Solucionario Introducción a la Termodinamica en Ingeniería Química: Smith, Van Ness & Abbott

Chapter 1 - Section A - Mathcad Solutions
1.4 The equation that relates deg F to deg C is: t(F) = 1.8 t(C) + 32. Solve this
equation by setting t(F) = t(C).
Guess solution:

t

t = 1.8t

Given

0

Find t
()

32

Ans.

40

P=

1.5 By definition:

F
A

F = mass g

A

P

3000bar

D

4mm

F

PA

g

9.807

m

D

4

F

mass

2

g

s

P=

1.6 By definition:

F
A

A

3000atm

D

0.17in

F

PA

g

32.174

ft

4

D

mass

2

sec

gh

13.535

1.8

gm
3

101.78kPa

13.535

g

9.832

mass

2

384.4 kg

Ans.

2

F
g

A

mass

2

0.023 in

1000.7 lbm

Ans.

gm
3

29.86in_Hg

m

h

2

56.38cm

s

Pabs

g

gh

32.243

Patm

ft

Pabs

h

2

176.808 kPa Ans.

25.62in

s

cm

Patm

12.566 mm

Patm

cm

Patm

A

F = mass g

P

1.7 Pabs =

2

Note: Pressures are in
gauge pressure.

Pabs

gh

1

Patm

Pabs

27.22 psia

Ans.

1.10 Assume the following:

13.5

gm

g

3

m

9.8

2

s

cm

P

1.11

P

h

400bar

h

g

Ans.

302.3 m

The force on a spring is described by: F = Ks x where Ks is the spring
constant. First calculate K based on the earth measurement then gMars
based on spring measurement on Mars.
On Earth:

F = mass g = K x

mass

g

0.40kg

9.81

m

x

2

1.08cm

s

F

F

mass g

F
x

Ks

3.924 N

Ks

363.333

N
m

On Mars:

x

0.40cm

gMars

1.12 Given:

FMars

gMars

FMars
mass

d
P=
dz

g

0.01

FMars

Kx

and:

mK

=

4

MP
RT

Substituting:

Separating variables and integrating:

1
dP =
P

Psea

1atm

ln

M

29

2

mK

PDenver
Psea

=

PDenver = Psea e
gm
mol

d
P=
dz

zDenver

Mg
zDenver
RT
Mg
zDenver
RT

g

9.8

MP
g
RT

Mg
dz
RT

0

Psea

Taking the exponential of both sides
and rearranging:

3

Ans.

kg

PDenver

After integrating:

10

m
2

s

3

R

82.06

cm atm

T

mol K

Mg
zDenver
RT
Mg
RT

(
10

zDenver

273.15)
K

0.194

zDenver

Psea e

PDenver

0.823 atm

Ans.

PDenver

PDenver

1 mi

0.834 bar

Ans.

1.13 The same proportionality applies as in Pb. 1.11.

gearth

ft

32.186

gmoon

2

5.32

M

lmoon

1.14

gearth
gmoon

costbulb

Ans.

18.767 lbf

Ans.

hr
0.1dollars
70W
10
day
kW hr

hr
5.00dollars
10
day
1000hr

costelec

dollars
yr

costelec

25.567

dollars
yr

costtotal

43.829

dollars
yr Ans.

18.262

costtotal

costbulb

D

18.76

113.498

113.498 lbm

wmoon

M gmoon

costbulb

1.15

learth

M

learth lbm

wmoon

lmoon

2

s

s

learth

ft

1.25ft

costelec

mass

250lbm

g

32.169

ft
2

s

3

Patm

(a) F

30.12in_Hg

Patm A

l

Work

1.7ft

D

D

2

3

10 lbf

Work

PE

mass g l

mass

1.227 ft

2

Ans.

Ans.

16.208 psia

F l

0.47m

A

2.8642

Pabs

PE

1.16

4

F

mass g

F
A

(b) Pabs

(c)

A

3
10 ft lbf Ans.

4.8691

Ans.

424.9 ft lbf

g

150kg

9.813

m
2

s

Patm

(a) F

Patm A

l

0.83m

EP

1.18

mass

EK

2

A
4

110.054 kPa

Work

EP

mass g l

u

1250kg

1
2
mass u
2

40

EK

m
s

Work

EK

Wdot =

200W

Ans.

1000 kJ

mass g h
0.91 0.92
time

Wdot

Ans.

1000 kJ

g

9.8

m
2

s

4

h

2

0.173 m

Ans.

10 N

F l

Work

1.19

1.909

Pabs

Work

D

4

F

mass g

F
A

(b) Pabs

(c)

A

101.57kPa

50m

Ans.

15.848 kJ Ans.

1.222 kJ

Ans.

Wdot

mdot

1.22
a) cost_coal

mdot

g h 0.91 0.92

0.488

25.00
ton

cost_coal

MJ
29
kg

cost_gasoline

0.95 GJ

kg
s

1

2.00
gal

37

GJ

cost_gasoline

Ans.

14.28 GJ

1

3

m

cost_electricity

0.1000
kW hr

cost_electricity

27.778 GJ

1

b) The electrical energy can directly be converted to other forms of energy
whereas the coal and gasoline would typically need to be converted to heat
and then into some other form of energy before being useful.
The obvious advantage of coal is that it is cheap if it is used as a heat
source. Otherwise it is messy to handle and bulky for tranport and
storage.
Gasoline is an important transportation fuel. It is more convenient to
transport and store than coal. It can be used to generate electricity by
burning it but the efficiency is limited. However, fuel cells are currently
being developed which will allow for the conversion of gasoline to electricity
by chemical means, a more efficient process.
Electricity has the most uses though it is expensive. It is easy to transport
but expensive to store. As a transportation fuel it is clean but batteries to
store it on-board have limited capacity and are heavy.

5

1.24 Use the Matcad genfit function to fit the data to Antoine's equation.
The genfit function requires the first derivatives of the function with
respect to the parameters being fitted.
B
T C

A

Function being fit:

f ( A B C)
T

e

First derivative of the function with respect to parameter A

d
f ( A B C)
T
dA

B

exp A

T

C

First derivative of the function with respect to parameter B

d
f ( A B C)
T
dB

1
T

C

B

exp A

T

C

First derivative of the function with respect to parameter C

d
f ( A B C)
T
dC

B
(
T

2

exp A

C)

18.5

3.18

9.5

5.48

0.2
11.8
t

9.45
16.9

23.1
32.7

Psat

28.2
41.9

44.4

66.6

52.1

89.5

63.3

129

75.5

187

6

B
T

C

T

t

273.15

lnPsat

ln ( )
Psat

Array of functions used by Mathcad. In this case, a0 = A, a1 = B and a2 = C.

exp a0
exp a0

F ( a)
T

a1
T

a2
a1

T

1
exp a0
T a2
a1
T

a2

2

Guess values of parameters

15

a2

guess

a1
T

exp a0

3000
50

a2
a1
T

a2

Apply the genfit function

A

13.421

A

B

B

genfit ( Psat guess F)
T

C

2.29

C

3

10

Ans.

69.053

Compare fit with data.
200

150
Psat
f ( A B C)
T

100

50

0
240

260

280

300

320

340

360

T

To find the normal boiling point, find the value of T for which Psat = 1 atm.

7

Psat

1atm

B

Tnb
A

Tnb
1.25
a) t1

1970

C2

C1 ( 1

t2
i)

273.15K

2000

dollars
gal

C1

0.35

1.513

329.154 K

Ans.

56.004 degC

C2

t2 t1

Tnb

C K

Psat
ln
kPa

i

5%

dollars
gal

The increase in price of gasoline over this period kept pace with the rate of
inflation.
b) t1

1970
Given

t2
C2
C1

= (1

2000
i)

C1

t2 t1

i

16000

dollars
yr

Find ( i)

i

C2

80000

dollars
yr

5.511 %

The salary of a Ph. D. engineer over this period increased at a rate of 5.5%,
slightly higher than the rate of inflation.
c) This is an open-ended problem. The strategy depends on age of the child,
and on such unpredictable items as possible financial aid, monies earned
by the child, and length of time spent in earning a degree.

8

2.1 (a)

Mwt

g

35 kg

9.8

m

z

2

5m

s

Work

(b)

Work

Mwt g z

Utotal

Utotal

Work

(c) By Eqs. (2.14) and (2.21):

dU

1.715 kJ Ans.

1.715 kJ

Ans.

d ( )= CP dT
PV

Since P is constant, this can be written:

MH2O CP dT = MH2O dU

MH2O P dV

Take Cp and V constant and integrate: MH2O CP t2 t1 = Utotal
kJ
MH2O 30 kg
CP
4.18
t1 20 degC
kg degC
t2

t1

Utotal
MH2O CP

t2

20.014 degC Ans.

(d) For the restoration process, the change in internal energy is equal but of
opposite sign to that of the initial process. Thus

Q

(e)

Utotal

Q

1.715 kJ

Ans.

In all cases the total internal energy change of the universe is zero.

2.2 Similar to Pb. 2.1 with mass of water = 30 kg.
Answers are:

(a) W = 1.715 kJ
(b) Internal energy change of
the water = 1.429 kJ
(c) Final temp. = 20.014 deg C
(d) Q = -1.715 kJ
9

2.4

The electric power supplied to the motor must equal the work done by the
motor plus the heat generated by the motor.

E

i 9.7amp

Wdotelect

Qdot

2.5

i
E

Wdotelect
t

Eq. (2.3):

U = Q

Step 1 to 2:

Ut12

Wdotelect

1.067

3

134.875 W

Ans.

W

W12

200J

Ut12

Q34

Q12

800J

Ut34

1.25hp

10 W

Qdot

Wdotmech

Q12

Step 3 to 4:

Wdotmech

110V

W12

6000J

W34

Q34

Ut34

W34

3

5.8

Ans.

10 J

300J

Ans.

500 J
t

t

Step 1 to 2 to 3 to 4 to 1: Since U is a state function, U for a series of steps
that leads back to the initial state must be zero. Therefore, the sum of the
t

U values for all of the steps must sum to zero.

Ut41

4700J

Ut23
Step 2 to 3:

Ut23

W23

Ut34

Ut41

Ans.

4000 J

Ut23

Ut12

4

Ut23

3

Q23

Q23

3800J

W23

10 J

200 J

Ans.

For a series of steps, the total work done is the sum of the work done for each
step.

W12341

1400J
10

W41

W12341

Step 4 to 1:

W12

W23

Ut41

Ut41

Ans.

10 J
3

4.5

Q41

W41

3

4.5

W41

4700J

Q41

Note:

W41

W34

10 J

200 J

Ans.

Q12341 = W12341

2.11 The enthalpy change of the water = work done.

M

Wdot

2.12 Q

CP

20 kg

4.18

kJ
kg degC

t

10 degC

M CP t

0.25 kW

0.929 hr

Wdot

Ans.

U

Q

12 kJ

W

U

19.5 kJ

Ans.

Q

12 kJ

U

W

U

7.5 kJ

Q

12 kJ

Ans.

2.13Subscripts: c, casting; w, water; t, tank. Then

mc Uc

mw Uw

mt Ut = 0

Let C represent specific heat,

C = CP = CV

Then by Eq. (2.18)

mc Cc tc

mw Cw tw

mt Ct tt = 0

mc

2 kg

mw

Cc

0.50

kJ
kg degC

Ct

mt

40 kg

tc

500 degC

Given

t1

mc Cc t2

t2

0.5

5 kg

kJ
kg degC

Cw

25 degC

t2

30 degC

tc = mw Cw

mt Ct

t2

Find t2
11

t2

4.18

kJ
kg degC

(guess)

t1

27.78 degC

Ans.

2.15

mass

(a)

(b)

T

g

CV

1 kg

Ut

1K

9.8

m

Ut

mass CV T

EP

2

kJ
kg K

4.18

Ans.

4.18 kJ

Ut

s

z

(c)

2.17

EP

EK

z

z

mass g

426.531 m Ans.

EK

u

Ut

1
mass
2

1000

50m

u

kg

5

A

3

A

2m

mdot

Wdot

uA

mdot

mdot g z Wdot

4

D

2

(b)

2

4 kg

1.571 10

7.697

s
3

10 kW

U1

P1

1002.7 kPa

U1

H1

763.131

U2

kJ
2784.4
kg

H2

U2

P1 V 1

kJ
kg

kJ
kg

cm
1.128
gm

V1

Ans.
3

P2

U

P2 V 2

2022.4

Ans.
3

kJ
762.0
kg

U

Ans.

3.142 m

H1

2.18 (a)

m
s

m
s

u

m

D

91.433

Ans.

1500 kPa

U2

U1

H

12

cm
169.7
gm

V2

H

2275.8

kJ
kg

H2

H1

Ans.

2.22

D1

(a)

u1

2.5cm

m
s

D2

5cm

For an incompressible fluid, =constant. By a mass balance,
mdot = constant = u 1A1 = u2A2

u2
(b)

D1

u1

u2

D2
1

EK

2

2

2

u2

1
2

2.23 Energy balance:

Mass balance:

u1

2

mdot3 H3

mdot1

1.0

Qdot

30

Qdot

H1

T1

CP

4.18

s

mdot1

mdot2 H2 = Qdot

H2 = Qdot

mdot2 CP T3

25degC

kJ

Ans.

kg

mdot2 H3

T1

kg
s

J

mdot2 = 0

mdot2 = Qdot

mdot1 CP T1

Ans.

1.875

mdot1

mdot Cp T3

T3 CP mdot1

m
s

mdot1 H1

mdot1 H3

or

0.5

EK

mdot3

Therefore:

T3

2

T2 = Qdot

mdot1 CP T1

mdot2 CP T2

mdot2

0.8

kg
s

T2

75degC

kJ
kg K

mdot2 CP T2

mdot2 CP

T3

43.235 degC

2

2.25By Eq. (2.32a):

By continuity,
incompressibility

H

u
= 0
2
A1

u2 = u1

13

A2

H = CP T

CP

4.18

kJ
kg degC

Ans.

2
u = u1

u1

u1

2

2 CP

D2

2
u = u1

1

m
s

14

D1

D1

2

1

A2

SI units:

T

2

A1

2

4

1

D2

D1

D2

2.5 cm

3.8 cm

4

T

0.019 degC

Ans.

T

D2

0.023 degC

Ans.

7.5cm

u1

T

2

2 CP

D1

1

4

D2

Maximum T change occurrs for infinite D2:

D2

cm

u1

T

2.26 T1

2 CP

300K

Wsdot

H

2

D1

1

T

D2

T2

u1

520K

ndot

98.8kW

CP T2

4

50

H

T1

10

0.023 degC

m
s

u2

kmol
hr

3.5

10

molwt

29

kg
kmol

7
R
2

CP

6.402

m
s

Ans.

3 kJ

kmol

By Eq. (2.30):

Qdot

H

2

u2
2

2

u1
2

molwt ndot

Wsdot Qdot

2

2.27By Eq. (2.32b):

By continunity,
constant area

H=

u2 = u1

u

2 gc

V2

u2 = u1

V1
14

also

T 2 P1
T 1 P2

Ans.

9.904 kW

V2
V1

=
2

T 2 P1
T 1 P2
2

u = u2

u1

2

2
u = u1

P1

R

T2

T2

P2

molwt

mol rankine

7
2

u1

20 psi

ft lbf

28

20

7
R T2
2

ft
s

T1

T1

579.67 rankine

H2

2726.5

kJ
kg

kJ
kg

Ans.

gm
mol

(guess)

578 rankine

Given

H = CP T =

1

T 1 P2

100 psi

3.407

2

T 2 P1

2

2

R T2

T1 =
T2

Find T2

u1
2

2

T 2 P1
T 1 P2

1 molwt

Ans.

578.9 rankine

(
119.15 degF)

2.28 u1

3

m
s

u2

200

m
s

H1
2

By Eq. (2.32a):

2.29 u1

u2

m
s
m
500
s

30

By Eq. (2.32a):

Q

H2

H1

3112.5

u1

u2

H1

334.9

kJ
kg

2

Q

2

kJ

H2

kg

2411.6

2945.7

(guess)
2

H2

Given

u2

578.36

m
s

H1 =

u2

u1

2

5 cm

V1

388.61

u2

2

Ans.
3

3

D1

kJ
kg

cm

gm
15

V2

667.75

cm

gm

Find u2

Continuity:

2.30 (a)

t1

D2

t2

30 degC

CV

u1 V2
u2 V1

D1

D2

Ans.

1.493 cm

n

250 degC

3 mol

J
mol degC

20.8

By Eq. (2.19):

Q

n CV t2

Q

t1

Ans.

13.728 kJ

Take into account the heat capacity of the vessel; then

cv

mv

Q

(b)

100 kg

mv cv

t1

200 degC

CP

29.1

n CV

CV

Q

Q

Ans.

11014 kJ

n

40 degC

4 mol

joule
mol degC
Q

n CP t2

t2

70 degF

5

kJ
kg degC

t1

t2

By Eq. (2.23):

2.31 (a) t1

t2

0.5

BTU
mol degF

n CV t2

Q

t1

18.62 kJ

n

350 degF

3 mol

By Eq. (2.19):

Q

t1

4200 BTU

Ans.

Take account of the heat capacity of the vessel:

mv

Q

mv cv

cv

200 lbm

(b) t1

400 degF

n CV

t2

0.12

BTU
lbm degF

Q

t1

t2

150 degF

16

10920 BTU

n

Ans.

4 mol

Ans.

CP

7

BTU
mol degF

Q

H1

2.33

n CP t2

1322.6

V1

Q

t1

BTU

BTU
lbm

1148.6

3

V2

lbm

78.14

ft

4

mdot

mdot

V2

10

22.997

u2

Wdot

2.34

H1

V1

H2

307

9.25

BTU
lbm

ft

H2

330

V2

u1

lbm

ft

0.28

173.99

BTU
lb

Ans.

39.52 hp

BTU

3

lbm

Ws

2

Wdot

Ws mdot

u1

u2

H1

20

ft
s

molwt

44

3

D1

lbm

D2

4 in

1 in

2

mdot

u2

4

D1 u1

mdot

V1

V2

mdot

u2

679.263

9.686

2

4

D2

2

Eq. (2.32a): Q

H2

H1

10 in

sec

2

D2

Eq. (2.32a): Ws

D2

ft
sec

2

4

3 in

4 lb

3.463

V1

mdot

10

D1

lbm

2

u2

ft
s

u1

3

2

D 1 u1

Ans.

