This document discusses thermodynamic properties of fluids, including: 1) Derivations of equations relating the primary thermodynamic properties of pressure, volume, temperature, internal energy, and entropy for homogeneous phases and fluids. 2) Calculations of changes in enthalpy, entropy, and internal energy based on changes in pressure and temperature. 3) The thermodynamic properties of Gibbs energy and residual properties. 4) An example problem calculating the enthalpy and entropy of saturated isobutane vapor at a specified temperature and pressure using compressibility factor and ideal gas heat capacity data.

1. Advanced Thermodynamics
Note 5
Thermodynamic Properties of Fluids
Lecturer: 郭修伯

2. Property relations for homogeneous phases
First law for a closed system d(nU) = dQ + dW
a special reversible process
rev rev d(nU) = dQ + dW
dW Pd(nV) rev = - dQ Td(nS) rev =
d(nU) = Td(nS) - Pd(nV )
Only properties of system are involved:
It can be applied to any process in a closed system (not necessarily
reversible processes).
The change occurs between equilibrium states.

3. The primary thermodynamic properties: P,V, T,U, and S
The enthalpy: H ºU + PV
The Helmholtz energy: A ºU -TS
The Gibbs energy: G º H -TS
For one mol of homogeneous fluid of constant composition:
dU = TdS - PdV
d(nU) = Td(nS) - Pd(nV )
dH = TdS +VdP
dA = -PdV - SdT
P
= - ¶ ÷ø
ö çè
÷ø
æ
¶
S V S
T
¶
æ
ö ¶
V
çèV
= ¶ ÷ø
ö çè
÷ø
æ
¶
S P S
T
¶
æ
ö ¶
P
çè
S
= ¶ ÷ø
ö çè
÷ø
æ
¶
¶
ö çè
V T V
P
æ
¶
T
S
V
= - ¶ ÷ø
ö çè
¶
dG =VdP - SdT Maxwell’s equations
P T P
T
÷ø
æ
¶
æ
ö ¶
çè

4. Enthalpy, entropy and internal energy change calculations f (P,T)
= ¶ dP
dP
dT H
+ ¶ ÷ø
æ
ö ¶
P
çè
dH H
T
÷ø
P T
æ
ö ¶
çè
dT S
+ ¶ ÷ø
æ
ö ¶
P
çè
dS = ¶
S
T
÷ø
P T
æ
ö ¶
çè
dH = TdS +VdP
P
H = ÷ø
P
C
æ
ö ¶
T
çè
¶
V
T S
= ¶ ÷ø
æ
ö ¶
P
çè
H
P
+ ÷ø
T T
¶
æ
ö ¶
çè
¶
æ
H
= ¶ ÷ø
= + ¶
T S
ö çè
P ÷ ÷ø
V dP
dH C dT T S
P
T
ö
ç çè
+ ÷ø
æ
¶
S
= - ¶ ÷ø
ö çè
P T P
V
T
÷ø
æ
¶
¶
æ
ö ¶
çè
dP
æ
dH = C dT + V - T ¶
V
ö çè
P ÷ ÷ø
T
P
ö
ç çè
÷ø
æ
¶
S C
P
T
¶
T
= ÷ø
P
æ
ö ¶
çè
= - ¶ ÷ø
ö çè
dP
V
V
¶
= - ¶
æ
ö ¶
T
çè
dS C dT
T
P
P ÷ø
P P T
T
÷øö çè
æ
¶
æ
ö ¶
çè
S
P T P
T
÷ø
æ
¶
æ
ö ¶
çè
H ºU + PV
V
P V
+ ¶ ÷ø
æ
ö ¶
P
çè
U
= ¶ ÷ø
æ
ö ¶
P
çè
H
P
+ ÷ø
T T T
¶
æ
ö ¶
çè
P V
T V
- ¶ ÷ø
ö çè
= - ¶ ÷ø
P
P
T T T U
P
÷ø
æ
¶
æ
ö ¶
çè
¶
æ
ö ¶
çè

