7/30/2019 Chapter14 B

1/12

Chapter 14 - Section B - Non-Numerical Solutions

14.2 Start with the equation immediately following Eq. (14.49), which can be modified slightly to read:

ln i =(nGR/RT)

ni (n Z)

ni+ n ln Z

ni+ 1

where the partial derivatives written here and in the following development without subscripts are

understood to be at constant T, n/ (or / n), and nj . Equation (6.61) after multiplication by n can

be written:nGR

RT= 2n(n B)

n

+

3

2n2(nC)

n

2 n ln Z

Differentiate:

(nGR/RT)

ni= 2

n (n B + nBi ) +

3

2

n 2

(2n2C + n2Ci ) n ln Z

ni ln Z

or(nG R/RT)

ni= 2 (B + Bi ) +

3

22(2C + Ci ) n

ln Z

ni ln Z

By definition, Bi

(n B)

ni

T,nj

and Ci

(nC)

ni

T,nj

The equation of state, Eq. (3.40), can be written:

Z = 1 + B + C2 or n Z = n + n(n B)

n

+ n2(nC)

n

2

Differentiate: (n Z)ni

= 1 +

n

(n B + n Bi ) +

n

2

(2n2C + n2Ci )

or (n Z)

ni= 1 + (B + Bi ) +

2(2C + Ci )

When combined with the two underlined equations, the initial equation reduces to:

ln i = 1 + (B + Bi ) +12

2(2C + Ci )

The two mixing rules are:

B = y2

1B11 + 2y1y2B12 + y2

2B22

C = y31 C111 + 3y21y2C112 + 3y1y

22 C122 + y

32 C222

Application of the definitions of Bi and Ci to these mixing rules yields:

B1 = y1(2 y1)B11 + 2y22B12 y

22B22

C1 = y21 (3 2y1)C111 + 6y1y

22 C112 + 3y

22 (1 2y1)C122 2y

32 C222

B2 = y21B11 + 2y

21B12 + y2(2 y2)B22

C2 = 2y31 C111 + 3y

21 (1 2y2)C112 + 6y1y

22 C122 + 2y

22 (3 2y2)C222

709

7/30/2019 Chapter14 B

2/12

In combination with the mixing rules, these give:

B + B1 = 2(y1B11 + y2B12)

2C + C1 = 3(y21 C111 + 2y1y2C112 + y

22 C122)

B + B2 = 2(y2B22 + y1B12)

2C + C2 = 3(y22 C222 + 2y1y2C122 + y21 C112)

In combination with the boxed equation these expressions along with Eq. (3.40) allow calculation of

ln 1 and ln 2.

14.11 For the case described, Eqs. (14.1) and (14.2) combine to give: yi P = xi Psat

i

sati

i

If the vapor phase is assumed an ideal solution, i = i , and yi P = xi Psat

i

sati

i

When Eq. (3.38) is valid, the fugacity coefficient of pure species i is given by Eq. (11.36):

ln i =Bii P

RTand sati =

Bii P sati

RT

Therefore, ln sati

i= ln sati ln i =

Bii Psat

i

RT

Bii P

RT=

Bii (Psat

i P)

RT

For small values of the final term, this becomes approximately:

sati

i= 1 +

Bii (Psat

i P)

RT

Whence, yi P = xi Psat

i 1 +Bii (P

sati P)

RT

or yi P xi Psat

i =xi P

sati Bii (P

sati P)

RT

Write this equation for species 1 and 2 of a binary mixture, and sum. This yields on the left the

difference between the actual pressure and the pressure given by Raoults law:

P P(RL) =x1B11 P

sat1 (P

sat1 P) + x2B22 P

sat2 (P

sat2 P)

RT

Because deviations from Raoults law are presumably small, P on the right side may be replaced by

its Raoults-law value. For the two terms,

P sat1 P = Psat

1 x1 Psat

1 x2 Psat

2 = Psat

1 (1 x2)Psat

1 x2 Psat

2 = x2(Psat

1 Psat

2 )

P sat2 P = Psat

2 x1 Psat

1 x2 Psat

2 = Psat

2 x1 Psat

1 (1 x1)Psat

2 = x1(Psat

2 Psat

1 )

