1) Estimate the fugacity of isobutylene as a gas at 280

Tc = 417.9 K

Pc = 40 bar

ω = 0.194

Thus,

Tr = T/Tc (since T= TcT

= (280+273.15) K/ 417.9 K =1.32364

Pr = P/Pc (since P= PcP

= 20 bar / 40 bar = 0.5

For pure gas, using Generalized correlation for fugacity coefficient;

Therefore:

6.1

0 422.0083.0

rTB −=

2.4

1 172.0139.0

rTB −=

32364.1

422.0083.0 −=

18645.0−=

32364.1

172.0139.0 −=

086022.0=

06413.0−=

( )1ln BBT

P

r

ri ωφ += °

( )1ln BBT

P

r

ri ωφ += °

[ 18645.032364.1

5.0−=

937886.0=iφ

UNIVERSITI MALAYSIA PERLIS

Pusat Pengajian Kejuruteraan Bioproses

ERT 206: Thermodynamics

Assignment -Solutions

Estimate the fugacity of isobutylene as a gas at 280 °C and 20 bar

Tr)

= (280+273.15) K/ 417.9 K =1.32364

Pr)

= 20 bar / 40 bar = 0.5

For pure gas, using Generalized correlation for fugacity coefficient;

6.132364

422

2.432364

172

( )( )]086022.0194.018645 +

For pure gas, using Generalized correlation for fugacity coefficient;

Fugacity of isobutylene at 280 °C and 20 bar is:

OR

Fugacity of pure gas can be also estimated using the following equation:

0φ and 1φ can be obtained in Lee/Kesler Generalized-correlation Table (Appendix E).

By interpolation, find 0φ and 1φ at Pr = 0.5 and Tr = 1.32364.

Calculation for 1φ :

At Tr =1.3

0328.11 =φ

At Tr =1.4

03875.11 =φ

At Tr =1.32364

0342.11 =φ

Calculation for 0φ :

At Tr =1.3

928.00 =φ

Pf ii φ=( )bar20937886.0=

bar76.18=

( )ωφφφ 10=i

0257.10399.1

0399.1

4.06.0

5.06.0 1

−−

=−− φ

0304.10471.1

0471.1

4.06.0

5.06.0 1

−−

=−− φ

0328.103875.1

03875.1

3.14.1

32364.14.1 1

−−

=−

− φ

9419.09141.0

9141.0

4.06.0

5.06.0 0

−−

=−− φ

At Tr =1.4

94415.00 =φ

At Tr =1.32364

931818.00 =φ

955.09333.0

9333.0

4.06.0

5.06.0 0

−−

=−− φ

928.094415.0

94415.0

3.14.1

32364.14.1 0

−−

=−

− φ

( ) ( )( ) 9379.00342.1931818.0194.010 ===

ωφφφi

2) For the system ethylene(1)/propylene(2) as a gas, estimate 121ˆ,ˆ,ˆ φff and 2φ̂ at T =

150°C, P = 30 bar, and y1 = 0.35.

Step 1: From Table B1, the data for ethylene and propylene are extracted as follows:

For ethylene,

Pc11 = 50.4 bar

Tc11 = 282.3 K

Vc11

= 131 cm3/mol

Zc11 = 0.281

ω11 = 0.087

For propylene,

Pc22 = 46.65 bar

Tc22 = 365.6 K

Vc22 = 188.4 cm

3/mol

Zc22 = 0.289

ω22 = 0.140

Step 2: Find Pc12,

Tc

12, Vc

12, Zc

12 and ω

12 using the following equations:

For chemically similar species, kij=0

Step 3: Find Tr12,

B

0, B

1, ijB̂ and Bij using the following equations:

ij Tcij /K Pcij/bar Vcij/cm3mol

-

1

Zcij ωij

11 282.3 50.4 131 0.281 0.087

22 365.6 46.65 188.4 0.289 0.140

12 321.26 48.19 157.966 0.285 0.1135

( ) ( )ijcjcicij kTTT −= 1

2/1

cij

cijcij

cijV

RTZP =

2

cjci

cij

ZZZ

+=

2

cjci

cij

ωωω

+=

33/13/1

2

+= cjci

cij

VVV

cij

rijT

TT =

6.1

0 422.0083.0

ijrTB −=

2.4

1 172.0139.0

ijrTB −=

cij

cijij

ijP

RTBB

ˆ=10ˆ BBB ijij ω+=

Step 4: Find , , using the following equations:

Step 5: Find and

For gas mixture,

Thus,

ij rijT /K 0B

1B ijB̂ ijB /cm

3mol

-1

11 1.499 -0.137816 0.107583 -0.128456 -59.819

22 1.1574 -0.25099 0.045911 -0.24456 -159.349

12 1.31715 -0.188579 0.084917 -0.178941 -99.181

( )12

2

2111ˆln δφ yB

RT

P+= ( )12

2

1222ˆln δφ yB

RT

P+=22111212 2 BBB −−=δ

2112ˆ,ˆ, φφδ

( ) ( ) ( )349.159819.59181.99212 −−−−−=δ

806.20=

( )( ) ( )[ ]806.2065.0819.59

15.27315014.83

30ˆln2

131 +−+

=− KmolKbarcm

barφ

043514.0−=

( )( ) ( )[ ]806.2035.0349.159

15.27315014.83

30ˆln2

132 +−+

=− KmolKbarcm

barφ

1337.0−=

8748.0ˆ2 =φ

9574.0ˆ1 =φ

1f̂ 2f̂

Py

f

i

ii

ˆˆ =φ

Pyf iii φ̂ˆ =

Pyf 111ˆˆ φ=

Pyf 222ˆˆ φ=

( )( )bar309574.035.0=

bar053.10=

( )( )bar308748.065.0=bar059.17=

3) Estimate 1φ̂ & 2φ̂ for an equimolar mixture of methyl ethyl ketone (1)/toluene(2) at