5. Determine the enthalpy and entropy changes of liquid water for a
change of state from 1 bar and 25°C to 1000 bar and 50°C. The data
for water are given.
H1 and S1 at 1 bar, 25°C
H2 and S2 at 1 bar, 50°C H3 and S3 at 1000 bar, 50°C
dP
æ
dH = C dT + V - T ¶
V
ö çè
P ÷ ÷ø
T
P
ö
ç çè
÷ø
æ
¶
( H 75.310(323.15 298.15) 1 (513 10 -
)(323.15) ) (18.204)(1000 1)
3400J /mol
dP
V
= - ¶
æ
ö ¶
T
çè
dS C dT
T
P
P ÷ø
dH C dT ( T )VdP P = + 1-b
( ) (1 ) ( ) 2 1 2 2 1 H C T T T V P P P D = - + - b -
V
V
T
b = ÷ø
P
¶
æ
ö ¶
çè
volume expansivity
S C T P D = - b -
2 V P P
T
ln ( ) 2 1
1
10
6
D = - + - ´ - =
( 513 10 -
) S (18.204)(1000 1)
5.13J /mol
10
75.310ln 323.15
298.15
6
D = - ´ - =

6. Internal energy and entropy change calculations f (V,T)
= ¶ dV
dV
dT U
+ ¶ ÷ø
æ
ö ¶
V
çè
dU U
ö T
çè
÷ø
V T
æ
¶
dT S
+ ¶ ÷ø
æ
ö ¶
V
çè
dS = ¶
S
ö T
çè
÷ø
V T
æ
¶
V
U = ÷ø
ö çè
V
C
æ
¶
T
¶
T S
U
= ¶ ÷ø
æ
ö ¶
çè
¶
V ÷ ÷ø ö
- ÷ø
P dV
æ
dU = C dT + T ¶
S
ö V
çè
T
ç çè
- ÷ø
æ
¶
P
= - ¶ ÷ø
ö çè
T V T
S
¶
V
÷ø
æ
¶
æ
ö ¶
çè
æ
= + ¶
ö çè
V ÷ ÷ø
P dV
dU C dT T P
T
V
ö
ç çè
- ÷ø
æ
¶
T S
= ¶ ÷ø
ö çè
S C
V
T
¶
ö T
çè
= ÷ø
V
æ
¶
= - ¶ ÷ø
ö çè
dV
S
P
¶
= + ¶
ö çè
V ÷ø
T
dS C dT
T
V
æ
¶
P
V
V
T T
æ
ö ¶
çè
V V T
U
T
÷ø
æ
¶
¶
æ
ö ¶
çè
P
T V T
V
÷ø
æ
¶
æ
ö ¶
çè

7. Gibbs energy G = G (P,T)
dG =VdP - SdT
dT
d G dG G
2
RT
º 1 - ÷ø
ö çè
RT RT
æ
Thermodynamic property of great potential utility
dG =VdP - SdT
dT
÷ø
çè
d æ
G ö º V
dP - H
RT
RT
RT
2 G º H -TS
dT
dP G RT
+ ¶ ÷ø
æ ( / ) ( / )
ö T
çè
G RT
P
d G
RT
÷ø
æ
¶
T P
æ
ö ¶
çè
= ¶ ÷ø
ö çè
G RT
= ¶( / )
ö çè
T P
V
RT
÷ø
æ
¶
T G RT
= - ¶( / )
ö çè
P T
H
RT
÷ø
æ
¶
G/RT = g (P,T)
The Gibbs energy serves as a generating function
for the other thermodynamic properties.

8. Residual properties
• The definition for the generic residual property:
M R º M -Mig
– M and Mig are the actual and ideal-gas properties,
respectively.
– M is the molar value of any extensive thermodynamic
properties, e.g., V, U, H, S, or G.
• The residual Gibbs energy serves as a generating
function for the other residual properties:
dT
ö
R R R
dP H
RT
V
RT
æ
d G
RT
2 - = ÷ ÷ø
ç çè

9. R R
= ÷ ÷ø
æ
æ const T dP
dT
ö
R R R
dP H
RT
V
RT
d G
RT
2 - = ÷ ÷ø
ç çè
V
RT
d G
RT
ö
ç çè
= òR P R
dP
V
RT
G
RT
0
RT
P
V R =V -V ig = ZRT -
P
= - ¶( / )
ö çè
( 1) P T
= ò - R P
Z dP
P
G
RT
0
T G RT
H
RT
÷ø
æ
¶
T Z
= - ¶ P
æ
ö ¶
çè
ò ÷ø
P
R
dP
P
T
H
RT
0
dP
T Z
= - ¶ P P
æ
ö ¶
çè
Z dP
ò ò - - ÷ø
P
R
P
P
T
S
R
0 0
( 1)
S R R G
R
RT
H
RT
R
= -
const T
const T
Z = PV/RT: experimental measurement .Given PVT data or an appropriate equation
of state, we can evaluate HR and SR and hence all other residual properties.