Combine the three preceding equations:

P P(RL) =x1x2B11(P

sat1 P

sat2 )P

sat1 x1x2B22(P

sat1 P

sat2 )P

sat2

RT

=x1x2(P

sat1 P

sat2 )

RT(B11 P

sat1 B22 P

sat2 )

710

7/30/2019 Chapter14 B

3/12

Rearrangement yields the following:

P P(RL) =x1x2(P

sat1 P

sat2 )

2

RT

B11 P

sat1 B22 P

sat2

P sat1 Psat

2

= x1x2(Psat

1 Psat

2 )

2

RT

B11 + (B11 B22)P

sat

2

P sat1 Psat

2

=x1x2(P

sat1 P

sat2 )

2

RT(B11)

1 +

1

B22

B11

P sat2

P sat1 Psat

2

Clearly, when B22 = B11, the term in square brackets equals 1, and the pressure deviation from the

Raoults-law value has the sign of B11; this is normally negative. When the virial coefficients are not

equal, a reasonable assumption is that species 2, taken here as the heavier species (the one with

the smaller vapor pressure) has the more negative second virial coefficient. This has the effect of

making the quantity in parentheses negative and the quantity in square brackets < 1. However, if this

latter quantity remains positive (the most likely case), the sign of B11 still determines the sign of the

deviations.

14.13 By Eq. (11.90), the definition ofi , ln i = ln fi lnxi ln fi

Whence,dln i

d xi=

dln fi

d xi

1

xi=

1

fi

d fi

d xi

1

xi

Combination of this expression with Eq. (14.71) yields:1

fi

d fi

d xi> 0

Because fi 0,

d fi

d xi > 0 (const T, P)

By Eq. (11.46), the definition of fi ,di

d xi= RT

dln fi

d xi=

RT

fi

d fi

d xi

Combination with Eq. (14.72) yields:di

d xi> 0 (const T, P)

14.14 Stability requires that G < 0 (see Pg. 575). The limiting case obtains when G = 0, in which

event Eq. (12.30) becomes:GE = RT

i

xi lnxi

For an equimolar solution xi = 1/N where N is the number of species. Therefore,

GE(max) = RTi

1

Nln

1

N= RT

i

1

Nln N = RT ln N

For the special case of a binary solution, N = 2, and GE(max) = RT ln 2

711

7/30/2019 Chapter14 B

4/12

14.17 According to Pb. 11.35, GE = 12 P y1y2 orGE

RT=

12 P

RTy1y2

This equation has the form:GE

RT= Ax 1x2

for which it is shown in Examples 14.5 and 14.6 that phase-splitting occurs for A > 2. Thus, theformation of two immiscible vapor phases requires: 12 P/RT > 2.

Suppose T = 300 K and P = 5 bar. The preceding condition then requires: 12 > 9977 cm3 mol1

for vapor-phase immiscibility. Such large positive values for 12 are unknown for real mixtures.

(Examples of gas/gas equilibria are known, but at conditions outside the range of applicability of the

two-term virial EOS.)

14.19 Consider a quadratic mixture, described by:GE

RT= Ax 1x2

It is shown in Example 14.5 that phase splitting occurs for such a mixture if A > 2; the value of

A = 2 corresponds to a consolute point, at x1 = x2 = 0.5. Thus, for a quadratic mixture,

phase-splitting obtains if:

GE > 2 1

2

1

2 RT = 0.5RT

This is a model-dependentresult. Many liquid mixtures are known which are stable as single phases,

even though GE > 0.5RT for equimolar composition.

