1 + Cx 21

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or ln 2 = x21 A + B2

(1 + 2 x2) +C 3

(1 + 2 x2 + 3 x22 )

The result of application of Eq. (12.6) reduces to equations of various forms; e.g.:

G E

RT = x1 x2 A +

B2

(1 + x2) +C 3

(1 + x2 + x22 )

12.40 (a ) As shown on page 458, x1 =1

1 + nand H = H (1 + n )

Eliminating 1 + n gives: H = H

x1( A)

Differentiation yields:d H

d n=

1 x1

d H d n

H

x21

dx1d n

=1

x1

d H dx1

H

x21

d x1d n

where dx1d n

= 1(1 + n )2

= x21

Whence,d H

d n= H x1

d H dx1

= H E x1d H E

dx1

Comparison with Eq. (11.16) written with M H E , H E 2 = H E x1

d H E

dx1

shows thatd H

d n= H E 2

(b) By geometry, with reference to the following gure,d H

d n=

H I n

Combining this with the result of Part ( a ) gives: H E 2 = H I

n

From which, I = H n H E 2

Substitute: H = H

x1=

H E

x1and n =

x2 x1

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Whence, I = H E

x1

x2 x1

H E 2 = H E x2 H E 2

x1

However, by the summability equation, H E x2 H E 2 = x1 H E 1

Then, I = H E 1

12.41 Combine the given equation with Eq. ( A) of the preceding problem:

H = x2( A21 x1 + A12 x2)

With x2 = 1 x1 and x1 = 1/( 1 + n ) (page 458): x2 =n

1 + n

The preceding equations combine to give:

H =n

1 + nA

211 + n +

A12

n1 + n

(a ) It follows immediately from the preceding equation that: limn 0

H = 0

(b) Because n /( 1 + n ) 1 for n , it follows that: limn

H = A12

(c) Analogous to Eq. (12.10b), page 438, we write: H E 2 = x21 [ A21 + 2( A12 A21 ) x2]

Eliminate the mole fractions in favor of n :

H E 2 =1

1 + n

2

A21 + 2( A12 A21 )n

1 + n

In the limit as n 0, this reduces to A21 . From the result of Part ( a ) of the preceding problem,it follows that

limn 0

d H d n

= A21

12.42 By Eq. (12.29) with M H , H = H i xi H i . Differentiate:

H t P , x

= H t P , x

i

xi H it P , x

With H t P , x

C P , this becomes H

t P , x= C P

i xi C Pi = C P

Therefore, H H 0 d ( H ) = t t 0 C P dt H = H 0 + t t 0 C P dt 697

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12.61 (a ) From the denition of M : M E = x1 x2M ( A)

Differentiate:d M E

dx1= M ( x2 x1) + x1 x2

d M

dx1( B)

Substitution of Eqs. ( A) & ( B) into Eqs. (11.15) & (11.16), written for excess properties, yieldsthe required result.

(b) The requested plots are found in Section A.

12.63 In this application the microscopic state of a particle is its species identity , i.e., 1, 2, 3, . . . . Byassumption, this label is the only thing distinguishing one particle from another. For mixing,

S t = S t mixed S t unmixed = S

t mixed

iS t i

where the total emtropies are given by Eq. (5.42). Thus, for an unmixed species i , and for the mixedsystem of particles,

S t i = k ln i = k ln

N i ! N i ! = 0 S

t mixed = k ln

N ! N 1! N 2! N 3!

Combining the last three equations gives: S t = k ln N !

N 1! N 2! N 3!

From which:S

R=

S t

R( N / N A)=

S t

k N =

1 N

ln N !

N 1! N 2! N 3! =

1 N

(ln N ! i

ln N i !)

ln N ! N ln N N and ln N i ! N i ln N i N i

S R

1

N ( N ln N N

i N i ln N i +

i N i ) =

1 N

( N ln N i

xi N ln xi N )

= 1 N

( N ln N i

xi N ln xi i

xi N ln N ) = i

xi ln x1

12.66 Isobaric data reduction is complicated by the fact that both composition and temperature vary frompoint to point, whereas for isothermal data composition is the only signicant variable. (The effectof pressure on liquid-phase properties is assumed negligible.) Because the activity coefcients arestrong functions of both liquid composition and T , which are correlated, it is quite impossible withoutadditional information to separate the effect of composition from that of T . Moreover, the P sati valuesdepend strongly on T , and one must have accurate vapor-pressure data over a temperature range.