10. { }
þ ý ü î í ì
H = H + H = H + C dT + - RT ¶
Z
ö çè
÷ø
dP
0
H R ICPH T T A B C D RT Z
= + ´ + - ¶
{ }
{ ( )}
{ ( )}
þ ý ü
î í ì
( , ; , , , )
H C T T RT Z
= + - + - ¶
ö çè
dP
dP
0
2
0 0
H R MCPH T T A B C D T T RT Z
= + ´ - + - ¶
ö çè
÷ø
æ
¶
þ ý ü
î í ì
÷ø
æ
¶
þ ý ü
î í ì
÷øö çè
æ
¶
æ
¶
ò
ò
ò
ò ò
P
P
ig
P
P
H
ig
P
ig
P
P
ig
P
P
T
T
ig
P
ig R ig
dP
P
T
P
T
P
T
P
T
0
2
0 0 0
0
2
0 0
0
2
0
( , ; , , , )
þ ý ü
R T Z
+ - ¶
ö çè
dP
R P
S S S S C dT
ò ò ò
ln ( 1)
R T Z
+ - ¶
ö çè
dP
Z dP
ò ò
0
( , ; , , , ) ln ( 1)
ln ln ( 1)
î í ì
Z dP
R T Z
dP
+ - ¶
ö çè
Z dP
Z dP
dP
- - ÷ø
æ
¶
þ ý ü
S R ICPS T T A B C D R P
R T Z
+ - ¶
ö çè
R P
S C T
î í ì
î í ì = + ´ -
þ ý ü
- - ÷ø
æ
¶
þ ý ü
î í ì
= + -
þ ý ü
î í ì
- - ÷ø
æ
¶
þ ý ü
î í ì
= + ´ -
þ ý ü
î í ì
- - ÷ø
æ
¶
þ ý ü
î í ì
= + = + -
ò ò
ò ò
P P
P
S ig
R MCPS T T A B C D T
P P
P
S
ig
P
ig
P P
P
ig
P P
P
T
T
ig
P
ig R ig
P
P
T
R P
P
T
P
P
T
P
T
P
P
T
P
P
P
T
P
T
0 0
0 0
0 0
0 0
0 0
0
0 0
0
0 0
0 0
0
0
( , ; , , , ) ln ln ( 1)

11. Calculate the enthalpy and entropy of saturated isobutane vapor at 360
K from the following information: (1) compressibility-factor for
isobutane vapor; (2) the vapor pressure of isobutane at 360 K is 15.41
bar; (3) at 300K and 1 bar, Hig = 18115 J /mol Sig = 295.976 J /mol ×K 0 0
(4) the
ideal-gas heat capacity of isobutane vapor:
Cig / R =1.7765 + 33.037´10-3
T
P
T Z
= - ¶ P
æ
ö ¶
çè
ò ÷ø
P
R
dP
P
T
H
RT
dP
T Z
= - ¶ P P
æ
ö ¶
çè
Z dP
S
0 ò ò - - ÷ø
P
R
P
P
T
R
0 0
( 1)
Z
¶
Graphical integration requires plots of P and (Z-1)/P vs. P.
T
/ ÷ø
P
æ
ö ¶
çè
dP
T Z
= - ¶ - òP
(360)(26.37 10 4 ) 0.9493
0
ö çè
- = ´ - = ÷ø
æ
¶
P
R
P
T
H
RT
dP
T Z
= - ò ¶ òP P
ö çè
Z dP
- = - - - = - - ÷ø
( 1) 0.9493 ( 0.02596) 0.6897
æ
¶
0 0
P
R
P
P
T
S
R
H R = -2841.3J /mol
S R = -5.734J /mol ×K
H = { Hig + R ´ ICPH(300,360;1.7765,33.037E - 3,0.0,0.0) } + H R =
21598.5 J 0 mol
{ } mol K
S Sig R ICPS E R S R J
×
(300,360;1.7765,33.037 3,0.0,0.0) ln 15.41 0
= + ´ - - + = 286.676
1

12. Residual properties by equation
of state
• If Z = f (P,T):
T Z
= - ¶ P
æ
ö ¶
çè
ò ÷ø
dP
T Z
= - ¶ P P
æ
ö ¶
çè
R
Z dP
S
÷ø
ò - ò - 0 P
R
dP
P
T
H
RT
0 0
P
P
P
T
R
( 1)
• The calculation of residual properties for gases
and vapors through use of the virial equations and
cubic equation of state.

