Chapter 12 - Section B - Non-Numerical Solutions

12.2 Equation (12.1) may be written: yi P = xi γi Pi sat .

Summing for i = 1, 2 gives: P = x1 γ1 P1sat + x2 γ2 P2sat .

dP dγ1 dγ2



sat sat
Differentiate at constant T : = P1 x1 + γ1 + P2 x2 − γ2
d x1 d x1 d x1

Apply this equation to the limiting conditions:

dγ2
For x1 = 0 : x2 = 1 γ1 = γ1∞ γ2 = 1 =0
d x1
dγ1
For x1 = 1 : x2 = 0 γ1 = 1 γ2 = γ2∞ =0
d x1
Then,
dP dP



= P1sat γ1∞ − P2sat or + P2sat = P1sat γ1∞
d x1 x1 =0 d x1 x1 =0
dP dP



= P1sat − P2sat γ2∞ or − P1sat = −P2sat γ2∞
d x1 x1 =1 d x1 x1 =1
Since both Pi sat and γi∞ are always positive definite, it follows that:

dP dP



≥ −P2sat and ≤ P1sat
d x1 x1 =0 d x1 x1 =1

12.4 By Eqs. (12.15), ln γ1 = Ax22 and ln γ2 = Ax12

γ1
Therefore, ln = A(x22 − x12 ) = A(x2 − x1 ) = A(1 − 2x1 )
γ2

y1 x2 P2sat

sat 
y1 /x1 P2


γ1
By Eq. (12.1), = = = α12 r
γ2 y2 x1 P1sat y2 /x2 P1sat

Whence, ln(α12 r ) = A(1 − 2x1 )

If an azeotrope exists, α12 = 1 at 0 ≤ x1az ≤ 1. At this value of x1 , ln r = A(1 − 2x1az )

The quantity A(1 − 2x1 ) is linear in x1 , and there are two possible relationships, depending on the
sign 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 γ1 is accompanied by the
opposite extremum in ln γ2 . Thus the difference ln γ1 − ln γ2 is also an extremum, and Eq. (12.8)
becomes useful:
γ1 d(G E/RT
ln γ1 − ln γ2 = ln =
γ2 d x1
Thus, given an expression for G E/RT = g(x1 ), we locate an extremum through:
d 2 (G E/RT ) d ln(γ1 /γ2 )
2
= =0
d x1 d x1

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

GE x1 x2
= A12 A21 where A ≡ A12 x1 + A21 x2
RT A
dA d 2A
Moreover, = A12 − A21 and =0
d x1 d x12

d(G E/RT ) x2 − x1 x1 x2 d A


Then, = A12 A21 − 2
d x1 A A d x1

d 2 (G E/RT ) 2 x2 − x1 d A x1 x2 d 2A dA 2x1 x2 d A x2 − x1


= A12 A21 − − − 2 − − +
d x12 A A2 d x 1 A d x12 d x1 A3 d x 1 A2

2x1 x2 d A 2
 
2 2(x2 − x1 ) d A


= A12 A21 − − +
A A2 d x1 A3 d x1
 2 
2A12 A21 d A d A


= −A2 − (x2 − x1 )A + x1 x2
A3 d x1 d x1

2A12 A21 dA dA



= A + x2 x1 −A
A3 d x1 d x1

This equation has a zero value if either A12 or A21 is zero. However, this makes G E/RT everywhere
zero, and no extremum is possible. If either quantity in parentheses is zero, substitution for A and
d A/d x1 reduces the expression to A12 = 0 or A21 = 0, again making G E/RT everywhere zero. We
conclude that no values of the parameters exist that provide for an extremum in ln(γ1 /γ2 ).
The Margules equation is given by Eq. (12.9b), here written:

GE dA d 2A
= Ax1 x2 where A = A21 x1 + A12 x2 = A21 − A12 =0
RT d x1 d x12

d(G E/RT ) dA
Then, = A(x2 − x1 ) + x1 x2
d x1 d x1

d 2 (G E/RT ) dA dA d 2A
= −2A + (x 2 − x 1 ) + (x 2 − x 1 ) + x x
1 2
d x12 d x1 d x1 d x12
dA dA
 