7000 BTU

H2

lbm

ft

3.058

By Eq. (2.23):

ft
sec

u1

u2

2
17

lb
hr

2

Ws

Ws
molwt

5360

Q

BTU
lbmol

98.82

BTU
lbm

gm
mol

Qdot

2.36

T1

Qdot

mdot Q

P

300 K

67128

BTU
hr

Ans.

1 kg

n

1 bar

28.9

V1

n

gm

34.602 mol

mol
3

3
bar cm T1
83.14
mol K P

cm
24942
mol

V1

V2

P dV = n P V 1

W= n

V2 = n P V1

3 V1

V1

Whence

W

Given:

V2
V1

joule

29

Q

n H

U

mol K

Q

= T1 3

Whence

T1

Q

W

U

n

CP T2

602.08 kJ

12.41

kJ
mol

Ans.

172.61 kJ

H

T2 = T1

CP

W

n P 2 V1

T2
H

3 T1

17.4

kJ
mol

Ans.

Ans.

Ans.

2.37 Work exactly like Ex. 2.10: 2 steps, (a) & (b). A value is required for PV/T,
namely R.

R

8.314

T1

(a) Cool at const V1 to P2
(b) Heat at const P2 to T2

Ta2

T1

P2
P1

Ta2

293.15 K

T2

333.15 K

P1

J
mol K

1000 kPa

P2

100 kPa

7
R
2

CP

29.315 K
18

CV

5
R
2

Tb

T2

Hb

C P Tb

Hb

Ua

CV Ta

Ua

R T1

V1

P1

Ta2

Tb

V1

303.835 K

2.437

8.841

Ta

10

5.484

10

mol

Ha

Ua

V 1 P2

P1

Ha

Ub

Hb

P2 V 2

V1

Ub

U

Ub

U

0.831

H

2.39

Ua
Ha

Hb

H

1.164

996

kg

9.0 10

3

D

5
2

kJ
mol

ms

1
cm

u

5

1 m
5 s
5
22133

Re

D

u

Re

Ta

263.835 K

mol
3 J

mol
V2

3

R T2

V2

P2

7.677

6.315

0.028

m

mol

3 J

10

mol

3 J

10

mol

Ans.

mol

4 kg

m
2

kJ

T1

3 J

3
3 m

10

Ta2

55333
110667
276667

19

Ans.

D

0.0001 Note: D = /D
in this solution

0.00635
fF

0.3305 ln 0.27 D

7
Re

0.9

2

0.00517

fF

0.00452
0.0039

0.313

mdot

u

4

D

2

mdot

1.956 kg
1.565 s

Ans.

9.778

0.632
P L

2
fF
D

2

P L

u

0.206

kPa
11.254 m

Ans.

3.88

2.42 mdot

4.5

kg
s

H1

761.1

kJ
kg

H2

536.9

kJ
kg

Assume that the compressor is adiabatic (Qdot = 0). Neglect changes in
KE and PE.
Wdot
Cost

mdot H2
15200

H1

Wdot

Wdot

1.009

3

10 kW

0.573

Cost

kW

20

799924 dollars Ans.

3.1

1 d

=

1

=

dT

d
dP

P

T

At constant T, the 2nd equation can be written:

d

=

2

ln

dP

6

44.1810

P

=

1

bar

2 = 1.01

1

ln ( )
1.01

P

P

Ans.

P2 = 226.2 bar

225.2 bar
3

3.4 b

cm
0.125
gm

c

2700 bar

P1

P2

1 bar

500 bar

V2

Since

Work =

a bit of algebra leads to

P dV
V1

P2

Work

c

P
P

b

dP

Work

0.516

Work

0.516

P1

J

Ans.

gm

Alternatively, formal integration leads to

Work

3.5

= a

P1

c P2

P1

a

bP

P2

1 atm

b ln

P2

b

P1

b

3.9 10

6

atm

1

b

V

3000 atm

0.1 10

1 ft

3

9

J
gm

Ans.

atm

2

(assume const.)

Combine Eqs. (1.3) and (3.3) for const. T:
P2

Work

(
a

V

Work

b P)P dP

P1
21

16.65 atm ft

3

Ans.

1

3.6

1.2 10

V1

3

degC

1

CP

0.84

kJ
kg degC

M

5 kg

t2

20 degC

3

m

1

P

1590 kg

t1

1 bar

0 degC

With beta independent of T and with P=constant,

dV
=
V

V2

dT

Vtotal

Vtotal

M V

Work

V1 exp

P Vtotal

M CP t2

P2

(a) Constant V:

T2

T1

U

H

R T1 ln

(c) Adiabatic:

T1

P1

Ans.

84 kJ

84 kJ

Ans.

83.99 kJ

Ans.

T

T1

U

10.91

15.28

H= 0

Q= 0
22

kJ
mol

kJ
mol

Work

and

CV

U = Q = CV T

T2

and

7
R
2

CP

600 K

and

Q

Ans.

Utotal

U=

P2

7.638 joule

Work

H

CP T

(b) Constant T:

Work

Q

1 bar

V1

Ans.

Q

Q and

CV T

m

t1

T

P1

3

V2

Htotal

W= 0

P2

10

Q

Utotal

8 bar

5

V

Work

7.638

Htotal

P1

t1

(Const. P)

Q

3.8

t2

525 K

Ans.

Ans.

and
10.37

Q= W
kJ
mol

Ans.

U = W = CV T

5
2

R

1

CP

T2

CV

P2

T1

U

and

U

5.586

3.9 P4

2bar

CP

H

kJ

7
R
2

10bar

T1

Step 41: Adiabatic

T

331.227 K

Ans.

mol

T4

R T1
P1

V1

P4

T1

kJ
mol

Ans.

5
R
2

CV

600K

7.821

T2

CP T

H

CV T

W

P1

T2

P1

V1

4.988

T4

3
3 m

10

378.831 K

mol

R
CP

P1

3 J

U41

CV T1

T4

U41

4.597

10

H41

CP T1

T4

H41

6.436

10

Q41

3bar

T2

Step 12: Isothermal

Q41

U41

W41

4.597

mol

J
0
mol

W41

P2

J
0
mol

mol
3 J

V2

600K

U12

0

J
mol

H12

0

J
mol

23

R T2
P2

3 J

10
3

V2

m
0.017
mol

U12

0

J
mol

H12

0

J
mol

mol

T1

Q12
W12

P3

2bar

V3

Q12

Q12

P1

W12

P3 V 3

T3

V2

Step 23: Isochoric

P2

R T1 ln

T3

R

U23

CV T3

H23

CP T3

T2

Q23

2 bar

T4

CV T3

W23

P4

0

T2

R T4

V4

3 J

10

mol
3 J

6.006

10

400 K
3 J

4.157

10

P4

3

V4

m
0.016
mol

U34

CV T4

T3

U34

439.997

H34

CP T4

T3

H34

615.996

Q34

CP T4

T3

Q34

W34

R T4

T3

W34

3.10 For all parts of this problem: T2 = T1

mol

mol
3 J
H23
5.82 10
mol
3 J
Q23
4.157 10
mol
J
W23 0
mol

J
mol

378.831 K

Step 34: Isobaric

U23

T2

6.006

615.996

175.999

J
mol
J

mol

J
mol
J

mol

and

Also
Q = Work and all that remains is
U= H= 0
to calculate Work. Symbol V is used for total volume in this problem.

P1

1 bar

P2

V1

12 bar
24

3

12 m

V2

3

1m

(a)

P2

Work = n R T ln

Work

P1
Work

P1 V1 ln

P2
P1
Ans.

2982 kJ

(b) Step 1: adiabatic compression to P2
1

5
3

Vi

P2 V i

W1

V1

P1
P2

(intermediate V)

P1 V 1

Vi

W1

3063 kJ

W2

1

3

2.702 m

2042 kJ

Step 2: cool at const P2 to V2

W2

P2 V 2

Work

W1

Vi

W2

Work

5106 kJ

Ans.

(c) Step 1: adiabatic compression to V2
Pi

W1

P1

V1

(intermediate P)

V2

Pi V 2

1

Step 2: No work.

Work

(d) Step 1: heat at const V1 to P2

62.898 bar

W1

P1 V 1

Pi

7635 kJ

W1

Work

7635 kJ

Ans.

W2

Work

13200 kJ

Ans.

W1 = 0

Step 2: cool at const P2 to V2

W2

P2 V 2

V1

Work

(e) Step 1: cool at const P1 to V2

W1

P1 V 2

W1

V1
25

1100 kJ

Step 2: heat at const V2 to P2

Work

3.17(a)

W2 = 0

Work

W1

No work is done; no heat is transferred.
t

U =

(b)

T= 0

Not reversible

T2 = T1 = 100 degC

The gas is returned to its initial state by isothermal compression.

Work = n R T ln
3

V1

3.18 (a) P1

P2 V2 ln

7
2

but

V2
4
3

3

P2

V1

6 bar

878.9 kJ Ans.

Work

V2

T1

500 kPa

303.15 K

CP

5
R
2

CV

R

n R T = P2 V 2

m

P2

100 kPa

CP

V1

V2

4m

Work

CV

Adiabatic compression from point 1 to point 2:

Q12

0

kJ
mol

1

U12 = W12 = CV T12

U12

CV T2

U12

3.679

T1

kJ
mol

T2

H12

CP T2

W12

H12

5.15

T1

kJ

W12

mol

T1

3.679

P2
P1
U12

kJ
mol

Cool at P2 from point 2 to point 3:

T3

Ans.

1100 kJ

H23

T1

U23

CV T3

CP T3

W23

T2

26

Q23

T2

U23

Q23

H23

Ans.

H23

5.15

Q23

5.15

kJ

U23

mol

kJ

W23

mol

kJ
mol

3.679

Ans.

kJ

1.471

Ans.

mol

Isothermal expansion from point 3 to point 1:

U31 =

H31 = 0

Q31

4.056

W31

P2

P1

R T3 ln

P3

W31

W31

P3

kJ
mol

FOR THE CYCLE:

Q

Q

Q12

1.094

Q23

Q31

U=

4.056

kJ
mol

Ans.

H= 0

Work

kJ
mol

W12

Work

Q31

1.094

W23

W31

kJ
mol

(b) If each step that is 80% efficient accomplishes the same change of state,
all property values are unchanged, and the delta H and delta U values
are the same as in part (a). However, the Q and W values change.
Step 12:

W12

Q12

Step 23:

W23

Q23

Step 31:

W31

Q31

W12

W12

W23
0.8

U23

4.598

Q12

0.92

kJ
mol

W23

0.8

U12

kJ

W12

1.839

kJ
mol

mol

W31 0.8

W31

Q23

5.518

kJ
mol

W31

W23

3.245

kJ
mol

Q31
27

3.245

kJ
mol

FOR THE CYCLE:

Q

Q12

Q

Q23

3.192

Work

kJ
mol

W12

Work

Q31

3.192

W23

W31

kJ
mol

3.19Here, V represents total volume.

P1

CP

21

CV

joule
mol K

CP

T2

208.96 K

P1 V 1

n

P1 V 1

T2

Work
n CV

P2

P1

V2

Ans.

T1

P2 V 2
P1 V 1

69.65 kPa

Ans.

Work

Work =

Work

Pext V2

994.4 kJ Ans,

U = Pext V

V1

Work

U = n CV T

R T1

V1

Ans.

1609 kJ

V2

P2

P1

200 kPa

T2

1

100 kPa

V2

V1

P1

(c) Restrained adiabatic:

Pext

P2

Work

P2

P2 V 2

V1

P2

600 K

T2

Work

CV

V2

(b) Adiabatic:

600 K

CP

V1

P1 V1 ln

T1

5 V1

R

Work = n R T1 ln

T1

Work

V2

1m

(a) Isothermal:
T2

3

V1

1000 kPa

T2

V 1 T2
V 2 T1
28

442.71 K

Ans.

P2

T1

147.57 kPa

Ans.

400 kJ Ans.

3.20

T1

CP

P1

423.15 K

7
2

Step 12:

If

V1
V2

=

2.502

Step 23:

V1
V3

kJ
mol
0

Process:

2.079

U12

mol

mol

R T1 ln r
()

Q12

2.502

CV T3

CP T3

T2

2.079

W23

kJ
mol

H23

Work

Q

Q23

2.502

0.424

2.91

kJ
mol

kJ
mol

kJ
mol

H

H12

H23

H

2.91

U

U12

U23

U

2.079

29

kJ
mol

T2

H23

W12

Q12

kJ

U23

U23

Q

323.15 K

W12

W12

U23

Work

0

T 3 P1

kJ
mol

kJ
mol

T3

T 1 P3

r

Q12

W23

T1

kJ

Then

Q23

Q23

0

3 bar

T2

R

2

H12

r=

W12

5

CV

R

P3

8bar

kJ
mol

kJ
mol

Ans.

Ans.

Ans.

Ans.

3.21 By Eq. (2.32a), unit-mass basis:

But

CP

u1

2.5

t2

3.22

28

Whence

H = CP T

R
7
2 molwt

molwt

u2

1
2

H

u2

T=

m
s

u2

t1

gm
mol
2

2

u1

2 CP

50

m
s

t1

u1

t2

2 CP

7
R
2

CV

5
R
2

P1

1 bar

P3

148.8 degC

Ans.

10 bar

CV T3

U

2.079

T1

mol

CP T3

H

Ans.

T3

303.15 K

H

T1

kJ

2.91

T1

kJ
mol

Each part consists of two steps, 12 & 23.
(a) T2

W23

T3

R T2 ln

P2

P3
P2

P1

150 degC

2

2

CP

U

2

u = 0

T2
T1
Work

W23

Work

6.762

Q

U

Q

4.684

30

kJ
mol

Ans.

Work

kJ
mol

Ans.

Ans.

403.15 K

(b) P2

T2

P1

H12

CP T2

W12

U12

W23

R T2 ln

Work

W12

Q

(c)

U

T2

U12

T3

CV T2

Q12

Q12

P3

H12

W12

T1

0.831

W23

P2

Work

W23

Q

Work

P2

T1

P3

T1

W12

kJ
mol

7.718

6.886

4.808

kJ
mol

kJ
mol

kJ
mol

CP T3

T2

Q23

CV T3

T2

W23

U23

Work

4.972

P1

H23

U23

Ans.

P2

R T1 ln

H23

Ans.

Work

Q

W12

U

W23

Q

Work

2.894

Q23

kJ
mol

kJ
mol

For the second set of heat-capacity values, answers are (kJ/mol):

U = 2.079

U = 1.247

(a)

Work = 6.762

Q = 5.515

(b)

Work = 6.886

Q = 5.639

(c)

Work = 4.972

Q = 3.725

31

Ans.

Ans.

3.23

T1

303.15 K

T2

T1

T3

393.15 K

P1

1 bar

P3

12 bar

CP

7
R
2

For the process:

U

U

Step 12:

W12

CV T3

1.871

P2

5.608

kJ
mol

mol

For the process:

Work

W12

Q

3.24

Work

5.608

W12 = 0

Work = W23 = P2 V3

But

T3 = T1

Also

W = R T1 ln

ln

P

T2 T1
P
=
P1
T1

P1 exp

T2

T2

R T1 ln

5.608

P
P1

Ans.

P2

U

P1

kJ
mol

W23

kJ

Q

mol

3.737

V 2 = R T3

kJ
mol

Ans.

T2

Work = R T2

So...

mol

Q23

kJ
mol

W23

Q23

kJ

Q12

W12

Step 23:

Q12

2.619

T1

W12

T3

0

CP T3

H

kJ

T1

P3

Q12

H

T1

5
R
2

CV

T1

Therefore

T1

T1

32

T1

800 K

P

350 K

2.279 bar

P1

Ans.

4 bar

3.25

3

VA

P

Define:

256 cm

P1

r

0.0639

VB

3750.3 cm

CV

= r

CP

Assume ideal gas; let V represent total volume:

P1 V B = P 2 V A

P
P1

3.26

T1

=

VA
VA

VA ( 1)
r

VB

VB

300 K

From this one finds:

VB

P1

r

1 atm

CP

7
R
2

3

Ans.

CP

R

CV

The process occurring in section B is a reversible, adiabatic compression. Let

TA ( )= TA
final

P ( )= P2
final

TB ( )= TB
final

Since the total volume is constant,

nA = nB

nA R TA
2 nA R T1
=
P2
P1

TB

TA TB
2 T1
=
P2
P1

or

(1)

1

(a)

P2
TA

2 T1

TB

1.25 atm
P2
P1

Q = nA

TB

Define

q=

TB

TA

319.75 K

T1

Q
nA

q

430.25 K

33

P2

(2)

P1
UA

CV TA

q

UB

TB

2 T1

3.118

kJ
mol

(3)

Ans.

(b)

Combine Eqs. (1) & (2) to eliminate the ratio of pressures:

(guess)

425 K

TB

300 K

TB

Find TB

TB

TA

319.02 K

Ans.

P2

1.24 atm

Ans.

kJ
mol

Ans.

1

TB = T1

Given

P2

P1

q

(c)

TA

P1

TA

2 T1

P2

P2

(1)

TB

P1

kJ
mol

2.993

1

T1

CV TA

3

q

2 T1

By Eq. (2),

TB

Eliminate

q

(1)

TB

TB

P2

(d)

2 T1

TB

325 K

q

TB

2 T1

CV TA

TB

TA

Ans.

TA

469 K

Ans.

q

2 T1

TA

1.323 atm

TB

q P1
2 T1 C V

4.032

kJ
mol

Ans.

from Eqs. (1) & (3):

P2

1.241 atm

Ans.

(2)

P2

TB

319.06 K

Ans.

(1)

TA

425.28 K

Ans.

P1

1

TB

T1

TA

2 T1

P2
P1
P2
P1

TB

34

6

3

3.30 B

P1

B'

cm

242.5

C

25200

mol

mol

P2

7.817

T

B

C

10

3 1

bar

B
2

373.15 K

2

55 bar

B'

1 bar

RT

C'

cm

R T

2

C'

2

3.492

10

5 1
2

bar

(a) Solve virial eqn. for initial V.

RT
P1

Guess:

V1

Given

P1 V 1
= 1
RT

B
V1

3

C

V1

2

V1

V2

Find V1

cm
30780
mol

cm
241.33
mol

V1

Solve virial eqn. for final V.
Guess:

RT
P2

V2

P2 V 2

Given

RT

= 1

B
V2

3

C

V2

2
V2

Find V2

Eliminate P from Eq. (1.3) by the virial equation:
V2

Work

RT

1

B
V

C
V

2

1
dV
V

Work

12.62

kJ
mol

V1

(b)

Eliminate dV from Eq. (1.3) by the virial equation in P:
P2

1

dV = R T
P

C' dP

W

1
P

RT

2
P1

W
35

12.596

kJ
mol

Ans.