13. Use of the virial equation of state
Z -1 = BP
RT
= ò - R P
Z dP
P
G
RT
0
( 1)
two-term virial equation
BP
RT
GR
RT
=
T Z
= - ¶ P
æ
ö ¶
çè
ò ÷ø
P
R
dP
P
T
H
RT
= - ¶ P
BP
æ +
ö ¶
çè
H
Z -1 = BP
0 ò ÷ø
P
R
dP
P
RT
T
T
RT
0
(1 )
ö
T æ
P
ö çè
B
dB
1
ö çè
ò ÷ ÷ø
ç çè
÷ø
æ - ÷ø
= - æ P
P
R
dP
P
T
dT
R T
H
RT
0 2
RT
dB
B
ö çè
÷ø
P
= æ -
dT
T
R
H R
RT
S R R G
R
RT
H
RT
R
= -
P
dB
R
dT
S R
RT
= -

14. Z -1 = Br +Cr 2
Three-term virial equation
= ò - R P
Z dP
P
G
RT
0
( 1)
= 2 r + 3 r 2 -
B C Z
GR
RT
ln
2
T Z
= - ¶ P
æ
ö ¶
çè
ò ÷ø
P
R
dP
P
T
H
RT
ù
T é
B
dC
C
r 1 r
dB
= æ - 2
ö çè
ö çè
H R
0 úû
êë
÷ø
æ - + ÷ø
2
dT
T
dT
T
RT
S R R G
R
RT
H
RT
R
= -
P = ZrRT
Application up to moderate pressure

15. Use of the cubic equation of state
• The generic cubic equation of state:
a T
( )
V b V b
( )( )
P RT
V b
+e +s
-
-
=
1
q r
b
b b
b
r + +
1 (1 )(1 )
Z
er sr
-
-
=
q º a(T)
bRT
ò r (Z -1) d r
= -ln(1- r
b) - qI
0
r
I
Z = - ÷ø
d dq
dT
¶ òr
æ
ö ¶
T
çè
r
r r
0
const T
I ò d ( r
b)
+ +
b b
º r
er sr
0 (1 )(1 )
qI
é
Z d T
1 lna ( )
= - + -1
d T
H
RT
r
r
R
ù
úû
êë
ln
é
Z d T
ù
ln b lna ( )
( ) qI
= - + -1
d T
S
R
r
r
R
úû
êë
ln
b º bP
RT

16. Find values for the residual enthalpy HR and the residual entropy SR
for n-butane gas at 500 K and 50 bar as given by Redlich/Kwong
equation.
T = 500 = 1.317
r 1.176
425.1
= 50 = r P = W = 0.09703
37.96
b P ( ) = 3.8689
r
T
r
q = Y
a T
r
T
W
r
Z q Z
= + - -
b b b
Z Z
+ +
( )( )
1
eb sb
e = 0
s =1 Z = 0.685
qI
I ò d ( r
b)
+ +
é
Z d T
1 lna ( )
= - + -1
d T
H
RT
r
r
R
ù
úû
êë
ln
é
Z d T
ù
ln b lna ( )
( ) qI
= - + -1
d T
S
R
r
r
R
úû
êë
ln
const T
b b
º r
er sr
0 (1 )(1 )
ö
æ
+
sb
I Z
13247 . 0 ln 1 = ÷ ÷ø
ç çè
+
-
=
eb
s e Z
H R = -4505 J
mol
S R J
mol ×
K
= -6.546

17. Two-phase systems
• Whenever a phase transition at constant
temperature and pressure occurs,
– The molar or specific volume, internal energy,
enthalpy, and entropy changes abruptly. The exception
is the molar or specific Gibbs energy, which for a pure
species does not change during a phase transition.
– For two phases α and β of a pure species coexisting at
equilibrium:
Ga = Gb
where Gα and Gβ are the molar or specific Gibbs energies of the individual phases

