Koretsky ch (7) (1)

Chapter 7 Solutions

Engineering and Chemical Thermodynamics 2e

Milo Koretsky

Wyatt Tenhaeff

School of Chemical, Biological, and Environmental Engineering

Oregon State University

[email protected]

mailto:[email protected]

7.1

The fugacity in the liquid is greater. At equilibrium, the fugacities are equal. For this to be the

case, the air in the room would have to be saturated (100% relative humidity). Since the air

contains less water than saturation, water will spontaneously evaporate, and the fugacity of the

vapor is smaller.

7.2

hmix = 0. The molecular basis for the Lewis fugacity rule is that all the intermolecular

interactions are the same. Therefore, the energy of the mixture is equal to that of the sum of the

pure species.

7.3

Mixture A. The molecular basis for the Lewis fugacity rule is that all the intermolecular

interactions are the same. n-pentane and n-hexane both have dispersion interactions and they are

approximately the same size (polarizability is similar)

7.4 The van der Waals parameter b approximates repulsive interactions with a hard sphere model

that is determined by the size of the molecules. Hence it is the weighted average of the size of

each of the species in the mixture. That can be seen in the form for a binary mixture:

1 1 2 2b y b y b

The parameter a represents van der Waals attractive interactions and is a “two-body”

interactions. Thus, you must sum together all the possible pair-wise interactions. That can be

seen in the form for a binary mixture:

1 1 1 1 2 12 2 1 21 2 2 2

2 2

1 1 1 2 12 2 22

a y y a y y a y y a y y a

y a y y a y a

7.5

There are many ways to approach this problem. One approach is shown below. If we assume an

ideal solution, for water (species 1), we get:

1 1 1

saty P x P

Solving for x1 at equilibrium

1

1

1

0.9eq

sat

y Px

P

Since we are at 90% RH. If we calculate the mole fraction of water for 96.55 mass % water, we

get x1 = 0.99 in sweat. Since 11

eqx x , water will have a tendency to evaporate so the fugacity of

water in the liquid is greater than the fugacity of water in the vapor.

7.6

(a)

The magnitude of the Henry’s Law constant is governed by the unlike (1-2) interactions. In the

case of acetone and water, strong intermolecular attraction exists due to dipole-dipole

interactions and hydrogen bonding. With methane and water, only dispersion is present.

Therefore, the interactions between methane and water are weaker, and the fugacity is greater.

There is a greater partial pressure for the methane-water system. The Henry’s Law constant is

greater for this system.

(b)

The Henry’s Law constant describes the unlike interactions. Since the unlike interactions result

in a fugacity that is equal to the pure fugacity of a, the unlike and like interactions are equal in

magnitude. Therefore, the solution is ideal for the entire composition range.

0

0.2

0.4

0.6

0.8

1

1.2

0 0.2 0.4 0.6 0.8 1

g a

xa

Activity Coefficient vs. Composition

7.7

(a)

First, we must realize that

mixE hh

Since

02

,

T

h

T

Tg

mix

nP

E

i

we see that T

g E is independent of temperature when the pressure and number of moles is held

constant. Therefore,

ii

nP

ba

nP

E

T

xAx

T

g

,,

is independent of temperature, which implies

inPT

A

,

is independent of temperature at constant pressure and moles. TA ~

(b)

Equation 7.24 provides

mixE

nT

E

vvP

g

i

,

Substituting the two suffix Margules equation in for the excess Gibbs energy, we find

0,

inT

ba

P

xAx

Therefore, we can see that the A parameter is independent of pressure at constant temperature

and moles.

0

50

100

150

200

0 0.5 1

[bar]

xa

0

0.4

0.8

1.2

1.6

2

0 0.5 1

lngb

xa

7.9

(a) See graph below. Since the Henry’s law constant which represents a-b interactions is less

than the pure species fugacity, the “tendency to escape” of a-b is lower and the a-b interactions

are stronger.

(b) Both lnga and lngb go to 1 and xi goes to 0. Therefore the Henry’s law reference state is being

used. Since gi> 0 the tendency to escape for some a-a or b-b interactions is greater than all a-a

interactions. Therefore the a-b interactions are stronger.

ˆa a af x

ˆa a af x f

7.10

7.11

200 oC and 1.56 MPa (Pi

sat = 1.55 MPa) ( 1sat

i ) so 1i

sat

i

f

P

70 oC and 1 bar (Pi

sat = 31.2 kPa) ideal 1i

sat

i

f

P

70 oC and 1.56 MPa (Pi

sat = 31.2 kPa) Poynting correction 1i

sat

i

f

P

7.12

Initially the system contains water and nitrogen in vapor liquid equilibrium at 1 atm. The

liquid is mostly water and the vapor contains a mole fraction of water where the fugacity of

the vapor equals the fugacity of the liquid. When the third component is added to the liquid,

the mole fraction of water in the liquid decreases, so its fugacity also decreases. To maintain

equilibrium, the fugacity of water in the vapor must also decrease to the point where it equals

the fugacity of water in the liquid. Therefore, some water will condense and the number of

moles of water in the vapor will decrease.

7.13

ˆ ˆ 2.34 kPal v sat

i i i if f y P P

7.14

(a)

Use Equation 7.7:

low

vi

ii P

fRTgg ln¼

Assume kPa 10lowP . From the steam tables at 500 ºC

kPa 10lowP : º kJ kJ kJˆ 3489.0 773.15K 9.8977 4163.4

kg kg K kgig

MPa 2P : kJ kJ kJ

ˆ 3467.6 773.15K 7.4316 2278.1 kg kg K kg

ig

Therefore,

kJ kJ2278.1 4163.4 1000 J/kJ 0.0180148 kg/mol

kg kg10 kPa exp

J8.314 773.15 K

mol K

v

if

MPa 1.97kPa 1970 vif

985.0i

(b)

For 500 ºC and 50 MPa, we obtain the following from the steam tables

kJ kJ kJ

ˆ 2720.1 773.15K 5.1725 1279.0 kg kg K kg

ig

Using data from Part (a), we can calculate the fugacity and fugacity coefficient:

kJ kJ1279.0 4163.4 1000 J/kJ 0.0180148 kg/mol

kg kg10 kPa exp

J8.314 773.15 K

mol K

v

if

32.4 MPav

if

and

648.0i

7.15

(a)

Equation 7.8 states

P

P

ilow

vi

low

dPvP

fRT ln

However, the Berthelot equation is not explicit in molar volume, so the integral must be

transformed.

dvTv

a

bv

RTdP

ii

32

2

Substituting this result into the integral, we get

i

low

v

P

RT ii

i

low

vi dv

Tv

a

bv

RTv

P

fRT

22

2ln

To determine the integral of the first term above, we use decomposition by partial fractions:

v

v b 2

1

v b

b

v b 2

so

v i

v i b 2

dvRT

Plow

vi

ln vi b b

v i b

RT

Plow

vi

and

i

low

v

P

RTi

iilow

vi

vRT

abv

bv

b

P

f

2

2)ln(ln

RT

P

vRT

a

bP

RT

bv

bP

RTbvb

P

f low

i

low

i

low

ilow

vi 12

ln11

ln2

Since bP

RT

low

, the expression simplifies to

RT

P

vRT

a

RT

bvP

RT

P

bvb

P

f low

i

ilowlow

ilow

vi 12

ln1

ln2

If we add lowPln to both sides and let 0lowP , we obtain

ln f i

v b

vi b ln

vi b RT

2a

RT 2vi

Therefore,

iii

vi

vRT

a

bv

b

bv

RTf

2

2exp

To obtain an expression for vi , we divide our expression for fugacity by total pressure:

i

v f i

v

P

RT

P vi b exp

b

vi b

2a

RT 2vi

(b)

From Problem 4.29, we got

vc 3b 3RTc

8Pc

and

a 9

8vcRTc

2

If we substitute into the definition for fugacity coefficient above, we get

i

v f i

v

P

8Tr

Pr 3vi,r 1 exp

1

3vi,r 1

9

4Tr

2vi,r

7.16

Equation 7.8:

low

vi

P

P

iP

fRTdPv

low

ln

For the Redlich-Kwong EOS

P RT

v b

a

T1/ 2v v b

so

dP RT

v b 2

a

T1/ 2v 2 v b

a

T1/ 2v v b 2

dv

Therefore,

vRT

v b 2

a

T1/ 2v v b

a

T1/ 2 v b 2

dv

RT

Plow

v

RT lnf i

v

Plow

To determine the integral of the first two terms term above, we use decomposition by partial

fractions. For the first term,

v

v b 2

1

v b

b

v b 2

so

v

v b 2

dvRT

Plow

v

ln v b b

v b

RT

Plow

v

For the second term:

1

v v b

1

b

1

v

1

v b

so

1

v v b dv

RT

Plow

v

1

bln

v

v b

RT

Plow

v

Thus, we get

RT lnf i

v

Plow

RT ln v b RT

b

v b

a

T1/ 2bln

v

v b

a

T1/ 2 v b RT

Plow

v

If we note that bP

RT

low

and let 0lowP , we obtain

bvT

a

bv

v

bT

a

bv

bRT

bv

RTRTfRT v

i

2/12/1lnlnln

Therefore,

bvRT

a

bv

v

bRT

a

bv

b

bv

RTf vi 2/32/3

lnexp

and

i

v f i

v

P

RT

P v b exp

b

v b

a

RT 3 / 2bln

v

v b

a

RT 3 / 2 v b

7.17

Equation 7.8:

low

vi

P

P

iP

fRTdPv

low

ln

The Peng-Robinson EOS can be written

P RT

v b

a(T)

v 2 2vb b2

Thus,

dP RT

v b 2

2a(T) v b

v 2 2vb b2 2

dv

Therefore,

v RT

v b 2

2a(T) v b

v 2 2vb b2 2

dv

RT

Plow

v

RT lnf i

v

Plow

Simplifying, we get

2 2

2 2

2 ( )ln

2low low

v vv

i

RT RTlow

P P

v v bf v a Tdv dv

P RTv b v vb b

(1)

To determine each of the integrals in Equation 1, we use decomposition by partial fractions. For

the first integral:

v

v b 2

1

v b

b

v b 2

so

v

v b 2

dvRT

Plow

v

ln v b b

v b

RT

Plow

v

(2)

For the second integral, decomposition leads to:

v v b

v2 2vb b2 2

1

v2 2vb b2 b

v b

v2 2vb b2 2

so

v v b

v2 2vb b2 2

dvRT

Plow

v

1

v2 2vb b2dv

RT

Plow

v

bv b

v 2 2vb b2 2

dvRT

Plow

v

(3)

We again use partial fractions. For the first term in Equation (3):

bvbvbbvbv 21

1

21

1

22

1

2

1

22

so

v

P

RT

v

P

RTlow

low

bv

bv

bdv

bvbv 21

21ln

22

1

2

1

22

(4)

For the second term in Equation (3):

v b

v2 2vb b2 2

v b

v2 2vb b2 2

2b

v2 2vb b2 2

(5)

Equation (5) can be substituted into Equation (3) to give two terms. The first term gives:

bv b

v 2 2vb b2 2

RT

Plow

v

dv b

2 v 2 2vb b2 RT

Plow

v

(6)

The second term is somewhat more problematic. Again decomposition gives:

2

222 21

1

21

1

4

1

2

2

bvbvbbvbv

b

or

22

222 21

1

21

1

21

12

21

1

4

1

2

2

bvbvbvbvbbvbv

b

(7)

Intergrating the first and third term in Equation (7) gives:

v

P

RT

v

P

RTlow

low

bvdv

bv 214

1

21

1

4

12

(8)

and

v

P

RT

v

P

RTlow

low

bvdv

bv 214

1

21

1

4

12

(9)

From above the second term gives:

v

P

RT

v

P

RTlow

low

bv

bv

bbvbv 21

21ln

24

1

21

1

21

1

2

1

(10)

Substituting Equations (4). (6). (8), (9) and (10) into Equation (3) gives:

v

P

RT

v

P

RT

v

P

RT

v

P

RT

v

P

RT

v

P

RT

lowlowlow

lowlowlow

bv

bv

bbvbv

bvbv

b

bv

bv

bdv

bvbv

bvv

21

21ln

24

1

214

1

214

1

2221

21ln

22

1

222222

(11)

