Advanced Thermodynamics
Note 2
Volumetric Properties of Pure Fluids
Lecturer: 郭修伯
A pressure-temperature diagram
• the sublimation curve
• the fusion curve
• the vaporization curve
• the triple point
• the critical point
Fig 3.1
A pressure-volume diagram
• The isotherms
– the subcooled-liquid and the
superheated-vapor regions
– isotherms in the subcooledliquid regions are steep
because liquid volumes change
little with large change in
pressure
• The two-phase coexist region
• The triple point is the horizontal
line
• The critical point
Fig 3.2
• An equation of state exists relating pressure, molar or specific
volume, and temperature for any pure homogeneous fluid in
equilibrium states.
• An equation of state may be solved for any one of the three
quantities P, V, or T as a function of the other two.
• Example:
V
V




dV  
 dT  
 dP
 T  P
 P T
1  V 
Volume expansivity:   

V  T  P
1  V 
Isothermal compressibility:    

V  P T
dV
 dT  dP
V
– For incompressible fluid, both β andκ are zero.
– For liquids β is almost positive (liquid water between 0°C and 4°C is
an exception), and κ is necessarily positive.
– At conditions not close to the critical point, β andκ can be assumed
constant:
V2
ln
V1
  (T2  T1 )   ( P2  P1 )
Virial equations of state
• PV along an isotherm:
• PV  a  bP  cP 2  a(1  BP  CP 2  DP3  ...)
– The limiting value of PV as P →0 for all the gases:
– PV *  a  f (T )
– PV *  a  RT , with R as the proportionally constant.
– Assign the value of 273.16 K to the temperature of the triple point
of water: PV *  R  273.16
• Ideal gas:
t
– the pressure ~ 0; the molecules are separated by infinite distance;
the intermolecular forces approaches zero.
–
PV *t
cm 3 bar
R
 83.1447
273.16
mol K
The virial equations
• The compressibility factor:
PV
Z
RT
• the virial expansion: Z  (1  BP  CP 2  DP3  ...)
– the parameters B’, C’, D’, etc. are virial coefficients,
accounting of interactions between molecules.
– the only equation of state proposed for gases having a
firm basis in theory.
– The methods of statistical mechanics allow derivation
of the virial equations and provide physical significance
to the virial coefficients.
Ideal gas
• No interactions between molecules.
• gases at pressure up to a few bars may often be considered
ideal and simple equations then apply
• the internal energy of gas depends on temperature only.
– Z = 1; PV = RT
– U = U (T)
– C   U   dU  C (T ) C   H   dH  dU  R  C  R  C (T )
V
V
P
V
P
 T V
dT
 T  P
dT
dT
U   CV dT H   C P dT
– Mechanically reversible closed-system process, for a unit mass or a
mole:
CV
CP
dQ  dW  CV dT
dQ 
VdP 
PdV
R
R
dW  PdV
• For ideal gas with constant heat capacities undergoing a
mechanically reversible adiabatic process:
dQ  dW  CV dT
T2  P2 
  
T1  P1 
R
CP
PV   const .
dW  PdV
dT
R dV

T
CV V
T2  V1 
  
T1  V2 
R
CV

CP
CV
– for monatomic gases,   1.67
– for diatomic gases,   1.4
– for simple polyatomic gases, such as CO2, SO2, NH3, and CH4,   1.3
• The work of an irreversible process is calculated:
– First, the work is determined for a mechanically reversible process.
– Second, the result is multiple or divided by an efficiency to give
the actual work.
Air is compressed from an initial condition of 1 bar and 25°C to a final state of 5
bar and 25 °C by three different mechanically reversible processes in a closed
system. (1) heating at constant volume followed by cooling at constant pressure; (2)
isothermal compression; (3) adiabatic compression followed by cooling at constant
volume. Assume air to be an ideal gas with the constant heat capacities, CV =
(5/2)R and CP = (7/2)R. Calculate the work required, heat transferred, and the
changes in internal energy and enthalpy of the air in each process.
Fig 3.7
Choose the system as 1 mol of air, contained in an imaginary frictionless piston
/cylinder arrangement.
For R = 8.314 J/mol.K, CV = 20.785, CP = 29.099 J/mol.K
The initial and final molar volumes are: V1 = 0.02479 m3 and V2 = 0.004958 m3
The initial and final temperatures are identical: ΔU = ΔH = 0
(1) Q = CVΔT + CPΔT = -9915 J; W = ΔU - Q = 9915 J
 P1 
(2) Q  W  RT ln    3990 J
 P2 
 V1 
T