14.21 Comparison of the Wilson equation, Eq. (12.18) with the modified Wilson equation shows that

(GE/RT)m = C(GE/RT), where subscript m distinguishes the modified Wilson equation from

the original Wilson equation. To simplify, define g (GE/RT); then

gm = Cg ngm = C ng(ngm)

n1= C

(ng)

n1ln(1)m = Cln 1

where the final equality follows from Eq. (11.96). Addition and subtraction of ln x1 on the left side

of this equation and ofClnx1 on the right side yields:

ln(x11)m lnx1 = Cln(x11) Clnx1

or ln(x11)m = Cln(x11) (C 1) lnx1

Differentiate:dln(x11)m

d x1= C

dln(x11)

d x1

C 1

x1

As shown in Example 14.7, the derivative on the right side of this equation is always positive. How-

ever, for C sufficiently greater than unity, the contribution of the second term on the right can make

dln(x11)M

d x1< 0

over part of the composition range, thus violating the stability condition of Eq. (14.71) and implying

the formation of two liquid phases.

14.23 (a) Refer to the stability requirement of Eq. (14.70). For instability, i.e., for the formation of two

liquid phases,

d2(GE/RT)

d x 21<

1

x1x2

712

7/30/2019 Chapter14 B

5/12

over part of the composition range. The second derivative of GE must be sufficiently negative so

as to satisfy this condition for some range of x1. Negative curvature is the norm for mixtures for

which GE is positive; see, e.g., the sketches of GE vs. x1 for systems (a), (b), (d), (e), and (f) in

Fig. 11.4. Such systems are candidates for liquid/liquid phase splitting, although it does not in

fact occur for the cases shown. Rather large values of GE are usually required.

(b) Nothing in principle precludes phase-splitting in mixtures for which GE < 0; one merely re-quires that the curvature be sufficiently negative over part of the composition range. However,

positive curvature is the norm for such mixtures. We know of no examples of liquid/liquid phase-

splitting in systems exhibiting negative deviations from ideal-solution behavior.

14.29 The analogy is Raoults law, Eq. (10.1), applied at constant P (see Fig. 10.12): yi P = xi Psat

i

If the vapor phase in VLE is ideal and the liquid molar volumes are negligible (assumptions inherent

in Raoults law), then the Clausius/Clapeyron equation applies (see Ex. 6.5):

dln P sati

d T=

Hlvi

RT2

Integration from the boiling temperature Tbi at pressure P (where P sati = P) to the actual temperatureT (where P sati = P

sati ) gives:

lnP sati

P=

TTbi

Hlvi

RT2d T

Combination with Eq. (10.1) yields:

yi = xi exp

TTbi

Hlvi

RT2d T

which is an analog of the Case I SLE equations.

14.30 Consider binary (two-species) equilibrium between two phases of the same kind. Equation (14.74)applies:

x i

i = x

i

i (i = 1, 2)

If phase is pure species 1 and phase is pure species 2, then x

1 =

1 = 1 and x2 =

2 = 1.

Hence, x 1

1 = x

1

1 = 1 and x2

2 = x

2

2 = 1

The reasoning applies generally to (degenerate) N-phase equilibrium involving N mutually immis-

cible species. Whence the cited result for solids.

14.31 The rules of thumb are based on Case II binary SLE behavior. For concreteness, let the solid be pure

species 1 and the solvent be liquid species 2. Then Eqs. (14.93) and (14.92a) apply:

x1 = 1 = expHsl1RTm1

T Tm1

T

(a) Differentiate:d x1

d T= 1

Hsl1RT2

Thus d x1/d T is necessarily positive: the solid solubility x1 increases with increasing T.

(b) Equation (14.92a) contains no information about species 2. Thus, to the extent that Eqs. (14.93)

and (14.92a) are valid, the solid solubility x1 is independent of the identity of species 2.

713

7/30/2019 Chapter14 B

6/12

(c) Denote the two solid phases by subscripts A and B. Then, by Eqs. (14.93) and (14.92a), the

solubilities xA and xB are related by:

xA

xB= exp

Hsl (TmB TmA )

RTmA TmB

where by assumption, HslA = Hsl

B Hsl

Accordingly, xA/xB > 1 if and only if TA < TB , thus validating the rule of thumb.

(d) Identify the solid species as in Part (c). Then xA and xB are related by:

xA

xB= exp

(HslB H

slA )(Tm T)

RTm T

where by assumption, TmA = TmB Tm

Notice that Tm > T (see Fig. 14.21b). Then xA/xB > 1 if and only if Hsl

A < Hsl

B , in

accord with the rule of thumb.