Simplifying Equation (11) gives

v v b

v 2 2vb b2 2

dvRT

Plow

v

1

4 2b2ln

v 1 2 bv 1 2 b

v

2 v 2 2vb b2

RT

Plow

v

(12)

Substituting Equations (12) and (2) into Equation (1) gives

lnf i

v

Plow

RT ln v b RT

b

v b

RT

Plow

v

2a(T)

RT

1

4 2bln

v 1 2 bv 1 2 b

v

2 v 2 2vb b2

RT

Plow

v

If we note that bP

RT

low

and let 0lowP , we obtain

ln f i

v lnRT

v b

b

v b

a(T)

RT

1

2 2bln

v 1 2 bv 1 2 b

v

v2 2vb b2

and

ln i

v lnRT

P v b

b

v b

a(T)

RT

1

2 2bln

v 1 2 bv 1 2 b

v

v2 2vb b2

7.18

(a)

We can calculate the fugacity from the steam tables using the following equations

low

vio

iiP

fRTgg ln

iii Tshg

We can take the reference state to be 374 ºC and 10 kPa. Using enthalpy and entropy values

from the steam tables, interpolation gives:

kJ

ˆ 2934.5 kg

o

ig

kJ

ˆ 890.9 kg

ig

Therefore,

mol

J 52864o

ig

mol

J 16049ig

Now, the fugacity can be calculated:

kPa 9382

K647Kmol

J 314.8

mol

J 28645

mol

J 16049

expkPa10

vif

MPa38.9vif

(b)

Following the development in Example 7.2, we can use the following equation to calculate the

fugacity from the van der Waals EOS,

i

i

i

vi

RTv

a

RT

bv

bv

bf

2lnexp

The “a” and “b” parameters for water are

mol

mJ 554.0

3

a

mol

m 1005.3

35b

The molar volume can be found from the van der Waals equation using the “solver” function on

a calculator.

mol

m 1080.3

34

iv

Substituting these values into the expression for fugacity, we get

MPa77.9vif

(c)

The reduced temperature and pressure can be calculated from data in Table A.1.2:

1rT 52.0rP 344.0

By interpolation of the data in Tables C.7 and C.8, we obtain

0log 0.08 1

log 0.0152

Calculate :

0 1log log log

82.0

Therefore,

MPa 47.982.0atm 114 Pf vi

The agreement in values is good for the three methods. However, Part (a) likely provides the

most accurate value since it is based directly on measured data for water.

7.19

The solution method will be illustrated for parts A and F only. The answers only will be given

for the remaining parts. The generalized correlation tables can be used to answer each part,

except Part F, for which we can use the van der Waals EOS.

(a) CH4

The reduced temperature and pressure can be found using data from Appendix A:

06.4rT 26.3rP

Also from Appendix A,

008.0

Using reduced temperature and pressure, the fugacity coefficient can be found from Tables C.7

and C.8:

0181.00614.0008.00176.0logloglog10

iivi

04.1 vi

Calculate fugacity:

bar 156vif

(b) C2H6

02.1vi

bar 153vif

(c) NH3

98.0vi

bar 147vif

(d) (CH3)2CO

842.0vi

bar 3.126vif

(e) C6H12

803.0vi

bar 5.120vif

(f) CO

Following the development in Example 7.2, we can use the following equation to calculate the

fugactity from the van der Waals EOS

i

i

i

vi

RTv

a

RT

bv

bv

bf

2lnexp

The “a” and “b” parameters for CO are (see Chapter 4 for equations)

mol

mJ 147.0

3

a

mol

m 1095.3

35b

The molar volume can be found from the van der Waals equation using the “solver” function on

a calculator.

mol

m 1048.4

34

iv

Substituting these values into the expression for fugacity, we get

bar5.156vif

Calculate the fugacity coefficient:

04.1P

f viv

i

As the strength of intermolecular forces between molecules increases, the fugacity coefficient

and fugacity decreases.

7.20

Use the virial equation expanded in pressure to express z as a function of pressure.

PBz '1

Calculate B’:

5 0.9 1 ' 30 10 Paz B

8 -1' 3.33 10 PaB

Find an expression for v:

PBP

RTv '1

Substitute into Equation 7.8 and integrate to obtain

low

vi

lowlow P

fPPB

P

Pln'ln

Simplifying and allowing Plow to go to zero results in

PBPf vi 'exp

Therefore,

bar 1.27vif

and

903.0bar 30

bar 1.27v

i

7.21

(a)

Equation 7.8:

P

P

ilow

i

low

dPvP

fRT ln

Manipulate the fugacity expression given in the problem statement to obtain:

CPP

P

P

f

P

f low

low

ii

lnlnln

Rearrange the above equation and substitute it into Equation 7.8.

P

P

ilowlow

low

dPvPPPT

RTP

PCPRT lnln

30065.0ln

Differentiate both sides with respect to P:

TPRTvi

30065.0

1

(b)

Substitute numerical values:

bar 30

1

K 15.353

30065.0K 15.353

Kmol

barm 10314.8

35

iv

mol

m 1093.3

34

iv

7.22

Equation 7.8:

P

P

ilow

vi

low

dPvP

fRT ln

The equation from the problem statement can be rearranged to yield

6.1

6.1422.0083.0

1

T

T

TP

T

PRTv c

c

ci

Substitute the above expression into Equation 7.8 and integrate (constant T):

lowc

c

c

lowlow

vi PP

T

T

TP

T

P

P

P

f

6.1

6.1422.0083.0lnln

or

lnf i

v

P

Tc

PcT0.083

0.422Tc

1.6

T1.6

P Plow

Let Plow go to zero gives:

f i

v PexpTcP

PcT0.083

0.422Tc

1.6

T1.6

Pexp

Pr

Tr

0.0830.422

Tr

1.6

and

i

v expTcP

PcT0.083

0.422Tc

1.6

T1.6

exp

Pr

Tr

0.0830.422

Tr

1.6

From Appendix A.1:

K2.373cT

bar 37.89cP

At 300 K and 20 bar, the expressions for fugacity and the fugacity coefficient provide

bar 33.17vif

867.0vi

7.23

We are given the Schrieber volume-explicit equation of state:

2RT kP c

v bP T

Starting with Equation 7.8, we substitute the EOS:

2

ln

low low

P Pv

ii

low P P

f RT kP cRT v dP b dP

P P T

Carrying out the integration, we find:

3

ln ln3

low

Pv

i

low P

f kP cPRT RT P bP

P T T

Expand the logarithm terms, so that we can cancel the ln lowP terms.

ln lnv

i lowRT f RT P ln ln lowRT P RT P 3

3low

P

P

kP cPbP

T T

Collect the log terms on the left-hand side, and use the definition of the fugacity coefficient for a

pure species:

3

ln ln3

low

Pv

vii

P

f kP cPRT RT bP

P T T

Now, since we are free to choose an arbitrary pressure for Plow, we will choose a pressure

vanishingly close to 0. Thus, as 0lowP ,

3

ln3

v

i

kP cPRT bP

T T

Rearranging this equation gives us an expression for the fugacity coefficient:

31

3

2expv

i

kP cP Pb

RT RT

7.24

For an ideal gas reference state

ln

low

Pvo i

i i i

low P

fg g RT v dP

P

From the equation of state

6 8

2 3

6.70 10 4.83 10RTP

v v v

So

6 8

2 3 4

13.40 10 14.49 10RTdP dv

v v v

And

6 8

2 3

13.40 10 14.29 10ln

i

low

vv

i

RTlow

P

f RTRT dv

P v v v

Integrating:

6 6 8 8

22

13.40 10 13.40 10 7.245 10 7.245 10ln ln

v

i i

RT RT

RTlow i iP Plow low

Plow

f vRT RT

P v v

Canceling out ln(Plow) subtracting ln(P), and letting Plow ->0, we get

6 8

2

1

13.40 10 7.245 10ln ln ln

v

i ii

i

f PvRT RT RT

P RT v v

6 8 6 8

2 2

1

6.70 10 4.83 10 13.40 10 7.245 10ln ln 1i

i i iRTv RTv RTv RTv

Substituting in numbers

0.79i and 15.2 bari if P

7.25

We need to pick an equation of state. We will use the virial expansion in pressure:

2' '1iPvz B P C P

RT

For the two states we have

v P T R z

m3/mol Pa K J/mol K

State 1 1.86 10-3 1.50 106 373.15 8.314 0.90 State 2 6.12 10-4 4.00 106 373.15 8.314 0.79

From this we can solve simultaneous equations for B’ and C

’:

2' '0.1 15 15B C

2' '0.21 40 40B C

Solving we get,

3' 7.58 10 barB and 5 2' 5.76 10 barC

We next solve for fugacity and fugacity coefficient using this equation of state:

ln

low

Pv

ii

low P

fRT v dP

P

2 2'1 ' ' 'ln ln

2low

Pv

ilow low

low lowP

f P CB C P dP B P P P P

P P P


2'

'ln ln2

v

ii

f CB P P

P

Substituting in numerical values gives:

0.736i

and

36.8 barv

if

7.26

To determine the fugacity of pure methane at 220 K and 69 bar accurately, we can use the

generalized correlations. The reduced temperature and pressure can be found using data from

Appendix A:

220

1.15190.6

r

c

TT

T

691.5

46r

c

PP

P


0.008

Using reduced temperature and pressure, the fugacity coefficient can be found from Tables

C.7 and C.8:

0 1log log log 0.16 0.008 0.034 0.16v

i i i

0.69v

i

Calculate fugacity:

1 1 47.7 barvf P

7.27

The data in the problem of the first printing are incorrect and should read

P [bar] v [m3/mol]

1.0 2.45 x10-2

5.1 4.78 x10-3

10.1 2.32 x10-3

15.2 1.50 x10-3

20.2 1.08 x10-3

25.3 8.34 x10-4

30.3 6.66 x10-4

35.4 5.44 x10-4

40.4 4.52 x10-4

45.5 3.78 x10-4

50.5 3.17 x10-4

Equation 7.9:

P

P

ivi

ideal

dPP

z 1ln

The following graph has been created with data above:

Integrate the data numerically using the Trapezoid Rule. We obtain

lni

v 0.3

Therefore,

i

v 0.74 and

fi

v i

vP 37.5 bar

-8

-7

-6

-5

-4

-3

-2

-1

0

0.0 10.0 20.0 30.0 40.0 50.0 60.0

(z-1

)/P

x 1

00

0 [

ba

r]

P (bar)

7.28

Rearrangement of Equation 7.9 yields

ln i P

zi 1

P

We can approximate the derivative at 500 bar by drawing a tangent line to the plot provided in

the problem statement and calculating the slope.

lni P

0.001 bar -1

Solving for the compressibility factor gives,

1001.0 PRT

Pvz ii

Therefore, the molar volume is:

v i

8.314 J

mol K

373.15 K 0.001 bar -1 500 bar 1

500 105 Pa

or

mol

m101.3

35

iv

7.29

If a gas obeys the Lewis fugacity rule, all the intermolecular interactions are the same. Therefore,

0mixh

So

1 1 2 2

Jln ln 1,730

molmix mixg T s RT x x x x

7.30

We want to calculate the fugacity coefficient of pure n-butane. The reduced temperature and

pressure can be found using data from Appendix A:

318.9

0.75425.2

r

c

TT

T

3.790.1

37.9 r

c

PP

P


0.193

Using reduced temperature and pressure, the fugacity coefficient can be found from Tables

C.7 and C.8:

0 1log log log 0.035 0.193 0.030 .041 v

i i i

0.91 v

i

Calculate fugacity:

1 1 1ˆ 0.41 barvf y P

7.31

(a)

1 1

vf P


190.6

1.0190.6

rT 32.2

0.7046.00

rP


0.008


and C.8:

0 1log log log 0.113 0.008 0.022 0.113v

i i i

0.77v

i

Calculate fugacity:

24.8 barv

if

(b)

1 1 1 19.8 barvf y P

7.32

(a)

Pure Species:

11ln

low

Pv

low P

fRT v dP

P

To acquire an expression for v1, set the mole fraction of species 1 equal to unity.