T
(3) adiabatic compression: 2 1  
 V2 
 1

 567.57 K
W  CV T  5600 J
cooling at constant V, W = 0.
overall, W = 5600 J, Q = ΔU - W = -5600 J.
V 
P2  P1  1   9.52 bar
 V2 
An ideal gas undergoes the following sequence of mechanically reversible
processes in a closed system:
(1) From an initial state of 70°C and 1 bar, it is compressed adiabatically to 150 °C.
(2) It is then cooled from 150 to 70 °C at constant pressure.
(3) Finally, it is expanded isothermally to its original state.
Calculate W, Q, ΔU, and ΔH for each of the three processes and for the entire cycle.
Take CV = (3/2)R and CP = (5/2)R. If these processes are carried out irreversibly but
so as to accomplish exactly the same changes of state (i.e. the same changes in P, T,
U, and H), then different values of Q and W result. Calculate Q and W if each step
is carried out with an efficiency of 80%.
Fig 3.8
Choose the system as 1 mol of air, contained in an imaginary frictionless piston
/cylinder arrangement. For R = 8.314 J/mol.K, CV = 12.471, CP = 20.785 J/mol.K
(1) For an ideal gas undergoing adiabatic compression, Q = 0
ΔU = W = CVΔT = 12.471(150 – 70) = 998 J
ΔH = CPΔT = 20.785(150 – 70) = 1663 J

 T2  ( 1)
P2  P1  
 1.689 bar
 T1 
(2) For the constant-pressure process:
Q = ΔH = CPΔT = 20.785(70 – 150) = -1663 J
ΔU = CVΔT = 12.471(70 – 150) = -998 J
W = ΔU – Q = 665 J
(3) Isotherm process, ΔU and ΔH are zero:
 P3 
Q  W  RT ln    1495 J
 P1 
(4) Overall: Q = 0 – 1663 + 1495 = -168 J
W = 998 + 665 – 1495 = 168 J
ΔU = 0
ΔH = 0
Irreversible processes:
(1) For 80% efficiency:
W(irreversible) = W(reversible) / 0.8 = 1248 J
ΔU(irreversible) = ΔU(reversible) = 998 J
Q(irreversible) = ΔU – W = -250 J
(2) For 80% efficiency:
W(irreversible) = W(reversible) / 0.8 = 831 J
ΔU = CVΔT = 12.471(70 – 150) = -998 J
Q = ΔU – W = -998 – 831 = -1829 J
(3) Isotherm process, ΔU and ΔH are zero:
W(irreversible) = W(reversible) x 0.8 = -1196 J
Q = ΔU – W = 1196 J
(4) Overall: Q = -250 – 1829 + 1196 = -883 J
W = 1248 + 831 – 1196 = 883 J
ΔU = 0
ΔH = 0
The total work required when the cycle consists of three irreversible steps is
more than 5 times the total work required when the steps are mechanically
reversible, even though each irreversible step is assumed 80% efficient.
A 400g mass of nitrogen at 27 °C is held in a vertical cylinder by a frictionless
piston. The weight of the piston makes the pressure of the nitrogen 0.35 bar higher
than that of the surrounding atmosphere, which is at 1 bar and 27°C. Take CV =
(5/2)R and CP = (7/2)R. Consider the following sequence of processes:
(1) Immersed in an ice/water bath and comes to equilibrium
(2) Compressed reversibly at the constant temperature of 0°C until the gas volume
reaches one-half the value at the end of step (1) and fixed the piston by latches
(3) Removed from the ice/water bath and comes to equilibrium to thermal
equilibrium with the surrounding atmosphere
(4) Remove the latches and the apparatus return to complete equilibrium with its
surroundings.
Nitrogen may be considered an ideal gas. Calculate W, Q, ΔU, and ΔH for each step
of the cycle.
The steps:
P
(1) 27 C, 1.35bar const