C' P dP

Ans.

Note: The answers to (a) & (b) differ because the relations between the two
sets of parameters are exact only for infinite series.
3.32 Tc

282.3 K

T

298.15 K

Pc

50.4 bar

P

12 bar

T
Tc

Tr

Pr

Tr

Pr

P
Pc
6

3

B

Given

cm

140

C

mol

PV
= 1
RT

7200

cm

mol

B
V

V

(b)

B0

V

Find ( )
V

Tr

V

Tr

Z

1

(c)

B0

cm
1919
mol

B1

B1

4.2

Tr

Z

PV
RT

Z

B0

1.6

Pr

3

V

2066

cm

mol

2

0.172

0.139

RT
P

C

0.422

0.083

V

2

3

B1

0.238

(guess)

0.087
(a)

1.056

Ans.

Z

0.929

V

cm
Ans.
1924
mol

0.304

2.262

0.932

10

V

3

3

ZRT
P

For Redlich/Kwong EOS:

0

1

( )
Tr

T r Pr

Tr

0.5

Pr
Tr

Table 3.1

Eq. (3.53)

36

Table 3.1

0.42748

0.08664

q Tr

Tr
Tr

Eq. (3.54)

Calculate Z

Guess:

Given

T r Pr

0.9

Eq. (3.52)

Z= 1

Z

Z

Z

Find Z)
(

(d)

q Tr

Z

T r Pr

Z

T r Pr

Z

T r Pr
3

Z RT

V

0.928

T r Pr

V

P

1916.5

cm

mol

Ans.

For SRK EOS:

2

1

Tr

1

0.480
Tr

q Tr

(e)

1

Tr

Guess:

Z

Table 3.1

2

Pr

T r Pr

Tr

T r Pr

q Tr

Z

Find Z)
(

0.9

Z

T r Pr

Z

V

0.928

T r Pr

T r Pr

Z

T r Pr
3

Z RT

V

P

1918

1

2

0.45724

0.07779

2

1

q Tr

cm

mol

Ans.

For Peng/Robinson EOS:

1

Tr

Eq. (3.53)

Tr

Eq. (3.52)

Given

Z

0.176

Eq. (3.54)

Calculate Z

Z= 1

1.574

2

Table 3.1

0.42748

0.08664

0

1

1

0.37464
Tr

1.54226

0.26992

Eq. (3.54)

Tr
37

2

T r Pr

1

Tr
Pr
Tr

2

Table 3.1

2

Table 3.1

Eq. (3.53)

Calculate Z

Guess:

0.9

Eq. (3.52)

Given

Z= 1

T r Pr

Z

Z

Find ( Z)

q Tr

Z

Z

T r Pr

Z

T r Pr
3

V

1900.6

cm

mol

305.3 K

Pc

T

323.15 K

Tr

T
Tc

Tr

P

15 bar

Pr

P
Pc

Pr

0.308

(guess)

0.100
6

3

(a)

cm
156.7
mol

B

Given

PV
= 1
RT

B
V

C

9650

cm

mol

V

(b)

B0

V

Find ( V)

Tr

B1

V

Tr

(c)

B0

cm
1625
mol

B1

B1

4.2

Tr

Z

PV

Z

B0

1.6

Pr

3

V

cm
1791
mol

2

0.172

0.139

RT
P

C

0.422

0.083

V

2

3

1

Ans.

1.058

48.72 bar

3.33 Tc

Z

Z

T r Pr

ZRT
P

V

0.92

T r Pr

Ans.

Z

RT

0.907

V

cm
Ans.
1634
mol

0.302

3.517

0.912

V

10

3

ZRT
P

3


1

0

0.08664
38

0.42748

Table 3.1

( )
Tr

0.5

Tr

Table 3.1

Pr

T r Pr

Tr

q Tr

Eq. (3.54)

Tr

Eq. (3.53)

Tr

Calculate Z

Guess:

Given

T r Pr

0.9

Eq. (3.52)

Z= 1

Z

Z

Z

Find Z)
(

(d)

q Tr

Z

T r Pr

Z

Z

T r Pr

cm
1622.7
mol

V

P

Ans.

For SRK EOS:

0.42748

0.08664

0

1

Tr

1

0.480
Tr

q Tr

0.176

Eq. (3.54)

Guess:

2

1

Tr

T r Pr

Tr

Calculate Z

Z

Table 3.1

2

Pr

Eq. (3.53)

Tr

0.9

Eq. (3.52)

Given

Z= 1

Z

1.574

Table 3.1

2

1

(e)

T r Pr
3

Z RT

V

0.906

T r Pr

T r Pr

Find Z)
(

q Tr

Z

T r Pr

Z
Z

V

0.907

T r Pr

Z RT
P

T r Pr
Z

T r Pr
3

V

1624.8

cm

mol

Ans.


1

2

1

0.07779

2
39

0.45724

Table 3.1

1

Tr

1

0.37464
Tr

q Tr

1.54226

0.26992

Eq. (3.54)

1

Guess:

Z

Tr

Table 3.1

2

Pr

T r Pr

Tr

Calculate Z

2

2

Eq. (3.53)

Tr

0.9

Eq. (3.52)

Given

Z= 1

T r Pr

q Tr

Z

Z

T r Pr

Z

T r Pr

Z

T r Pr
3

ZRT

V

0.896

T r Pr

cm
1605.5
mol

V

Z

Find ( Z)

3.34 Tc

318.7 K

T

348.15 K

Tr

T
Tc

Tr

Pc

37.6 bar

P

15 bar

Pr

P
Pc

Pr

Ans.

1.092

0.399

P

0.286
(guess)
6

3

(a)

B

Given

194

cm

C

mol

PV
= 1
RT

15300

cm

mol

B

V

V

cm
1722
mol

Find ( V)

(b)

B0

0.083

V

1930

mol

2

3

V

3

cm

C

V

V

2

RT
P

0.422
Tr

1.6

B0

40

Z

0.283

PV
RT

Z

0.893

Ans.

B1

0.172

0.139

Tr

Z

1

B0

(c)

B1

4.2

Pr

B1

Z

Tr

0.02

V

0.899

3

Z RT

V

P

1734

cm

mol

Ans.


( )
Tr

Tr

0.5

Table 3.1

Pr

T r Pr

Table 3.1

0.42748

0.08664

0

1

Tr

q Tr

Eq. (3.54)

Tr

Eq. (3.53)

Tr

Calculate Z

Guess:

Given

T r Pr

0.9

Eq. (3.52)

Z= 1

Z

Z

Find Z)
(

(d)

q Tr

Z

Z

T r Pr

Z

T r Pr

Z

T r Pr
3

Z RT

V

0.888

T r Pr

cm
1714.1
mol

V

P

Ans.

For SRK EOS:

0.42748

0.08664

0

1

1

Tr
q Tr

1

0.480
Tr

1.574

0.176

Eq. (3.54)

Tr

41

2

1

T r Pr

Tr

Table 3.1

2

Table 3.1

2

Pr
Tr

Eq. (3.53)

Calculate Z
Guess:
Eq. (3.52)
Given

Z= 1

T r Pr

Z

(e)

q Tr

Z

Find ( Z)

Z

0.9

Z

T r Pr

Z

Z

T r Pr

T r Pr
3

ZRT
P

V

0.895

T r Pr

V

1726.9

Ans.

mol


1

1

2

0.45724

0.07779

2

1

Tr

1

0.37464
Tr

q Tr

T r Pr

Guess:

Z

T r Pr

q Tr

Z

Find ( Z)

Tr

2

Table 3.1

2

Pr

Eq. (3.53)

Tr

0.9

523.15 K

Z

T r Pr

Z

ZRT
P

V

0.882

P

T r Pr

cm
152.5
mol

B

Given

Z

PV
RT

T r Pr
Z

T r Pr
3

cm
1701.5
mol

V

Ans.

1800 kPa
6

3

(a)

1

Table 3.1

Eq. (3.52)

Z= 1

T

2

0.26992

Tr

Given

Z

1.54226

Eq. (3.54)

Calculate Z

3.35

cm

C

mol

C

B
V

PV
= 1
RT

V

2

2

V

RT
(guess)
P

V

5800

cm

Find ( V)

3

V

2250
42

cm

mol

Z

0.931

Ans.

(b) Tc

Tr

Tr

B1

647.1 K

Pc

T
Tc

Pr

0.139

0.172
Tr

(c) Table F.2:

B0

Pc

4.2

0.422

0.083

Tr

B0

0.082

B1

0.51

Z

0.281

1.6

1

B0

B1

Pr
Tr

3

Z RT
P

V

P

Pr

0.808

0.345

220.55 bar

Z

molwt

V

V

cm
molwt
124.99
gm

V

0.939

cm
2268
mol

cm
2252
mol

Ans.
3

gm
18.015
mol

3

or

cm
53.4
mol

B

T

Zi

Z1i

C

2620

cm

mol

2

D

5000

cm

mol

3

n

mol

273.15 K

PV

Given

i

9

6

3

3.37

Ans.

RT

0 10

Pi

fPi Vi Pi
RT

1

B Pi
RT

= 1

10

B
V
10

C
V

2

D
V

fP V)
(

3

RT
Pi

Vi

20 i bar

Find V)
(

(guess)

Eq. (3.12)

Eq. (3.38)

43

Z2i

1
2

1
4

B Pi
RT

Eq. (3.39)

1·10 -10

Z2i

Z1i

Zi

20

1

1

1

40

0.953

0.953

0.951

60

0.906

0.906

0.895

0.861

0.859

0.83
0.749

80

Pi

100

bar

0.819

0.812

120

0.784

0.765

0.622

140

0.757

0.718

0.5+0.179i

160

0.74

0.671

0.5+0.281i

180

0.733

0.624

0.5+0.355i

200

0.735

0.577

0.5+0.416i

0.743

0.53

0.5+0.469i

Note that values of Z from Eq. (3.39) are not physically meaningful for
pressures above 100 bar.
1

0.9
Zi

0.8

Z1 i
Z2 i

0.7

0.6

0.5

0

50

100
Pi bar

44

150
1

200

3.38 (a) Propane:

Tc

Tc

313.15 K

P

Tr

T

Pc

T

Tr

369.8 K

0.847

Pr

0.152

42.48 bar

13.71 bar

P
Pc

Pr

0.323


( )
Tr

Tr

0.5

Table 3.1

Pr

T r Pr

Table 3.1

0.42748

0.08664

0

1

Tr

q Tr

Eq. (3.54)

Tr

Eq. (3.53)

Tr

Calculate Z for liquid by Eq. (3.56) Guess:

Z

0.01

Given

Z=

Z

T r Pr

Z

Find Z) Z
(

T r Pr

0.057

Z

Calculate Z for vapor by Eq. (3.52)

Z

T r Pr

q Tr

T r Pr
3

Z RT
P

V

1

T r Pr

V

Guess:

108.1

cm

Ans.

mol

Z

0.9

V

cm
1499.2
mol

Given

Z= 1

Z

T r Pr

Find Z)
(

q Tr

Z

T r Pr

Z
Z Z

V

0.789

45

T r Pr
T r Pr

Z RT
P

3

Ans.

Rackett equation for saturated liquid:

T
Tc

Tr

Tr

0.847

3

Vc

V

200.0

V c Zc

cm

Zc

mol

1 Tr

0.276

3

0.2857

V

94.17

cm

Ans.

mol

For saturated vapor, use Pitzer correlation:

B0

0.083

0.422
Tr

B1

0.139

V

0.468

B1

0.207

0.172
Tr

RT
P

B0

1.6

4.2

R B0

B1

Tc

V

Pc

46

1.538

3
3 cm

10

mol

Ans.

Parts (b) through (t) are worked exactly the same way. All results are
summarized as follows. Volume units are cu.cm./mole.
R/K, Liq. R/K, Vap. Rackett Pitzer
(a) 108.1
1499.2
94.2 1537.8
(b) 114.5

1174.7

98.1

1228.7

(c) 122.7

920.3

102.8

990.4

(d) 133.6

717.0

109.0

805.0

(e) 148.9

1516.2

125.4

1577.0

(f) 158.3

1216.1

130.7

1296.8

(g) 170.4

971.1

137.4

1074.0

(h) 187.1

768.8

146.4

896.0

(i) 153.2

1330.3

133.9

1405.7

(j) 164.2

1057.9

140.3

1154.3

(k) 179.1

835.3

148.6

955.4

(l) 201.4

645.8

160.6

795.8

(m)

61.7

1252.5

53.5

1276.9

(n)

64.1

1006.9

55.1

1038.5

(o)

66.9

814.5

57.0

853.4

(p)

70.3

661.2

59.1

707.8

(q)

64.4

1318.7

54.6

1319.0

(r)

67.4

1046.6

56.3

1057.2

(s)

70.8

835.6

58.3

856.4

(t)

74.8

669.5

60.6

700.5
47

3.39 (a) Propane

T

Tc

273.15)K

T

Tr

(
40

T
Tc

Tr

Pc

369.8 K

Pr

0.847

0.152

42.48 bar

P

P
Pc

13.71 bar

Pr

313.15 K

0.323

From Table 3.1 for SRK:

0.42748

0.08664

0

1

2

1

Tr

1

0.480

Tr

q Tr

1.574

0.176

Eq. (3.54)

2

1

2

Tr

Pr

T r Pr

Tr


Eq. (3.53)

Tr
Z

0.01

Given

Z=

Z

T r Pr

Find Z)
(

Z

Z

T r Pr

0.055

Z


Z

T r Pr

q Tr

T r Pr
3

Z RT
P

V

1

T r Pr

cm
104.7
mol

V

Guess:

Z

Ans.

0.9

Given

Z= 1

Z

T r Pr

Find Z)
(

q Tr

Z

0.78

Z

T r Pr

Z

V

48

T r Pr
Z

T r Pr

Z RT
P

T r Pr
3

V

1480.7

cm

mol

Ans.

SRK, Liq. SRK, Vap. Rackett Pitzer
(a) 104.7
1480.7
94.2 1537.8
(b) 110.6

1157.8

98.1

1228.7

(c) 118.2

904.9

102.8

990.4

(d) 128.5

703.3

109.0

805.0

(e) 142.1

1487.1

125.4

1577.0

(f) 150.7

1189.9

130.7

1296.8

(g) 161.8

947.8

137.4

1074.0

(h) 177.1

747.8

146.4

896.0

(i) 146.7

1305.3

133.9

1405.7

(j) 156.9

1035.2

140.3

1154.3

(k) 170.7

815.1

148.6

955.4

(l) 191.3

628.5

160.6

795.8

(m)

61.2

1248.9

53.5

1276.9

(n)

63.5

1003.2

55.1

1038.5

(o)

66.3

810.7

57.0

853.4

(p)

69.5

657.4

59.1

707.8

(q)

61.4

1296.8

54.6

1319.0

(r)

63.9

1026.3

56.3

1057.2

(s)

66.9

817.0

58.3

856.4

(t)

70.5

652.5

60.6

700.5
49

3.40 (a) Propane

T

(
40

Tr

Tc

T
Tc

Pc

369.8 K

T

273.15)K

Tr

P

P
Pc

13.71 bar

Pr

313.15 K

Pr

0.847

0.152

42.48 bar

0.323

From Table 3.1 for PR:
2

1

Tr

1
1

0.37464
1

2

0.26992

2

Eq. (3.54)

1

2

Pr

T r Pr

Tr

Tr

0.45724

0.07779

2

Tr

q Tr

1.54226


Eq. (3.53)

Tr
Z

0.01

Given

Z=

Z

T r Pr

Find Z)
(

Z

Z

T r Pr

0.049

Z


Z

T r Pr

q Tr

T r Pr
3

Z RT
P

V

1

T r Pr

cm
92.2
mol

V

Guess:

Z

Ans.

0.6

Given

Z= 1

Z

T r Pr

Find Z)
(

q Tr

Z

0.766

Z

T r Pr

Z

V

50

T r Pr
Z

T r Pr

Z RT
P

T r Pr
3

V

1454.5

cm

mol

Ans.

PR, Liq. PR, Vap. Rackett Pitzer
(a) 92.2
1454.5
94.2 1537.8
(b)

97.6

1131.8

98.1

1228.7

(c) 104.4

879.2

102.8

990.4

(d) 113.7

678.1

109.0

805.0

(e) 125.2

1453.5

125.4

1577.0

(f) 132.9

1156.3

130.7

1296.8

(g) 143.0

915.0

137.4

1074.0

(h) 157.1

715.8

146.4

896.0

(i) 129.4

1271.9

133.9

1405.7

(j) 138.6

1002.3

140.3

1154.3

(k) 151.2

782.8

148.6

955.4

(l) 170.2

597.3

160.6

795.8

(m)

54.0

1233.0

(n)

56.0

987.3

55.1

1038.5

(o)

58.4

794.8

57.0

853.4

(p)

61.4

641.6

59.1

707.8

(q)

54.1

1280.2

54.6

1319.0

(r)

56.3

1009.7

56.3

1057.2

(s)

58.9

800.5

58.3

856.4

(t)

62.2

636.1

60.6

700.5

53.5

1276.9

51

3.41 (a) For ethylene,

molwt

28.054

gm
Tc
mol

282.3 K

Pc

50.40 bar

0.087

T

Tr

Tc

328.15 K

P

35 bar

P
Pc

Pr

T

Tr

1.162

Pr

0.694

From Tables E.1 & E.2:

Z

Z0

Z1

Z

Z0

Z1

0.838

0.033

0.841

Vtotal

ZnRT
P

Vtotal

0.421 m

P

115 bar

Vtotal

0.25 m

Tr

1.145

Pr

P
Pc

From Tables E.3 & E.4: Z0

0.482

Z1

0.126

18 kg

n

molwt

(b) T

323.15 K

T
Tc

Tr

Z

Z0

Z

Z1

mass

Pr

ZRT

mass

n molwt

Ans.

3

P Vtotal

n

0.493

3

n

2.282

2171 mol

Ans.