RT

BAP

PRTv

11

Substituting this expression into the integral and integrating, we get

22lnln

221 low

lowlow

v PP

RT

BA

P

P

P

f

By simplifying and letting 0lowP ,

ln f1

v ln P A B RT

P2

2

Therefore,

ln 1

v lnf1

v

P

A B

RT

P2

2

2

50100.3100.9exp

2exp

255

2

1P

RT

BAv

93.01

v

Species in a mixture:

11

1

ˆln

low

Pv

low P

fRT V dP

y P

The expression for the extensive volume is

32121321 nnnBnnAPP

RTnnnV

Therefore,

BAPP

RTV 1

Substituting this result into the above integral and integrating, we obtain the following after

simplifiction

2

1ˆln

2

v P A B

RT RT

2

1ˆ exp

2

v P A B

RT RT

2

5 5

1 2 2

50 atm 1 1ˆ exp 9.0 10 3.0 10 2 atm atm

v

1ˆ 0.93v

(b)

When the vapor and liquid are in equilibrium,

1 1ˆ ˆv lf f

Hence,

1ˆ 15 atmvf

We also know the value of the fugacity coefficient from part (a). This can be used to calculate

the mole fraction in the vapor.

11

1

ˆˆ 0.93

vv f

y P

322.0

atm5093.0

atm151 y

7.33

(a)

The Lewis fugacity rule:

1 1ˆv v

From Example 7.2

2

ln lniv

i

i i

v b Pb a

v b RT RTv

From the van der Waals EOS,

mol

m 1015.1

33

1v

Now, the fugacity coefficient can be computed by substituting values.

89.01

v

The fugacity is calculated using the fugacity coefficient.

1ˆ 0.2 0.89 30 bar 5.34 barvf

(b)

The truncated virial form of the van der Waals equation for the mixture is

RT

anbnP

RTnV mixTmixTT

1

The expressions for mixa and mixb are

2332133112213232

221

21 222 ayyayyayyayayayamix

12 1 2 13 1 3 23 2 3, , a a a a a a a a a

332211 bybybybmix

Rewriting the volume in terms of moles, we get

23321

3213

321

3112

321

213

321

23

2321

22

1321

21

33221121

222

1

annn

nna

nnn

nna

nnn

nna

nnn

n

annn

na

nnn

n

RTbnbnbn

P

RTnnV

To calculate fugacity, we use the following expression

11

1

ˆln

low

Pv

low P

fRT V dP

y P

Find 1V :

mix

nnPT

aayayayRT

bP

RT

dn

VV

13322111

,,,11 222

1

32

Substituting this expression into the integral and integrating we get,

1 11 1 2 2 3 132

1

ˆ 1ln ln 2 2 2

v

low mix low

low low

f bPP P y a y a y a a P P

y P P RT RT

By subtracting lowPln and Pln from both sides and then letting 0lowP , we obtain

1 11 1 1 2 2 3 132

1

ˆ 1ˆln ln 2 2 2v

v

mix

f by a y a y a a P

y P RT RT

Substitute numerical values and evaluate:

1ˆln 0.10v

1ˆ 0.906v

Calculate fugacity:

1 1 1ˆ ˆ 0.2 0.906 30 barv vf y P

1ˆ 5.44 barvf

7.34

(a)

Equation 7.14:

, ,

ˆln

i a

low

Vv

a

nRTa low a T V nP

f PRT dV

y P n

The equation of state provided in the problem statement can be rewritten as

2

222 222

V

BnnBnnBnnBnBnBn

V

nnnRTP

bccbaccaabbacccbbbaaacba

Therefore,

2,,

2221

V

BnBnBn

VRT

n

P accabbaaa

nVTaai

Substitute the above equation into Equation 7.14 and integrate

ˆ 2 2 22 2 2

ln ln lnv

a aa b ab c ac lowa a aa b ab c ac

a low low

n B n B n B Pf n B n B n B nRTV

y P V P nRT

Canceling the lowPln from both sides and then allowing lowP to go to zero, we obtain

ˆ 2 2 2 2 2 2ln ln ln ln

v

a a aa b ab c ac a aa b ab c ac

a

f n B n B n B n B n B n BnRTV nRT

y V V V

Now subtract Pln from both sides:

ˆ 2 2 2ˆln ln ln

vva a aa b ab c aca

a

f n B n B n BnRT

y P PV V

2 2 21ˆln lnv a aa b ab c ac

a

n B n B n B

z V

2 2 21ˆ expv a aa b ab c ac

a

y B y B y B

z v

2 2 2ˆ expv a a aa b ab c ac

a

y P y B y B y Bf

z v

(b)

For this system

2.0ay

3.0by

5.0cy

Using the virial coefficient data:

mol

m 10392.2

34

mixB

Therefore,

v

B

RT

Pvz mix 1

mol

m 1036.1

33v

(Note: There are two solutions to the equation, but the other value is not sensible.)

Now, calculate the fugacity coefficient by substituting the appropriate values:

ˆ 1.04v

a

ˆ 0.2 15 bar 1.04 3.12 barv

af

(c)

The EOS reduces to the following expression for pure methane:

Pva

RT1

Baa

va

For this expression, we can write

dP RT1

va

2

2Baa

va

3

dva

Develop an expression for the fugacity coefficient similar to the method in Example 7.2

vadPRT

Plow

va

RT lnfa

v

Plow

Substitute the expression for dP into the above integral:

2

21ln

low

v v

aa aa

RT a a low

P

B fRT dv RT

v v P

Integrating and evaluating the limits, we obtain

lnfa

v

Plow

lnva

2Baa

va

ln

RT

Plow

2Baa

RT

Plow

Cancel the logarithmic terms containing Plow and then let Plow go to zero:

ln fa

v 2Baa

va

lnva

RT

Therefore,

fa

v RT

va

exp2Baa

va

a

v 1

za

exp2Baa

va

Since we are employing the Lewis fugacity rule

ˆ av a

v 1

za

exp2Baa

va

ˆ f av ya fa

v yaRT

va

exp2Baa

va

To calculate the fugacity and fugacity coefficient, we will need to find the pure species molar

volume:

va 1.61103 m3

mol

Substituting values into the above expressions, we obtain

ˆ av 0.975

ˆ f av 2.92 bar

7.37

Calculate fugacity and fugacity coefficient of phenol in a mixture of 20 mole % phenol (1) and

80 mole % oxygen (2) at 694.2 K and 24.52 bar using the following:

(a) Ideal gas law

For an ideal gas, we assume the fugacity coefficient is 1.

Next fugacity is calculated by:

1 1 1ˆ ˆ 0.2 1 24.52 barv vf y P

1ˆ 4.904 barvf

(b) The Lewis Fugacity Rule (choose the method that gives you as accurate an answer as

possible).

Options:

1) Ideal gas – definitely not the most accurate!

2) E.O.S – The van der Waals E.O.S is more accurate than the ideal gas assumption,

however, the “van der Waals equation is not as accurate as more modern cubic equations

of state.” (Text pg 310)

3) Generalized Correlations – as noted in class on 2-6-08, the Lee-Kesler tables (utilized in

the generalized correlations) are more accurate than the van der Waals E.O.S.

From the above information, we can choose a more modern cubic equation of state (as will be

demonstrated with the Redlich-Kwong E.O.S. in part c) or we can choose the Generalized

Correlations. For variety, let’s use the Generalized Correlations.

To use the tables in the book, we utilize the form of the generalized correlations:

(0) (1)

1log( ) log log

r

c

PP

P

r

c

TT

T

From Appendix A.1 we find Pc and Tc for phenol, which we can then use to get Pr and Tr.

Pc 61.3

bar Pr 0.4

Tc 694.2

K Tr 1

ω 0.440

Now we can use these values to find (0) (1)& . (0)log 0.061 (1)log 0.0122

1log( ) ( 0.061) 0.440 ( 0.0122)

We are using the Lewis Fugacity Rule, so the pure species fugacity coefficient is equal to the

mixture fugacity coefficient

1 1ˆ 0.86

1 1 1ˆ 0.2 0.86 24.52 barv vf y P

1ˆ 4.2 barvf

We can check our answer for fugacity coefficient with ThermoSolver:

We can also look at results for fugacity coefficient using the Peng Robinson E.O.S. via

ThermoSolver:

Note: For the Peng Robinson E.O.S. we would use the pure species fugacity coefficient given in

the right column with the Lewis Fugacity Rule, but ThermoSolver gives both the pure species

fugacity coefficient as well as the fugacity coefficient of species i in the mixture.

(c) The Redlich-Kwong Equation of State, See Table 7.1

From Table 7.1 we get

11 1 11.5 1.5

1 1ˆln( ) ln ln 1 2 ln 1( )

v b aRT b a bb a a

Pv v v b bRT v b bRT b v

Note: In this equation a represents amix and b represents bmix.

We need to find a1, a2, a, b1, b2, b and v

2 2

1 1 1 2 1 2 2 22a y a y y a a y a

1 1 2 2b y b y b

1 0.2y

2 0.8y

Note: The Redlich-Kwong parameters a and b are different from those for the van der Waals

equation and cannot be interchanged 2 2.50.42748 c

c

R Ta

P

0.08664 c

c

RTb

P

Phenol (subscript 1) Oxygen (subscript 2) Mixture (no subscript)

Tc [K] 694.2 154.6

Pc [bar] 61.3 50.46

a [J m3K

0.5/mol

2] 61.2 1.740 6.86

b [m3/mol] 8.16E-05 2.207E-05 3.40E-05

Next we can use the Redlich-Kwong E.O.S. to find v.

1/2

RT aP

v b T v v b

v can be found by using a solver function: 3m

0.00234 mol

v

Or by using an approximation as demonstrated below:

v(i1) RT

P b

a v(i) b PT1/ 2v(i) v(i) b

Start with the ideal gas law:

3

(0) 3 m2.35 10

mol

RTv

P

then 3

(1) 3 m2.34 10

mol

RTv

P

: and 3

(2) 3 m2.34 10

mol

RTv

P

Now use the equation from Table 7.1 to find the fugacity coefficient.

11 1 11.5 1.5

1 1ˆln( ) ln ln 1 2 ln 1( )

v b aRT b a bb a a


1ˆ 0.943v

1 1 1ˆ ˆ 0.2 0.94 24.52 barv vf y P

1ˆ 4.63 barvf

7.38

(a)

11ln

low

Pv

low P

fRT v dP

P

,1

1

,18

r

r

PRT RTv

P P T

,11

,1

ln8

low

Pvc

low cP

RTf RTRT dP

P P P

,11

,1

ln ln8

vc

low

low low c

RTf PRT RT P P

P P P

,111

,1

ln ln8

vrv

r

Pf

P T

2190.6

365rT

27.720.6

46.2rP

1

1ln

8

v

1 1.13v

1 1 1ˆ 7.85 barv vf y P

(b)

11

1

ˆln

low

Pv

low P

fRT V dP

y P

1 ,1 2 ,2

1 2

,1 ,28 8

r r

r r

n P n PRT RTV n n

P P T T

2 3

,1

1

1 ,1, , ,8

c

cT P n n

RTV RTV

dn P P

,11

1 ,1

ˆln

8low

Pvc

low cP

RTf RTRT dP

y P P P

,11

1 ,1

ˆln ln

8

vc

low

low low c

RTf PRT RT P P

y P P P

,111

1 ,1

ˆˆln ln

8

vrv

r

Pf

y P T

2190.6

365rT

27.720.6

46.2rP

1

1ˆln

8

v

1ˆ 1.13v

1 1 1ˆ ˆ 7.85 barv vf y P

(c)

Values are the same so all the intermolecular interactions are the same

7.39

(a)

1 2 2 ' ' 2 '

1 11 1 2 12 2 22

1 2

2n n RT RT

V n B n n B n BP n n

2

2 ' ' 2 ' ' '

1 1 11 1 2 12 2 22 1 11 2 12

1 , ,

2 2 2

T P n

V RTV RT y B y y B y B RT y B y B

n P

2

' ' '

1 1 11 2 12

1 , ,

2 2

T P n

V RTV RT B RT y B y B

n P

' ' '11 1 11 2 12

1

' ' '

1 11 2 12

ˆ 1ln 2 2

ln 2 2

low low

P Pv

low P P

low

low

fRT V dP RT B y B y B dP

y P P

Py B y B B P P

P

' ' '11 1 11 2 12

1

ˆˆln ln 2 2

vvf

y B y B B Py P

(b)

y1 = 0.4 '

11B - 1.9 x 10-7

[Pa-1

]

'

12B - 3.6 x 10-8

[Pa-1

]

'

22B - 2.0 x 10-9

[Pa-1

]

So

1̂ 0.84v

And

1 1 1ˆ ˆ 4.0 barv vf y P

(c)

Since '

11B and '

11B are very different, we do not expect the Lewis rule to be a good approximation.