 0 C, 1.35bar
(2)
(3)
(4)
1
T
0 C , V2 const

 0 C , V3  V2
2
V
0 C , V3 const

 27  C , V4  V3
4 T1
27 C, P4 T
 27 C, 1.35bar
n
m
 14.286 mol
M
Fig 3.9
(1) W  n PdV  nPV  nRT  3207 J Q  nH  nC T  11224 J
1
1
P
1

nU1  Q1  W1  11224  3207  8017 J
(2) U  H  0
2
2
Q2  W2  nRT ln
V3
 22487 J
V2
(3) W  0 Q  nU  nC T  8017 J nH  nC T  11224 J
3
3
V
3
P
3
(4) the oscillation of the piston
U 4  H 4  0 Q4  W4
Air flows at a steady rate through a horizontal insulated pipe which
contains a partly closed valve. The conditions of the air upstream
from the valve are 20°C and 6 bar, and the downstream pressure is 3
bar. The line leaving the valve is enough larger than the entrance line
so that the kinetic-energy change as it flows through the valve is
negligible. If air is regarded as an ideal gas, what is the temperature
of the air some distance downstream from the valve?
Flow through a partly closed valve is known as a throttling process.
For steady flow system:

d (mU )cv
1
 
   Q  W
   H  u 2  zg m
dt
2
  fs

Ideal gas:
H   C P dT
The result that ΔH = 0 is general for a throttling process.
H  0
T2  T1
If the flow rate of the air is 1 mol/s and if the pipe has an inner diameter of 5 cm,
both upstream and downstream from the valve, what is the kinetic-energy change
of the air and what is its temperature change? For air, CP = (7/2)R and the molar
mass is M = 29 g/mol.
Upstream molar volume:
RT1 83.14  293.15
1
V
6
3 m 2
V1 

10  4.062 10
u

n

n
 2.069 m
1
mol
s
P1
6
A
A
Downstream molar volume:
V2  2V1 u2  2u1  4.138 m
s
The rate of the change in kinetic energy:
2
2
1 2
1 2
3 (4.138  2.069 )
  u   nM  u   (1 29 10 )
m
 0.186 J
s
2
2 
2 

d (mU )cv
1
 
   Q  W
   H  u 2  zg m
dt
2
  fs

C
 1 
m  P T   u 2    0
 2 
M
T  0.0064 K
Application of the virial equations
• Differentiation:
 Z 
2
   B  2CP  3DP  ...
 P T
 Z 
 B


 P T ; P 0
• the virial equation truncated to two terms satisfactorily
represent the PVT behavior up to about 5 bar
Z  1  BP
Z  1
BP
B
 1
RT
V
• the virial equation truncated to three terms provides good
results for pressure range above 5 bar but below the critical
pressure
B C
2
Z

1

 2
Z  1  BP  C P
V V
Reported values for the virial coefficients of isopropanol vapor at 200°C are:
B = -388 cm3/mol and C = -26000 cm6/mol2. Calculate V and Z for isopropanol
vapor at 200 °C and 10 bar by (1) the ideal gas equation; (2) two-term virial
equation; (3) three-term virial equation.
(1) For an ideal gas, Z = 1:
V
3
RT 83.14  473.15

 3934 cm
mol
P
10
(2) two-term virial equation:
V
3
RT
 B  3934  388  3546 cm
mol
P
Z
PV
 0.9014
RT
(3) three-term virial equation:
Vi 1 
RT
P
 B C 

388  26000 
cm3
1   2   39341 



3539
1st iteration
2 
mol
 3934 (3934) 
 Vi Vi 
Ideal gas value
...
3
After 5 iterations V4 ~ V5  3488 cm
mol
Z
PV
 0.8866
RT
Cubic equations of state
• Simple equation capable of representing both
liquid and vapor behavior.
RT

• The van del Waals equation of state: P 
a
V b V 2
– a and b are positive constants
– unrealistic behavior in the two-phase region. In reality,
two, within the two-phase region, saturated liquid and
saturated vapor coexist in varying proportions at the
saturation or vapor pressure.
– Three volume roots, of which two may be complex.
– Physically meaningful values of V are always real,
positive, and greater than constant b.
Fig 3.12
A generic cubic equation of state
• General form:
P
RT
 (V   )