= −2A + 2(x2 − x1 ) = 2 (x1 − x2 ) −A
d x1 d x1

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

dA
(x2 − x1 ) − A = (x2 − x1 )(A21 − A12 ) − A21 x1 − A12 x2 = A21 x2 + A12 x1 − 2(A21 x1 + A12 x2 ) = 0
d x1
Substituting x2 = 1 − x1 and solving for x1 yields:

A21 − 2A12 (r − 2) A21

x1 = or x1 = r≡
3(A21 − A12 ) 3(r − 1) A12

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 When r = 2, x1 = 0, and the extrema in ln γ1 and ln γ2 occur at the left edge of a diagram such
as those of Fig. 12.9. For values of r > 2, the extrema shift to the right, reaching a limiting value for
r = ∞ at x1 = 1/3. For positive values of the parameters, in all of these cases A21 > A12 , and the
intercepts of the ln γ2 curves at x1 = 1 are larger than the intercepts of the ln γ1 curves at x1 = 0.
When r = 1/2, x1 = 1, and the extrema in ln γ1 and ln γ2 occur at the right edge of a diagram
such as those of Fig. 12.9. For values of r < 1/2, the extrema shift to the left, reaching a limiting
value 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 γ1 curves at x1 = 0 are larger than the intercepts of the ln γ2 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:

GE d(G E/RT ) GE d(G E/RT )

ln γ1 = + x2 and ln γ2 = − x1
RT d x1 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:

GE x1 x2
= A12 A21 where D ≡ A12 x1 + A21 x2
RT D

d(G E/RT ) x2 − x1 x1 x2
 
Then, = A12 A21 − 2 (A12 − A21 )
d x1 D D
x1 x2 x2 − x1 x1 x2


and ln γ1 = A12 A21 + x2 − 2 (A12 − A21 )
D D D
A12 A21 x1 x22
 
2
= x1 x2 + x2 − x1 x2 − (A12 − A21 )
D D
A12 A21 x22 A12 A21 x22
= (D − A 12 x 1 + A 21 x 1 ) = (A21 x2 + A21 x1 )
D2 D2
A12 A221 x22 A21 x2 2
−2
D A12 x1 + A21 x2 −2




= = A 12 = A 12 = A 12
D2 D A21 x2 A21 x2

A12 x1 −2


ln γ1 = A12 1 +
A21 x2

The equation for ln γ2 is derived in analogous fashion.

∂(nG E/RT )
 
(b) With the understanding that T and P are constant, ln γ1 =
∂n 1 n2

and Eq. (12.16) may be written:

nG E A12 A21 n 1 n 2
= where n D = A12 n 1 + A21 n 2
RT nD

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 Differentiation in accord with the first equation gives:
  
1 n1 ∂(n D)


ln γ1 = A12 A21 n 2 −
n D (n D)2 ∂n 1 n 2

A12 A21 n 2  n1 A12 A21 x2 A12 x1



ln γ1 = 1− A12 = 1−
nD nD D D

A12 A21 x2 A12 A21 x2 A12 A221 x22

= (D − A x
12 1 ) = A x
21 2 =
D2 D2 D2
The remainder of the derivation is the same as in Part (a).

12.10 This behavior requires positive deviations from Raoult’s law over part of the composition range and
negative 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 values
over the remainder, with an intermediate crossing of the x1 axis. Because these deviations are usually
quite small, the vapor pressures P1sat and P2sat must not be too different, otherwise the dewpoint and
bubblepoint curves cannot exhibit extrema.