60.898 kg

3.42 Assume validity of Eq. (3.38).
3

P1

1bar

Z1

P1 V 1
R T1

T1

Z1

300K

0.922

cm
23000
mol

V1

R T1
Z1
P1

B

With this B, recalculate at P2

Z2

1

B P2
R T1

Z2

0.611

P2

V2
52

R T1 Z2
P2

B

1

3
3 cm

1.942 10

mol

5bar

V2

3.046

10

3
3 cm

mol

Ans.

3.43 T

753.15 K

Tc

513.9 K

Tr

T
Tc

Tr

1.466

P

6000 kPa

Pc

61.48 bar

Pr

P
Pc

Pr

0.976

0.645

B0

0.083

0.422
Tr

B1

0.172

0.139

Tr

RT
P

V

B0

For an ideal gas:

3.44 T

320 K

0.104

3

V

cm
989
mol

V

Pc

cm
1044
mol

Ans.
3

RT
P

V

P

0.146

B1

4.2

Tc

B1 R

B0

1.6

Tc

369.8 K

Pc

Zc

16 bar

42.48 bar

0.276

molwt

3

Vc

0.152

Tr

Vliq

Vtank

cm
200
mol

T
Tc

Tr

V c Zc

1 Tr

3

B0

0.083

Tr

B1

0.139

1.6

0.172
Tr

4.2

Pr

gm
mol

0.377
3

cm

Vliq

0.8 Vtank
Vliq
molwt

0.422

P
Pc

0.2857

mliq

0.35 m

Pr

0.865

44.097

B0

0.449

B1

0.177

53

96.769

mliq

127.594 kg

mol

Ans.

RT
P

Vvap

B0

B1 R

Tc

3
3 cm

Vvap

1.318

mvap

Pc

10

2.341 kg

mol

0.2 Vtank

mvap

Vvap

Ans.

molwt

3.45 T

B0

Tc

425.1 K

2.43 bar

Pc

37.96 bar

Pr

Vvap

0.083

0.422
Tr

B1

0.139

RT
P

1.6

0.172
Tr

V

T

Tr

0.200

P

298.15 K

4.2

B0

3

gm
mol

0.624

Tc

B1 R

V

Pc

Vvap

mvap

0.064

0.661

B1

58.123

0.701

Pr

Pc

molwt

16 m

B0

Tr

Tc
P

mvap

V
molwt

9.469

98.213 kg

3
3 cm

10

mol

Ans.

P

333.15 K

Tc

305.3 K

Tr

T
Tc

Tr

1.091

14000 kPa

Pc

48.72 bar

Pr

P
Pc

Pr

2.874

0.100

3.46 (a) T

Vtotal

From tables E.3 & E.4: Z0

3

molwt

0.15 m

0.463
54

Z1

0.037

30.07

gm
mol

Z

Z0

Vtotal

methane

(b)

Z

Z1

methane

V
molwt

Vtotal

V

P

40 kg

or

Tr =

Whence

V

0.459

Z

Tr =

PV

29.548

R Tc

at

90.87

mol

P V = Z R T = Z R Tr Tc

20000 kPa

0.889
Z

V

P

cm

Ans.

49.64 kg

where

3

Z RT

Pr

P
Pc

Pr

mol
kg

4.105

This equation giving Tr as a function of Z and Eq. (3.57) in conjunction with
Tables E.3 & E.4 are two relations in the same variables which must be
satisfied at the given reduced pressure. The intersection of these two
relations can be found by one means or another to occur at about:

Tr

Whence

T

V

or

T

or

3

0.15 m

Vtotal

T

298.15 K

molwt

0.087

50.40 bar
P V = P r Pc V = Z R T

40 kg
molwt

Whence

0.693

118.5 degC Ans.

Pc

282.3 K

Pr =

Z

Tr Tc

391.7 K

3.47 Vtotal

Tc

and

1.283

Z

RT
Pc V

where

Pr = 4.675 Z

at

Tr

55

T
Tc

4.675

Tr

1.056

28.054

gm
mol

This equation giving Pr as a function of Z and Eq. (3.57) in conjunction with
Tables E.3 & E.4 are two relations in the same variables which must be
satisfied at the given reduced temperature. The intersection of these two
relations can be found by one means or another to occur at about:

Pr

and

1.582

3.48 mwater

Z

P

Pc Pr

3

Vtotal

15 kg

P

0.338

V

0.4 m

Interpolate in Table F.2 at 400 degC to find:

298.15 K

Tc

305.3 K

Tr1

P1

2200 kPa

Pc

48.72 bar

Pr1

3

T2

Z0

mwater

T1
Tc
P1
Pc

3

V

26.667

Z

Z1

Z1

0.422
Tr2

B1

0.139

T2
Tc

Z R T1
P1

Tr2

0.806

B0

1.6

0.172
Tr2

Tr1

0.977

Pr1

0.452

1.615

3

V1

B1

4.2

0.113

0.116

R T2

P2
V1

B0

gm

0.0479

V1

.8105

Tr2

493.15 K

0.083

cm

Ans.

Assume Eq. (3.38) applies at the final state.

B0

Ans.

0.100

0.35 m

From Tables E.1 & E.2: Z0

Z

Vtotal

P = 9920 kPa

3.49 T1

Vtotal

79.73 bar

B1 R

P2

Tc
Pc
56

42.68 bar

Ans.

cm
908
mol

3.50 T

B0

0.5 m

0.083

0.139

73.83 bar

0.422
Tr

B1

304.2 K

Pc

3

Vtotal

B0

4.2

V

10 kg
molwt

0.224

0.997

molwt

44.01

0.036

Vtotal

V

Tr

Tc

0.341

B1

1.6

0.172
Tr

T

Tr

Tc

303.15 K

3
3 cm

2.2

10

mol

RT

P
V

B0

B1 R

P

Tc

10.863 bar

Ans.

Pc

3.51 Basis: 1 mole of LIQUID nitrogen

P

B0

Tc

126.2 K

Tr

1 atm

Pc

34.0 bar

Pr

0.038

Tn

molwt

77.3 K

0.083

0.139

1

B0

4.2

B1

Pr
Tr

0.842

B1

0.172
Tr

Z

mol

B0

1.6

1.209

Z

0.957

57

Tr

P
Pc

Vliq

0.613

Pr

Tc

gm

0.422
Tr

B1

28.014

Tn

0.03

3

34.7 cm

gm
mol

P Vliq

nvapor

nvapor

Z R Tn

5.718

10

3

mol

Final conditions:

ntotal

T

1 mol

V

nvapor

RT

Pig

Pig

V

V

T
Tc

69.005

Tr

ntotal

Tr

298.15 K

3

2 Vliq

cm

2.363

mol

359.2 bar

Use Redlich/Kwong at so high a P.

2

2

a

Tr R Tc
Pc

a

3
3 bar cm
0.901 m
2

b

Eq. (3.42)

RT
V b

Tr

.5

R Tc

Tr

0.651

Eq. (3.43)

Pc
3

b

mol

P

( )
Tr

0.42748

0.08664

a
V ( b)
V

Eq. (3.44)

cm
26.737
mol

P

450.1 bar

Ans.

3

3.52 For isobutane:

T1

Tr1

Tr1

300 K

T1
Tc

0.735

Tc

408.1 K

Pc

36.48 bar

V1

1.824

P1

4 bar

T2

415 K

P2

cm

75 bar

Pr1

Pr1

P1

Tr2

Pc

Tr2

0.11

58

T2
Tc

1.017

Pr2

Pr2

P2
Pc

2.056

gm

From Fig. (3.17):

r1

2.45

The final T > Tc, and Fig. 3.16 probably should not be used. One can easily
show that
r=

P Vc

with Z from Eq. (3.57) and
Tables E.3 and E.4. Thus

Z RT
3

Vc

Z

262.7

Z0

cm

Z0

0.181

mol

Z

Z1

Eq. (3.75):

V1

r2

Tc

3

r1

cm
2.519
gm

V2

Pc

469.7 K

1.774

r2

Z R T2

r2

3.53 For n-pentane:

0.0756

P2 V c

0.322

V2

Z1

0.3356

33.7 bar

Ans.

0.63

1

gm
3

cm

T1

Tr1
Tr1

P1

291.15 K

T1

Pr1

Tc

T2

1 bar
P1

Tr2

Pc

T2

0.03

Tr2

2.69

r2

r2
2

2

1

0.532

r1

3.54 For ethanol: Tc

Pc

Pc

3.561

gm
3

Ans.

cm

513.9 K

T

453.15 K

Tr

T
Tc

Tr

0.882

61.48 bar

P

200 bar

Pr

P
Pc

Pr

3.253

3

Vc

P2

2.27

From Fig. (3.16):

By Eq. (3.75),

Pr2

0.88

r1

120 bar

Pr2

Tc

Pr1

0.62

P2

413.15 K

167

cm

mol

molwt
59

46.069

gm
mol

From Fig. 3.16:

r

2.28

=

r

r

c=

0.629

Vc

r

Vc

gm

Ans.

3

cm

molwt

3.55 For ammonia:

Tc

405.7 K

T

293.15 K

Tr

T
Tc

Tr

0.723

Pc

112.8 bar

P

857 kPa

Pr

P
Pc

Pr

0.076

Vc

cm
72.5
mol

Zc

0.242

3

Eq. (3.72):

B0

Vliquid

0.083

0.139

Vvapor

cm
27.11
mol

Vliquid

B0

4.2

B0

B1 R

0.627

B1

1.6

0.172
Tr

RT
P

3

0.2857

0.422
Tr

B1

V c Zc

1 Tr

0.253

0.534

Tc
Pc

3

Vvapor

cm
2616
mol
3

V

Vvapor

Vliquid

V

60

cm
2589
mol

Ans.

Alternatively, use Tables E.1 & E.2 to get the vapor volume:

Z0

Z1

0.929

Vvapor

Z

0.071

Z0

Z

Z1

0.911

3

ZRT
P

Vvapor

cm

2591

mol
3

V

Vvapor

V

Vliquid

cm

2564

Ans.

mol

3.5810 gal. of gasoline is equivalent to 1400 cu ft. of methane at 60 degF and 1
atm. Assume at these conditions that methane is an ideal gas:
3

R

ft atm
0.7302
lbmol rankine

V

1400 ft

T

P

1 atm

n

3

519.67 rankine

PV
RT

n

3.689 lbmol

For methane at 3000 psi and 60 degF:

Tc

190.6 1.8 rankine

T

519.67 rankine

Tr

T
Tc

Tr

1.515

Pc

45.99 bar

P

3000 psi

Pr

P
Pc

Pr

4.498

Z

0.822

0.012
From Tables E.3 & E.4:

Z0

0.819

Vtank

ZnRT
P

Z1

Z

0.234

Vtank

61

5.636 ft

Z0
3

Z1
Ans.

3.59 T

P

25K

3.213bar

Calculate the effective critical parameters for hydrogen by equations (3.58)
and (3.56)

43.6

Tc
1

K

1

30.435 K

Pc

10.922 bar

2.016T

20.5

Pc

Tc

bar

21.8K

44.2K
2.016T

0

P
Pc

Pr

Pr

Initial guess of volume:

T

Tr

0.294

Tr

cm
646.903
mol

3

RT
P

V

0.821

V

Tc

Use the generalized Pitzer correlation

B0

0.083

0.422
Tr

Z

1

B0

1.6

B0

B1

0.495

B1

0.139

0.172
Tr

Pr
Tr

Z

0.823

Ans.

4.2

B1

0.254

Experimental: Z = 0.7757


0

1

Tr

T r Pr

Tr

0.5

Pr
Tr

Table 3.1

Eq. (3.53)

62

Table 3.1

0.42748

0.08664

q Tr

Tr
Tr

Eq. (3.54)

Calculate Z

Guess:

Given

T r Pr

0.9

Eq. (3.52)

Z= 1

Z

Z

q Tr

Z

Find Z)
(

3.61For methane:

Z Z

Tc

0.012

T

T r Pr
T r Pr

Ans.

0.791

At standard condition:

Z

T r Pr

Experimental: Z = 0.7757

Pc

190.6K

(
60

32)

5
9

45.99bar

273.15 K

T

P

Tr

T
Tc

B0

0.083

Z0

Z

1

Tr

0.422
Tr
Pr

B0

Z0

1.6

Pitzer correlations:

T
Tc

Tr

0.083

0.422
Tr

B1

0.139

1.6

0.172
Tr

4.2

0.134

B1

0.139

0.172

T
T

Tr

Z RT
P

(
50

32)

5
9

0.109

Tr

0.00158
3

V1

273.15 K

m
0.024
mol
P

300psi

Pr

0.45

283.15 K

1.486

B0

B1

B1

V1

0.998

B1

Pr

Z1

0.998

4.2

0.022

Z1

P
Pc

Tr

Z

Z1

(a) At actual condition:

B0

B0

Z0

Tr

Pr

1.515

1atm

Pr

Pitzer correlations:

288.706 K

0.141

0.106

63

Pr

P
Pc

Z0

Z

1

B0

Z0

Pr
Tr

Z1

6 ft

q1

150 10

(b) n1

(c) D

u

Z0

0.957

Z

Z1

0.958

q1

n1

7.485

q2

V1

q1
V1

22.624in
q2
A

A

u

4

D

8.738

2

m
s

64

3 kmol

10

hr

A

Ans.

0.0322
3

V2

P

V2

q2

day

Z1

Tr

ZRT

V2

3

Pr

B1

6.915

Ans.

2

0.259 m

0.00109

10

6 ft

m

mol

3

day

Ans.

3.62

0.012

0.286

0.087

0.281

0.1

0.279

0.140

0.289

0.152

0.276

0.181

0.282

0.187

0.271

0.19

0.267

0.191

0.277

0.194

0.275

0.196

0.273

0.2

0.274

0.205

0.273

0.21

0.273

0.21

ZC

Use the first 29 components in Table B.1
sorted so that values are in ascending
order. This is required for the Mathcad
slope and intercept functions.

0.271
m

slope

b

intercept

r

corr

0.212

0.272

0.218

0.275

0.23

0.272

0.235

0.269

0.252

0.264

0.28

0.265

ZC

0.297

0.256

m

0.301

0.266

0.302

0.266

0.303

0.263

0.31

0.263

0.322

0.26

0.326

0.261

( 0.091)

0.27

0.262

ZC

ZC

ZC

(
0.291)
2

( 0.878) r

0.771

0.29
0.28

b 0.27
0.26
0.25

0

0.1

0.2

The equation of the line is:
Zc = 0.291 0.091
65

0.3

0.4

Ans.

5

7
R
2

Cv

298.15K

P1

P2

T1

T3

5bar

P3

Cp

1bar

5bar

3.65 Cp

T1

2

R

1.4

Cv

Step 1->2 Adiabatic compression
1

T2

T1

P2

T2

P1

472.216 K

U12

Cv T2

T1

U12

3.618

H12

Cp T2

T1

H12

5.065

Q12

W12

0

kJ
mol
kJ

mol

kJ
mol

kJ
mol

Q12

0

U12

W12

3.618

Ans.

Ans.

Ans.
kJ
mol

Ans.

Step 2->3 Isobaric cooling

U23

Cv T3

T2

U23

3.618

H23

Cp T3

T2

H23

5.065

Q23

W23

R T3

Q23

H23

W23

T2

5.065

1.447

kJ
mol
kJ

mol
kJ

mol
kJ

mol

Ans.

Ans.

Ans.

Ans.

Step 3->1 Isothermal expansion

U31

Cv T1

T3

U31

0

H31

Cp T1

T3

H31

0

66

kJ
mol
kJ

mol

Ans.
Ans.

Q31

R T3 ln

W31

P1

Q31

Q31

P3

W31

kJ
mol
kJ
3.99
mol

3.99

Ans.

Ans.

For the cycle

Qcycle

Q12

Wcycle

Q23

W12

Qcycle

Q31

W23

Wcycle

W31

1.076

1.076

kJ
Ans.
mol

kJ
Ans.
mol

Now assume that each step is irreversible with efficiency:

80%

Step 1->2 Adiabatic compression

W12

Q12

W12

U12

W12

Q12

W12

4.522

kJ
mol

kJ
mol

0.904

Ans.

Ans.

Step 2->3 Isobaric cooling

W23

Q23

W23

U23

W23

1.809

kJ
mol

kJ
mol

Q23

5.427

W31

W23

3.192

Ans.

Ans.

Step 3->1 Isothermal expansion

W31

Q31

W31

U31

Q31

W31

3.192

kJ
mol
kJ

mol

Ans.

Ans.

For the cycle

Qcycle

Q12

Wcycle

W12

Q23

W23

Qcycle

Q31

Wcycle

W31

67

kJ
Ans.
mol
kJ
3.14
mol Ans.
3.14

3.67 a) PV data are taken from Table F.2 at pressures above 1atm.

125
150

1757.0

175
P

2109.7

1505.1

200

1316.2

cm

1169.2

gm

V

kPa

225
250

300
Z

955.45

T

(
300

273.15)
K

M

1051.6

275

3

18.01

gm
mol

875.29

1
VM

PVM
RT

i

0 7

If a linear equation is fit to the points then the value of B is the y-intercept.
Use the Mathcad intercept function to find the y-intercept and hence, the
value of B
Yi

Zi

1

3

Xi

B

intercept ( Y)
X

A

i

slope ( Y)
X

Ans.

6
5 cm
10
2

i

A

cm
128.42
mol

1.567

B

mol

X

0

mol
3

cm

10

5 mol
3

cm

8 10

5 mol
3

cm

Below is a plot of the data along with the linear fit and the extrapolation to
the y-intercept.
68

115

(Z-1)/p

120

125

130
0

2 10

5

4 10

5

6 10

p

5

8 10

5

(Z-1)/p
Linear fit

b) Repeat part a) for T = 350 C
PV data are taken from Table F.2 at pressures above 1atm.

125
150

1912.2

175
P

2295.6

1638.3

200

1432.8

225

V

kPa

1273.1

250

300
Z

1040.7

T

( 350

273.15)K

M

1145.2

275

3

cm
gm

18.01

gm
mol

953.52

1
VM

PVM
RT

i

0 7

If a linear equation is fit to the points then the value of B is the y-intercept.
Use the Mathcad intercept function to find the y-intercept and hence, the
value of B
Yi

Zi

1

3

Xi

i

B

intercept ( X Y)

i
69

B

105.899

cm

mol

Ans.

A

A

slope ( X Y)

6
5 cm
10
2

1.784

mol

X

0

mol
3

10

cm

5 mol
3

8 10

5 mol
3

cm

cm

the y-intercept.
90

(Z-1)/p

95

100

105

110
0

2 10

5

4 10

5

6 10

p

5

8 10

5

(Z-1)/p
Linear fit

c) Repeat part a) for T = 400 C
PV data are taken from Table F.2 at pressures above 1atm.