18. Ga = Gb dGa = dGb
dG =VdP - SdT
ab
a b
S
S S
dPsat
= D
= -
Va dPsat - Sa dT =V b dPsat - Sb dT a b
ab
V
V V
dT
D
-
dH = TdS +VdP DHab = TDSab
The latent heat of phase transition
ab
ab
H
= D
T V
dPsat
dT
D
The Clapeyron equation
DVab =V = RT
sat
v
P
Ideal gas, and Vl << Vg
H R d P
sat
ln
d T
(1/ )
D lv = -
The Clausius/Clapeyron equation

19. ab
ab
H
= D
T V
dPsat
dT
D
The Clapeyron equation: a vital connection between the properties of different phases.
P = f (T)
ln Psat = A- B For the entire temperature range from the triple point
T
to the critical point
Psat A B
ln = - The Antoine equation, for a specific temperature range
T +
C
= + + +
t t t t
-
t
1
ln
A B 1.5 C 3 D 6 Psat
r
The Wagner equation, over a wide
temperature range.
r t =1-T

20. Two-phase liquid/vapor systems
• When a system consists of saturated-liquid and
saturated-vapor phases coexisting in equilibrium,
the total value of any extensive property of the
two-phase system is the sum of the total properties
of the phases:
M = (1- xv )Ml + xvMv
– M represents V, U, H, S, etc.
– e.g., V = (1- xv )V v + xvV v

21. Thermodynamic diagrams and tables
• A thermodynamic diagram represents the temperature,
pressure, volume, enthalpy, and entropy of a substance on
a single plot. Common diagrams are:
– TS diagram
– PH diagram (ln P vs. H)
– HS diagram (Mollier diagram)
• In many instances, thermodynamic properties are reported
in tables. The steam tables are the most thorough
compilation of properties for a single material.

22. Fig. 6.2

23. Fig. 6.3

24. Fig. 6.4

25. Superheated steam originally at P1 and T1 expands through a nozzle to an
exhaust pressure P2. Assuming the process is reversible and adiabatic, determine
the downstream state of the steam and ΔH for the following conditions:
(a) P1 = 1000 kPa, T1 = 250°C, and P2 = 200 kPa.
(b) P1 = 150 psia, T1 = 500°F, and P2 = 50 psia.
Since the process is both reversible and adiabatic, the entropy change
of the steam is zero.
(a) From the steam stable and the use of interpolation,
At P1 = 1000 kPa, T1 = 250°C:
H1 = 2942.9 kJ/kg, S1 = 6.9252 kJ/kg.K
At P2 = 200 kPa,
S2 = S1= 6.9252 kJ/kg.K < Ssaturated vapor,
the final state is in the two-phase liquid-vapor region:
S xv Sl xvSv 2 2 2 2 2 = (1- ) + xv xv2 2 6.9252 =1.5301(1- ) + 7.1268
0.9640 2 xv =
H H H 315.9 kJ 2 1 D = - = -
(1 0.964)(504.7) (0.964)(2706.3) 2627.0 2 kg H = - + =

26. (a) From the steam stable and the use of interpolation,
At P1 = 150 psia, T1 = 500°F:
H1 = 1274.3 Btu/lbm, S1 = 1.6602 Btu/lbm.R
At P2 = 50 psia,
S2 = S1= 1.6602 Btu/lbm.K > Ssaturated vapor,
the final state is in the superheat region:
T 283.28 F 2 =
H H H 99.0 Btu 2 1 D = - = -
lbm
H 1175.3 Btu / lbm 2 =

27. A 1.5 m3 tank contains 500 kg of liquid water in equilibrium with pure water
vapor, which fills the remainder of the tank. The temperature and pressure are
100°C, and 101.33 kPa. From a water line at a constant temperature of 70°C and
a constant pressure somewhat above 101.33 kPa, 750 kg of liquid is bled into
the tank. If the temperature and pressure in the tank are not to change as a result
of the process, how much energy as heat must be transferred to the tank?
d mU Energy balance: ( )cv =  + ¢ ¢
Q H m
dt
d mU ( )cv =  + ¢ cv
Q H dm
dt
dt
cv cv Q = D(mU) - H¢Dm
H =U + PV
cv cv cv Q = D(mH) - D(PmV ) - H¢Dm
cv cv cv Q = (m H ) - (m H ) - H¢Dm 2 2 1 1
At the end of the process, the tank still contains saturated liquid and
saturated vapor in equilibrium at 100°C and 101.33 kPa.