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

GE GE 1
= x1 x2 (A21 x1 + A12 x2 ) and (equimolar) = (A12 + A21 )
RT RT 8

But [see page 438, just below Eq. (12.10b)]: A12 = ln γ1∞ A21 = ln γ2∞

GE 1 GE 1
(equimolar) = (ln γ1∞ + ln γ2∞ ) or (equimolar) = ln(γ1∞ γ2∞ )
RT 8 RT 8

GE
12.24 (a) By Eq. (12.6): = x1 ln γ1 + x2 ln γ2
RT
= x1 x22 (0.273 + 0.096 x1 ) + x2 x12 (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 )

GE
= 0.273 x1 x2
RT
(b) The preceding equation is of the form from which Eqs. (12.15) are derived. From these,

ln γ1 = 0.273 x22 and ln γ2 = 0.273 x12

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

12.25 Write Eq. (11.100) for a binary system, and divide through by d x1 :
d ln γ1 d ln γ2 d ln γ2 x1 d ln γ1 x1 d ln γ1
x1 + x2 =0 whence =− =
d x1 d x1 d x1 x2 d x1 x2 d x2

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Integrate, recalling that ln γ2 = 1 for x1 = 0:
x1 x1
x1 d ln γ1 x1 d ln γ1
ln γ2 = ln(1) + d x1 = d x1
0 x2 d x2 0 x2 d x2
d ln γ1
(a) For ln γ1 = Ax22 , = 2Ax2
d x2
x1
Whence ln γ2 = 2A x1 d x1 or ln γ2 = Ax12
0

GE
By Eq. (12.6), = Ax1 x2
RT

(b) For ln γ1 = x22 (A + Bx2 ),

d ln γ1
= 2x2 (A + Bx2 ) + x22 B = 2Ax2 + 3Bx22 = 2Ax2 + 3Bx2 (1 − x1 )
d x2
x1 x1 x1
Whence ln γ2 = 2A x1 d x1 + 3B x1 d x1 − 3B x12 d x1
0 0 0

3B 2 3B B

  
ln γ2 = Ax12 + x − Bx13 or ln γ2 = x12 A+ 2
− Bx1 = x1 A + (1 + 2x2 )
2 1 2 2

GE 3B
Apply Eq. (12.6): = x1 x22 (A + Bx2 ) + x2 x12 (A + − Bx1 )
RT 2
Algebraic reduction can lead to various forms of this equation; e.g.,

GE B
 
= x1 x2 A + (1 + x2 )
RT 2

(c) For ln γ1 = x22 (A + Bx2 + C x22 ),

d ln γ1
= 2x2 (A + Bx2 + C x22 ) + x22 (B + 2C x2 ) = 2Ax2 + 3Bx22 + 4C x23
d x2
= 2Ax2 + 3Bx2 (1 − x1 ) + 4C x2 (1 − x1 )2
x1 x1 x1
Whence ln γ2 = 2A x1 d x1 + 3B x1 (1 − x1 )d x1 + 4C x1 (1 − x1 )2 d x1
0 0 0

x1 x1 x1
or ln γ2 = (2A + 3B + 4C) x1 d x1 − (3B + 8C) x12 d x1 + 4C x13 d x1
0 0 0

2A + 3B + 4C 3B + 8C



ln γ2 = x12 − x13 + C x14
2 3

3B 8C

 
2
ln γ2 = x12 A+ + 2C − B + x1 + C x1
2 3

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B C
or ln γ2 = x12 A + (1 + 2x2 ) + (1 + 2x2 + 3x22 )
2 3

The result of application of Eq. (12.6) reduces to equations of various forms; e.g.:
 