125
150

2066.9

175
P

2481.2

1771.1

200

1549.2

225

kPa

V

1376.6

250

1238.5

275

1125.5

300

1031.4
70

3

cm
gm

T

( 400

273.15)K

M

18.01

gm
mol

1
VM

PV M
RT

Z

i

0 7

If a linear equation is fit to the points then the value of B is the
y-intercept.
Use the Mathcad intercept function to find the y-intercept and hence,
the value of B

Yi

Zi

1

3

Xi

B

i

intercept ( X Y)

B

i

A

A

slope ( X Y)

89.902

cm

mol

Ans.

6
5 cm
10
2

2.044

mol

X

0

mol
3

cm

10

5 mol
3

8 10

5 mol
3

cm

cm

the y-intercept.
70

(Z-1)/p

75

80

85

90
0

2 10

5

4 10
p

(Z-1)/p
Linear fit

71

5

6 10

5

8 10

5

3.70

Create a plot of

(Z

1) Z Tr

vs

Pr

Pr
Z Tr

Data from Appendix E at Tr = 1
0.01

0.9967

0.05

0.9832

0.10

0.9659
Z

0.20

Pr

0.9300

0.40

X

0.7574

0.80

1

0.8509

0.60

Tr

0.6355

Pr

(Z

Y

Z Tr

1) Z Tr
Pr

Create a linear fit of Y vs X
Slope

slope ( X Y)

Intercept
Rsquare

Slope

intercept ( X Y)

0.033

Intercept

corr ( X Y)

Rsquare

0.332
0.9965

0.28
0.3

Y
Slope X Intercept

0.32
0.34

0

0.2

0.4

0.6

0.8

1

1.2

1.4

X

The second virial coefficient (Bhat) is the value when X -> 0
Bhat

Intercept

By Eqns. (3.65) and (3.66)

Bhat
B0

0.083

0.422
Tr

These values differ by 2%.
72

1.6

B0

0.332
0.339

Ans.
Ans.

3.71 Use the SRK equation to calculate Z

T

T
273.15)Kr

Tc

150.9 K

T

(
30

Pc

48.98 bar

P

300 bar

Tr

Pr

Pc

2.009

Pr

Tc
P

6.125

0.0

0.42748 Table 3.1

0.08664

0

1

2

1

Tr

1

0.480

Tr

q Tr

0.176

Eq. (3.54)

1

Tr

Table 3.1

2

Pr

T r Pr

Tr

Calculate Z

Guess:

Z

Eq. (3.53)

Tr

0.9

Eq. (3.52)

Given

Z= 1

Z

1.574

2

T r Pr

q Tr

Z

Find Z)
(

Z

T r Pr

Z

V

1.025

T r Pr

T r Pr

Z

T r Pr
3

Z RT
P

cm
Ans.
86.1
mol

V

This volume is within 2.5% of the ideal gas value.

3.72 After the reaction is complete, there will be 5 moles of C 2H2 and 5 moles of
Ca(OH)2.
First calculate the volume available for the gas.

n

5mol

V

Vt

3

0.4 1800 cm

3

cm
5 mol 33.0
mol

Vt
n

Vt

3

555 cm

3

V

73

111

cm

mol

Use SRK equation to calculate pressure.

T

Tc

308.3 K

Pc

( 125

61.39 bar

T
Tc

Tr

273.15) K

Tr

0.0

0.42748 Table 3.1

0.08664

0

1

2

1

Tr

1

1.574

0.176

2

2

2

Pc

Eq. (3.45)

3
3 bar cm
3.995 m
2

RT
V b

3.73 mass

b

Tc

0.152

( 10

Vc

Pc

369.8K

200.0

197.8 bar

Ans.

273.15)K

3

0.276

cm
36.175
mol

P

b)
T

35000kg

Eq. (3.46)

Pc
3

a
V (V

R Tc

b

mol

Zc

Tr

Eq. (3.54)

R Tc

Tr

P

1

Table 3.1

2

Tr

a

a

0.480

Tr

q Tr

1.291

cm

mol

n

M

42.48bar

mass
M

n

44.097
7.937

gm
mol
5

10 mol

a) Estimate the volume of gas using the truncated virial equation

Tr

B0

T

Tr

Tc

0.083

0.422
Tr

B0

1.6

0.766

P

Eq. (3-65)

B1

Pr

1atm

0.139

0.172
Tr

B1

0.564
74

0.389

4.2

P
Pc

Eq. (3-66)

Z

1

B0

B1

Pr

Z

Tr

ZnRT

Vt

0.981

Vt

P

3

This would require a very large tank. If the
D
tank were spherical the diameter would be:

7 3

2.379
6

Vt

3

10 m m

D

32.565 m

b) Calculate the molar volume of the liquid with the Rackett equation(3.72)
Vliq

V c Zc

P

1

0.2857

Vvap

3

Vliq

6.294atm

Z

1 Tr

P
Pc

Pr

B0

B1

85.444

Pr

0.878

Pr
Tr

ZRT
P

Vvap

Guess:

Vtank

Given

90%

mol

0.15

Z

cm

3
3 cm

3.24 10

mol

90% Vliq n

Vtank

10%

Vtank

= n

Vvap

Vtank

Find Vtank

Vtank

Vliq

75.133 m

This would require a small tank. If the tank
D
were spherical, the diameter would be:

3

3

6

Vtank

D

5.235 m

Although the tank is smaller, it would need to accomodate a pressure of 6.294
atm (92.5 psi). Also, refrigeration would be required to liquify the gaseous
propane stream.

75

4.1 (a) T0

T

473.15 K

For SO2: A

B

5.699

n

1373.15 K

0.801 10

H

47.007

C

5

D

0.0

1.015 10

R ICPH T0 T A B C D

H

3

10 mol

kJ

T

523.15 K

For propane:A

28.785 10

H

161.834

3

Ans.

12 mol

C

6

8.824 10

D

0

R ICPH T0 T A B C 0.0

H

470.073 kJ

n

1473.15 K

B

1.213

n H

Q
(b) T0

Q

mol

kJ
mol

n

473.15 K

For ethylene:

A

2 (guess)

Q= nR

A T0

3

10 kJ Ans.

800 kJ
3

14.394 10
K

B

B
2
T0
2

1

n

A

1.967

2

T

2.905

533.15 K

For 1-butene:

1.424

1.942

6

4.392 10

C

K

2

Given

Find

(b) T0

Q

10 mol

n H

Q

4.2 (a) T0

Q

C
3
T0
3

1

T

T0

15 mol

Q

31.630 10
K

B

76

3

1

Ans.

1374.5 K

2500 kJ
3

C

9.873 10
K

2

6

(guess)

3

Q= nR

Given

A T0

2

n

500 degF

C
3
T0
3

1

T

2.652

Find

(c) T0

B
2
T0
2

1

3

T

T0

1

Ans.

1413.8 K
6

Q

40 lbmol

10 BTU

Values converted to SI units

T0

n

533.15K

For ethylene:

A

Q= nR

Q

10 mol

B

14.394 10
K

1.424

2 (guess)

4

1.814

6

1.055 10 kJ

3

4.392 10

C

K

Given

A T0

1

Find

B
2
T0
2

2

T

2.256

C
3
T0
3

1

T

T0

3

6

2

1

1202.8 K
Ans.

T = 1705.4degF
4.3 Assume air at the given conditions an ideal gas. Basis of calculation is 1 second.
P

T0

1 atm

(
T

T

PV
R T0

T0

773.15 K

n

For air:

H

32degF) 273.15K

T0

n

A

T

250 ft

V

Convert given values to SI units

T

3

V

122 degF

932 degF

7.079 m

3

T0

32degF

273.15K

323.15 K

266.985 mol

3.355

B

0.575 10

R ICPH T0 T A B C D
77

3

C

0.0

D

0.016 10

5

kJ

13.707

4.4 molwt

Q

gm
mol

100.1

T0

10000 kg
molwt

n

n

For CaCO3: A

323.15 K

B

10 mol

2.637 10

H

9.441

3

C

D

0.0

3.120 10

5

R ICPH T0 T A B C D

H

1153.15 K

3
10 BTU Ans.

4

9.99

12.572

3.469

T

mol

n H

Q

H

4 J

10

Q

mol

n H

6

Q

Ans.

9.4315 10 kJ

4.7 Let step 12 represent the initial reversible adiabatic expansion, and step 23
the final constant-volume heating.

T1

298.15 K

T3

298.15 K

P1

121.3 kPa

P2

101.3 kPa

P3

104.0 kPa

T2

T3

CP

30

T2

290.41 K

Given

J
mol K

T2 = T1

4.9a) Acetone: Tc

Hn

(guess)

29.10

P2

P2
P3

R
CP

CP

P1
Pc

508.2K

kJ

47.01bar

Tn

Trn

mol

CP

Find CP

Tc

56.95

Tn

329.4K

Trn

J
Ans.
mol K

0.648

Use Eq. (4.12) to calculate H at Tn ( Hncalc)

1.092 ln
Hncalc

R Tn

Pc

1.013

bar

0.930

Trn

78

Hncalc

30.108

kJ
mol

Ans.

To compare with the value listed in Table B.2, calculate the % error.

%error

Hncalc

Hn

%error

Hn

3.464 %

Values for other components in Table B.2 are given below. Except for
acetic acid, acetonitrile. methanol and nitromethane, agreement is within
5% of the reported value.
Acetone
Acetic Acid
Acetonitrile
Benz
ene
iso-Butane
n-Butane
1-Butanol
Carbon tetrachloride
Chlorobenz
ene
Chloroform
Cyclohexane
Cyclopentane
n-Decane
Dichloromethane
Diethyl ether
Ethanol
Ethylbenz
ene
Ethylene glycol
n-Heptane
n-Hexane
M ethanol
M ethyl acetate
M ethyl ethyl k
etone
Nitromethane
n-Nonane
iso-Octane
n-Octane
n-Pentane
Phenol
1-Propanol
2-Propanol
Toluene
W ater
o-Xylene
m-Xylene
p-Xylene

Hn (k mol)
J/
30.1
40.1
33.0
30.6
21.1
22.5
41.7
29.6
35.5
29.6
29.7
27.2
40.1
27.8
26.6
40.2
35.8
51.5
32.0
29.0
38
.3
30.6
32.0
36.3
37.2
30.7
34.8
25.9
46.6
41.1
39.8
33.4
42.0
36.9
36.5
36.3

79

% error
3.4%
69.4%
9.3%
-0.5%
-0.7%
0.3%
-3.6%
-0.8
%
0.8
%
1.1%
-0.9%
-0.2%
3.6%
-1.0%
0.3%
4.3%
0.7%
1.5%
0.7%
0.5%
8
.7%
1.1%
2.3%
6.7%
0.8
%
-0.2%
1.2%
0.3%
1.0%
-0.9%
-0.1%
0.8
%
3.3%
1.9%
2.3%
1.6%

b)

33.70

469.7
Tc

507.6
562.2

30.25

Pc

K

48.98

36.0
68.7

H25

273.15 K

80.0

366.1

72.150
J

M

433.3 gm
392.5

Tn

Tr2

Tc

(
25

273.15)
K
Tc

H2

H25 M

0.673

H2

0.628

H1

31.533

Hn

31.549

kJ
33.847 mol

32.242
1

Tr2

1

0.38

Tr1

Eq. (4.13) %error

26.448
H2calc

H1

26.429

0.631
H2calc

86.177 gm
78.114 mol
82.145

0.658
Tr1

kJ
30.72 mol

29.97

366.3

80.7

Tr1

28.85

Hn

bar

43.50

560.4

Tn

25.79

kJ
Ans.
33.571 mol

32.816

31.549

kJ
33.847 mol

32.242

H2

H2
0.072

26.429
H2

H2calc

%error

0.052
0.814

%

1.781

The values calculated with Eq. (4.13) are within 2% of the handbook values.

4.10 The ln P vs. 1/T relation over a short range is very nearly linear. Our
procedure is therefore to take 5 points, including the point at the
temperature of interest and two points on either side, and to do a linear
least-squares fit, from which the required derivative in Eq. (4.11) can be
found. Temperatures are in rankines, pressures in psia, volumes in cu
ft/lbm, and enthalpies in Btu/lbm. The molar mass M of tetrafluoroethane is
102.04. The factor 5.4039 converts energy units from (psia)(cu ft) to Btu.

80

(a) T

459.67

V

5

1.934

18.787

P

0

23.767

t

5

26.617

dPdT

T

H

2

ti

1
459.67

yi

ln Pi

15

slope ( x y) slope

( P) 3

xi

10

29.726
slope

1 5

5

21.162
Data:

i

0.012

4952

slopedPdT

T V dPdT
5.4039

H

0.545

90.078

Ans.

The remaining parts of the problem are worked in exactly the same
way. All answers are as follows, with the Table 9.1 value in ( ):
(a)

H = 90.078

( 90.111)

(b)

H = 85.817

( 85.834)

(c)

H = 81.034

( 81.136)

(d)

H = 76.007

( 75.902)

(e)

H = 69.863

( 69.969)

119.377

4.11

M

32.042
153.822

H is the value at
0 degC.

536.4
mol

Tc

334.3

512.6 K Pc

80.97 bar Tn

337.9 K

556.4

gm

54.72

45.60

349.8

Hexp is the given
value at the normal
boiling point.
81

Tr1

273.15K
Tc

Tr2

Tn
Tc

270.9
H

1189.5

0.509

246.9
J
gm

Hexp

217.8

J
gm

1099.5

Hn

Hn

Hexp
Hexp

Hn

PCE

0.38

Tr1

0.52

Hexp

1195.3

1.092 ln

R Tn

Hn

Hexp

4.03 %

M

Pc

1.013

bar

0.930

Tr2

100%

0.34

247.7
Hn

1

H

Tr2

0.77

J
1055.2
gm
193.2

(b) By Eq. (4.12):

PCE

1

0.629

This is the % error

100%

245
Hn

0.659

Tr2

0.491

194.2

(a) By Eq. (4.13)

PCE

0.533

Tr1

0.623

J
gm

PCE

8.72

%

0.96

192.3

4.12 Acetone
0.307

Tc

508.2K

Pc

Tn

329.4K

P

1atm

Pr

P
Pc

47.01bar

Zc

0.233

3

Vc
Tr

cm
209
mol
Tn
Tc

Tr

0.648

82

Hn

29.1
Pr

kJ
mol

0.022

Generalized Correlations to estimate volumes
Vapor Volume
B0

0.422

0.083

Tr
B1

Tr

4.2

Pr

Z

1

V

B1

Pr

Z R Tn
P

Tr

Tr

0.762

Eq. (3.65)

B1

0.172

0.139

B0

B0

1.6

0.924

Eq. (3.66)

Z

V

(Pg. 102)

0.965

3
4 cm

2.609

10

mol

Liquid Volume
2

Vsat

V c Zc

1 Tr

3

7

Eq. (3.72)

Vsat

Combining the Clapyeron equation (4.11)

gives

A

Vsat

14.3145

V
B

2.602

A

B

H= T V

V

2

e

C)

3
4 cm

10

d
Psat
dT

T C

Psat = e

(
T

V

H= T V

B

A

with Antoine's Equation

cm
70.917
mol

mol

2756.22

C

83

228.060

B
( C)
T

B

A

Tn 273.15K

Hcalc

B

Tn V
Tn

C

K

e
2

273.15K

kPa
K

C
K

Hcalc

29.662

kJ
Ans.
mol

%error

Hcalc

Hn

%error

1.9 %

Hn

The table below shows the values for other components in Table B.2. Values
agree within 5% except for acetic acid.
Acetone
Acetic Acid
Acetonitrile
Benz
ene
isoButane
nButane
1Butanol
Carbon tetrachloride
Chlorobenz
ene
Chloroform
Cyclohexane
Cyclopentane
nDecane
Dichloromethane
Diethyl ether
Ethanol
Ethylbenz
ene
Ethylene glycol
nHeptane
nHexane
M ethanol
M ethyl acetate
M ethyl ethyl k
etone
Nitromethane
nNonane
isoOctane
nOctane
nPentane
Phenol
1- panol
Pro

Hn ( J/
k mol)
29
.7
37.6
31.3
30.8
21.2
22.4
43.5
29
.9
35.3
29
.3
29
.9
27.4
39
.6
28
.1
26
.8
39
.6
35.7
53.2
31.9
29
.0
36
.5
30.4
31.7
34.9
37.2
30.8
34.6
25.9
45.9
41.9
84

% error
1.9
%
58
.7%
3.5%
0.2%
0.7%
0.0%
0.6
%
0.3%
0.3%
0.1%
0.1%
0.4%
2.2%
0.2%
0.9
%
2.8
%
0.5%
4.9
%
0.4%
0.4%
3.6
%
0.2%
1.3%
2.6
%
0.7%
0.1%
0.6
%
0.2%
- %
0.6
1.1%

p
2-Propanol
Toluene
W ater
o-Xylene
m-Xylene
p-Xylene

40.5
33.3
41.5
36.7
36.2
35.9

1.7%
0.5%
2.0%
1.2%
1.4%
0.8%

4.13 Let P represent the vapor pressure.

T

ln

Given

P

P

348.15 K

P
= 48.157543
kPa

Find ( ) dPdT
P

P

5622.7 K
T

P

Clapeyron equation:

B

dPdT =

Pc

AL

13.431

BL

2.211

0.029

bar
K
3

cm
96.49
mol

Vliq

H
T dPdT

1

512.6K

AV

dPdT

3

4.14 (a) Methanol: Tc

AL

4.70504
T

T
K

4.70504 ln

H
T V Vliq

Vliq

PV
V
RT

CPL ( )
T

5622.7 K
T

joule
31600
mol

V = vapor molar volume. V

Eq. (3.39)

2

H

87.396 kPa

(guess)

100 kPa

BL
K
BV
85

cm
1369.5
mol

B

Tn

80.97bar

51.28 10

T

CL
K

2

T

2

3

Ans.