7.40

Using the Lewis fugacity rule for n-pentane (1):

1 1 1ˆ vf y P


495

1.0469.6

rT 18

0.7038.74

rP


0.251


and C.8. After interpolation, we get:

0 1log log log 0.086 0.251 0.019 0.0907v

i i i

0.81v

i

Calculate fugacity:

1ˆ 0.137 0.81 18 2.00 barvf

7.41

From the definition of fugacity:

2 31 1 , , ,

ˆln

Tlow

low

P Vv

ii

n RTlow P T V n n

P

f PRT V dP dV

y P n

Putting the EOS in terms of T, V, n1, and, n2:

2 21 2 1 1 2 2 1 2 12

2

1 1 2 2

2n n RT n a n a n n aP

V V n c n c

Differentiating

2 3

2 2

1 1 1 2 2 1 2 121 1 2 12

2 3

1 , , , 1 1 2 2 1 1 2 2

2 22 2

T V n n

c n a n a n n an a n aP RT

n V V n c n c V n c n c

Into the equation above:

2 2

1 1 1 2 2 1 2 121 1 2 12

2 3

1 1 1 2 2 1 1 2 2

ˆ 2 22 2ln

T T T

low low low

V V Vv

i

n RT n RT n RTlow

P P P

c n a n a n n af n a n aRTRT dV dV dV

y P V V n c n c V n c n c

Integrating we get

1 1 2 12

1 1 1 2 21 1 2 2

2 2

1 1 1 2 2 1 2 12 2 2

1 1 2 2

1 1 2 2

ˆ 1 1ln ln 2 2

1 12

v

i

T Tlow

low low

T

low

f VRT RT n a n a

n RT n RTy P V n c n cn c n c

P P

c n a n a n n aV n c n c n RT

n c n cP

Taking and 1 1 2 2T

low

n RTn c n c

P

1 1 2 12

1 1 1 2 2

2 2

1 1 1 2 2 1 2 12 2 2

1 1 2 2

ˆ 1 1ln ln 2 2

1 12

v

i

T Tlow

low low

T

low

f VRT RT n a n a

n RT n RTy P V n c n c

P P

c n a n a n n aV n c n c n RT

P


1 1 2 12

1 1 1 2 2

2 2

1 1 1 2 2 1 2 12 2

1 1 2 2

ˆ 1ln ln 2 2

12

v

i

low

fRT RT z n a n a

y P V n c n c

c n a n a n n aV n c n c

Simplifying:

1 1 2 12 11 2

ˆln ln 2y a y a c a

RT RT zv c v c

7.42

From the definition of fugacity:

2 31 1 , , ,

ˆln

Tlow

low

P Vv

ii

n RTlow P T V n n

P

f PRT V dP dV

y P n

Putting the EOS in terms of T, V, n1, and, n2:

2 2 21 2 3 1 1 2 2 3 3 1 2 12 1 3 13 2 3 23

2

1 1 2 2 3 3 1 1 2 2 3 3

2 2 2n n n RT n a n a n a n n a n n a n n aP

V n b n b n b V n c n c n c

Differentiating

2 3

1 1 2 3

2

1 1 1 2 2 3 3, , , 1 1 2 2 3 3

2 2 2

1 1 1 2 2 3 3 1 2 12 1 3 13 2 3 231 1 2 12 3 13

2 3

1 1 2 2 3 3 1 1 2 2 3 3

2 2 2 22 2 2

T V n n

b n n n RTP RT

n V n b n b n b V n b n b n b

c n a n a n a n n a n n a n n an a n a n a

V n c n c n c V n c n c n c

1 1 1 2 2 3 3

1 1 2 3

2

1 1 2 2 3 3

1 1 2 12 3 13

2

1 1 2 2 3 3

2 2 2

1 1 1 2 2 3 3 1 2 1

ˆln

2 2 2

2 2

T

low

T

low

T

low

Vv

i

n RTlow

P

V

n RT

P

V

n RT

P

f RTRT dV

y P V n b n b n b

b n n n RTdV

V n b n b n b

n a n a n adV

V n c n c n c

c n a n a n a n n a

2 1 3 13 2 3 23

3

1 1 2 2 3 3

2 2

T

low

V

n RT

P

n n a n n adV

V n c n c n c

Integrating we get

1 1 2 2 3 3

11 1 2 2 3 3

1 1 2 3

1 1 2 2 3 31 1 2 2 3 3

1 1 2 12 3 13

1 1 2 2 3 31 1

ˆln ln

1 1

1 12 2 2

v

i

Tlow

low

T

low

T

low

V n b n b n bfRT RT

n RTy Pn b n b n b

P

RTb n n nn RTV n b n b n b

n b n b n bP

n a n a n an RTV n c n c n c

n c nP

2 2 3 3

2 2 2

1 1 1 2 2 3 3 1 2 12 1 3 13 2 3 23

2 2

1 1 2 2 3 3

1 1 2 2 3 3

2 2 2

1 1

T

low

c n c

c n a n a n a n n a n n a n n a

V n c n c n c n RTn c n c n c

P

Taking 1 1 2 2 3 3T

low

n RTn b n b n b

P and 1 1 2 2 3 3

T

low

n RTn c n c n c

P

1 1 2 2 3 3

1

1 1 2 3

1 1 2 2 3 3

1 1 2 12 3 13

1 1 2 2 3 3

2 2 2

1 1 1 2 2 3 3 1

ˆln ln

1 1

1 12 2 2

2

v

i

Tlow

low

T

low

T

low

V n b n b n bfRT RT

n RTy P

P

RTb n n nn RTV n b n b n b

P

n a n a n an RTV n c n c n c

P

c n a n a n a n n

2 12 1 3 13 2 3 23

2 2

1 1 2 2 3 3

2 2

1 1

T

low

a n n a n n a

V n c n c n c n RT

P


1 1 2 2 3 3

1

1 1 2 3

1 1 2 2 3 3

1 1 2 12 3 13

1 1 2 2 3 3

2 2 2

1 1 1 2 2 3 3 1 2 12 1 3 13 2 3 23 2

1 1 2 2 3 3

ˆln ln

1

12 2 2

12 2 2

v

i

T

PV P n b n b n bfRT RT

y P n RT

RTb n n nV n b n b n b

n a n a n aV n c n c n c

c n a n a n a n n a n n a n n aV n c n c n c

Simplifying:

1 1 2 12 3 131 11 2

ˆln ln 2y a y a y aRTb c abP

RT RT zRT v b v c v c

7.43

Let “a” refer to methane and “b” refer to hydrogen sulfide. From Example 7.4

2

ˆln lna a b a bmixv a

a

mix

y a y a aP v b b

RT v b RTv

In the above expression, v and mixb depend on mole fractions. First, calculate aa , ba , ab , and

bb :

c

c

P

RTa

2

64

27

c

c

P

RTb

8

2

3

mol

mJ 230.0aa

mol

m 1031.4

35

ab

2

3

mol

mJ 454.0ba

mol

m 1034.4

35

bb

To calculate v, we need amix and bmix:

bbbabaaamix ayaayyaya 22 2

bbaamix bybyb

Using these expressions we can find the molar volume with the van der Waals EOS.

2v

a

bv

RTP mix

mix

Once the molar volume is calculated, the molar fugacity coefficient can be found using the

expression from Example 7.4. The following table can be created:

ya yb amix bmix v ln(a) a

0 1 0.454 4.34 x 10-5

0.000437 0.001 1.001

0.1 0.9 0.428 4.34 x 10-5

0.000447 -0.007 0.993

0.2 0.8 0.403 4.33 x 10-5

0.000456 -0.013 0.987

0.3 0.7 0.379 4.33 x 10-5

0.000465 -0.018 0.982

0.4 0.6 0.355 4.33 x 10-5

0.000473 -0.023 0.978

0.5 0.5 0.333 4.33 x 10-5

0.000481 -0.026 0.974

0.6 0.4 0.311 4.32 x 10-5

0.000488 -0.029 0.972

0.7 0.3 0.289 4.32 x 10-5

0.000494 -0.031 0.970

0.8 0.2 0.269 4.32 x 10-5

0.000500 -0.032 0.968

0.9 0.1 0.249 4.31 x 10-5

0.000506 -0.033 0.967

1 0 0.230 4.31 x 10-5

0.000512 -0.033 0.967

From ThermoSolver using the Peng-Robinson EOS:

ya yb aö

0 1 1.033

0.1 0.9 1.02

0.2 0.8 1.008

0.3 0.7 1.000

0.4 0.6 0.992

0.5 0.5 0.986

0.6 0.4 0.982

0.7 0.3 0.978

0.8 0.2 0.977

0.9 0.1 0.975

1 0 0.975

Plot the activity coefficient versus the mole fraction of methane for both methods on the same

graph:

The values agree relatively well.

a vs. ya From the van der Waals EOS

Compared to Thermosolver Results

0.950

0.975

1.000

1.025

1.050

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

ya (mole fraction)

a

van der Waals EOS

Thermosolver Results

7.44

Let “a” refer to methane and “b” refer to hydrogen sulfide. From Problem 7.5:

1.5 1.5

1 1ˆln ln ln 1 2 ln 1v aa a a

b aRT b a bb a a


(Note: a is short for amix and b is short for bmix)

In the above expression, v , amix, and mixb depend on mole fractions. First, calculate aa , ba , ab ,

and bb :

c

c

P

TRa

5.22

42748.0 c

c

P

RTb 08664.0

2

1/23

mol

KmJ 22.3aa

mol

m 1099.2

35

ab

2

1/23

mol

KmJ 90.8ba

mol

m 1001.3

35

bb

To calculate the v, we need amix and bmix:

bbbabaaamix ayaayyaya 22 2

bbaamix bybyb

Using these expressions we can find the molar volume with the Redlich-Kwong EOS:

mix

mix

mix bvvT

a

bv

RTP

2/1

Once the molar volume is calculated, the fugacity coefficient can be found by substituting the

appropriate values into the expression from Problem 7.5. The following table can be created:

ya yb amix bmix v aöln aö

0 1 8.900 3.01 x 10-5

0.000437 0.036 1.037

0.1 0.9 8.205 3.01 x 10-5

0.000449 0.024 1.024

0.2 0.8 7.538 3.01 x 10-5

0.000460 0.013 1.013

0.3 0.7 6.899 3.00 x 10-5

0.000470 0.005 1.005

0.4 0.6 6.289 3.00 x 10-5

0.000479 -0.002 0.998

0.5 0.5 5.707 3.00 x 10-5

0.000487 -0.007 0.993

0.6 0.4 5.153 3.00 x 10-5

0.000495 -0.011 0.989

0.7 0.3 4.627 3.00 x 10-5

0.000502 -0.014 0.986

0.8 0.2 4.130 2.99 x 10-5

0.000508 -0.016 0.984

0.9 0.1 3.661 2.99 x 10-5

0.000514 -0.018 0.982

1 0 3.220 2.99 x 10-5

0.000520 -0.018 0.982

We have from the Peng-Robinson EOS using ThermoSolver:

ya yb aö

0 1 1.033

0.1 0.9 1.02

0.2 0.8 1.008

0.3 0.7 1.000

0.4 0.6 0.992

0.5 0.5 0.986

0.6 0.4 0.982

0.7 0.3 0.978

0.8 0.2 0.977

0.9 0.1 0.975

1 0 0.975

The data plotted on the same graph reveal:

Clearly, the results from the two solution methods agree well. The Redlich-Kwong EOS

provides a fugacity coefficient that is slightly larger over the entire composition range.

a vs. y a From the Redlich-Kwong EOS

Compared to Thermosolver Results

0.950

0.975

1.000

1.025

1.050

0 0.2 0.4 0.6 0.8 1ya

a Redlich-Kwong EOS

Thermosolver

7.45

Let “a” refer to methane and “b” refer to hydrogen sulfide.