14.34 The shape of the solubility curve is characterized in part by the behavior of the derivative d yi /d P

(constant T). A general expression is found from Eq. (14.98), y1 = Psat

1 P/F1, where the enhance-

ment factor F1 depends (at constant T) on P and y1. Thus,

d y1

d P=

P sat1P 2

F1 +P sat1

P

F1

P

y1

+

F1

y1

P

d y1

d P

= y1

P+ y1

ln F1

P

y1

+

ln F1

y1

P

d y1

d P

Whence,d y1

d P=

y1

ln F1

P

y1

1

P

1 y1

ln F1

y1

P

(A)

This is a general result. An expression for F1 is given by Eq. (14.99):

F1 sat1

1exp

Vs1 (P Psat

1 )

RT

From this, after some reduction:

ln F1

P

y1

=

ln 1

P

y1

+Vs1

RTand

ln F1

y1

P

=

ln 1

y1

P

Whence, by Eq. (A),d y1

d P=

y1

ln 1

P

y1

+Vs1

RT

1

P

1 + y1

ln 1

y1

P

(B)

714

7/30/2019 Chapter14 B

7/12

This too is a general result. If the two-term virial equation in pressure applies, then ln 1 is given by

Eq. (11.63a), from which: ln 1

P

y1

=1

RT(B11 + y

22 12) and

ln 1

y1

P

= 2y212 P

RT

Whence, by Eq. (B),d y1

d P=

y1

Vs1 B11 y

22 12

RT

1

P

1 2y1y212 P

RT

The denominator of this equation is positive at any pressure level for which Eq. (3.38) is likely to be

valid. Hence, the sign of d y1/d P is determined by the sign of the group in parentheses. For very low

pressures the 1/ P term dominates and d y1/d P is negative. For very high pressures, 1/ P is small,

and d y1/d P can be positive. If this is the case, then d y1/d P is zero for some intermediate pressure,

and the solubility y1 exhibits a minimum with respect to pressure. Qualitatively, these features are

consistent with the behavior illustrated by Fig. 14.23. However, the two-term virial equation is onlyvalid for low to moderate pressures, and is unable to mimic the change in curvature and flattening

of the y1 vs. P curve observed for high pressures for the naphthalene/CO2 system.

14.35 (a) Rewrite the UNILAN equation:

n =m

2s

ln(c + Pes ) ln(c + Pes )

(A)

As s 0, this expression becomes indeterminate. Application of lHopitals rule gives:

lims0

n = lims0

m

2

Pes

c + Pes+

Pes

c + Pes

= m2

P

c + P+ P

c + P

or lims0 n =m P

c + P

which is the Langmuir isotherm.

(b) Henrys constant, by definition: k limP0

dn

d P

Differentiate Eq. (A):dn

d P =m

2s es

c + Pes es

c + Pes

Whence, k =m

2s

es

c

es

c

=

m

cs

es es

2

or k =

m

cssinh s

(c) All derivatives ofn with respect to P are well-behaved in the zero-pressure limit:

limP0

dn

d P=

m

cssinh s

715

7/30/2019 Chapter14 B

8/12

limP0

d2n

d P 2=

m

c2ssinh 2s

limP0

d3n

d P 3=

2m

c3ssinh 3s

Etc.

Numerical studies show that the UNILAN equation, although providing excellent overall corre-

lation of adsorption data at low-to-moderate surface coverage, tends to underestimate Henrys

constant.

14.36 Start with Eq. (14.109), written as:

ln(P/n) = ln k +

n0

(z 1)dn

n+ z 1

With z = 1 + Bn + Cn2 + , this becomes:

ln(P/n) = ln k + 2Bn +3

2C n2 +

Thus a plot of ln(P/n) vs. n produces ln k as the intercept and 2B as the limiting slope (for

n 0). Alternatively, a polynomial curve fit of ln(P/n) in n yields ln k and 2B as the first

two coefficients.