28. Q = (m H ) - m H - H¢Dm 2 2 cv () 1 1 cv cv H¢ = 293.0 kJ saturated liquid @70
C
kg
saturated liquid C
kJ Hlcv
= 419.1 @100
kg
saturated vapor C
Hv kJ
kg
cv
= 2676.0 @100
From the steam table, the specific volumes of saturated liquid and saturated
vapor at 100°C are 0.001044 and 1.673 m3/kg , respectively.
( ) 500(419.1) 1.5 (500)(0.001044) 1 1 1 1 1 1 = + = + - =
m H mlHl mvHv kJ
cv (2676.0) 211616
1.673
= 500 + 1.5 - (500)(0.001044) + = + mv ml2 2 1.5 =1.673 + 0.001044
m ml mv2 2 2 750
1.673
m H mlHl mvHv kJ
( )cv 524458 2 2 2 2 2 2 = + =
Q m H m H H m kJ ( )cv ( )cv cv 524458 211616 (750)(293.0) 93092 2 2 1 1 = - - ¢D = - - =

29. c r P = P P c r T = T T
T Z
= - ¶ P
æ
ö ¶
çè
ò ÷ø
P
R
dP
P
T
H
RT
0
dP
T Z
= - ¶ P P
æ
ö ¶
çè
Z dP
ò ò - - ÷ø
P
R
P
P
T
S
R
0 0
( 1)
ö
æ
¶
T Z
= - ¶ r
ò ÷ ÷ø
ç çè
r
P
dP
r
r
r P
r
R
c
P
T
H
RT
0
2
ö
æ
¶
dP
T Z
= - ¶ r r
Z dP
P r
ò ò - - ÷ ÷ø
ç çè
r
P
r
r
r
r P
r
R
P
P
T
S
R
0 0
( 1)
Z = Z 0 +wZ1
ù
ú ú
û
é
T Z
dP
T Z
= - ò ¶ ò r
- - ¶
ê ê
ë
ö
÷ ÷ø
æ
¶
ç çè
ù
ú ú û
é
ê ê
ë
ö
÷ ÷ø
æ
¶
ç çè
r
r
r
P
dP
r
r
r P
r
P
r
r
r P
r
R
c
P
T
P
T
H
RT
0
1
2
0
0
2 w
ù
ú ú
û
é
dP
T Z
Z dP
dP
T Z
= - ò ¶ ò - - ò r ¶
ò r
ê ê
ë
ö
Z dP
P r
- - ÷ ÷ø
æ
¶
ç çè
ù
ú ú
û
é
ê ê
ë
ö
P r
- - ÷ ÷ø
æ
¶
ç çè
r
r r
r
P
r
r
r
r P
r
P
r
r
r
r P
r
R
P
P
T
P
P
T
S
R
0
1
0
1
0
0
0
0
( 1) w ( 1)
( ) ( ) ( , , )
é
H
0 H
R
1
= +w
HRB TR PR OMEGA
RT
RT
( ) ( ) ( , , )
H
R
RT
c
c
R
c
=
ù
ú úû
ê êë
é
0 1
= +w SRB TR PR OMEGA
Table E5 ~ E12
S R R S
R
R
S
R
R
=
ù
ú úû
ê êë

30. T R
T
0 = + ò +
H Hig 2
C ig
dT H2 0 P
2
T R
T
0 = + ò +
H Hig 1
C ig
dT H1 0 P
1
T R R
T
D = ò + -
H 2
C ig
dT H HP 2 1
1
S S C dT R P
2
T R
T
= + ò - ln +
ig S
ig
P
2
P
T
0
2 0
2
0
S S C dT R P
1
T R
T
= + ò - ln +
ig S
ig
P
1
P
T
0
1 0
1
0
ig
P H C T T H H2 1 2 1 D = ( - ) + -
R R
S C dT R P
2 1
T R R
T
D = ò - ln + -
ig
P S S
2 2
1
P
T
1
H
S C ig
P S S
P 2 1
R R
D = ln 2 + -
S
P
1