(a)

From Example 7.4,

2

ˆln lna a b a bmixv a

a

mix

x a x a aP v b b

RT v b RTv

First, calculate aa , ba , ab , and bb :

c

c

P

RTa

2

64

27

c

c

P

RTb

8

2

3

mol

mJ 230.0aa

mol

m 1031.4

35

ab

2

3

mol

mJ 454.0ba

mol

m 1034.4

35

bb

To calculate v, we need amix and bmix:

2

322

mol

mJ 333.02 bbbabaaamix ayaayyaya

mol

m 10325.4

35

bbaamix bybyb

Using these expressions we can find the molar volume with the van der Waals EOS:

2v

a

bv

RTP mix

mix

so

v 4.81104 m3

mol

Substituting these values into the equation for the fugacity coefficient, we get

ˆ 0.974v

a

(b)

From Problem 7.5:

1.5 1.5

1 1ˆln ln ln 1 2 ln 1v aa a a

b aRT b a bb a a


(Note: a is short for amix and b is short for bmix)

First, calculate aa , ba , ab , and bb :

c

c

P

TRa

5.22

42748.0 c

c

P

RTb 08664.0

2

3

mol

mJ 22.3aa

mol

m 1099.2

35

ab

2

3

mol

mJ 90.8ba

mol

m 1001.3

34

bb

To calculate the v, we need amix and bmix:

707.52 22 bbbabaaamix ayaayyaya

mol

m 100.3

35

bbaamix bybyb

Using these expressions we can find the molar volume with the Redlich-Kwong EOS:

mix

mix

mix bvvT

a

bv

RTP

2/1

mol

m 10871.4

34v

Substituting these values into the expression for the fugacity coefficient, we get

ˆ 0.993v

a

(c)

Using Kay’s mixing rules, we have the following expressions:

bcbacapc TyTyT ,,

bcbacapc PyPyP ,,

bbaapc yyw

Substituting values from Appendix A, we get

K 9.281pcT

bar 69.67pcP

054.0pc

Therefore,

58.1rT

03.1rP

From the generalized correlation tables

0336.0log 0

0371.0log 1

Therefore,

logv 0.0336 0.054 0.0371

v 0.930

(d)

Using the Peng-Robinson EOS in ThermoSolver, we obtain:

ˆ 0.986v

a

A summary of the results for each solution method is provided in the following table. The

percent differences are based on the fugacity coefficient found using ThermoSolver.

Solution Method vaö % Difference

(a) van der Waals 0.974 1.22

(b) Redlich-Kwong 0.993 0.71

(c) Generalized Correlations – Kay’s 0.930 5.68

(d) ThermoSolver / Peng-Robinson 0.986 0

Clearly, all of the solution methods agree reasonably well with the Peng-Robinson EOS. The

fugacity coefficient calculated with the Redlich-Kwong EOS agrees the best.

7.46

For the virial equation, we have

RT

PBz mix 1

where

bbbabbaaaamix ByByyByB 22 2

At 127 oC, the second virial coefficients are

mol

cm 16

3

aaB and

mol

cm 101

3

bbB

Solve for volume:

ba

bbbabbaaaabamixT

T

nn

BnBnnBn

P

RTnnBn

P

RTnV

22 2

To get the partial molar volume, we differentiate with repect to na.

2

22

,,

222

ba

bbbabbaaaa

ba

abbaaa

nPTaa

nn

BnBnnBn

nn

BnBn

P

RT

n

VV

b

or

2 22 2 2a a aa b ab a aa a b ab b bb

RTV y B y B y B y y B y B

P

We must now plug this into Equation 7.13 and integrate:

2 2

,

ˆ2 2 2 = ln

low

P v

aa aa b ab a aa a b ab b bb

a lowP

fRTy B y B y B y y B y B dP RT

P p

so

low

P

2 2

P

ˆln 2 2 2 = ln

v

aa aa b ab a aa a b ab b bb

low a low

fPRT y B y B y B y y B y B dP RT

P y P

Rearranging

low

P

2 2

P

ˆˆ2 2 2 = ln ln

vva

a aa b ab a aa a b ab b bb a

a

fy B y B y B y y B y B dP RT RT

y P

Integrating, we get

2 2 ˆ2 2 2 ln v

a aa b ab a aa a b ab b bb ay B y B y B y y B y B P RT

Setting ya=0 and yb=1:

ˆ2 lnab bb aB B P RT

3ˆln

cm34.5

2 mol

a bb

ab

RTB

PB

Compare the value to the geometric average

mol

cm 2.40

3

bbaaab BBB

This problem can also be solved using the form of the virial equation:

z 1Bmix

v

In that case, the solution becomes:

Bab v

2ln

ˆ avP

RT

55.9

cm3

mol

7.47

(a)

We can start with Equation 7.14 to find the fugacity coefficient:

V

P

nRT nVTlow

v

low

dVn

P

Py

fRT

2,,11

1ö

ln

Rewrite the equation of state to include extensive volume and moles:

2/12/3

2/121221121

2/12/3 TV

nnanan

V

RTnn

Tv

a

v

RTP mix

Differentiate:

2/121

22112/12112/12/3

,,1 2

1

2nn

anannna

TVV

RT

n

P

nVT

Substitute this expression into Equation 7.14 and integrate to obtain

1/2 1 1 2 211 1 2 1/21/2 1/2

1 1 2

1/21/2 1 1 2 2

1 1 2 1/21/2

1 2

ˆ 2ln ln

2

2

2

v

low

low

low

n a n aVPfRT RT a n n

y P nRT V T n n

n a n aPa n n

nRT T n n

This expression can be simplified by canceling the terms containing Plow and then taking the

limit as Plow goes to zero. This results in

1/2 1 1 2 211 1 2 1/21/2 1/2

1 1 2

ˆ 2ln ln

2

v n a n af nRTRT RT a n n

y V V T n n

The above equation is equivalent to

11 1 1 2 2

1

ˆ 2 1ln ln

2

vf RTRT RT a y a y a

y v vT

If we subtract the natural log of pressure from both sides of the equation and rearrange, we

obtain

1 1 1 1 2 2

2 1ˆln ln2

v RTa y a y a

Pv RT vT

Now, find the numerical value of the fugacity coefficient by substituting values for all of the

variables.

1 6

8.314 500 2 1ˆ exp ln 800 0.33 800 0.66 50021.78 10 0.002 8.314 500 0.002 500

v

1ˆ 0.689v

Note: The pressure was calculated prior to substitution using the given EOS.

Pa1078.1

500002.0

50066.080033.0

002.0

500314.8 6

3

P

(b)

The Lewis fugacity rule states

1 1ˆv v

Start with Equation 7.26 to find v1

:

P

Plow

v

low

dPvP

fRT 1

1ln

For the given equation of state

12/51

2/1

1

21

2

3dv

vT

a

v

RTdP

Substitute the above expression into Equation 7.8 and change the limits of integration:

1

12/31

2/1

1

1

1

2

3ln

v

P

RTlow

v

low

dvvT

a

v

RT

P

fRT

Now perform the integration to obtain

RT

P

T

a

Tv

a

RT

PvRT

P

fRT lowlow

low

v1

1

111 33lnln

If we cancel the natural log terms containing Plow and then let Plow go to zero, the above equation

simplifies to

1

1

11

3lnln

TvRT

a

v

RTf v

Now subtract the natural log of P from both sides of the equation:

1

1

11

1 3lnlnln

TvRT

a

Pv

RT

P

f vv

v1 is the molar volume of species 1 at the temperature and pressure of the mixture in Part (a) We

can calculate it from the given EOS.

31

1

6

500

8005008.314Pa 1087.1

vv

mol

m00187.0

3

1v

Substitute values into the expression for the fugacity coefficient and evaluate:

1 6

8.314 500 3 800exp ln

1.78 10 0.00187 8.314 500 500 0.00187

v

687.01

v

The fugacity coefficient calculated using the Lewis fugacity rule is equal to the fugacity

coefficient in Part (a)

7.48

Gibbs energy can be written as

dPvdTsdg iii

Therefore,

iT

ii

T

i vP

PvTs

P

g

Equation 7.8 states

lowiilow

iii PRTfRTg

P

fRTgg lnlnln ¼¼

Differentiate:

º ln ln lni i low ii

TT T

g RT f RT P fgRT

P P P

Hence,

ln ii

i

TT

fgv RT

P P

7.49

It can be shown that

iT

i vP

fRT

ln

To solve this problem, we can assume the molar volume of liquid water at 300 ºC is independent

of pressure. Therefore,

P

P

li

f

f

isat

i

sat

i

dPvfdRT ln

lnl

sati i

sat

i

f vP P

f RT

We need to calculate sat

if , but before we can do that, we must choose a reference. Use

300 ºCT and kPa10P as the reference. From the steam tables

º kJ9.2812

kg Ks

º kJ3076.5

kgh

The Gibbs energy at the reference state:

º kJ kJ kJ J3076.5 573.15 K 9.2812 2243.0 40407.2

kg kg K kg molg

For MPa 5810.8satP and C¼ 300T , the steam tables allow us the calculate the Gibbs

energy:

mol

J 5.9378

kg

kJ 6.520

Kkg

kJ7044.5K 15.573

kg

kJ9.2748satg

Now use Equation 7.7 to find satf :

15.573314.8

2.404075.9378expkPa 10satf

kPa 6729satf

Once we find liv , we can calculate the fugacity. From the saturated steam tables,

mol

m 1053.2kg/mol0180148.0

kg

m 001404.0

35

3sativ

Therefore,

5

5 52.53 10 6729 kPa exp 300 10 Pa-85.18 10 Pa

8.314 573.15if

bar 75.4kPa 7542 if

7.50

(a)

From Equation 7.35 we have the following relationship

sat

ivi

li Pf

Since the system pressure is low (1 bar), the fugacity coefficient is unity. Calculate the

saturation pressure using Antoine Equation data in Appendix A.1:

42.34260

9.21540580.9expsat

iP

bar 610.0satiP

Therefore,

bar 610.0 sati

li Pf

(b)

Now that the system pressure is high, we can’t simplify the calculation as we did in Part (a).

Instead, we will use Equation 7.36.

P

P

isati

sati

li

sat

i

dPRT

vPf exp

Since bar 610.0satiP , the saturation fugacity coefficient of n-butane is unity. From the

problem statement, we can obtain vi:

3

4

3

0.058123 kg/mol m1.00 10

mol579 kg/m

ii

i

MWv

Since the density is constant with respect to pressure, we can substitute numerical values into

Equation 7.36 and integrate:

4 3 5 41.00 10 m /mol 200 10 Pa-6.10 10 Pa0.610 bar exp

J8.314 260 K

mol K

l

if

1.53 barl

if

7.51

We can use Equation 7.36 to calculate the fugacity of pure liquid acetone.

P

P

isati

sati

li

sat

i

dPRT

vPf exp

Calculate the saturation pressure using Antoine’s Equation data in Appendix A.1:

93.35382

46.29400311.10expsat

iP

bar 64.4satiP

Since the saturation pressure is 4.64 bar, we cannot assume the saturation fugacity coefficient to

be unity. We can use reduced generalized correlation tables to calculate the fugacity coefficient.