337.9K

CL

131.13 10

CV

3.450 10

6

R

12.216 10

3

6

CPV ( )
T

P

BV

AV

CV

T

K

K

Tsat

3bar

2

T

2

R

T1

368.0K

T2

300K

500K

Estimate Hv using Riedel equation (4.12) and Watson correction (4.13)

Tn

Trn

Trn

Tc

1.092 ln
Hn

Pc

Hv

Trsat

1

Trn

kJ
mol

35.645

kJ
mol

T2

CPL ( ) T
T d

100

38.301

Hv

Hv

CPV ( ) T
T d

T1

n

0.718

0.38

Tsat

H

Trsat

Hn

Tc

R Tn

Trn

1

Hn

Tsat

1.013

bar

0.930

Trsat

0.659

kmol
hr

H

49.38

T sat

Q

n H

Q

1.372

3

10 kW

kJ
mol
Ans.

(b) Benzene:

Hv = 28.273

kJ
mol

H = 55.296

kJ
mol

Q = 1.536 10 kW

(c) Toluene

Hv = 30.625

kJ
mol

H = 65.586

kJ
mol

Q = 1.822 10 kW

4.15 Benzene

Tc

T1sat

562.2K

451.7K

Pc

48.98bar

T2sat

86

358.7K

3

3

Tn

353.2K

Cp

162

J
mol K

Estimate Hv using Riedel equation (4.12) and Watson correction (4.13)

Trn

Tn

Trn

Tc

Hn

Hv

Hn

30.588

30.28

R Tn

Trn

Tr2sat

1

0.638

Hn

Tc

1.013

bar

0.930

1

Tr2sat

Hv

Pc

1.092 ln

T2sat

Tr2sat

0.628

0.38

Trn

kJ
mol

kJ
mol

Assume the throttling process is adiabatic and isenthalpic.
Guess vapor fraction (x): x

Given

Cp T1sat

0.5

4.16 (a) For acetylene:

Tc

Tn

ln
Hn

R Tn 1.092
1

Tr

1

Hn

Hf

227480

0.498

Tn

61.39 bar

Tr

0.614

Pc
bar

J
mol

T
Tc

Tr

1.013
Trn

16.91

Hv

0.930

Hn

6.638

0.38

Trn

Hv

x

Find ( x)

Ans.

189.4 K

298.15 K

Trn

Tc

Pc

308.3 K

T

Trn

x

T2sat = x Hv

H298

Hf

87

Hv

0.967

kJ
mol
kJ
mol

H298

220.8

kJ
mol

Ans.

kJ
mol

(b) For 1,3-butadiene:

H298 = 88.5

(c) For ethylbenzene:

H298 = 12.3

(d) For n-hexane:

H298 = 198.6

(e) For styrene:

H298 = 103.9

4.17

1st law:

dQ = dU

Ideal gas:

Since

mol

kJ
mol

and

V dP = R dT

V dP = P

P

(B)
1

V

R dT

which reduces to

or

dQ =

P dV =

R

1

R dT
1

dQ = CV dT

R dT
1

R dT
1

dQ = CP dT

CP

dV = V dP

dV

Combines with (A) to give:

dQ = CP dT

V dP = R dT

P dV

Combines with (B) to yield:

or

(A)

P dV

P dV

then

P V = const

from which

kJ

dW = CV dT

PV= RT

Whence

kJ
mol

R dT

1

R dT

(C)

Since CP is linear in T, the mean heat capacity is the value of
CP at the arithmetic mean temperature. Thus
Tam

88

675

CPm

R 3.85

0.57 10

Integrate (C):

P1

4.18

R

950 K

P2

P1

T1

400 K

1.55

J

Q

6477.5

P2

11.45 bar

H298 = 4 058 910 J

R T2

1

1 bar

Tam

T2

CPm

Q

3

Ans.

Ans.

T1

mol

1

T2
T1

Ans.

For the combustion of methanol:
CH3OH(g) + (3/2)O2(g) = CO2(g) + 2H2O(g)
H298

393509

H298

676485

For 6 MeOH:

2 ( 241818) ( 200660)

For the combustion of 1-hexene:
C6H12(g) + 9O2(g) = 6CO2(g) + 6H2O(g)
H298
H298

6 ( 393509) 6 ( 241818) ( 41950)
3770012

H298 = 3 770 012 J

Ans.

Comparison is on the basis of equal numbers of C atoms.

4.19

C2H4 + 3O2 = 2CO2 + 2H2O(g)
H298

[ ( 241818) 2 ( 393509) 52510]
2

J
mol

Parts (a) - (d) can be worked exactly as Example 4.7. However, with
Mathcad capable of doing the iteration, it is simpler to proceed differently.
89

Index the product species with the numbers:
1 = oxygen
2 = carbon dioxide
3 = water (g)
4 = nitrogen
(a) For the product species, no excess air:

0

3.639

0.506

2

5.457

1.045

n

A

10

B

2

3.470
3.280

3

1.157

10 K

0.593

ni Ai B

A

1 4

A

0.121
0.040

D

ni Bi

ni D i
i

i

i

B

54.872

5 2

D

K

1.450

11.286
i

0.227

0.012

1
K

5 2

D

1.621 10 K

T0

298.15K

T

For the products,

CP

HP = R

dT

R
T0

The integral is given by Eq. (4.7). Moreover, by an energy balance,

H298

HP = 0

(guess)

2

H298 = R A T0

Given

8.497

Find

1

T

T0

B
2
T0
2

2

T

1

D
T0

2533.5 K

1

Ans.

Parts (b), (c), and (d) are worked the same way, the only change being in the
numbers of moles of products.
(b)

nO = 0.75

nn = 14.107

T = 2198.6 K

Ans.

(c)

nO = 1.5

nn = 16.929

T = 1950.9 K

Ans.

(d)

nO = 3.0

nn = 22.571

T = 1609.2 K

Ans.

2

2

2

2

2

2

90

(e) 50% xs air preheated to 500 degC. For this process,
Hair

H298

HP = 0

Hair = MCPH (
298.15

773.15)

For one mole of air:
3

MCPH 773.15 298.15 3.355 0.575 10

0.0

0.016 10

5

= 3.65606

For 4.5/0.21 = 21.429 moles of air:
Hair = n R MCPH T
Hair
Hair

21.429 8.314 3.65606 (
298.15
309399

J
mol

J
mol

The energy balance here gives:

H298

1.5
n

773.15)

3.639
5.457

HP = 0

0.227

0.506

2

Hair

1.045

A

2

B

3.470

1.450

3.280

16.929

A

3

ni Ai

B

B

5

10 K

0.121

2

0.040

ni Bi

D

i

78.84

1.157

D

0.593

i

A

10
K

ni D i
i

1
0.016
K

D

5 2

1.735

10 K

2 (guess)

H298

Find

Hair = R A T0

1

D
T0

Given

B
2
T0
2

2

1

1

7.656

T
91

T0 K

T

2282.5 K K

Ans.

4.20

n-C5H12 + 8O2 = 5CO2 + 6H2O(l)
By Eq. (4.15) with data from Table C.4:

H298

5 ( 393509) 6 ( 285830) ( 146760)

H298 = 3 535 765 J
4.21

Ans.

The following answers are found by application of Eq. (4.15) with
data from Table C.4.
(a) -92,220 J

(n) 180,500 J

(b) -905,468 J

(o) 178,321 J

(c) -71,660 J

(p) -132,439 J

(d) -61,980 J

(q) -44,370 J

(e) -367,582 J

(r) -68,910 J

(f) -2,732,016 J

(s) -492,640 J

(g) -105,140 J

(t) 109,780 J

(h) -38,292 J

(u) 235,030 J

(i) 164,647 J

(v) -132,038 J

(j) -48,969 J

(w) -1,807,968 J

(k) -149,728 J

(x) 42,720 J

(l) -1,036,036 J

(y) 117,440 J

(m) 207,436 J

(z) 175,305 J

92

4.22

The solution to each of these problems is exactly like that shown in
Example 4.6. In each case the value of Ho298 is calculated in Problem
4.21. Results are given in the following table. In the first column the

letter in ( ) indicates the part of problem 4.21 appropriate to the
value.

T/
K

(a)
(b
)
(f
)
(i
)
(j
)
(l
)
(m)
(n
)
(o
)
(r
)
(t
)
(u
)
(v
)
(w)
(x
)
(y
)

4.23

A

873.15
773.15
923.15
973.15
583.15
683.15
850.00
1350.00
1073.15
723.15
733.15
750.00
900.00
673.15
648.15
1083.15

-5.871
1.861
6.048
9.811
-9.523
-0.441
4.575
-0.145
-1.011
-1.424
4.016
7.297
2.418
2.586
0.060
4.175

103 B
4.181
-3.394
-9.779
-9.248
11.355
0.004
-2.323
0.159
-1.149
1.601
-4.422
-9.285
-3.647
-4.189
0.173
-4.766

106 C
0.000
0.000
0.000
2.106
-3.450
0.000
0.000
0.000
0.000
0.156
0.991
2.520
0.991
0.000
0.000
1.814

10-5 D
-0.661
2.661
7.972
-1.067
1.029
-0.643
-0.776
0.215
0.916
-0.083
0.083
0.166
0.235
1.586
-0.191
0.083

IDCPH/
J

-17,575
4,729
15,635
25,229
-10,949
-2,416
13,467
345
-9,743
-2,127
7,424
12,172
3,534
4,184
125
12,188

Ho298

J
HoT/
-109,795
-900,739
-2,716,381
189,876
-59,918
-1,038,452
220,903
180,845
168,578
-71,037
117,204
247,202
-128,504
-1,803,784
42,845
129,628

This is a simple application of a combination of Eqs. (4.18) & (4.19) with
evaluated parameters. In each case the value of Ho298 is calculated in Pb.

4.21. The values of A, B, C and D are given for all cases except for
Parts (e), (g), (h), (k), and (z) in the preceding table. Those missing are as
follows:
Part No.
A
103 B 106 C 10-5 D
(e)
-7.425 20.778
0.000
3.737
(g)
-3.629
8.816 -4.904
0.114
(h)
-9.987 20.061 -9.296
1.178
(k)
1.704 -3.997
1.573
0.234
(z)
-3.858 -1.042
0.180
0.919

93

4.24 q

6 ft

150 10

3

T

day

(
60

5
32) K
9

T

273.15K

288.71 K

P

1atm

The higher heating value is the negative of the heat of combustion with water
as liquid product.
Calculate methane standard heat of combustion with water as liquid product:
CH4 + 2O2 --> CO2 +2H2O
Standard Heats of Formation:

J
mol

HfCH4

74520

HfCO2

393509

Hc

HfCO2

HfO2

J
mol

J
mol

HfH2Oliq

2 HfH2Oliq

HigherHeatingValue

0

285830

HfCH4

J
mol

2 HfO2
5 J

Hc

Hc

8.906 10

mol

Assuming methane is an ideal gas at standard conditions:
n

q

P
RT

n HigherHeatingValue
4.25

8 mol

n

1.793 10
5dollar
GJ

day
5 dollar

7.985 10

day

Ans.

Calculate methane standard heat of combustion with water as liquid product
Standard Heats of Formation: CH4 + 2O2 --> CO2 +2H2O
J
mol

HfCH4

74520

HfCO2

393509

HcCH4

HfCO2

HcCH4

890649

J
mol

HfO2

0

J
mol

HfH2Oliq

2 HfH2Oliq
J
mol
94

HfCH4

285830

J
mol

2 HfO2

Calculate ethane standard heat of combustion with water as liquid product:
Standard Heats of Formation:C2H6 + 7/2O2 --> 2CO2 +3H2O

J
mol

HfC2H6

83820

HcC2H6

2 HfCO2

HcC2H6

1560688

3 HfH2Oliq

HfC2H6

7
2

HfO2

J
mol

Calculate propane standard heat of combustion with water as liquid product
Standard Heats of Formation:C3H8 + 5O2 --> 3CO2 +4H2O

J
mol

HfC3H8

104680

HcC3H8

3 HfCO2

HcC3H8

2219.167

4 HfH2Oliq

HfC3H8

5 HfO2

kJ
mol

Calculate the standard heat of combustion for the mixtures

a) 0.95 HcCH4

0.02 HcC2H6

0.02 HcC3H8

921.714

kJ
mol

b) 0.90 HcCH4

0.05 HcC2H6

0.03 HcC3H8

946.194

kJ
mol

c) 0.85 HcCH4

0.07 HcC2H6

0.03 HcC3H8

932.875

kJ
mol

Gas b) has the highest standard heat of combustion.
4.26

2H2 + O2 = 2H2O(l)

Hf1

C + O2 = CO2(g)

Hf2

N2(g)+2H2O(l)+CO2(g)=(NH2)2CO(s)+3/2O2

H

Ans.

2 ( 285830) J

393509 J

631660 J

.
N2(g)+2H2(g)+C(s)+1/2O2(g)=(NH2)2CO(s)

H298

Hf1

Hf2

H298

H
95

333509 J

Ans.

4.28

On the basis of 1 mole of C10H18
(molar mass = 162.27)
Q

43960 162.27 J

6

Q

7.133 10 J

This value is for the constant-volume reaction:
C10H18(l) + 14.5O2(g) = 10CO2(g) + 9H2O(l)
Assuming ideal gases and with symbols representing total properties,
Q=
T

U=

H

( )=
PV

298.15 K

H

Q

H

R T ngas

ngas

R T ngas

(
10

14.5)mol
6

H

7.145 10 J

This value is for the constant-V reaction, whereas the STANDARD
reaction is at const. P.However, for ideal gases H = f(T), and for liquids H
is a very weak function of P. We therefore take the above value as the
standard value, and for the specified reaction:

C10H18(l) + 14.5O2(g) = 10CO2(g) + 9H2O(l)

H

9H2O(l) = 9H2O(g)
Hvap 9 44012 J
___________________________________________________
C10H18(l) + 14.5O2(g) = 10CO2(g) + 9H2O(g)
H298
4.29

H

Hvap

H298

6748436 J

Ans.

FURNACE: Basis is 1 mole of methane burned with 30% excess air.
CH4 + 2O2 = CO2 + 2H2O(g)
Entering:

Moles methane

n1

1

Moles oxygen

n2

2 1.3

Moles nitrogen

n3

2.6

96

79
21

n2

2.6

n3

9.781

Total moles of dry gases entering

n

n1

n2

n

n3

13.381

At 30 degC the vapor pressure of water is
4.241 kPa. Moles of water vapor entering:

n4

Leaving:

4.241
13.381
101.325 4.241

n4

CO2 -- 1 mol
H2O -- 2.585 mol
O2 -- 2.6 - 2 = 0.6 mol
N2 -- 9.781 mol

0.585

(1)
(2)
(3)
(4)

By an energy balance on the furnace:

Q=

H=

H298

For evaluation of

1
A

0.6
9.781

i

HP we number species as above.
5.457

2.585

n

HP

3.470
3.639

1.045
1.450

B

R

8.314

B

48.692

0.121

D

0.227
0.040

D

ni Bi

i

i

10.896983 10

3

ni Di

C

D

0

4

5.892 10

The TOTAL value for MCPH of the product stream:

HP

R M CPH ( 303.15K 1773.15K A B C D) ( 1773.15

HP

732.013

kJ
mol

From Example 4.7:

Q

HP

H298

5

10

J
mol K

i

A

3

0.593

ni Ai B

A

10

0.506

3.280

1 4

1.157

H298

802625

Q = 70 612 J
97

J
mol

Ans.

303.15)K

HEAT EXCHANGER: Flue gases cool from 1500 degC to
50 degC. The partial pressure of the water in the flue gases leaving the
furnace (in kPa) is
pp

n2
n1

n2

n3

n4

101.325

pp

18.754

The vapor pressure of water at 50 degC (exit of heat exchanger) is 12.34
kPa, and water must condense to lower its partial pressure to this value.
Moles of dry flue gases:

n

n1

n3

n4

n

11.381

Moles of water vapor leaving the heat exchanger:
n2

12.34
n
101.325 12.34

n2

Moles water condensing:

1.578

n

2.585

1.578

Latent heat of water at 50 degC in J/mol:
H50

2382.918.015

J
mol

Sensible heat of cooling the flue gases to 50 degC with all the water as
vapor (we assumed condensation at 50 degC):
Q

R MCPH (
323.15 K 1773.15 K A B C D)(
323.15

1773.15)
K

Q = 766 677 J

4.30

4NH3(g) + 5O2(g) = 4NO(g) + 6H2O(g)
BASIS: 4 moles ammonia entering reactor
Moles O2 entering = (5)(1.3) = 6.5
Moles N2 entering = (6.5)(79/21) = 24.45
Moles NH3 reacting = moles NO formed = (4)(0.8) = 3.2
Moles O2 reacting = (5)(0.8) = 4.0
Moles water formed = (6)(0.8) = 4.8
98

n H50
Ans.

ENERGY BALANCE:

H=

HR

H298

HP = 0

REACTANTS: 1=NH3; 2=O2; 3=N2

4
n

3.578

6.5

A

3.639

24.45
i

3.020
B

0.506 10

3.280

A

1 3

118.161

ni Ai

B

3

5

D

0.227 10

0.593

B

0.040

D

ni Bi

ni Di
i

i

i

A

0.186

C

0.02987

5

D

0.0

1.242 10

TOTAL mean heat capacity of reactant stream:

HR

HR

R MCPH ( 348.15K 298.15K A B C D) ( 298.15K

52.635

348.15K)

kJ
mol

The result of Pb. 4.21(b) is used to get
J
H298 0.8 ( 905468)
mol
PRODUCTS: =NH3; 2=O2; 3=NO; 4=H2O; 5=N2
1

0.8

3.020

2.5
n

3.578
3.639

0.506

3.2

A

3.387

B

0.629

0.186
10
K

0.227

3

D

0.014

4.8

1.450

3.280

0.593

2

0.121

24.45
i

3.470

5

10 K

0.040

1 5

A

ni Ai

B

119.65

D

B

1
0.027
K

By the energy balance and Eq. (4.7), we can write:
(guess)
2
T0 298.15K

99

ni Di
i

i

i

A

ni Bi

D

4

8.873 10 K

2

H298

B
2
T0
2

1

2

1

1

T

3.283

Find

4.31

H R = R A T0

D
T0

Given

T

T0

Ans.

978.9 K

C2H4(g) + H2O(g) = C2H5OH(l)

n

BASIS: 1 mole ethanol produced

H= Q=

Energy balance:

H298

[ 277690

HR

(
52510

1mol

H298

241818)
]

J
mol

4 J

H298

8.838 10

mol

Reactant stream consists of 1 mole each of C2H4 and H2O.