50°C and 25kPa. Set all kij=0

ij Tcij /K Pcij/bar Vcij/cm3mol

-1 Zcij ωij

11 535.5 41.5 267 0.249 0.323

22 591.8 41.1 316 0.264 0.262

STEP 1: Find Tcij, Vcij, Zcij, Pcij and ωcij

STEP 2: Find riiT and

STEP 3: Find and

( ) ( )12

2/1

221112 1 kTTTc −=

( ) ( )018.5915.5352/1 −×=

K563=

12

121212

c

ccc

V

RTZP =

( )( )13

3

291

563/14.83256.0−⋅

⋅=

molcm

KKmolbarcm

bar3.41=

33/1

12

3/1

1112

2

+= cc

c

VVV

33/13/1

2

316267

+=

13291 −= molcm

264.02

264.0249.0=

+=

2

262.0323.0 += 293.0=

rijT

11

11

c

rT

TT =

12

12

c

rT

TT =

22

22

c

rT

TT =

5.535

15.2735011

+=rT

0.563

15.2735012

+=rT

8.591

15.2735022

+=rT

603.0= 546.0= 574.0=

0B

6.1

11

0

11

422.0083.0

rTB −=

6.1

22

0

22

422.0083.0

rTB −=

6.1

12

0

12

422.0083.0

rTB −=

6.1603.0

422.0083.0 −=

6.1546.0

422.0083.0 −=

6.1574.0

422.0083.0 −=

865.0−= 028.1−= 943.0−=

1B

2

121112

ccc

ZZZ

+=

2

121112

ccc

ωωω

+=

STEP 4: Calculate , , and

2.4

11

1 172.0139.011

rTB −=

2.4

22

1 172.0139.022

rTB −=

2.4

12

1 172.0139.012

rTB −=

2.4603.0

172.0139.0 −=

2.4546.0

172.0139.0 −= 2.4574.0

172.0139.0 −=

3.1−= 045.2−= 632.1−=

1

1111

0

1111ˆ BBB ω+=

11

111111

ˆ

c

c

P

RTBB =

1

2222

0

2222ˆ BBB ω+=

22

222222

ˆ

c

c

P

RTBB =

( )3.1323.0865.0 −+−=

( )045.2262.0028.1 −+−=

( )( )bar

KKmolbarcm

5.41

5.535/14.832849.1 3 ⋅−=

2849.1−=

56379.1−=

( )( )bar

KKmolbarcm

1.41

8.591/14.8356379.1 3 ⋅−=

molcm /1378 3−=

molcm /1872 3−=

iiB̂ijB̂

iiB ijB

12

121212

ˆ

c

c

P

RTBB =

1

1212

0

1212ˆ BBB ω+=

( )632.1293.0943.0 −+−=

421176.1−=

( )( )bar

KKmolbarcm

3.41

563/14.83421176.1 3 ⋅−=

molcm /1611 3−=

STEP 5: Calculate and

( )12

2

2111ˆln δφ yB

RT

P+=

( )12

2

1222ˆln δφ yB

RT

P+=

22111212 2 BBB −−=δ

( ) 13281872137816112 −=++−= molcm

( )( ) ( )[ ]285.01378

15.3238314

25 2+−= 0128.0−=

( )( ) ( )[ ]285.01872

15.3238314

25 2+−= 0173.0−=

987.0ˆ1 =φ

983.0ˆ2 =φ

2φ̂1φ̂

4) A system formed initially of 2 mol CO2, 5 mol H2, and 1 mol CO undergoes the

reactions:

Develop expressions for the mole fractions of the reacting species as functions of the

reaction coordinates for the two reactions.

Step 1: List down the stoichiometric number of each species and find the total

stoichiometic number for both reactions:

Step 2: Find the total no of moles initially present;

Step 3: Express the composition of each species in terms of reaction coordinate

according to the following equation:

i CO2 H2 CH3OH H2O CO

j vj

1 -1 -3 1 1 0 -2

2 -1 -1 0 1 1 0

( ) ( ) ( ) ( )gOHgOHCHgHgCO 2322 3 +→+

( ) ( ) ( ) ( )gOHgCOgHgCO 222 +→+

∑∑

+

+=

j jj

j jjii

ivn

vny

ε

ε

0

,0

81520 =++=n

1

21

28

22 ε

εε−

−−=COy

1

21

28

352 ε

εε−

−−=Hy

1

1

283 εε−

=OHCHy

1

21

282 εεε

−

−=OHy

1

2

28

1

εε

−

+=COy