From data in Appendix A.1,

75.0K 508.1

K 382rT 099.0

bar 47.01

bar .644sat

rP 309.0

Use these values in Tables C.7 and C.8 to determine the fugacity coefficient:

0438.00297.0309.00346.0log sati

904.0sati



5 3 5 57.34 10 m /mol 100 10 Pa-4.64 10 Pa0.904 4.64 bar exp

J8.314 382 K

mol K

l

if

5.23 barl

if

7.52


P

P

isati

sati

li

sat

i

dPRT

vPf exp

Calculate the saturation pressure using Antoine’s Equation data in Appendix A.1:

1872.46

exp 9.1058333 25.16

sat

iP

20.56 barsat

iP

Since the saturation pressure is 20.56 bar, we cannot assume the saturation fugacity coefficient to

be unity. We can use reduced generalized correlation tables to calculate the fugacity coefficient.

From data in Appendix A.1,

333 K

0.9370 K

rT 20.53 bar

0.4844.24 bar

sat

rP 0.152

Use these values in Tables C.7 and C.8 to determine the fugacity coefficient:

log 0.107 0.152 0.051 0.115sat

i

0.767sat

i



Taking the Poynting correction to be negligible, we get

15.77 barl sat sat

i i if P

7.53


P

P

isati

sati

li

sat

i

dPRT

vPf exp

Using values from the steam tables:

3.169 kPasat

iP

so

1sat

i

and

3

5 m1.8 10

moliv

Taking vi to be constant

exp 4557 kPal sat sat satii i i i

vf P P P

RT

7.54

(a)

Begin by drawing a line tangent to the plot for ˆ l

af as 0ax . Then, extend the line to 1ax .

The Henry’s constant is equal to the intercept of the vertical gridline when 1ax . Therefore,

a 19.5 kPaH

(b)

The activity coefficient is defined as

ˆ

ˆ

l

aa o

a a

f

x fg

Using a Henry’s law reference state, this equation becomes

ˆ l

aa

a a

f

xg

H

From the provided graph,

ˆ 0.4 12 kPal

a af x

ˆ 0.8 31 kPal

a af x

Therefore,

12 kPa

0.4 1.540.4 19.5 kPa

a axg

31 kPa

0.8 1.990.8 19.5 kPa

a axg

(c)

From the Gibbs-Duhem Equation, we know

0lnln

a

bb

a

aa

dx

dx

dx

dx

gg

Therefore,

a

b

dx

d gln must be negative. Since the activity coefficient is based on a Lewis-Randall

reference state, 0ln bg as 0ax . Therefore, at 4.0ax , bgln is negative and 1bg .

(d)

Since 1/ RLb

g , the a-b interactions are stronger than the pure species interactions.

(e)

When a gas is in equilibrium with a liquid, the fugacities of each phase are equal. For an ideal

gas, ˆ v

a af y P . Setting fugacities equal, we get

ˆ ˆv l

a a ay P f f

Therefore,

ˆ 12 kPa

20 kPa

l

aa

fy

P

6.0ay

7.55

(a)

Both species are based on the Lewis-Randall reference state because

0ln ag as 1ax

0ln bg as 1bx

(b)

The Gibbs-Duhem equation states

0lnln bbaa dxdx gg

This can be differentiated to provide

0lnln

a

bb

a

aa

dx

dx

dx

dx

gg

By drawing tangent lines to the activity coefficient lines at 6.0ax , we obtain:

33.1ln

a

a

dx

d g

2ln

a

b

dx

d g

Therefore,

0002.024.033.16.0

(c)

We can’t use the two-suffix Margules equation because the lines for the activity coefficients

aren’t symmetric. Therefore, we will use the three-suffix Margules. Infinite dilution data from

the graph:

5.2ln ag

5.1ln bg

From Table 7.2:

BART a

5.2K 300

Kmol

J 314.8lng

BART b

5.1K 300

Kmol

J 314.8lng

Solve simultaneously:

A 4988.4 J mol-1

B 1247.1 J mol-1

Therefore,

gE xa xb 4988.4 1247.1 xa xb J mol-1

(d)

The mixture will not separate into two phases. The system is more stable as a mixture as shown

by the activity coefficients being less than one. Furthermore, mixg is always negative because

gE is always negative. (This becomes apparent by examining the magnitude of the A and B

parameters in the three-suffix Margules equation.) Therefore, it is thermodynamically favorable

for the system to mix.

7.56

(a)

The activity coefficient for species A is based on the Lewis-Randall reference state. For the

Lewis-Randall reference state, the natural log of the activity coefficient of species A goes to zero

as the mole fraction of A goes to one since all interactions are a-a interactions.

(b)

Equation 7.84 can be used to calculate the fugacity of species a. It states

P

P

lasat

asata

la

sat

a

dPRT

vPf exp

However, sata is assumed equal to one since the saturation pressure is low. The Poynting

correction can also be assumed equal to one since the system pressure is low. Therefore,

sata

la Pf

kPa 80laf

(c)

The value of aH can be calculated using Equation 7.75. It states

a

aa

f

Hg

From the graph provided in the problem statement, we find

ln 2.5ay

Using the value from Part (b), we can calculate Ha:

5.2expkPa 80aH

kPa 975aH

(d)

From Table 7.1,

2lnba AxRT g

As 1bx , 5.2ln ag . Therefore,

-1 12.52.5 8.314 J mol K 300 K

1

RTA

A 6235.5 J mol-1

(e)

We first calculate the mole fraction of a in the liquid mixture.

4.0mol 5

mol 2

tot

aa

n

nx

Now, we can obtain the activity coefficient from the graph.

875.0ln ag

40.2 ag

This allows us to fugacity of species a in the liquid which is equal to the fugacity of species a in

the vapor phase.

ˆ ˆ 0.4 2.40 80 kPav l o

a a a a af f x fg

ˆ 76.8 kPav

af

The vapor mole fraction can be determined from the definition of fugacity in the vapor phase if

we assume ideal gas behavior, which is reasonable at 1 bar.

ˆ v

a af y P

768.0kPa 100

kPa 8.76ay

(f)

In part D, the Margules parameter was calculated for the mixture based on a Lewis-Randall

reference state. From Equation 7.56:

ba RTAx gln2

2-1

-1 -1

6235.5 J mol 0.4exp 1.49

8.314 J mol K 300 Kbg

7.57

(a)

Assume the two-suffix Margules equation represents the excess Gibbs energy data well.

Therefore, for any of the data points in the provided graph, the following relationship should be

true

21

21

xxR

A

R

g

xAxg

E

E

When 5.02 x :

K 1285.05.0

K 32

R

A

To calculate the activity coefficients of cyclohexane, we can use the following equations

T

xR

A

RTAx

21

2

221

exp

ln

g

g

(i).

23.1

K 343

75.0K 128exp

2

2

2

g

g

(ii).

45.1

K 343

1K 128exp

2

2

2

g

g

(b)

From Equation 7.75:

2

22

f

Hg

where

satPf

22

Calculate the saturation pressure of cyclohexane using data from Appendix A.1:

bar722.0

50.50343

63.27661325.9exp

2

2

sat

sat

P

P

Therefore,

bar 05.1

bar 722.045.1

2

2

H

H

(c)

Since A>0, gE is always greater than one. Therefore, like interactions are stronger.

7.58

To find the activity coefficients, we can use the following two equations:

Eaa GRT gln

Ebb GRT gln

The derivations of expressions for ag will be shown below. The method for finding bg is

analogous, and the final expressions can be found in Table 7.2.

(a)

The excess Gibbs energy:

ba

ba

ba

baE

nn

nnBA

nn

nnG

Find EaG :

2

,,

2 bababab

nVTa

EEa xBxxxBAxxx

dn

GdG

b

Substitute ba xx 1 :

32 43bb

Ea BxxBAG

Therefore,

RT

BxxBAbb

a

32 43expg

(b)

Rewrite excess Gibbs energy as

ba

baE

BnAn

nnABG

Find EaG :

2

22

,, ba

b

nVTa

EEa

BnAn

nBA

dn

GdG

b

Divide the numerator and denominator of the right-hand side of the above expression by ba nn

:

2

22

ba

bEa

BxAx

xBAG

Therefore,

2

22

exp

ba

ba

BxAx

xB

RT

Ag

(c)

Rewrite excess Gibbs energy as

ba

ababb

ba

babaa

E

nn

nnn

nn

nnnRTG lnln

Find EaG :

baab

baabba

abba

abbaababa

Ea

xx

xx

xx

xxxxxRTG

1ln

This can be rewritten as

baab

bab

abba

abbbbaba

Ea

xx

x

xx

xxxxRTG

111ln

2

baab

bab

abba

abbbbaba

Ea

xx

x

xx

xxxxRTG

111ln

After some manipulation, the following is obtained

baab

ba

abba

abbbaba

Ea

xxxxxxxRTG ln

baab

ba

abba

abbbabaa

xxxxxxxlnexpg

which is consistent with the expression in Table 7.2.

(d)

Rewrite the excess Gibbs energy as

abab

baabab

baba

bababa

E

nn

nn

nn

nnRTG

GG

GG

Find EaG :

2

2

2

22ba

,,G

G

G

G

abab

babab

baba

bba

nVTa

EEa

nn

n

nn

nRT

dn

GdG

b

Dividing the numerators and denominators of the above expression by ba nn , we obtain

22

2ba2

G

G

G

G

abab

abab

baba

bab

Ea

xxxxRTxG

Therefore,

22

2ba2

G

G

G

Gexp

abab

abab

baba

baba

xxxxx

g

7.59

Three-suffix Margules:

At infinite dilution:

BARTRT a 2lnlng

BARTRT b 2/3lnlng

Solving simultaneously,

3lnRTA 2/3lnRTB

For an equimolar solution:

RT lnga RT ln 3 3ln 3 /2 0.5 2

4ln 3 /2 0.5 3

RT lngb RT ln 3 3ln 3 /2 0.5 2

4ln 3 /2 0.5 3

Solving, we obtain

ga 1.11

gb 1.19

Van Laar:


RT lnga

RT ln 2 A

RT lngb

RT ln 3/2 B


2

5.02/3ln5.02ln

5.02/3ln2lnln

RTRT ag

RT lngb RT ln 3/2 ln 2 0.5

ln 2 0.5 ln 3/2 0.5

2

Solving, we obtain

ga 1.10

gb 1.18

Wilson:


RT lnga

RT ln 2 RT ln ab ba 1

RT lngb

RT ln 3/2 RT ln ba ab 1


21.1

407.0

ba

ab


lnga ln 0.5 0.5 0.407 0.51.21

0.5 0.5 1.21

0.407

0.5 0.5 0.407

lngb ln 0.5 0.5 1.21 0.50.407

0.5 0.5 0.407

1.21

0.5 0.5 1.21

Solving, we obtain

ga 1.10

gb 1.17

7.60

The problem statement provides the following information:

9.0ax 2.0

ax 1.0b

x 8.0b

x

At equilibrium:

gg aaaa xx

gg aa

a

a

x

xlnlnln

ggbbbb

xx

ggbb

b

b

x

xlnlnln

Use the composition data provided in the problem statement and the expressions from Table 7.1:

2 2 3 30.9 1ln 3 0.8 0.1 4 0.8 0.1

0.2 8.314 493.15A B B

2 2 3 30.1 1ln 3 0.2 0.9 4 0.2 0.9

0.8 8.314 493.15A B B


A 10100 J mol-1 B 1300 J mol-1

7.61

Let the subscript “a” represent water and “b” represent ethanol. Since the activity coefficients at

infinite dilution are different for water and ethanol, the two suffix Margules equation cannot be

used. Instead, employ the three suffix Margules equation:

babaE xxBAxxg

From Table 7.2, we have

32 43lnbba BxxBART g

32 43ln aab BxxBART g

Substituting infinite dilution data, we obtain a system of equations that can be solved for A and

B:

2 3J

ln 8.314 343.15 K ln 2.62 3 1 4 1mol K

aRT A B Bg

2 3J

ln 8.314 343.15 K ln 7.24 3 1 4 1mol K

bRT A B Bg

A 4200 J mol-1 and

B 1450 J mol-1

To obtain the fugacity of liquid water, we use the following equation

ˆ l o

a a a af x fg

Since the activity coefficient for water as 0ax is greater than one, the activity coefficient is

based on the Lewis-Randall reference state. Therefore, oaf is the fugacity of pure water at 70

ºC. The following relationship also holds since the saturation pressure at 70 ºC is so low that the

water vapor behaves as an ideal gas and the Poynting correction can be neglected.

sata

oa Pf

From the steam tables,

31.2 kPasat

aP (70 ºC)

Using the values of A and B calculated above, we can calculate the coefficient for a mixture of

40 mole % water and 60 mole % ethanol at 70 ºC:

2 3

-1 1

4198 3 1450 0.6 4 1450 0.6exp

8.314 J mol K 343.15 K

1.90

a

a

g

g

Now, we can calculate the fugacity.