14.37 For species i in a real-gas mixture, Eqs. (11.46) and (11.52) give:

g

i = i (T) + RT ln yi i P

At constant temperature, dg

i = RT dln yi i P

With di = dg

i , Eq. (14.105) then becomes:

a

RTd + dln P +

i

xi dln yi i = 0 (const T)

For pure-gas adsorption, this simplifies to:

a

RTd = dln P + dln (const T) (A)

which is the real-gas analog of Eq. (14.107). On the left side of Eq. (A), introduce the adsorbate

compressibility factor z through z a/RT = A/n RT:

a

RTd = d z + z

dn

n(B)

where n is moles adsorbed. On the right side of Eq. (A), make the substitution:

dln = (Z 1)d P

P(C)

which follows from Eq. (11.35). Combination of Eqs. (A), (B), and (C) gives on rearrangement (see

Sec. 14.8):

dlnn

P= (1 z)

dn

n dz + (Z 1)

d P

P

which yields on integration and rearrangement:

n = k P exp

P0

(Z 1)d P

P exp

n0

(1 z)dn

n+ 1 z

This equation is the real-gas analog of Eq. (14.109).

716

7/30/2019 Chapter14 B

9/12

14.39 & 14.40 Start with Eq. (14.109). With z = (1 bm)1, one obtains the isotherm:

n = k P(1 bn) exp

bn

1 bn

(A)

For bn sufficiently small, expbn

1 bn 1

bn

1 bn

Whence, by Eq. (A), n k P(1 2bn) or n k P

1 + 2bk P

which is the Langmuir isotherm.

With z = 1 + n, the adsorption isotherm is: n = k P exp(2n)

from which, for n sufficiently small, the Langmuir isotherm is again recovered.

14.41 By Eq. (14.107) with a = A/n,Ad

RT= n

d P

P

The definition of and its derivative are:

A

RTand d =

A d

RT

Whence, d = nd P

P(A)

By Eq. (14.128), the Raoults law analogy, xi = yi P/ P

i . Summation for given P yields:

i

xi = P

i

yi

P i(B)

By general differentiation,

di xi = P d

i

yi

P i +

i

yi

P id P (C)

The equation,

i xi = 1, is an approximation that becomes increasingly accurate as the solution

procedure converges. Thus, by rearrangement of Eq. (B),

i

yi

P i=

i

xi

P=

1

P

With P fixed, Eq. (C) can now be written in the simple but approximate form:

di

xi =d P

P

Equation (A) then becomes:

d = n di

xi or = n

i

xi

where we have replaced differentials by deviations. The deviation in

i xi is known, since the true

value must be unity. Therefore,

i

xi = P

i

yi

P i 1

717

7/30/2019 Chapter14 B

10/12

By Eq. (14.132), n =1

i

(xi /ni )

Combine the three preceding equations:

=

P

i

yi

P i 1

i

(xi /ni )

When xi = yi P/ P

i , the Raoults law analogy, is substituted the required equation is reproduced:

=

P

i

yi

P i 1

P

iyi

P i ni

14.42 Multiply the given equation for GE/RT by n and convert all mole fractions to mole numbers:

nGE

RT= A12

n1n2

n+ A13

n1n3

n+ A23

n2n3

n

Apply Eq. (11.96) for i = 1:

ln 1 = A12n2

1

n

n1

n2

+ A13n3

1

n

n1

n2

A23

n2n3

n2

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

Introduce solute-free mole fractions:

x 2 x2

x2 + x3=

x2

1 x1and x 3 =

x3

1 x1

Whence, ln 1 = A12x2(1 x1)

2 + A13x3(1 x1)

2 A23x2x

3(1 x1)

2

For x1 0, ln

1 = A12x2 + A13x

3 A23x

2x

3

Apply this equation to the special case of species 1 infinitely dilute in pure solvent 2. In this case,

x 2 = 1, x3 = 0, and

ln 1,2 = A12 Also ln

1,3 = A

13

Whence, ln 1 = x2 ln

1,2 + x

3 ln

1,3 A23x

2x

3

In logarithmic form the equation immediately following Eq. (14.24) on page 552 may be applied to

the several infinite-dilution cases:

lnH1 = ln f1 + ln

1 lnH1,2 = ln f1 + ln

1,2 lnH1,3 = ln f1 + ln

1,3

Whence, lnH1 ln f1 = x2(lnH1,2 ln f1) + x

3(lnH1,3 ln f1) A23x

2x

3

or lnH1 = x2 lnH1,2 + x

3 lnH1,3 A23x

2x

3

718

7/30/2019 Chapter14 B

11/12

14.43 For the situation described, Figure 14.12 would have two regions like the one shown from to ,

probably one on either side of the minimum in curve II.