1 2 n

i

1.424

A

3.470

1
1
14.394

B

1.450

ni Ai B

A

B

HR

Q

4.392

C

10

0.0

C

6

D

C

0.01584

0.0

ni Di
i

4.392 10

6

D

R M CP (
H 298.15K 593.15K A B C D)(
298.15K

4

1.21 10

593.15K)

4 J

2.727 10

HR

mol

Q

H298 1mol

100

5

10

0.121

D

ni Ci
i

i

4.894

HR

3

ni Bi

i

A

10

115653 J

Ans.

4.32 One way to proceed is as in Example 4.8 with the alternative pair of reactions:
CH4 + H2O = CO + 3H2

H298a

205813

CH4 + 2H2O = CO2 + 4H2

H298b

164647

BASIS: 1 mole of product gases containing 0.0275 mol CO2; 0.1725 mol CO;
& H2O 0.6275 mol H2
Entering gas, by carbon & oxygen balances:
0.0275 + 0.1725 = 0.2000 mol CH4
0.1725 + 0.1725 + 2(0.0275) = 0.4000 mol H2O
J
H298
0.1725 H298a 0.0275 H298b
mol
The energy balance is written

Q=

HR

H298

1.702
3.470

9.081

B

1.450

ni Ai B

A

HR

HR

10

i

3

B

2.164

C

0.0

0.4
10

6

D

2.396 10

3

C

0.0

5

10

0.121

D

ni Ci

ni Di
i

i

i

1.728

mol

0.2

n

1 2

C

ni Bi

i

A

4.003 10

HP

REACTANTS: 1=CH4; 2=H2O

A

4 J

H298

4.328 10

7

D

3

4.84 10

RI
CPH ( 773.15K 298.15K A B C D)
4 J

1.145 10

mol

PRODUCTS: 1=CO2; 2=CO; 3=H2O; 4=H2

0.0275
n

0.1725
0.1725
0.6275

5.457
A

3.376
3.470

1.045
B

3.249

0.557
1.450
0.422

101

1.157
10

3

D

0.031
0.121
0.083

5

10

i

A

1 4

B

ni Ai

HP

B

3.37

ni Di
i

i

i

A

D

ni Bi

6.397 10

4

C

3

D

0.0

3.579 10

RI
CPH (
298.15K 1123.15K A B C D)
4 J
2.63 10
mol

HP

Q

HR

H298

Q

HP mol

4.33 CH4 + 2O2 = CO2 + 2H2O(g)

Ans.

54881 J

H298a

H298b

C2H6 + 3.5O2 = 2CO2 + 3H2O(g)

802625

1428652

BASIS: 1 mole fuel (0.75 mol CH4; 0.25 mol C2H6) burned completely with
80% xs. air.
O2 in = 1.8[(0.75)(2) + (0.25)(3.5)] = 4.275 mol
N2 in = 4.275(79/21) = 16.082 mol
Product gases: CO2 = 0.75 + 2(0.25) = 1.25 mol
H2O = 2(0.75) + 3(0.25) = 2.25 mol
O2 = (0.8/1.8)(4.275) = 1.9 mol
N2 = 16.082 mol

H298

0.75 H298a

0.25 H298b

J
mol

Energy balance:
Q = H = H298
HP
PRODUCTS: 1=CO2; 2=H2O; 3=O2; 4=N2

1.25
n

5.457
3.470

1.450

A

1.9

1 4

B

3.639

16.082
i

8 10

HP = Q

1.045

2.25

5

Q

3.280

74.292

10
K

3

D

B

102

0.227

D

ni Bi

0.015

0.121

5

10 K

0.040

ni Di
i

i

i

H298
1.157

0.593

ni Ai B

A

A

0.506

J
mol

1
K

C

0.0

D

4 2

9.62 10 K

2

By the energy balance and Eq. (4.7), we can write:
T0

303.15K

Q

2

(guess)

H298 = R A T0

2

1

1

1.788

4.34

B
2
T0
2

1

D
T0

Given

T

T0

T

Find

Ans.

542.2 K

BASIS: 1 mole of entering gases containing 0.15 mol SO2; 0.20 mol
O2; 0.65 mol N2

SO2 + 0.5O2 = SO3

Conversion = 86%

SO2 reacted = SO3 formed = (0.15)(0.86) = 0.129 mol

O2 reacted = (0.5)(0.129) = 0.0645 mol

Energy balance:

H773 =

HR

H298

HP

Since HR and HP cancel for the gas that passes through the converter
unreacted, we need consider only those species that react or are formed.
Moreover, the reactants and products experience the same temperature
change, and can therefore be considered together. We simply take the
number of moles of reactants as being negative. The energy balance is
then written: H773 = H298
Hnet
H298

[ 395720

( 296830) ]0.129

J
mol

1: SO2; 2: O2; 3: SO3

0.129
n

0.0645

5.699
A

0.129

i

1 3

3.639

B

8.060

A

A

0.06985

0.506 10

3

5

D

0.227 10
2.028

1.056

ni Ai B
i

1.015

0.801

ni Bi

D

ni Di

i

B

2.58 10
103

i
7

C

0

D

4

1.16 10

Hnet

R M CPH (
298.15K 773.15K A B C D)(
773.15K

Hnet

77.617

H773

298.15K)

J
mol

H298

H773

Hnet

12679

J
mol

Ans.

4.35 CO(g) + H2O(g) = CO2(g) + H2(g)
BASIS: 1 mole of feed consisting of 0.5 mol CO and 0.5 mol H2O.
Moles CO reacted = moles H2O reacted = moles CO2 formed = moles H2
formed = (0.6)(0.5) = 0.3
Product stream:

moles CO = moles H2O = 0.2
moles CO2 = moles H2 = 0.3

Energy balance:

Q=

H298

0.3 [ 393509

H=

HR

H298

J
214818)
]
mol

( 110525

HP
4 J

H298

2.045 10

mol

Reactants: 1: CO 2: H2O

0.5

n
i

0.5

3.376

A

3.470

1.450

ni Ai B

A

1 2

0.557

B

HR

HR

0.031

D

B

ni Di
i

i

3.423

5

10

0.121

D

1.004 10

3

0 D

C

R M CPH (
298.15K 398.15K A B C D)(
298.15K

3

4.5 10

398.15K)

3 J

3.168 10

Products:

0.2
0.3
0.3

mol

1: CO 2: H2O 3: CO2 4: H2

0.2
n

3

ni Bi

i

A

10

3.376
A

3.470
5.457

0.557
B

3.249

1.450
1.045
0.422

104

0.031
10

3

D

0.121
1.157
0.083

5

10

i

1 4 A

ni Ai

B

i

A

ni Bi

D

i

3.981

B

8.415 10

i
4

C

4

0 D

HP

R MCPH ( 298.15K 698.15K A B C D) ( 698.15K

HP

1.415 10

Q

ni Di
3.042 10

298.15K)

4 J

HR

mol

H298

HP mol

Q

9470 J Ans.

4.36 BASIS: 100 lbmol DRY flue gases containing 3.00 lbmol CO2 and 11.80
lbmol CO x lbmol O2 and 100-(14.8-x)= 85.2-x lbmol N2. The oil therefore
contains 14.80 lbmol carbon;a carbon balance gives the mass of oil burned:
14.8

12.011
lbm
0.85

209.133 lbm

The oil also contains H2O:
209.133 0.01
lbmol 0.116 lbmol
18.015
Also H2O is formed by combustion of H2 in the oil in the amount
209.133 0.12
lbmol
2.016

12.448 lbmol

Find amount of air entering by N2 & O2 balances.
N2 entering in oil:
209.133 0.02
lbmol
28.013

0.149 lbmol

lbmol N2 entering in the air=(85.2-x)-0.149 =85.051-x
lbmol O2 in flue gas entering with dry air =
3.00 + 11.8/2 + x + 12.448/2
=
15.124 + x lbmol
(CO2) (CO) (O2) (H2O from combustion)
Total dry air = N2 in air + O2 in air = 85.051 - x + 15.124 + x = 100.175 lbmol
105

Since air is 21 mol % O2,

0.21 =

15.124 x
100.175

x

(
0.21 100.175

15.124)lbmol

x

5.913 lbmol

O2 in air = 15.124 + x = 21.037 lbmols
N2 in air = 85.051 - x = 79.138 lbmoles
N2 in flue gas = 79.138 + 0.149 = 79.287 lbmols
[CHECK: Total dry flue gas
= 3.00 + 11.80 + 5.913 + 79.287
= 100.00 lbmol]
Humidity of entering air, sat. at 77 degF in lbmol H2O/lbmol dry air,
P(sat)=0.4594(psia)
0.4594
14.696 0.4594

0.03227

lbmol H2O entering in air:
0.03227 100.175 lbmol

3.233 lbmol

If y = lbmol H2O evaporated in the drier, then
lbmol H2O in flue gas = 0.116+12.448+3.233+y
= 15.797 + y
Entering the process are oil, moist air, and the wet material to be dried, all at
77 degF. The "products" at 400 degF consist of:
3.00 lbmol CO2
11.80 lbmol CO
5.913 lbmol O2
79.287 lbmol N2
(15.797 + y) lbmol H2O(g)
Energy balance:

Q=

H=

H298

HP

where Q = 30% of net heating value of the oil:

Q

0.3 19000

BTU
209.13 lbm
lbm

Reaction upon which net heating value is based:
106

Q

6

1.192 10 BTU

OIL + (21.024)O2 = (14.8)CO2 + (12.448 + 0.116)H2O(g) + (0.149)N2
H298a

19000 209.13 BTU

6

H298a

3.973 10 BTU

To get the "reaction" in the drier, we add to this the following:
(11.8)CO2 = (11.8)CO + (5.9)O2
H298b

11.8 ( 110525

(y)H2O(l) = (y)H2O(g)
H298c ( y)

393509) 0.42993 BTU
Guess: y

50

44012 0.42993 y BTU

[The factor 0.42993 converts from joules on the basis of moles to Btu on the
basis of lbmol.]
Addition of these three reactions gives the "reaction" in the drier, except for
some O2, N2, and H2O that pass through unchanged. Addition of the
corresponding delta H values gives the standard heat of reaction at 298 K:
H298 ( y)

H298a

H298b

H298c ( y)

For the product stream we need MCPH:
1: CO2 2: CO 3:O2 4: N2 5: H2O
T0

298.15

r

1.986

400

T

459.67
1.8

T

477.594

3

1.045

1.157

11.8
n y)
(

5.457
3.376

0.557

0.031

5.913

A

i

B

3

5

D

0.227 10

0.593

0.040

3.470

y

1 5 A ( y)

1.450

0.121

n y) i Ai B ( y)
(
i

T
T0

0.506 10

3.280

79.278
15.797

3.639

1.602

n y) i Bi D ( y)
(

n y) i Di
(

i

CP ( y)

107

r A ( y)

i

B ( y)
T0
2

1

D ( y)
T0

2

CP ()(
y 400

Given

y

y

H298 ()
y

Find y
()

(lbmol H2O evaporated)

49.782

y 18.015
209.13

Whence

77)BTU = Q

(lb H2O evap. per lb oil burned)
Ans.

4.288

4.37 BASIS: One mole of product gas containing 0.242 mol HCN, and
(1-0.242)/2 = 0.379 mol each of N2 and C2H2. The energy balance is

Q=

H=

H298

H298

HP

( 135100
2

227480)

0.242
J
2

3

H298

5.169 10 J

Products:

0.242
n

4.736

0.379

A

3.280

0.379

ni Ai

0.725
3

0.593 10

D

0.040

1.952

B

4.7133

B

D

ni Bi

1.2934 10

ni Di
i

3

4

0 D

C

HP

R M CPH (
298.15K 873.15K A B C D)(
873.15K

HP

6.526 10

2.495 10 J

Q

298.15K)mol

4

HP

H298

Q

HP

4

2.495 10 J

30124 J

Ans.

4.38 BASIS: 1 mole gas entering reactor, containing 0.6 mol HCl, 0.36 mol O2,
and 0.04 mol N2.
HCl reacted = (0.6)(0.75) = 0.45 mol
4HCl(g) + O2(g) = 2H2O(g) + 2Cl2(g)
108

5

10

1.299

i

i

A

B

6.132

1 3 A

i

1.359

For this reaction,

H298

[ ( 241818)
2

Evaluate

T0

4 ( 92307) ]

5 J

H298

1.144 10

mol

by Eq. (4.21) with

H823

T

298.15K

J
mol

823.15K

1: H2O 2: Cl2 3: HCl 4=O2

2

3.470

2

n

4.442

A

4

0.089

B

3.156

1
i

1.45

0.623

3.639

A

1 4

ni Ai

10

3

D

0.344
0.151

0.506

B

D

ni Di
i

i

0.439

H823

H298

MCPH T0 T

H823

117592

B

5

8 10
A

B

5

10

0.227

ni Bi

i

A

0.121

C

0

D R T

4

D

T0

C

8.23 10

J
mol

Heat transferred per mol of entering gas mixture:

Q

H823
4

Q

0.45 mol

4.39 CO2 + C = 2CO

Ans.

J
(a)
mol
J
(b)
221050
mol

H298a

2C + O2 = 2CO

13229 J

172459

H298b

Eq. (4.21) applies to each reaction:

For (a):

2
n

1
1

3.376
A

1.771

0.557
B

0.771 10

5.457

1.045
109

0.031
3

D

5

0.867 10
1.157

i

A

1 3

B

ni Ai

H1148a

B

0.476

i

4

7.02 10

H298a
R M CP 298.15K 1148.15K
H

H1148a

ni Di

i

i

A

D

ni Bi

A

C

B

C

D

0

5

1.962 10

D (
1148.15K

298.15K)

5 J

1.696 10

mol

For (b):

2
n

3.376

1

A

3.639

2
i

1 3

0.557
B

H1148b

ni Ai

5

D

0.227 10

B

0.867

ni Bi

D

ni Di

i

0.429

B

9.34 10

H298b
R M CP 298.15K 1148.15K
H

H1148b

3

0.771

i

A

0.506 10

1.771

A

0.031

i
4

C

A

B

0

C

D

D (
1148.15K

5 J

2.249 10

mol

The combined heats of reaction must be zero:
nCO

2

H1148a

nO

H1148b = 0

2

nCO

Define:

r=

2

nO

H1148b

r

H1148a

2

110

r

5

1.899 10

1.327

298.15K)

For 100 mol flue gas and x mol air, moles are:
Flue gas
Air
Feed mix
CO2

12.8

0

12.8

CO

3.7

0

3.7

O2

5.4

0.21x

5.4 + 0.21x

N2

78.1

0.79x

78.1 + 0.79x

Whence in the feed mix:

r=

12.5
5.4
r
mol
0.21

x

12.8
5.4 0.21 x

x

19.155 mol

100
19.155

Flue gas to air ratio =

Ans.

5.221

Product composition:

nCO

3.7

nN

78.1

2

2 ( 12.8

0.21 19.155)

0.79 19.155

nCO

Mole % N2 =

100

48.145
93.232

2

nCO

Mole % CO =

nCO
nN

5.4

nN

100

34.054
Ans.

2

34.054

65.946

H298a

802625

J
mol

H298b

4.40 CH4 + 2O2 = CO2 + 2H2O(g)

519641

J
mol

CH4 + (3/2)O2 = CO + 2H2O(g)

BASIS: 1 mole of fuel gas consisting of 0.94 mol CH4 and 0.06 mol N2
Air entering contains:

1.35 2 0.94

2.538

79
21

2.538

9.548

mol O2

mol N2
111

Moles CO2 formed by reaction =

0.94 0.7

0.658

Moles CO formed by reaction =

0.94 0.3

0.282

H298

0.658 H298a

5 J

H298

0.282 H298b

Moles H2O formed by reaction =

0.94 2.0

Moles O2 consumed by reaction =

2 0.658

6.747 10

mol

1.88

3
0.282
2

1.739

Product gases contain the following numbers of moles:
(1)
(2)
(3)
(4)
(5)

CO2: 0.658
CO: 0.282
H2O: 1.880
O2: 2.538 - 1.739 = 0.799
N2: 9.548 + 0.060 = 9.608

0.658

1.045

1.157

0.282
n

5.457
3.376

0.557

0.031

1.880

A

3.470

B

1.450 10

0.799

3.280

D

0.593

0.121

0.506

9.608
i

3.639

3

1 5 A

ni Ai

B

A

0.227
0.040

ni Bi

i

D

B

9.6725 10

i
3

C

R M CPH (
298.15K 483.15K A B C D)(
483.15K

HP

7.541 10

HH2O mdotH2O

4

0 D

HP

Energy balance:

ni Di

i

45.4881

5

10

3.396 10
298.15K)

4 J

mol
Hrx

H298

Hrx ndotfuel = 0

112

HP

Hrx

kJ
mol
kg
34.0
sec

599.252

mdotH2O

From Table C.1:

HH2O

( 398.0

104.8)

kJ
kg

HH2O mdotH2O

ndotfuel

ndotfuel

Hrx

16.635

mol
sec

Volumetric flow rate of fuel, assuming ideal gas:

ndotfuel R 298.15 K

V

3

m
0.407
sec

V

101325 Pa

4.41 C4H8(g) = C4H6(g) + H2(g)

Ans.

H298

109780

J
mol

BASIS: 1 mole C4H8 entering, of which 33% reacts.
The unreacted C4H8 and the diluent H2O pass throught the reactor
unchanged, and need not be included in the energy balance. Thus

T

T0

1

n

by Eq. (4.21):
Evaluate
H798
1: C4H6 2: H2 3: C4H8

1
2.734
A

3.249

26.786
B

0.422

1.967
i

298.15 K

798.15 K

1

8.882
10

3

C

0.0

0.0

31.630

6

10

D

9.873

5

0.083 10
0.0

1 3

ni Ai B

A

4.016

B

H298

H798

1.179 10

Q

3

C

9.91 10

MCPH 298.15K 798.15K
5 J

0.33 mol H798

mol

Q

38896 J
113

Ans.

ni Di
i

i

4.422 10

H798

D

ni Ci

i

i

A

C

ni Bi

A

B

7

D

C

3

8.3 10

D R T

T0

4.42Assume Ideal Gas and P = 1 atm

P

R

1atm

a) T0

(
70

T0

3 BTU

7.88 10

mol K

T

294.261 K

ICPH T0 T 3.355 0.575 10

3

T0

T

459.67)
rankine

BTU
sec

305.372 K
5

0

0.016 10

R ICPH T0 T 3.355 0.575 10

Vdot

b) T0

38.995 K

(
24

3

5

0

0.016 10

T

T0

ndot

39.051

Vdot

ft
33.298
sec
Q

13K

12

kJ
s

3

5

0

0.016 10

45.659 K

Q
R ICPH T0 T 3.355 0.575 10

3

0

5

0.016 10

Vdot

4.43Assume Ideal Gas and P = 1 atm

(
94

459.67)
rankine

1.61 10

3

Vdot

31.611

P

T

(
68

m
0.7707
s

1atm

459.67)
rankine

3

atm ft
mol rankine

3

ft
50
sec

ndot

mol
s

3

ndot R T0
P

a) T0

Ans.

kJ
mol K

ndot

Vdot

mol
s

3

m
0.943
s

Vdot

273.15)
K

8.314 10

3

3

ndot R T0
P

ICPH T0 T 3.355 0.575 10

R

12

Q

ndot

R

Q

20rankine

ndot

P Vdot
R T0
114

ndot

56.097

mol
s

Ans.