12.67 (a ) Written for G E , Eqs. (11.15) and (11.16) become:

G E 1 = G E + x2

dG E

dx1and G E 2 = G

E x1dG E

dx1

Divide through by RT ; dene G G E

RT ; note by Eq. (11.91) that

G E i RT

= ln i

Then ln 1 = G + x2d G

dx1and ln 2 = G x1

d G

d x1

Given:G E

x! x2 RT = A1/ k with A x1 Ak 21 + x2 A

k 12

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Whence: G = x1 x2 A1/ k andd G

d x1= x1 x2

d A1/ k

dx1+ A1/ k ( x2 x1)

d A1/ k

dx1=

1k

A(1/ k ) 1d Adx1

=1k

A1/ k

A( Ak 21 A

k 12 ) and

d G

dx1= x1 x2

A1/ k

k A( Ak 21 A

k 12 )+ A

1/ k ( x2 x1)

Finally, ln 1 = x22 A1/ k ( A

k 21 A

k 12 ) x1

k A+ 1

Similarly, ln 2 = x21 A1/ k 1

( Ak 21 Ak 12 ) x2

k A

(b) Appropriate substitition in the preceding equations of x1 = 1 and x1 = 0 yields:

ln 1 = A1/ k = ( Ak 12 )

1/ k = A12 ln 2 = A1/ k = ( Ak 21 )

1/ k = A21

(c) Let g G E

x1 x2 RT = A1/ k = ( x1 Ak 21 + x2 A

k 12 )

1/ k

If k = 1, g = x1 A21 + x2 A12 (Margules equation)

If k = 1, g = ( x1 A 121 + x2 A

112 )

1 = A21 A12

x1 A12 + x2 A21(van Laar equation)

For k = 0, , + , indeterminate forms appear, most easily resolved by working with thelogarithm:

ln g = ln( x1 Ak 21 + x2 Ak 12 )

1/ k =1k

ln x1 Ak 21 + x2 Ak 12

Apply lH opitals rule to the nal term:

d ln x1 Ak 21 + x2 Ak 12

dk =

x1 Ak 21 ln A21 + x2 Ak 12 ln A12

x1 Ak 21 + x2 Ak 12

( A)

Consider the limits of the quantity on the right as k approaches several limiting values.

For k 0, ln g x1 ln A21 + x2 ln A12 = ln A x121 + ln A

x212 and g = A

x121 A

x212

For k , Assume A12 / A21 > 1, and rewrite the right member of Eq. ( A) as

x1 ln A21 + x2( A12 / A21 )k ln A12 x

1+ x

2( A

12/ A

21)k

For k , limk

( A12 / A21 )k 0 and limk

ln g = ln A21

Whence g = A21 except at x1 = 0 where g = A12

For k + , limk

( A12 / A21 )k and limk

ln g = ln A12

Whence g = A12 except at x1 = 1 where g = A21

If A12 / A21 < 1 rewrite Eq. ( A) to display A21 / A12 .

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12.68 Assume that Eq. (12.1) is the appropriate equilibrium relation, written as

xe e P sate = xe

e Psat

e = ye P e EtOH

Because P is low, we have assumed ideal gases, and for small xe let e e . For volume fraction e in the vapor, the ideal-gas assumption provides v

e ye , and for the liquid phase, with xe small

le = xe V le

xe V le + xb V b

xe V le xb V b

xe V le

V bb blood

ThenV bV e

le

e Psat

e ve P

volume % EtOH in bloodvolume % EtOH in gas

V e P

V b e Psat

e

12.70 By Eq. (11.95), H E

RT = T

( G E / RT ) T P , x

G E

RT = x1 ln( x1 + x2 12 ) x2 ln( x2 + x1 21 ) (12.18)

( G E / RT ) T x

= x1 x2

d 12dT

x1 + x2 12

x2 x1d 21dT

x2 + x1 21

H E

RT = x1 x2T

d 12dT

x1 + x2 12+

d 21dT

x2 + x1 21

i j =V j

V iexp

a i j

RT (i = j ) (12.24)

d i jdT

=V jV i

exp a i j RT

a i j RT 2

= i ja i j

RT 2

H E = x1 x212 a 12

x1 + x2 12+ 21

a 21 x2 + x1 21

Because C E P = d H E / dT , differentiate the preceding expression and reduce to get:

C E P R

= x1 x2 x1 12 (a 12 / RT )2

( x1 + x2 12 )2+

x2 21 (a 21 / RT )2

( x2 + x1 21 )2

Because 12 and 21 must always be positive numbers, C E P must always be positive.

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