14.44 By Eq. (14.136) with V2 = V2:V2

RT= ln(x22)

Represent ln 2 by a Taylor series:

ln 2 = ln 2|x1=0 +dln 2

d x1

x1=0

x1 +1

2

d2 ln 2

d x 21

x1=0

x 21 +

But at x1 = 0 (x2 = 1), both ln 2 and its first derivative are zero. Therefore,

ln 2 =1

2

d2 ln 2

d x 21

x1=0

x 21 +

Also, lnx2 = ln(1 x1) = x1 x 21

2

x 31

3

x 41

4

Therefore, ln(x22) = + lnx2 + ln 2 = x1 12

1

1

2

d2 ln 2

d x 21

x1=0

x 21

andV2

x1RT= 1 +

1

2

1

1

2

d2 ln 2

d x 21

x1=0

x1 +

Comparison with the given equation shows that: B=1

2

1

1

2

d2 ln 2

d x 21

x1=0

14.47 Equation (11.95) applies:

(GE/RT)

T

P,x

= HE

RT2

For the partially miscible system GE/RT is necessarily large, and if it is to decrease with increasing

T, the derivative must be negative. This requires that HE be positive.

14.48 (a) In accord with Eqs. (14.1) and (14.2), yii

satiP = xi i P

sati Ki

yi

xi=

i Psat

i

P

sati

i

12 K1

K2

=1 P

sat1

2

P sat2

sat1

1

2

sat

2

(b) 12(x1 = 0) =1 P

sat1

P sat2

1(Psat

1 )

1 (Psat

2 )

2(Psat

2 )

2(Psat

2 )=

1 Psat

1

P sat2

1(Psat

1 )

1 (Psat

2 )

12(x1 = 1) =P sat1

2 Psat

2

1(P

sat1 )

1(Psat

1 )

2 (Psat

1 )

2(Psat

2 )=

P sat12 P

sat2

2 (P

sat1 )

2(Psat

2 )

The final fractions represent corrections to modified Raoults law for vapor nonidealities.

719

7/30/2019 Chapter14 B

12/12

(c) If the vapor phase is an ideal solution of gases, then i = i for all compositions.

14.49 Equation (11.98) applies:

ln i

T

P,x

= HEi

RT2

Assume that HE and HEi are functions of composition only. Then integration from Tk to T gives:

lni (x , T)

i (x, Tk)=

HEiR

TTk

d T

T2=

HEiR

1

T

1

Tk

=

HEiRT

T

Tk 1

i (x , T) = i (x , Tk) exp

HEiRT

T

Tk 1

14.52 (a) From Table 11.1, p. 415, find:

GE

T

P,x

= SE = 0 and GE is independent ofT.

Therefore

GE

RT =

FR (x )

RT

(b) By Eq. (11.95),

(GE/RT)

T

P,x

= HE

RT2= 0

GE

RT= FA(x )

(c) For solutions exhibiting LLE, GE/RT is generally positive and large. Thus and are positive

for LLE. For symmetrical behavior, the magic number is A = 2:

A < 2 homogeneous; A = 2 consolute point; A > 2 LLE

With respect to Eq. (A), increasing T makes GE/RT smaller. thus, the consolute point is an up-

per consolute point. Its value follows from:

RTU

= 2 TU = 2R

The shape of the solubility curve is as shown on Fig. 14.15.

14.53 Why? Because they are both nontoxic, relatively inexpensive, and readily available. For CO2, its

Tc is near room temperature, making it a suitable solvent for temperature-sensitive materials. It is

considereably more expensive than water, which is probably the cheapest possible solvent. However,

both Tc and Pc for water are high, which increases heating and pumping costs.

720