GE B C 2
= x1 x2 A + (1 + x2 ) + (1 + x2 + x2 )
RT 2 3

1
= H (1 + ñ)
12.40 (a) As shown on page 458, x1 = and H
1 + ñ

Eliminating 1 + ñ gives:
= H
H (A)
x1



d H 1 dH H d x1 1 dH H d x1
Differentiation yields: = − 2 = − 2
d ñ x1 d ñ x1 d ñ x1 d x1 x1 d ñ

d x1 −1
where = = −x12
d ñ (1 + ñ)2

d H dH dHE
Whence, = H − x1 = H E − x1
d ñ d x1 d x1

dHE
Comparison with Eq. (11.16) written with M ≡ H E , H̄2E = H E − x1
d x1

d H
shows that = H̄2E
d ñ

d H
−I
H
(b) By geometry, with reference to the following figure, =
d ñ ñ

−I
H
Combining this with the result of Part (a) gives: H̄2E =
ñ

From which,
− ñ H̄ E
I = H 2

E
Substitute:
= H = H
H and ñ =
x2
x1 x1 x1

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 HE x2 H E − x2 H̄2E
Whence, I = − H̄2E =
x1 x1 x1

However, by the summability equation, H E − x2 H̄2E = x1 H̄1E

Then, I = H̄1E

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

= x2 (A21 x1 + A12 x2 )
H

ñ
With x2 = 1 − x1 and x1 = 1/(1 + ñ) (page 458): x2 =
1 + ñ
The preceding equations combine to give:



= ñ A21 A12 ñ
H +
1 + ñ 1 + ñ 1 + ñ

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

=0
lim H
ñ→0

(b) Because ñ/(1 + ñ) → 1 for ñ → ∞, it follows that:

= A12
lim H
ñ→∞

(c) Analogous to Eq. (12.10b), page 438, we write: H̄2E = x12 [A21 + 2(A12 − A21 )x2 ]
Eliminate the mole fractions in favor of ñ:

2  
E 1 ñ
H̄2 = A21 + 2(A12 − A21 )
1 + ñ 1 + ñ

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

d H
lim = A21
ñ→0 d ñ


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


 

∂H ∂H ∂ Hi
= − xi
∂t P,x ∂t P,x i
∂t P,x



∂H ∂H
With ≡ CP, this becomes = CP −  xi C P i = C P
∂t P,x ∂t P,x i
t
H t
Therefore, d(H ) = C P dt H = H0 + t0 C P dt
H0 t0

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12.61 (a) From the definition of M: M E = x1 x2 M (A)

dME dM
Differentiate: = M(x2 − x1 ) + x1 x2 (B)
d x1 d x1

Substitution of Eqs. ( A) & (B) into Eqs. (11.15) & (11.16), written for excess properties, yields
the 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, . . . . By
assumption, this label is the only thing distinguishing one particle from another. For mixing,

S t = Smixed
t t
− Sunmixed t
= Smixed −  Sit
i

where the total emtropies are given by Eq. (5.42). Thus, for an unmixed species i, and for the mixed
system of particles,
Ni ! N!
Sit = k ln i = k ln =0 t
Smixed = k ln
Ni ! N1 ! N2 ! N3 ! · · ·
N!
Combining the last three equations gives: S t = k ln
N1 ! N2 ! N3 ! · · ·
S S t S t 1 N! 1
From which: = = = ln = (ln N ! −  ln Ni !)
R R(N /N A ) kN N N1 ! N2 ! N3 ! · · · N i

ln N ! ≈ N ln N − N and ln Ni ! ≈ Ni ln Ni − Ni
S 1 1
≈ (N ln N − N −  Ni ln Ni +  Ni ) = N (N ln N −  xi N ln xi N )
R N i i i

1
= (N ln N −  xi N ln xi −  xi N ln N ) = −  xi ln x1
N i i i

12.66 Isobaric data reduction is complicated by the fact that both composition and temperature vary from
point to point, whereas for isothermal data composition is the only significant variable. (The effect
of pressure on liquid-phase properties is assumed negligible.) Because the activity coefficients are
strong functions of both liquid composition and T , which are correlated, it is quite impossible without
additional information to separate the effect of composition from that of T . Moreover, the Pi sat values
depend 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:

dG E dG E
Ḡ 1E = G E + x2 and Ḡ 2E = G E − x1
d x1 d x1
GE Ḡ iE
Divide through by RT ; define G ≡ ; note by Eq. (11.91) that = ln γi
RT RT
dG dG
Then ln γ1 = G + x2 and ln γ2 = G − x1
d x1 d x1
GE
Given: = A1/k with A ≡ x1 Ak21 + x2 Ak12
x! x2 RT

