ITK-233
Termodinamika Teknik Kimia I
3 SKS
2 - PVT Behavior of Fluid, Equation of State
DICKY DERMAWAN
www.dickydermawan.net78.net
dickydermawan@gmail.com
P – T Diagram
P – v Diagram
T-v Diagram
Equation of State
For the regions of the diagram where a single phase exist:
f ( P, V , T )  0
This means an equation of state exist relating P, V, T. An EOS may be
solved for any one of the three quantities as a function of the other
two, viz.:
V = V(T,P)
 V 
 V 
 dV  
dT



 dP
 T  P
 P  T
1  V 
 

Volume expansivity:
V  T  P
Isothermal compressibility:   
1  V 


V  P  T

dV
  dT   dP
V
Example
For liquid acetone at 20oC & 1 bar:
Β = 1.487 x 10-3 oC-1 κ = 62 x x 10-6 bar-1
For aceton, find:
 P 
a. The value of  T 
V
V= 1.287 cm3 g-1
b. The pressure generated by heating at constant V from
20oC & 1 bar to 30oC
c. The change in volume for a change from 20oC & 1 bar to
0oC & 10 bar.
Problem 3.1
Express the volume expansivity β and isothermal
compressibility κ as functions of density ρ and its
partial derivatives.
The isothermal compressibility coefficient () of water
at 50oC and 1 bar is 44.18 x 10-6 bar-1. To what
pressure must water be compressed at 50oC to
change its density by 1%? Assume that  is
independent of P.
Problem 3.2 & 3.3
Generally, volume expansivity β and isothermal compressibility
κ depend on T and P.
Prove that
  
  
    
 P  T
 T  P
The Tait equation for liquids is written for an isotherm as:
AP 

V  V0  1 

 B P
where V is specific or molar volume, Vo is the hypothetical
molar or specific volume at P = 0 and A & B are positive
constant. Find an expression for the isothermal
compressibility consistent with this equation.
Problem 3.4
For liquid water the isothermal compressibility is
given by:
c

V ( P  b)
where c & b are functions of temperature only.
If 1 kg of water is compressed isothermally &
reversibly from 1 bar to 500 bar at 60oC, how much
work is required?
At 60oC, b=2700 bar and c = 0.125 cm3 g-1
Problem 3.5
Calculate the reversible work done in compressing
1 ft3 of mercury at a constant temperature of
32oF from 1 atm to 3000 atm. The isothermal
compressibility of mercury at 32oF is:
κ/atm-1 = 3.9 x 10-6 – 0.1 x 10-9 P (atm)
Problem 3.6
Five kilograms of liquid carbon tetrachloride undergo
a mechanically reversible, isobaric change of state at
1 bar during which the temperature change from 0oC
to 20oC. Determine ΔVt, W, Q, and ΔUt
The properties for liquid carbon tetrachloride at 1 bar
& ooC may be assumed independent of temperature:
β = 1.2 x 10-3 K-1
Cp = 0.84 kJ kg-1 K-1
ρ = 1590 kg m-3
Problem 3.7
A substance for which κ is a constant undergo a
mechanically reversible process from initial state
(P1,V1) to final state (P2,V2), where V is molar
volume.
a. Starting with the definition of κ, show that the path
of the process is described by
V  A(T) exp(  P)
b. Determine an exact expression which gives the
isothermal work done on 1 mol of this constant-κ
substance
P – v Diagram
Virial EOS & Ideal Gas
Virial expansion: PV  RT (1  B' P  C' P 2  D' P3  ...
Alternative form: PV  RT (1  B  C2  D3  ...
V
Compressibility factor
V
PV
Z
RT
For ideal gas: Z = 1, thus
PV  RT
V
The Ideal Gas
The internal energy of a real gas is a function of pressure and
temperature.
This pressure dependency is the result of forces between the
molecules.
In an ideal gas, such forces does not exist. No energy would be
required to alter the average intermolecular distance, and
therefore no energy would be required to bring about volume
& pressure changes in an ideal gas at constant temperature. In
other word, the internal energy of an ideal gas is a function of
temperature only.
U  U (T )
Implied Property Relations for an Ideal Gas
dU
 U 
CV  
 CV (T)
 