V  b (V  b)(V 2  V   )
– where b, θ, κ,λ and η are parameters depend on temperature and
(mixture) composition.
– Reduce to the van der Waals equation when η= b, θ= a, and κ=λ=
0.
– Set η= b, θ= a (T), κ= (ε+σ) b, λ = εσb2, we have:
P
RT
a(T )

V  b (V  b)(V  b)
• where ε and σ are pure numbers, the same for all substances, whereas
a(T) and b are substance dependent.
• Determination of the parameters:
– horizontal inflection at the critical point:
 2P 
 P 

   2   0
 V T ;cr  V T ;cr
• 5 parameters (Pc, Vc, Tc, a(Tc), b) with 3 equations, one has:
3 RTc
Vc 
8 Pc
Zc 
27 R 2Tc2
a
64 Pc
b
1 RTc
8 Pc
PcVc 3

RTc 8
•
• Unfortunately, it does not agree with the experiment. Each
chemical species has its own value of Zc.
• Similarly, one obtain a and b at different T.
RT
a(T )
P

V  b V (V  b)
a(T )  
 (Tr ) R 2Tc2
Pc
b
RTc
Pc
Two-parameter and three-parameter
theorems of corresponding states
• Two-parameter theorem: all fluids, when compared at the same
reduced temperature and reduced pressure, have approximately the
same compressibility factor, and all deviate from ideal-gas behavior to
about the same degree.
T
P
• Define reduced temperature and reduced pressure: Tr 
Pr 
Tc
Pc
• Not really enough to describe the state, a third corresponding-states
parameter is required.
– The most popular such parameter is the acentric factor (K.S. Pitzer, 1995):
  1.0  log Prsat T 0.7
r
• Three-parameter theorem: all fluids having the same value of ω, when
compared at the same reduced temperature and reduced pressure, and
all deviate from ideal-gas behavior to about the same degree.
• Vapor and vapor-like
V
q
RT
a(T )
V b
b
P
P (V  b)(V  b)
a (T )  (Tr )

bRT
Tr
Z  1    q

V starts with V(ideal-gas)
and then iteration
bP
P
 r
RT
Tr
Z 
( Z   )( Z   )
• Liquid and liquid-like
 RT  bP  VP 
V  b  (V  b)(V  b) 

a(T )


a (T )  (Tr )
bP
P
q


 r
bRT
Tr
RT
Tr
V starts with V = b and
then iteration
1   Z 

Z    ( Z   )( Z   )
 q 
Equations of state which express Z as a function of Tr and Pr are said to be
generalized, because of their general applicability of all gases and liquids.
2-parameter/3-parameter E.O.S.
• Express Z as functions of Tr and Pr only, yield 2parameter corresponding states correlations:
– The van der Waals equation
– The Redlich/Kwong equation
• The acentric factor enters through function α(Tr;ω)
as an additional paramter, yield 3-parameter
corresponding state correlations:
– The Soave/Redlich/Kwong (SRK) equation
– The Peng/Robinson (PR) equation
Table 3.1
Given that the vapor pressure of n-butane at 350K is 9.4573 bar, find the molar
volumes of (1) saturated-vapor and (2) saturated-liquid n-butane at these conditions
as given by the Redlich/Kwong equation.
350
Tr 
 0.823
425.1
q
 (Tr )
 6.6048
Tr
9.4573
Pr 
 0.2491
37.96
 
Pr
 0.026214
Tr
(1) The saturated vapor
Z  1    q
Z 
( Z   )( Z   )
Z starts at Z = 1 and converges on Z = 0.8305
ZRT
cm3
V
 2555
P
mol
(2) The saturated liquid
1   Z 