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 dG d A1/k
Whence: G = x1 x2 A1/k and = x1 x2 + A1/k (x2 − x1 )
d x1 d x1

d A1/k 1 dA 1 A1/k k dG A1/k k

= A(1/k)−1 = (A21 −Ak12 ) and = x1 x2 (A21 −Ak12 )+A1/k (x2 −x1 )
d x1 k d x1 k A d x1 kA

(Ak21 − Ak12 )x1

 
Finally, ln γ1 = x22 A1/k +1
kA

(Ak − Ak12 )x2

 
Similarly, ln γ2 = x12 A1/k 1 − 21
kA

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

ln γ1∞ = A1/k = (Ak12 )1/k = A12 ln γ2∞ = A1/k = (Ak21 )1/k = A21

GE
(c) Let g≡ = A1/k = (x1 Ak21 + x2 Ak12 )1/k
x1 x2 RT
If k = 1, g = x1 A21 + x2 A12 (Margules equation)
A21 A12
If k = −1, g = (x1 A−1
21 + x 2 A12 )
−1 −1
= (van Laar equation)
x1 A12 + x2 A21
For k = 0, −∞, +∞, indeterminate forms appear, most easily resolved by working with the
logarithm:
1 
ln g = ln(x1 Ak21 + x2 Ak12 )1/k = ln x1 Ak21 + x2 Ak12

k
Apply l’Hôpital’s rule to the final term:

d ln x1 Ak21 + x2 Ak12 x1 Ak21 ln A21 + x2 Ak12 ln A12

 
= (A)
dk x1 Ak21 + x2 Ak12

Consider the limits of the quantity on the right as k approaches several limiting values.
x1 x2 x1 x2
• For k → 0, ln g → x1 ln A21 + x2 ln A12 = ln A21 + ln A12 and g = A21 A12
• For k → ± ∞, Assume A12 /A21 > 1, and rewrite the right member of Eq. (A) as

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

x1 + x2 (A12 /A21 )k

• For k → −∞, lim (A12 /A21 )k → 0 and lim ln g = ln A21

k→−∞ k→−∞

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

• For k → +∞, lim (A12 /A21 )k → ∞ and lim ln g = ln A12

k→∞ k→∞

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 γe Pesat = xe γe∞ Pesat = ye P e ≡ EtOH

Because P is low, we have assumed ideal gases, and for small xe let γe ≈ γe∞ . For volume fraction
ξe in the vapor, the ideal-gas assumption provides ξev ≈ ye , and for the liquid phase, with xe small

xe Vel xe Vel xe Vel

ξel = ≈ ≈ b ≡ blood
xe Vel + xb Vb xb Vb Vb

Vb l ∞ sat volume % EtOH in blood Ve P

Then ξ γ P ≈ ξev P ≈
Ve e e e volume % EtOH in gas Vb γe∞ Pesat

HE ∂(G E /RT )


12.70 By Eq. (11.95), = −T
RT ∂T P,x
E
G
= −x1 ln(x1 + x2 12 ) − x2 ln(x2 + x1 21 ) (12.18)
RT
d12 d21


∂(G E /RT )
 x1 x2 x2 x1
=− dT − dT
∂T x x1 + x2 12 x2 + x1 21

d12 d21 
E
H dT dT
= x1 x2 T  +
 
RT x1 + x2 12 x2 + x1 21


Vj −ai j
i j = exp (i = j) (12.24)
Vi RT
di j Vj ai j ai j


−ai j
= exp 2
= i j
dT Vi RT RT RT 2

12 a12 21 a21


H E = x1 x2 +
x1 + x2 12 x2 + x1 21

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

C PE x1 12 (a12 /RT )2 x2 21 (a21 /RT )2

 
= x1 x2 +
R (x1 + x2 12 )2 (x2 + x1 21 )2

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

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