T0

307.594 K

T

ICPH T0 T 3.355 0.575 10

3

293.15 K
5

0

0.016 10

50.7 K

3 BTU

R

7.88 10

Q

R ICPH T0 T 3.355 0.575 10

mol K
3

0

5

0.016 10

ndot

Q

b) T0

R

( 35

273.15)K

T

273.15)K

mol K

3

m
1.5
sec

3

P Vdot
R T0

ndot

ICPH T0 T 3.355 0.575 10

3

0

ndot
5

0.016 10

59.325

8.314 10

Q

R ICPH T0 T 3.355 0.575 10

35.119 K

3

0

5

0.016 10

Q

ndot
17.3216

4.44 First calculate the standard heat of combustion of propane
C3H8 + 5O2 = 3CO2(g) + 4H2O (g)

H298

Cost

3

393509

J
mol

4

241818

6 J

2.043 10

2.20

mol
s

kJ
mol K

R

H298

BTU
Ans.
sec

3
5 atm m

8.205 10

Vdot

( 25

22.4121

dollars
gal

mol
80%

115

J
mol

104680

J
mol

kJ
Ans.
s

Estimate the density of propane using the Rackett equation
3

Tc
T

Vsat

369.8K
(
25

Zc

273.15)
K

V c Zc

Heating_cost

1 Tr

0.276
Tr

Vc

T
Tc

Tr

cm
200.0
mol
0.806

3

0.2857

Vsat

Vsat Cost

cm
89.373
mol

Heating_cost

H298

Heating_cost

0.032

dollars
MJ

33.528

Ans.

dollars
6

10 BTU
4.45 T0

(
25

a) Acetylene

273.15)
K

T

(
500

273.15)
K

Q

R ICPH T0 T 6.132 1.952 10

Q

2.612 10

3

0

5

1.299 10

4 J

mol

The calculations are repeated and the answers are in the following table:
J/mol
a) Acetylene
26,
120
b) Ammonia
20,
200
c) nbutane
71,
964
d) Carbon diox
ide
21,
779
e) Carbon monox
ide
14,
457
f) Ethane
3 420
8,
g) Hydrogen
13866
,
h) Hydrogen chloride
14,
040
i) M ethane
233
,18
j Nitric ox
)
ide
14, 0
73
k) Nitrogen
14,
276
l) Nitrogen diox
ide
20,
846
m) Nitrous ox
ide
22,
019
n) Ox
ygen
15,
052
o) Propylene
46,
147
116

4.46 T0

( 25

Q

273.15)K

30000

a) Acetylene

T

( 500

273.15)K

J
mol
Given Q = R ICPH T0 T 6.132 1.952 10

T

Find ( T)

T

835.369 K

T

3

273.15K

0

5

1.299 10

562.2 degC

The calculations are repeated and the answers are in the following table:

a) Acetylene
b) Ammonia
c) n-butane
d) Carbon dioxide
e) Carbon monoxide
f) Ethane
g) Hydrogen
h) Hydrogen chloride
i) Methane
j) Nitric oxide
k) Nitrogen
l) Nitrogen dioxide
m) Nitrous oxide
n) Oxygen
o) Propylene

4.47 T0

( 25

273.15)K

T(
K)
835.4
964.0
534.4
932.9
1248.0
690.2
1298.4
1277.0
877.3
1230.2
1259.7
959.4
927.2
1209.9
636.3

T

T(
C)
562.3
690.9
261.3
659.8
974.9
417.1
1025.3
1003.9
604.2
957.1
986.6
686.3
654.1
936.8
363.2

( 250

a) Guess mole fraction of methane:

y

273.15) K

Q

11500

0.5

Given
y ICPH T0 T 1.702 9.081 10
(1
y

3

2.164 10

y) ICPH T0 T 1.131 19.225 10
Find ( y)

y

0.637

Ans.
117

3

6

= Q

0 R

5.561 10

6

0 R

J
mol

b) T0

(
100

273.15)
K

T

(
400

Guess mole fraction of benzene

273.15)K

y

Q

54000

J
mol

0.5

Given
y ICPH T0 T
(
1

3

0.206 39.064 10

y)ICPH T0 T

y

Find y
()

y

c) T0

(
150

3

3.876 63.249 10

273.15)
K

6

13.301 10

= Q

0 R
6

20.928 10

0 R

Ans.

0.245
T

(
250

Guess mole fraction of toluene

273.15)K

y

Q

17500

J
mol

0.5

Given
y ICPH T0 T 0.290 47.052 10
(
1
y

3

15.716 10

y)ICPH T0 T 1.124 55.380 10
Find y
()

y

0.512

3

6

= Q

0 R

18.476 10

6

0 R

Ans.

4.48 Temperature profiles for the air and water are shown in the figures below.
There are two possible situations. In the first case the minimum
temperature difference, or "pinch" point occurs at an intermediate location
in the exchanger. In the second case, the pinch occurs at one end of the
exchanger. There is no way to know a priori which case applies.
Pi h a End
nc t

I e m e a e Pi h
nt r di t nc

TH1

Se ton I
ci I

Se ton I
ci
TH1

Se ton I
ci

THi

Se ton I
ci I

THi
T
TH2 TC1

TC1
TCi

TCi
TC2

118

TH2
T
TC2

To solve the problem, apply an energy balance around each section of the
exchanger.
T H1

Section I balance:

mdotC HC1

HCi = ndotH

CP dT
THi

T Hi

Section II balance:

HC2 = ndotH

mdotC HCi

CP dT
TH2

If the pinch is intermediate, then THi = TCi + T. If the pinch is at the end,

then TH2 = TC2 + T.

a) TH1

T

10degC

TC1

2676.0

TCi

For air from Table C.1:A

kJ
kg

3.355 B

100degC

TC2

25degC

HCi

100degC

HC1

1000degC

419.1

kJ
kg

HC2

104.8

0.575 10

3

C

0 D

kJ
kg
5

0.016 10

kmol
s

Assume as a basis ndot = 1 mol/s.

ndotH

Assume pinch at end:

TH2

TC2

THi

110degC

Guess:

mdotC

1

kg
s

1

T

Given

mdotC HC1

mdotC HCi

mdotC
THi
mdotC
ndotH

THi

TCi

HCi = ndotH R ICPH THi TH1 A B C D Energy balances
on Section I and
HC2 = ndotH R ICPH TH2 THi A B C D II

170.261 degC mdotC

Find mdotC THi THi

0.011

kg
mol

70.261 degC

Ans.

TH2
119

TC2

10 degC

11.255

kg
s

Since the intermediate temperature difference, THi - TCi is greater than
the temperature difference at the end point, TH2 - TC2, the assumption of a
pinch at the end is correct.
b) TH1

TC1

T

10degC

2676.0

TCi

kJ
kg

100degC

TC2

25degC

HCi

100degC

HC1

500degC

419.1

kJ
kg

HC2

104.8

1

kmol
s

THi

TCi

T

TH2

110degC

Assume as a basis ndot = 1 mol/s.

ndotH

Assume pinch is intermediate:

Guess:

mdotC

1

kg
s

kJ
kg

Given
mdotC HC1

mdotC HCi

mdotC
TH2

mdotC
ndotH

THi

TCi

HCi = ndotH R ICPH THi TH1 A B C D Energy balances
on Section I and
HC2 = ndotH R ICPH TH2 THi A B C D II
Find mdotC TH2 TH2

5.03 10

3 kg

10 degC

mol

48.695 degC

mdotC

5.03

kg
s

Ans.

TH2

TC2

23.695 degC

Since the intermediate temperature difference, THi - TCi is less than the
temperature difference at the end point, TH2 - TC2, the assumption of an
intermediate pinch is correct.
4.50a) C6H12O6(s) + 6 O2(g)= 6 CO2(g) + 6 H2O(l)
1 = C6H12O6 , 2 = O2 , 3 = CO2 , 4 = H2O

H0f1

1274.4

kJ
mol

H0f2
120

0

kJ
mol

M1

180

gm
mol

H0f3

H0r

393.509

6 H0f3

kJ
mol

6 H0f4

b) energy_per_kg

mass_glucose

H0f4

150

kJ
kg

H0f1

285.830

mass_person energy_per_kg
H0r

M3

H0r

6 H0f2

mass_person

kJ
mol

44

2801.634

gm
mol

kJ
mol

Ans.

57kg

mass_glucose

M1

0.549 kg Ans.

c) 6 moles of CO2 are produced for every mole of glucose consumed. Use
molecular mass to get ratio of mass CO2 produced per mass of glucose.
6

275 10 mass_glucose

6 M3
M1

8

Ans.

2.216 10 kg

4.51 Assume as a basis, 1 mole of fuel.
0.85 (CH4(g) + 2 O2(g) = CO2(g) + 2 H2O(g))
0.10(C2H6 (g) + 3.5 O2(g) = 2 CO2(g) + 3 H2O(g))
-----------------------------------------------------------------0.85 CH4(g) + 0.10 C2H6(g) + 2.05 O2(g) = 1.05 CO2(g) + 2 H2O(g)
1 = CH4, 2 = C2H6, 3 = O2, 4 = CO2, 5 = H2O 6 = N2

kJ
mol

H0f1

74.520

H0f4

393.509

a) H0c

H0c

825.096

kJ
mol

2 H0f5

83.820

H0f5

kJ
mol

1.05 H0f4

kJ
mol

H0f2

241.818

0.85 H0f1

H0f3

0

kJ
mol

kJ
mol

0.10 H0f2

1.05 H0f3

Ans.

b) For complete combustion of 1 mole of fuel and 50% excess air, the exit
gas will contain the following numbers of moles:

n3

n3

0.5 2.05mol
121

1.025 mol

Excess O2

n4

1.05mol

n5

2mol

n6

0.05mol

79
1.5 2.05mol
21

n6

11.618 mol

Total N2

Air and fuel enter at 25 C and combustion products leave at 600 C.

T1

A

B

C

D

Q

(
25

T2

273.15)
K

n3 3.639

n4 6.311

(
600

n5 3.470

273.15)
K

n6 3.280

mol

n3 0.506

n4 0.805

n5 1.450

n6 0.593 10

3

mol

n3 0

n4 0

n5 0

n6 0 10

6

mol

n3 ( 0.227) n4 ( 0.906) n5 0.121

5

n6 0.040 10

mol

H0c

ICPH T1 T2 A B C D R

122

Q

529.889

kJ
mol

Ans.

5.2 Let the symbols Q and Work represent rates in kJ/s. Then by Eq. (5.8)

Work

=

QH

TC

TC

= 1

TH

TH

323.15 K

Work

TC

QH 1

250

kJ
s

kJ
s

Work
or

148.78

Work

TH

By Eq. (5.1),

QH

798.15 K

148.78 kW which is the power. Ans.

QC

QH

QC

Work

101.22

kJ
s

Ans.

5.3 (a) Let symbols Q and Work represent rates in kJ/s

TH

TC

750 K

By Eq. (5.8):

But

TC

1
Work

=

QH

Work

300 K

0.6

TH

Whence

95000 kW

QH

Work

QH

QC
(b)

QH
QH

0.35

QC

5.4 (a) TC

303.15 K

Carnot

1

TC
TH

QC

Work
QH

TH

Work

6.333

QH

Work

5
1.583 10 kW Ans.

5
2.714 10 kW Ans.

QC

1.764

4
10 kW Ans.

5
10 kW Ans.

623.15 K

0.55
123

Carnot

0.282

Ans.

(b)

0.35

Carnot

By Eq. (5.8),

Carnot

0.55

TC

TH

1

0.636

833.66 K Ans.

TH

Carnot

5.7 Let the symbols represent rates where appropriate. Calculate mass rate of
LNG evaporation:
3

V

9000

molwt

m

P

s

17

gm

mLNG

mol

T

1.0133 bar

PV
molwt
RT

298.15 K

mLNG

6254

kg
s

Maximum power is generated by a Carnot engine, for which

Work

QH

=

QC

QC

TH

512

Work

QH

QC

TH
TC

QH

1=

QC

TH
TC

1

1

113.7 K

QC

kJ
mLNG
kg

QC

=

TC

303.15 K

QC

5.8

QC

3.202

Work

QH

Work

6

10 kW
6

10 kW

8.538

Ans.

6

5.336

Ans.

10 kW

Take the heat capacity of water to be constant at the valueCP

(a) T1

273.15 K

SH2O
Sres

CP ln
Q
T2

T2

T2
T1

373.15 K

Q

CP T2

SH2O

1.305

1.121

kJ
kg K

Q

kJ
kg K

Sres

T1

124

Ans.

4.184

418.4

kJ
kg K
kJ

kg

Stotal

SH2O

Sres

Stotal

0.184

kJ

Ans.

kgK

(b) The entropy change of the water is the same as in (a), and the total
heat transfer is the same, but divided into two halves.

Q

Sres

2

Stotal

1
373.15 K

1
323.15 K

Sres

Sres

Stotal

SH2O

0.097

1.208

kJ

kJ
kgK

Ans.

kgK

(c) The reversible heating of the water requires an infinite number of heat
reservoirs covering the range of temperatures from 273.15 to 373.15 K,
each one exchanging an infinitesimal quantity of heat with the water and
raising its temperature by a differential increment.

5.9

P1

T1

P1 V

n

R T1

(a) Const.-V heating;

T2

T1

Q
n CV

But

P2
P1

=

1.443 mol

U= Q

T2

By Eq. (5.18),

1

T1

Whence

3

0.06m

5

CV

2

R

W = Q = n CV T2

Q

15000 J

T1

3

10 K

S = n CP ln
T2

V

500 K

n

1 bar

S

T2
T1

R ln

n CV ln

T2
T1

P2
P1

S

20.794

(b) The entropy change of the gas is the same as in (a). The entropy
change of the surroundings is zero. Whence

Stotal = 10.794

J
K

Ans.

The stirring process is irreversible.
125

J
K

Ans.

5.10 (a) The temperature drop of the second stream (B) in either
case is the same as the temperature rise of the first stream CP
(A), i.e., 120 degC. The exit temperature of the second
stream is therefore 200 degC. In both cases we therefore
have:

SA
SA

8.726

463.15

CP ln

SB

343.15
J
mol K

473.15
593.15

CP ln

SB

7
R
2

6.577

J

Ans.

mol K

(b) For both cases:

Stotal

SA

Stotal

SB

2.149

J
mol K

Ans.

(c) In this case the final temperature of steam B is 80 degC, i.e., there is
a 10-degC driving force for heat transfer throughout the exchanger.
Now

SA
SA

8.726

463.15
343.15

CP ln

Stotal

SB

J
mol K

SA

SB

SB

dQ

Since dQ/T = dS,

8.512

Stotal

dW

5.16 By Eq. (5.8),

Ans.

J
mol K

dW = dQ

T

dW = dQ

J
mol K

0.214

T

= 1

353.15
473.15

CP ln

Ans.

T

dQ
T

T dS

Integration gives the required result.

T1

Q

CP T2

T2

600 K

T1

Q

5.82

T

400 K

126

3 J

10

mol

300 K

S

CP ln

Work

Q

Q

T2
T1
T

Q

5.17 TH1

11.799

Work

S

Q

Work

Q

Sreservoir

S

S

2280

3540

Sreservoir

T

J
mol K

Sreservoir

0

600 K

TC1

J
mol K

J
mol

Ans.

J
mol

11.8

Ans.

J

Ans.

mol K

Process is reversible.

300 K

TH2

For the Carnot engine, use Eq. (5.8):

W
QH1

TC2

300 K

=

TH1

250 K

TC1

TH1

The Carnot refrigerator is a reverse Carnot engine. W
TH2 TC2
=
Combine Eqs. (5.8) & (5.7) to get:
TC2
QC2
Equate the two work quantities and solve for the required ratio of the heat
quantities:
TC2 TH1 TC1
Ans.
r
r 2.5
TH1 TH2 TC2

5.18 (a) T1

H

S

(b)

C p T2

Cp ln

T2
T1

T2

1.2bar

T1
R ln

P2

mol

4.365

S

P1

3 J

H = 5.82 10

450K

H

P1

300K

1.582

S = 1.484

127

P2

6bar

3 J

10

mol

J
mol K

J
mol K

Ans.

Ans.

Cp

7
R
2

3 J

(c)

H = 3.118 10

(d)

H = 3.741 10

(e)

H = 6.651 10

S = 4.953

mol
3 J

S = 2.618

mol
3 J

J
mol K

J
mol K

S = 3.607

mol

J
mol K

5.19This cycle is the same as is shown in Fig. 8.10 on p. 305. The equivalent states
are A=3, B=4, C=1, and D=2. The efficiency is given by Eq. (A) on p. 305.
Temperature T4 is not given and must be calaculated. The following equations
are used to derive and expression for T4.
For adiabatic steps 1 to 2 and 3 to 4:

T1 V 1

1

= T2 V 2

1

1

T3 V 3

For constant-volume step 4 to 1:

P2
T2

=

P3
T3

Solving these 4 equations for T4 yields: T4 = T1
7
R
2

T1

(
200

T4

T1

T3

Eq. (A) p. 306

1

(
1000

T4

T2

T3

1.4

Cv

T2

273.15)
K

T2

Cp

5
R
2

Cv

1

V1 = V4

For isobaric step 2 to 3:

Cp

= T4 V 4

T3

(
1700

873.759 K

1

T4

T1

T3

T2

128

273.15)
K

0.591

Ans.

273.15)
K

Solucionario Introducción a la Termodinamica en Ingeniería Química: Smith, Van Ness & Abbott

Solucionario Introducción a la Termodinamica en Ingeniería Química: Smith, Van Ness & Abbott

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