 T  V dT
H  U  PV  U(T)  RT
Thus H is also a funcion of temperature only
dH
 H 
CP  
 CP (T)
 
 T  P dT
dH
CP 
 C P ( T )  C v (T )  R
dT
Isothermal Process
Closed System, Ideal Gas, Mechanically Reversible
U  0
PV = constant
H  0
RT
V2
P1
W   PdV  
dV  RT ln
 RT ln
V
V1
P2
QW
Isobaric Process
Closed System, Ideal Gas, Mechanically Reversible
U   CV dT
H   CP dT
W   PdV  P dV  P (V2  V1 )  R (T2  T1 )
Q  H
TV-1 = constant
Isochoric (Constant V) Process
Closed System, Ideal Gas, Mechanically Reversible
U   CV dT
H   CP dT
W   PdV  0
Q  U
TP-1 = constant
Adiabatic Process
Closed System, Ideal Gas, Mechanically Reversible
Q  0  dQ  0 
RT

dW  P dV C v dT  0  P dV  C v dT  
dV 
V

dU  C v dT 
 RT 
 C v dT   P d

W  PdV
 P 

U   C
H   C
V
P
T2  V1 
  
T1  V2 
T2  P2 
  
T1  P1 
R
Cv
R
CP

dT
P2  V1 

 
P1  V2 
dT
CP
Cv
T V  1  cons tan t
CP

CV
TP
1 

 cons tan t
PV   con tan t
Polytropic Process
Closed System, Ideal Gas, Mechanically Reversible
Polytropic process can be considered as general form
of process.
T V 1  cons tan t
TP
1 

 cons tan t
PV   con tan t
Isobaric process
Isothermal process
Adiabatic process
Isochoric process
:δ=0
:δ=1
:δ=γ
:δ=∞
Example 3.2
Air is compressed from an initial condition of 1 bar & 25oC to a final
state of 5 bar & 25oC by three different mechanically reversible
processes in a closed system:
a. Heating at constant volume followed by cooling at constant
pressure
b. Isothermal compression
c. Adiabatic compression followed by cooling at constant volume.
Assume air to be an ideal gas with the constant heat capacities: CV =
5/2 R, CP = 7/2 R.
Sketch the process in a PV diagram & calculate the work required,
heat transferred, and the changes in internal energy & entalphy of
the air for each processes
Example 3.3
An ideal gas undergoes the following sequence of
mechanically reversible processes in a closed system:
a. From an initial state of 70oC & 1 bar, it is compressed
adiabatically to 150oC
b. It is then cooled from 150oC to 70oC at constant pressure
c. Finally, it is expanded isothermally to its original state
Sketch the process in a PV diagram & calculate W, Q, U, and
H for each of the three processes and for the entire cycle.
Take CV = 3/2 R and CP = 5/2R
Problem 3.8
One mole of an ideal gas with CV = 5/2 R, CP = 7/2 R
expands from P1 = 8 bar & T1 = 600 K to P2 = 1
bar by each of the following path:
(a) Constant volume
(b) Contant temperature
(c) Adiabatically
Assuming mechanical reversibility, calculate W, Q,
U, and H for each of the three processes.
Sketch each path in a single PV diagram
Problem 3.9
An ideal gas initially at 600 K & 10 bar undergoes a four-step
mechanically reversible cycle in a closed system. In step 1-2
pressure decreases isothermally to 3 bar; in step 2-3
pressure decreases at constant volume to 2 bar; in step 3-4
volume decreases at constant pressure; and in step 4-1 the
gas returns adiabatically to its initial step.
a. Sketch the cycle on a PV diagram
b. Determine (where unknown) both T & P for states 1, 2, 3,
and 4
c. Calculate W, Q, U, and H for each step of the cycle
Data: CV = 5/2 R, CP = 7/2 R
Problem 3.10
An ideal gas, CV = 3/2 R & CP = 5/2 R is changed from P1 = 1 bar & Vt1
= 12 m3 to P2 = 12 bar & Vt2 = 1 m3 by the following mechanically
reversible processes:
a. Isothermal compression
b. Adiabatic compression followed by cooling at constant pressure
c. Adiabatic compression followed by cooling at constant volume
d. Heating at constant volume followed by cooling at constant
pressure
e. Cooling at constant pressure followed by heating at constant
volume
Calculate W, Q, Ut, and Ht for each of these processes, and sketch
the paths of all processes on a single PV diagram
Example 3.4
A 400 gram mass of nitrogen at 27oC 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 surroundings
atmosphere, which is at 1 bar & 27oC. Thus the nitrogen is initially at
a pressure of 1.35 bar, and is in mechanical & thermal equilibrium
with its surroundings. Consider the following sequence of process:
The apparatus is immersed in an ice/water bath and is allowed to
come to equilibrium
b. A variable force is slowly applied to the piston so that the nitrogen
is compressed reversibly at the constant temperature of 0oC until
the gas volume reaches one-half the value at the end of step a. At
this point the piston is held in place by latches.
a.
Example 3.4 (cont’)
The apparatus is removed from ice/water bath and comes
to thermal equilibrium with the surrounding atmosphere
at 27oC
d. The latches are removed, and the apparatus is allowed to
return to complete equilibrium with its surroundings
c.
Sketch the entire cycle on a PV diagram, and calculate Q, W,
U & H for the nitrogen for each step of the cycle. Nitrogen
may be considered an ideal gas for which CV = 5/2 R and CP
= 7/2 R
Cubic Equation of State:
van der Waals Equation (1873)
RT
a
P
 2
Vb V
27 R 2 TC
a
64 PC
2
1 RTC
b
8 PC
3 RTC
VC 
8 PC
PC VC 3
ZC 