Z    ( Z   )( Z   )
q



ZRT
cm3
V
 133.3
P
mol
Z starts at Z = β and converges on Z = 0.04331
Generalized correlations for gases
• Pitzer correlations for the compressibility factor:
Z  Z 0  Z 1
– Z0 = F0 (Tr, Pr)
– Simple linear relation between Z and ω for given values
of Tr and Pr.
– Of the Pitzer-type correlations available, the Lee/Kesler
correlation provides reliable results for gases which are
nonpolar or only slightly polar (App. E).
– Only tabular nature (disadvantage)
Pitzer correlations for the 2nd virial
coefficient
• Correlation:
BP
0 Pr
1 Pr
Z  1
 1 B
 B
RT
Tr
Tr
Z  Z 0  Z 1
Z 0  1 B 0
Pr 1
P
Z  B1 r
Tr
Tr
– Validity at low to moderate pressures
– For reduced temperatures greater than Tr ~ 3, there
appears to be no limitation on the pressure. 0
0.422
B  0.083  1.6
Tr
– Simple and recommended.
0.172
– Most accurate for nonpolar species.
B1  0.139 
Tr4.2
Determine the molar volume of n-butane at 510K and 25 bar by, (1) the ideal-gas
equation; (2) the generalized compressibility-factor correlation; (3) the generalized
virial-coefficient correlation.
(1) The ideal-gas equation
RT
cm3
V
 1696.1
P
mol
(2) The generalized compressibility-factor correlation
510
25
the acentric factor
 0.659
Tr 
 1.200 Pr 
  0.200
37.96
425.1
the Lee/Kesler correlation
Z  0.865
0
Z  0.038
1
ZRT
cm3
 1480.7
Z  Z  Z  0.873 V 
P
mol
0
1
(3) The generalized virial-coefficient correlation
0.172
0.422 1
510
0
B

0
.
139

B

0
.
083

Tr 
 1.200
Tr4.2
Tr1.6
425.1
Z  1  B0
3
Pr
P
 B1 r  0.879 V  ZRT  1489.1 cm
Tr
Tr
P
mol
What pressure is generated when 1 (lb mol) of methane is stored in a volume of 2
(ft)3 at 122°F using (1) the ideal-gas equation; (2) the Redlish/Kwong equation; (3)
a generalized correlation .
(1) The ideal-gas equation
P
RT 0.7302(122  459.67)

 212.4 atm
V
2
(2) The RK equation
RTc
581.67
 (Tr ) R 2Tc2
atm
3
b



0
.
4781
ft
a
(
T
)



453
.
94
Tr 
 1.695
Pc
Pc
ft 6
343.1
P
RT
a(T )

 187.49 atm
V  b V (V  b)
(3) The generalized compressibility-factor correlation is chosen (high pressure)
P
ZRT Z (0.7302)(122  459.67)

 212.4Z atm
V
2
Pr 
581.67
P
Z
Tr 
 1.695

343.1
45.4 0.2138
P  189.0 atm
Z starts at Z = 1 and
converges on Z = 0.890
A mass of 500 g of gases ammonia is contained in a 30000 cm3 vessel immersed in
a constant-temperature bath at 65°C. Calculate the pressure of the gas by (1) the
ideal-gas equation; (2) a generalized correlation .
Vt
cm3
V
 1021.2
n
mol
(1) The ideal-gas equation
P
RT
 27.53 bar
V
(2) The generalized virial-coefficient correlation is chosen (low pressure, Pr ~ 3 )
338.15
Tr 
 0.834
405.7
Pr ~
B 0  0.083 
27.53
 0.244
112.8

Z  1  B 0  B1
the acentric factor
 TP
r
r
P
ZRT
 23.76 bar
V
 1  0.541
Pr
Tr
0.422
Tr1.6
  0.253
B1  0.139 
0.172
Tr4.2
Generalized correlations for liquids
• The generalized cubic equation of state (low accuracy)
• The Lee/Kesler correlation includes data for subcooled
liquids
– Suitable for nonpolar and slightly polar fluids
• Estimation of molar volumes of saturated liquids
– Rackett, 1970: V sat  V Z (1Tr )0.2857
c c
• Generalized density correlation for liquid (Lydersen,
Greenkorn, and Hougen, 1955):
 Vc
r 
c

V

V2  V1 r1
r 2
Fig 3.17
For ammonia at 310 K, estimate the density of (1) the saturated liquid; (2) the
liquid at 100 bar
(1) Apply the Rackett equation at the reduced temperature
Tr 
310
 0.7641 Vc  72.47 Z c  0.242
405.7
V sat  Vc Z c(1Tr )
0.2857
cm3
 28.33
mol
(2) At 100 bar
100
Pr 
 0.887
112.8
Tr 
310
 0.7641
405.7
Fig 3.17
r  2.38
cm3
V
 30.45
r
mol
Vc
 r1,310K , saturatedliquid
 r1
2.34
cm 3
310K
V2  V1
V
 29.14
 28.65
r 2
 r 2,100bar
2.38
mol