31. Estimate V, U, H, and S for 1-butene vapor at 200°C and 70 bar if H and S are
set equal to zero for saturated liquid at 0°C. Assume that only data available are:
T K c = 420 P bar c = 40.43 w = 0.191 T 266.9K (nomal boiling pt.) n =
Cig / R 1.967 31.630 10 3T 9.837 10 6T 2
P
= + ´ - - ´ -
=1.127 r T =1.731 r P
0 1
Z = Z +wZ
= +
0.485 (0.191)(0.142)
0.512
=
cm
mol
V ZRT
P
3
= = 287.8
Assuming four steps:
(a) Vaporization at T1 and P1 = Psat
(b) Transition to the ideal-gas state at (T1, P1)
(c) Change to (T2, P2) in the ideal-gas state
(d) Transition to the actual final state at (T2, P2)
Fig 6.7

32. Fig 6.7

33. Step (a)
ln Psat = A- B
T
ln1.0133 = A- B
266.9
ln 40.43 = A- B
420
For = 273.15K,
Psat = 1.2771 bar
9.979
lv
n
H
D = -
1.092(ln 1.013) =
0.930
-
rn
c
n
T
P
RT
The latent heat at normal boiling point:
The latent heat at 273.15 K:
0.38
æ
-
= 1
-
1
ö
÷ ÷
ø
ç ç
è
H
D
D
T
r
rn
lv
lv
n
T
H
DHlv = 21810 J
mol
DHab = TDSab
Slv J
mol ×
K
D = 79.84
Step (b) = 0.650 r T = 0.0316 r P
( ) ( ) ( , , ) 0.0985
é
0 1
= + HRB TR PR OMEGA
= = -
ù
ú úû
ê êë
H
R
RT
H
RT
H
R
RT
c
c
R
c
w
( ) ( ) ( , , ) 0.1063
é
0 1
= + SRB TR PR OMEGA
= = -
ù
ú úû
S R R R
ê êë
S
R
S
R
R
w
H R 344 J 1 - =
mol
S R J
mol ×
K
- = 0.88 1

34. Step (c)
DHig = 8.314´ ICPH(273.15,473.15;1.967,31.630E -3,-9.837E - 6,0.0) = 20564 J
mol
8.314 (273.15,473.15;1.967,31.630 3, 9.837 6,0.0) 8.314ln 70
Sig ICPS E E
D = ´ - - - -
J
mol ×
K
=
22.18
1.2771
Step (d) =1.127 r T =1.731 r P
( ) ( ) 2.430
é
= +
0 1
= -
ù
ú úû
H w
ê êë
R
c
c
R
R
c
H
RT
H
RT
RT
( ) ( ) 1.705
é
= +
0 1
= -
ù
ú úû
S R R R
ê êë
S
R
S
R
R
w
H R 8485 J 2 = -
mol
S R J
mol ×
K
= -14.18 2
H = DH = 21810 - (-344) + 20564 -8485 = 34233 J
mol
S S J
mol ×
K
= D = 79.84 - (-0.88) + 22.18 -14.18 = 88.72
Total
U = H - PV= 34233- (70)(287.8) /10 = 32218 J
mol

35. Gas mixtures
• Simple linear mixing rules
w ºå
yw i i ºå
i
Pc i ci T y T ºå
i
Pc i ci P y P
i
Estimate V, HR, and SR for an equimolar mixture of carbon dioxide and propane
at 450K and 140 bar
T y T K
= (0.5)(304.2) + (0.5)(369.8) = 337
Pc ºå i ci i
P y P bar
Pc i ci ºå = (0.5)(73.83) + (0.5)(42.48) = 58.15 = 2.41 pr P
i
=1.335 pr T Z 0 = 0.697
Z1 = 0.205
188 . 0 ) 152 . 0 )( 5 . 0 ( ) 224 . 0 )( 5 . 0 ( = + = ºåi
i i w yw Z = Z 0 +wZ1 = 0.736
( ) ( ) = =196.7 cm
1.762
mol
V ZRT
P
3
H é
0 1
ù
( ) ( ) = +
w = -
1.029
ú úû
ê êë
R
pc
R
pc
R
pc
H
RT
H
RT
RT
é
= +
0 1
= -
ù
ú úû
S R R R
ê êë
S
R
S
R
R
w
H R = -4937 J /mol S R = -8.56 J /mol ×K