R TC 8
Theorem of Corresponding State;
Accentric Factor
All fluids, when compared at the same reduced
temperature & reduced pressure, have approximately
the same compressibility factor, and all deviate from
ideal gas behavior to about the sam degree
T
Tr 
TC
Accentric Factor
(Pitzer)
P
Pr 
PC
 
  1.0  log Pr
sat
Tr  0.7
Critical Constants & Accentric Factors:
Paraffins
Tc/K
Pc/bar
-6
3.
Vc/10 m mol
-1
Zc

Critical Constants & Accentric Factors:
Olefin & Miscellaneous Organics
Tc/K
Pc/bar
-6
3.
Vc/10 m mol
-1
Zc

Critical Constants & Accentric Factors:
Miscellaneous Organic Compounds
Tc/K
Pc/bar
-6
3.
Vc/10 m mol
-1
Zc

Critical Constants & Accentric Factors:
Elementary Gases
Tc/K
Pc/bar
-6
3.
Vc/10 m mol
-1
Zc

Critical Constants & Accentric Factors:
Miscellaneous Inorganic Compounds
Tc/K
Pc/bar
-6
3.
Vc/10 m mol
-1
Zc

Cubic Equation of State:
Redlich/Kwong Equation (1949)
RT
a (T )
P

V  b V ( V  b)
2
2
R 2 TC
 1 2 R TC
a (T)    (Tr ) 
 0.42748  Tr 
PC
PC
2
RTC
RTC
b
 0.08664
PC
PC
ZC 
PC VC 1

R TC 3
Cubic Equation of State:
Soave/Redlich/Kwong Equation (1972)
RT
a (T )
P

V  b V ( V  b)
2
R 2 TC
R 2 TC
a (T)    (Tr , ) 
 0.42748  (Tr , ) 
PC
PC
2

(Tr , )  1  (0.480  1.574   0.176 2 )  (1  Tr )
1/ 2

RTC
RTC
b
 0.08664
PC
PC
ZC 
PC VC 1

R TC 3
2
Cubic Equation of State:
Peng - Robinson Equation (1976)
RT
a (T )
P

V  b V ( V  b)
2
R 2 TC
R 2 TC
a (T)    (Tr , ) 
 0.45724  (Tr , ) 
PC
PC

2
(Tr , )  1  (0.37464  1.54226   0.26992 2 )  (1  Tr )
1/ 2

RTC
RTC
b
 0.07779
PC
PC
ZC 
PC VC
 0.30740
R TC
2
Example 3.8
Given that the vapor pressure of n-butane at 350 K &
9.4573 bar, find the molar volumes of saturated
vapor and saturated liquid n-butane at these
conditions as given by:
a. Van der Waals
b. Redlich/Kwong
c. Soave/Redlich/Kwong
d. Peng-Robinson
Application of The Virial Equations
At low to moderate pressure, it is common to use truncated virial
PV
BP
equation:
Z
 1
RT
RT
At pressure above the range of applicability of the above eqn, the
appropriate form is:
PV
B C
Z
 1  2
RT
V V
Benedict/Webb/Rubin equation:
RT B0 RT  A0  C0 / T 2 bRT  a a
c 
 