ˆ 0.40 1.90 31.2 kPal

af

ˆ 23.7 kPal

af

7.62

To find expressions for agln and bgln , we can use Equations 7.55 and 7.56:

bE

b

aEa

RTG

RTG

g

g

ln

ln

The expression for the excess molar Gibbs energy can be rewritten as

2bababa

baE xxxCxBxAx

xABxg

By multiplying the above expression by the total number of moles and rewriting mole fractions

in terms of moles, we obtain

32

bababa

ba

baE nnnnnCnBnAn

nABnG

Differentiation provides

561222

bbb

ba

bEa xxCx

BxAx

BxAG

561222

aaa

ba

aEb xxCx

BxAx

AxBG

Substitute these expressions into Equations 7.55 and 7.56:

5612

1ln 2

2

bbbba

ba xxCx

BxAx

BxA

RTg

5612

1ln 2

2

aaaba

ab xxCx

BxAx

AxB

RTg

7.63

Applying Equation E6.4E to the property k = (gE/RT)

gives:

1

, ,

i

i i

E E

EE mi

i

n

P n T n

g gRT RT Gg

d dT dP dxRT T P RT

Substituting the following equations into the above expression

gE

RTT

P ,n i

hmix

RT 2 (Equation 7.75)

gE

RTP

T ,n i

vmix

RT (Equation 7.74)

we obtain

dgE

RT

hmix

RT 2dT

vmix

RTdP lng idxi

ni 1

m

For isobaric binary data, the pressure is constant, and the expression can be reduced to

12112221121lnlnlnln xddxdT

RT

hdxdxdT

RT

h

RT

gd mixmix

E

gggg

12

1

2ln dxdT

RT

hmix

g

g

Now, integrate the above expression

1

2

1

2ln dxdT

RT

h

RT

gd mix

E

g

g

By the definition of excess Gibbs energy, we know RTg E / is zero when 01 x and 11 x .

Therefore,

1

0

12

11

02

1

1

1

1

ln 0

x

x

xT

xT

mix dxdTRT

h

g

g

1

02

1

0

12

11

1

1

1

ln

xT

xT

mixx

x

dTRT

hdx

g

g

We can also show the following using differentials

2

1

T

dT

Td

Hence,

1

1

01

1

02

1

0

12

1

1

1

1

1

1

1

1ln

xT

xT

mixxT

xT

mixx

xT

dR

hdT

RT

hdx

g

g

7.64

(a)

Since the activity coefficients are approximately equal, we can assume the two-suffix Margules

equation sufficiently models excess Gibbs energy. Calculate the A parameter for both activity

coefficients, and use the average value for subsequent calculations:

mol

J 4.60227.1lnK 15.303

Kmol

J 314.8ln aRTA g

mol

J 6.73734.1lnK 15.303

Kmol

J 314.8ln bRTA g

mol

J 670A

Equation 7.32:

ˆ l l

a a a af x fg

Calculate the fugacity of pure hexane (a) at 30 ºC and 1 bar assuming the saturated hexane vapor

acts ideally since its saturation pressure is low at 30 ºC:

sata

la Pf

bar 25.0laf (Used Antoine’s equation and data from Appendix A.)

Calculate ag when 8.0 ,2.0 ba xx :

19.1

15.303314.8

8.0670exp

2

ag

Therefore,

ˆ 0.2 1.19 0.25 barl

af

ˆ 0.06 barl

af

(b)


07.1

15.303314.8

5.0670exp

2

ag

Therefore,

ˆ 0.5 1.07 0.25 barl

af

ˆ 0.134 barl

af

(c)


00.1

15.303314.8

1.0670exp

2

ag

Therefore,

ˆ 0.9 1.00 0.25 barl

af

ˆ 0.23 barl

af

7.65

We are given the density and the Henry’s law constant for 1-propanol in water:

31

cm

g 80.0l , and bar 61.0aH

(a)

Noting that the system pressure is high (100 bar), we begin with Equation 7.85 from the text:

1

1

1 1 1

lsatv

P PRTl sat satf P e

The specific molar volume of 1-propanol is found from the given density:

mol

cm1.75

g/cm 80.0

g/mol 1.60 3

3

1

11

l

l MWv

The saturation pressure is found using the Antoine equation, and the appropriate constants given

in Appendix A:

15.80298

38.31669237.10ln 1

CT

BAPsat

bar 027.01 satP

Since the saturation pressure is very low, we can assume that 1-propanol acts ideally. Thus,

(ideal) 11 sat

And we can now find the fugacity of 1-propanol in the liquid:

11

36

5 5

1 1 3

m75.1 10

Jmol0.027 bar exp 100 10 0.027 10J m

8.314 298mol K

lsatv

P PRTl satf P e

K

Cranking the arithmetic, we get the pure-species fugacity for 1-propanol:

1 0.037 barlf

Now, we can calculate the activity coefficient from the pure-species fugacity and the given

Henry’s law constant.

Recall that (Equation 7.75):

11

1

0.61 bar16.70

0.037 barl

H

fg

Now we can use a model for the free Gibbs energy to determine the solution fugacity. Since the

two species are similar in size and polarity, we can expect the two-suffix Margules (2SM) model

to fit it adequately. From Table 7.1, we know that:

2

21ln AxRT g

Since at infinite dilution, x2 = 1, we can easily determine the 2SM parameter, A :

81.2ln 1 gRT

A

Now, use the same model to determine the activity coefficient in the given mixture (i.e. x2 = 0.6):

01.16.081.2ln22

21 xRT

Ag

75.21 g

And finally, we can find the solution fugacity of 1-isopropanol in the given mixture:

1 1 1 1ˆ 0.4 2.75 0.037 barlf x fg

bar 040.0ˆ1 lf

7.66

You have the parameters for this system from the van Laar (vL) equation:

mol

J 3000vLA , and

mol

J 5040vLB

(a)

Find the 3SM parameters A and B :

From Table 7.2, the van Laar parameters are related to the activity coefficients by:

2

ln

ba

b

aBxAx

BxART g , and

2

ln

ba

a

bBxAx

AxBRT g

From these values, we can calculate the infinite-dilution activity coefficients for the two species,

and calculate the 3SM parameters from these. Note that for species a at infinite dilution, species

b is pure (i.e. xa = 0, and xb = 1). The bracketed terms in the expressions above then equal one.

mol

J 3000ln ART ag , and

mol

J 5040ln BRT ag

Applying this same reasoning to the 3SM model,

mol

J 3000ln BART ag , and

mol

J 5040ln BART bg .

Solving the two equations with two unknowns, we find:

mol

J4020 A , and

mol

J1020 B

(b)

Calculate the fugacity of liquid benzene (a) in a 30 mole% mixture in 1-propanol at 81kPa.

Start with Equation 7.77 from the text:

ˆ l sat

a a a a a a af x f x Pg g (Note: the system pressure is low, so sat

a

l

a Pf )

Calculate the activity coefficient using either the vL or 3SM model. We will use the 3SM model,

the results are slightly different if you choose the vL model (can you explain why?)

From Table 7.2, we find the expression for the activity coefficient of a in the 3SM model:

mol

J 20704)3(ln 32 bba BxxBART g

Solving for the activity coefficient of benzene in the mixture,

04.2ag

Now, we need to find the saturation pressure of benzene. Turning to the Antoine equation and

the tables in Appendix A, we calculate:

147.0ln

CT

BAP sat

a

bar 86.0sat

aP

Substituting these values into the original equation, we find the fugacity of liquid benzene in the

mixture:

bar 86.004.23.0ˆ sat

aaa

l

a Pxf g

bar 53.0ˆ l

af

(c)

Determine the mole fraction of vapor in the mixture, assuming the system is in equilibrium.

Since the system is in equilibrium, we know that

bar 53.0ˆˆ l

a

v

aaa

v

aa fffxPy g

We can assume again that the vapor phase is ideal, so that

bar 53.0Pya

Solving for ya, and substituting the system pressure (0.81 bar) :

65.0bar 81.0

bar 53.0ay

7.67

We are asked to determine the fugacity of a, and the Henry’s Law constant Ha in a mixture of a

and b at 30 kPa and 20°C. The saturation pressure of a and an equation for the excess Gibbs

energy of the mixture is given:

kPa 50sat

aP , and

baba

E

xxxxRT

g5.025.0

(a)

Looking in Table 7.1, we find that the given equation for gE is a variant of the Margules 3-suffix

equation. If we rewrite the given equation and compare it to the Margules equation, the values

for the two coefficients are given by:

babababababa

E xxRTxRTxxAxAxxg 5.025.0

RTAab 5.0 and

RTAba 25.0

Reading from the same table (or explicitly evaluating the derivative), we find an expression for

the activity coefficient of a in the mixture:

aabbaabba xAAAxRT 2ln 2g

Substitute the values for Aab and Aba, and the mole fraction of each species (xa = 0.2, xb = 0.8):

aba RTxRTxRT 5.025.025.0ln 2 g

256.05.05.0ln 2 aba xxg

29.1 ag

Now use the relation for fugacity of a condensed phase to find the fugacity l

af̂ :

kPa 9.12ˆ sat

aaa

l

a Pxf g

(b)

Now calculate the Henry’s law constant. First, find the activity of a at infinite dilution:

5.05.05.0ln0

2

axaba xxg

68.1

ag

Now multiply the infinite-dilution activity coefficient by the saturation pressure:

kPa 84 sat

aaa PH g

7.68

From the expressions in Table 7.4, we have:

Acetone Water

*

1 1 2 2

i ii

x r

x r x r

0.545 0.455

1 1 2 2

i ii

x q

x q x q

0.417 0.583

''

' '

1 1 2 2

i ii

x q

x q x q

0.501 0.499

li -0.42 -2.32

For the combinatorial part of the activity coefficients, we get:

*

*1 1 11, 1 2 1 2*

1 1 2

ln ln ln 0.2362

combinatorial

z rq l l

x r

g

and

*

*2 2 22, 2 1 2 1*

2 2 1

ln ln ln 0.1162

combinatorial

z rq l l

x r

g

For the residual part of the activity coefficients, we calculate the energy parameters

1212 exp 0.204

a

T

and 21

21 exp 1.35a

T

With these values, we get:

' ' ' ' ' 21 121, 1 1 2 21 2 1 ' ' ' '

1 2 21 1 12 2

ln ln 0.568residual q q

g

and

' ' ' ' ' 12 212, 2 1 12 2 1 2 ' ' ' '

1 12 2 1 2 21


g

Summing the combinatorial and residual parts and taking the exponential gives

g1 = 2.23 and g2 = 1.24.

These values are 3 – 6% lower than the experimentally measured values of exp

1 2.30g and exp

2 1.32g .