  
P




1

exp


 2 
2
3
6
3 2
2
V
V
V
V
VT  V 
 V 
and its modifications are inspired by volume expansion virial eqn
and are used in the petroleum & natural gas industries for light
hydrocarbons.
Example 3.7
Reported values for the virial coefficients of isopropanol
vapor at 200oC are:
B = -388 cm3 mol-1 C = -26000 cm6 mol-2
Calculate V and Z for isopropanol at 200oC & 10 bar by:
a. The ideal gas equation
b. Two term truncated pressure expansion virial
equation
c. Three term truncated volume expansion virial
equation
Pitzer Correlation for
the Second Virial Coefficient
 BPC  Pr
BP

Z  1
 1  
RT
 RTC  Tr
BPC
 B0   B1
RTC
0.422
B  0.083  1.6
Tr
0
0.172
B  0.139  4.2
Tr
1
Generalized Correlations 4 Gases: Pitzer type:
Lee -Kessler
Z  Z0  Z1
Tr
Z0
Pr
Generalized Correlations 4 Gases: Pitzer type:
Lee -Kessler
Z  Z0  Z1
Tr
Z0
Pr
Generalized Correlations 4 Gases: Pitzer type:
Lee -Kessler
Z  Z0  Z1
Tr
Z1
Pr
Generalized Correlations 4 Gases: Pitzer type:
Lee -Kessler
Z  Z0  Z1
Tr
Z1
Pr
Real Gas, EOS
Calculate Z and V for steam at 250oC and 1,800 kPa by the
following:
a. Ideal gas equation
b. Truncated virial equation with the following
experimental values of virial coefficient:
B = -152.5 cm3 mol-1
C = -5,800 cm6 mol-2
c. Pitzer correlation for virial equation
d. Pitzer-type correlation of Lee – Kessler
e. Steam table
f. van der Waals equation
g. Redlich/Kwong equation
h. Soave/Redlich/Kwong equation
i. Peng-Robinson Equation
Correlation for Liquids: Rackett Equation
V
sat
 VC Z C
(1 Tr ) 0.2 85 7
Correlation for Liquids:
Lydersen, Greenkorn & Hougen
Problem 3.45
A 30 m3 tank contains 14 m3 of liquid n-butane in
equilibrium with its vapor at 25oC. Estimate the total
mass of n-butane in the tank. The vapor pressure if
n-butane at the given temperature is 2.43 bar.
Ideal Gas
A cylinder pressure vessel having an inside diameter of
50 cm and height of 1.25 cm. Amount of the
ammonia gas is confined in that a pressure vessel.
Measured pressure and temperature of gas are 30
bars and 200oC. What are the specific volume and
amount of the ammonia gas in pressure vessel (in SI
unit), assuming ammonia is an ideal gas?
Real Gas, EOS
Calculate Z and for steam at 250oC and 1,800 kPa by the
following:
The ideal gas equation
The truncated virial equation with the following experimental
values of virial coefficient:
B = -152.5 cm3 mol-1
C = -5,800 cm6mol-2
The van der Waals equation
The Redlich/Kwong equation
The steam table
Data: critical point of steam, TC = 647.1 K; PC = 220.55 bar;
= 55.9 cm3mol-1
1st law after diagram
1 kmole of ideal gas (Cv = 3R/2) undergoes a three-step
mechanically reversible cycle in a closed system.
Gas in initial state at 500 kPa and 27 oC is heated at constant
pressure
Followed by adiabatic expansion until the its pressure
become 120 kPa
Finally step, gas is isothermally compressed to initial
condition
From that process description:
Illustrate the cycle process at P-V diagram in T parameter!
b.
Calculate H, U, Q and W for each step of process and
total!