7.69


Ethanol Benzene

*

1 1 2 2

i ii

x r

x r x r

0.319 0.681

1 1 2 2

i ii

x q

x q x q

0.368 0.632

''

' '

1 1 2 2

i ii

x q

x q x q

0.214 0.786

li -0.41 1.76


*

*1 1 11, 1 2 1 2*

1 1 2

ln ln ln 0.0632

combinatorial

rzq l l

x r

g

and

*

*2 2 22, 2 1 2 1*

2 2 1

ln ln ln 0.0222

combinatorial

rzq l l

x r

g


1212 exp 1.266

a

T

and 21

21 exp 0.467a

T


' ' ' ' ' 21 121, 1 1 2 21 2 1 ' ' ' '

1 2 21 1 12 2


g

and

' ' ' ' ' 12 212, 2 1 12 2 1 2 ' ' ' '

1 12 2 1 2 21


g


g1 =1.32 and g2 =1.09.

exp 11

1 1

1.66sat

y P

x Pg

7.70


Acetone Chloroform

*

1 1 2 2

i ii

x r

x r x r

0.192 0.808

1 1 2 2

i ii

x q

x q x q

0.200 0.800

''

' '

1 1 2 2

i ii

x q

x q x q

0.200 0.800

li -0.42 0.1


*

*1 1 11, 1 2 1 2*

1 1 2

ln ln ln 0.0082

combinatorial

rzq l l

x r

g

and

*

*2 2 22, 2 1 2 1*

2 2 1

ln ln ln 0.0012

combinatorial

rzq l l

x r

g


1212 exp 1.745

a

T

and 21

21 exp 0.737a

T


' ' ' ' ' 21 121, 1 1 2 21 2 1 ' ' ' '

1 2 21 1 12 2


g

and

' ' ' ' ' 12 212, 2 1 12 2 1 2 ' ' ' '

1 12 2 1 2 21


g


g1 =0.585 and g2 =0.951.

exp 11

1 1

0.536sat

y P

x Pg

7.71

The fugacity of ethanol in solution is calculated with the following equation:

1 1 1 1ˆ l satf x Pg

To find the activity coefficient, we can use Equation 7.73:

333322311

313

233222211

212

133122111

111

1331221111 ln1ln

xxx

x

xxx

x

xxx

x

xxxg

Substituting the values given in the problem statement, we obtain

192.0ln 1 g

21.11 g

Now calculate the saturation pressure of ethanol at 60 ºC using Antoine’s Equation data in

Appendix A.1.

bar 468.0bar 68.4115.333

98.38032917.12exp

1

satP

Therefore, the fugacity of the ethanol in the liquid is

1ˆ 0.3 1.21 0.468 bar 0.17 barlf

7.72

First, we need to find the Wilson parameters at 8 ºC. We can use the following equations:

RT

ab

av

bvab

exp

RT

ba

vb

avba

exp

Find the energetic parameters (lowercase lambdas) at 60 ºC, and use them to calculate the

uppercase lambdas at 8 ºC. The molar volumes for ethanol and 1-propanol can be calculated

using the Rackett EOS. The saturated steam tables provide an estimate for water’s molar

volume. Using the above equations with the 60 ºC data, we obtain:

Pairing (J/mol)

12 1.13 x 102

21 6.83 x 102

13 1.39 x 103

31 3.52 x 103

23 4.73 x 103

32 5.04 x 103

Now, we can calculate the Wilson parameters at 8 ºC since the energetic parameters are less

sensitive to changes in temperature. We obtain

Pairing

12 1.214 21 0.586 13 0.204 31 0.602 23 0.038

32 0.400

We will find the fugacity of the ethanol with the following equation

1 1 1 1ˆ l satf x Pg

To find the activity coefficient, with the Wilson parameters at 8 ºC:

333322311

313

233222211

212

133122111

111

1331221111 ln1ln

xxx

x

xxx

x

xxx

x

xxxg

32.11 g

We can find the saturation pressure from Antoine’s Equation data in Appendix A.1.

bar 028.0bar 68.4115.281

98.38032917.12exp

1

satP

Therefore,

3

1ˆ 0.2 1.32 0.028 bar 7.39 10 barlf

7.73

Equation 7.81 provides

2

,

ln i ii

P x

H h

T RT

g

Equation 6.26 states that

mix i ii

H H h

Therefore,

2

,

ln mixi i

P x

H

T RT

g

For the expression given in the problem statement

2 3

2 1 2 1 2 11 2

2 3

1 2 1 2 1 2 1 2

3802 1200 1554447.8mix

n n n n n nn nH

n n n n n n n n

Thus, we find

2 3 2 2 3

2 1 1 2 1 2 21

28731.8 4156.6 13131.4 3708.2mixH x x x x x x x

For an equimolar solution

1

J1062.5

molmixH

Now calculate the activity coefficient at 100 ºC by integrating Equation 7.81:

15.333

1

15.373

1

314.8

5.1062

65.1ln 1g

58.11 g

7.74

Since we are assuming that the tin and cadmium form a regular solution,

mixE hg

Find the activity coefficient by using the following equation:

CdE

CdRTG gln

To find E

CdG , start with

SnCd

SnCdmix

E

nn

nnHG

13000

Differentiating provides

213000 Sn

ECd

XG

Therefore,

K 15.773Kmol

J 314.8

6.013000exp

2

Cdg

07.2Cdg

7.75

Select the isothermal ethanol/water experimental data set in the Models for gE – Parameter

Fitting menu. The temperature, 74.79 ºC, is automatically selected. We have the capability of

determining the activity coefficients with three different objective functions, but only the

coefficients found using the pressure objective function, OFP, are shown below.

ThermoSolver provides the following plot

We can look at the value of each objective function to determine which model fits the data best.

We want the objective function with the smallest value. The values are tabulated below for each

model.

Pressure Objective Function Values (bar2)

Two Suffix

Margules

Three Suffix

Margules

van Laar

Wilson NRTL

Part (a) Part (b) Part (c) Part (d) Part (e) Two

Suffix Margules

Three Suffix Margules

van Laar Wilson NRTL

A A B A B ab ba Gab Gba ab ba

3652.3 3521.0 -1102.3 5001.0 2692.8 0.167 0.869 0.979 0.523 0.055 1.683

1.34E-02 2.77E-04 5.45E-

05 1.40E-

04 5.39E-

05

The NRTL model best represents the data.

7.76

Select the isothermal pentane/acetone experimental data set in the Models for gE – Parameter

Fitting menu. The temperature, 25 ºC, is automatically selected. We have the capability of

determining the activity coefficients with three different objective functions, but only the

coefficients found using the pressure objective function, OFP, are shown below.




model.


Part (a) Part (b) Part (c) Part (d) Part (e) 2 Suffix

Margules Three Suffix

Margules van Laar Wilson NRTL


4371.1 4365.8 208.7 4150.2 4600.6 0.366 0.226 1.963 2.280 0.573 0.700

Two Suffix

Margules

Three Suffix

Margules

van Laar

Wilson NRTL

1.69E-03 1.11E-03 1.08E-

03 1.73E-

04 1.00E-

04


7.77

Select the isothermal chloroform/heptane experimental data set in the Models for gE –

Parameter Fitting menu. The temperature, 25 ºC, is automatically selected. We have the

capability of determining the activity coefficients with three different objective functions, but

only the coefficients found using the pressure objective function, OFP, are shown below.




model.


Two Suffix

Margules

Three Suffix

Margules

van Laar

Wilson NRTL

4.39E-05 1.33E-06 2.66E-

07 1.64E-

07 9.12E-

08

Part (a) Part (b) Part (c) Part (d) Part (e) 2 Suffix

Margules Three Suffix

Margules van Laar Wilson NRTL


1329.8 1235.5 261.6 1028.7 1548.1 1.109 0.478 0.559 0.856 0.519 0.138


7.78

We are given a plot of oxygen content vs. partial pressure of oxygen for an emulsion of 24%

(w/v) perflubron (C8F17Br) in water. Estimate the oxygen capacity of the emulsion, which might

be used as a blood substitute, in units of moles of oxygen per liter of emulsion. Compare this

result with the solubility of oxygen in pure water at the same conditions (25°C, atmospheric

pressure). Use these results to determine the value of the Henry’s Law constant for oxygen in

pure perflubron (not the emulsion).

a) First, estimate [O2] in the emulsion, as

moles O2/L solution.

We are given a graph showing the

concentration of oxygen in the emulsion at

25°C. We are given ambient pressure is 1.0

atm.

The standard atmosphere contains 21% O2,

and 79% N2 and other gases. Assuming air

acts as an ideal gas at 1 atm, the partial pressure of oxygen is:

2 2

159.6 mmHgO OP y P

At this partial pressure, the oxygen concentration of the emulsion is 1.4 mL/100 cm3. Since the

air acts as an ideal gas, the number of moles of oxygen in the emulsion is:

2

2

35 -6

3

5

2

J m1.013 10 1.4 mL 10

m mL5.726 10 moles O

J8.314 298K

mol K

O

O

P Vn

RT

If we assume that the gas dissolves completely (i.e. there is no volume change when the 1.4 mL

of oxygen is absorbed into the 100 cm3 of emlusion), then

4

22

moles O 5.73 10 moles

L solution L solutionO

(where 1 L = 10 x 100 cm

3)

b) Compare the result from part A with the oxygen capacity of pure water:

From Table 8.1, we can get the Henry’s Law constant 2

44,253.9 barO H for oxygen in water

at 25°C. Since pressure is low and the gas solution is dilute, we can assume ideal behavior, so

Equation 8.30 holds:

PO2

= 160 mmHg

O2 ≈ 1.4 mL/100 cm3

PO2

= 160 mmHg

O2 ≈ 1.4 mL/100 cm3

2

2

2

60.21 bar4.75 10

44,253.9 bar

O

O

O

y Px

H

Since 2 2

2

2 2 2

O O

O

O H O H O

n nx

n n n

, we can write

2 2 2O O H On x n . Assuming the solution volume is

approximately equal to the volume of pure water, we can define the molality of oxygen to be:

2 2

2

2

O O

solution H O

n nO

V V

Substituting the above relation for 2On , and noting that 2

2

2

H O

H O

H O

Vv

n ,

2

2

2

O

H O

xO

v

The molar volume of water can be found from the density at 25°C and molecular weight:

2

2

2

18 g/mole L0.018

1000 g/L mole

H O

H O

H O

Mv

Substituting this value, we calculate:

2

2

64 2

22 2

2

mole O4.75 102.64 10

liter H O liter H O0.018

mole H O

O

H O

xO

v

Since 2 2O H OV V , the volume of solution is essentially the same as that of the water:

4 22

mole O2.64 10

literO .

Thus, the perflubron emulsion can carry approximately 2.2 times as much oxygen as pure

water.

c) Find the value of the Henry’s Law constant for oxygen in pure perflubron (PFB).

Previously, we found the mole fraction of oxygen in the emulsion

and in water at the given partial pressure of oxygen. Using the

definition of the Henry’s Law constant, we can write (for each of

the two phases in the emulsion):

xa

Ha

yaP

Xa, yaP

xa

Ha

yaP

Xa, yaP

aa

a

y P

xH (see the figure at right)

To find xa, we will first find the number of moles of PFB and water in 100 cm3 of the emulsion.

The emulsion contains 24g of PFB per 100cm3 of solution. This is equivalent to:

1 mole PFB24g PFB 0.0481 mole PFB

498.96 g PFB

, and

331 cm PFB

24g PFB 12.435 cm PFB1.93 g PFB

.

Since 12.435 cm3 of the total 100 cm

3 is taken up by the PFB, the remaining 87.565 cm

3 must be

water (since this is an emulsion, we can assume that 0mixV ). Therefore, we have

3 2 22 23

2

1.00 g H O 1 mole H O87.565 cm H O 4.865 moles H O

cm 18 g H O

.

Oxygen must be dissolved in either the water or PFB. A simple mole balance on the total oxygen

taken up by the emulsion 2 ,O emulsionn allows us to determine the amount of O2 in the PFB phase:

2 2 2 2, , ,O PFB O emulsion O H On n n

We determined the molar uptake of the emulsion and pure water in Parts (a) and (b), above.

However, recall that we calculated 2 2,O H On as the moles of O2 taken up by 100 cm

3 of water.

Since we have less than 100 cm3 of water, we must scale

2 2,O H On appropriately:

2

35 5

, 3

86.565 cm5.726 10 moles 2.64 10 moles

100 cmO PFBn

2

-5

, 22.64 10 moles OO PFBn

Now find the mole fraction of O2 dissolved in the PFB in the 100 cm3 of PFB/H2O emulsion:

2 2

2

2

5

2,

3.44 10 moles O

0.0481 moles PFB

O O

O PFB

O PFB PFB

n nx

n n n

2

4

, 7.153 10O PFBx

Applying the definition of the Henry’s Law constant above, we can compute 2 ,O PFBH :

2

2

2

, 4

,

0.21 1.01325 bar297.5 bar

7.153 10

O

O PFB

O PFB

y P

x

H

2 , 300 barO PFB H (Note that a smaller Ha implies a larger xa for a given Pa)

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