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Ch.3 Properties of
Pure Substances
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Objectives
 Introduce the concept of a pure substance.
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Discuss the physics of phase-change processes.
Illustrate the P-v, T-v, and P-T property diagrams and P-v-T surfaces of
pure substances.
Demonstrate the procedures for determining thermodynamic properties of
pure substances from tables of property data.
Describe the hypothetical substance “ideal gas” and the ideal-gas equation
of state.
Apply the ideal-gas equation of state in the solution of typical problems.
Introduce the compressibility factor, which accounts for the deviation of
real gases from ideal-gas behavior.
Present some of the best-known equations of state.
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Pure Substance
 In Chemistry you defined a pure substance as an
Air
element or a compound
 Something that can not be separated
 In Thermodynamics we’ll define it as something
that has a fixed (same) chemical composition
throughout
Examples of Pure Substance
 Water, nitrogen, helium, and carbon dioxide, for
example, are all pure substances.
 A mixture of water liquid and water vapor, for
example, is a pure substance because both
phases have the same chemical composition.
N2
Water
vapor
Water
liquid
Pure substance
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Example of Non-Pure Substance
Air Gas
 A mixture of liquid air and gaseous air, however, is
not a pure substance.
Air
liquid
 This is because different air components condense at
different temperatures at a specified pressures and
thus the composition of liquid air and gas air will be
different..
Not a Pure
substance
Phases of Pure Substance
 A phase is identified as having a distinct molecular
arrangement.
 This molecular arrangement is homogeneous
throughout the system.
 The phase separated from the other phases by easily
identifiable boundary surfaces.
vapor
liquid
Solid
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Solid Phase of Pure Substance
 The molecules in a solid are arranged in a lattice
(network or pattern) that is repeated throughout.
 Three dimensional pattern
 Large attractive forces between atoms or
molecules
 Molecule at relatively fixed position in solid.
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Liquid Phase of Pure Substance
 When a solid reaches a sufficiently high
temperature the velocity (and thus the
momentum) of the molecules reach a
point where the intermolecular forces are
partially defeated and groups of the
molecules break away (melting point)
 In liquid the molecular spacing is not
much different from that of solids,
except that they can rotate and translate
freely (they are not at fixed positions
relative to each other)
 Distance between molecules increase
slightly as a solid turns to liquid
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Gas Phase of Pure Substance
 In the gas phase, the molecules are far
apart from each other, and a molecular
order is nonexistent.
 Molecules move about at random,
continually colliding with each other and
the walls of the container they are in
 High kinetic energy
 In order to liquefy, lots of that kinetic
energy must be released
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Liquid phase to Gas Phase
 Let us study what would happen when we heat a liquid phase of pure
substance at a constant pressure
Liquid Water
Pressure under the piston = Pressure above the piston
Piston cylinder
device –
maintains
constant
pressure
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T
5
2
3
Phase Change
Processes on a Tv diagram
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P is constant
1
Constant pressure
line
v
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Phase Change Processes on a T-v diagram
 Consider a piston-cylinder device with water
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inside at 20oC and 1 atm pressure .
At this P and T, water is called compressed
(or subcooled) liquid state.
Compressed liquid means that it is not about
to vaporize.
The system is heated and the piston is allowed
to float and thus the pressure will be constant.
T and v will increase until the system reaches
100 C at which any addition of heat will cause
some of the liquid to vaporize
The temperature at which a pure substance
changes phase is called the saturation
temperature, Tsat.
At Tsat, Liquid and vapor phases are in
equilibrium.
A liquid that is about
to vaporize is called
Saturated Liquid.
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 Adding more heat will cause
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boiling to start. Liquid gradually
evaporates (state 3) but
temperature will remain constant,
The only change is the increase in
the specific volume (v) until it
reaches state 4 (saturated vapor).
Heating the system further, will
increase both the temperature and
specific volume (state 5). This
single-phase state is called
“Superheated vapor”
Repeat this experiment for higher
pressures.
Similar curves will be obtained but
at higher sat. temperature.
Note that the sat. liquid specific
volume (vsat,l ) will increase while
the sat. vapor specific volume
(vsat,g ) will decrease
A substance between
saturated liquid (state 2) and
saturated vapor (state 4) is
called saturated liquid-vapor
mixture.
A vapor that is about to
condense is called
Saturated vapor.
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Latent Heat
 Def.: the amount of energy absorbed or released during a phase change
process.
 Latent heat of fusion: the amount of energy absorbed during melting.
Which is equal to the amount of energy released during freezing.
 Latent heat of vaporization : the amount of energy absorbed during
vaporization which is equal to the amount of energy released during
condensation.
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Saturation Temperature and Pressure

Water at a pressure of 101.325 kPa, Tsat is
100oC. Conversely, at a temperature of
100oC, Psat is 101.325 kPa.
 Latent heat: fusion and vaporization.
 The magnitude of the latent heats depend
on the temperature or pressure at which
the phase change occur.
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Saturation Temperature and Pressure

At a given pressure, the temperature at which a
pure substance changes phase is called the
saturation temperature. Likewise the pressure.
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Critical Point
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The critical point is defined as the
point at which the saturated liquid
and saturated vapor states are
identical.
At the critical pressure, there will be
no distinct phase change process.
Instead, the specific volume of the
substance will continually increase
and at all times there will be only
one phase present.
The saturated liquid states can be
connected by a line called the
saturated liquid line.
The saturated vapor states can be
connected by another line, called the
saturated vapor line, to form a phase
dome.
Three main regions can be
identified.
vsat,l and vsat,g will be the same and we
speak of Pcrit, Tcrt, and vcrit.
Critical properties in Table A-1
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Phase Change Processes on a P-v diagram
 Decrease P gradually but
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keep T constant.
State 1
T = 150 C
Water boils at Psat
The pressure at which a
pure substance changes
phase is called the
saturation pressure Psat.
At Psat, Liquid and vapor
phases are in equilibrium.
From State 2 to 4, no
weights are removed
(P=constant) and T is kept
constant but heating
causes liquid to vaporize.
State 2
T = 150 C
State 3
T = 150 C
State 4
T = 150 C
State 5
T = 150 C
1
2
3
4
5
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P-v diagrams with Solid Phase
 P – v diagram of substance that
contracts on freezing (most
substance)
 P – v diagram of substance that
expands on freezing (such as
water)
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Triple point
Constant T
Line
 Under some conditions all three
phases of substance coexist in
equilibrium at states along the
triple line.
 The states on the triple line of
substance have the same pressure
and temperature but different v.
 The triple line appears as a point
on the P-T diagram.
 The triple point of water occurs
at T= 0.01 C and P=0.6113 Kpa
vapor
liquid
Solid
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Property Diagrams
 P-T diagram
(or Phase diagram)
 The P-T diagram is often called
phase diagram since all three
phases are separated by three lines,
namely the sublimation line
(between solid and vapor regions),
the vaporization line (between
liquid and vapor regions), and the
melting line (between solid and
liquid).
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T
P-v-T surfaces
You can plot P, T, V on a
three dimensional graph
Top view
v
e
tur
a
r
pe
Tem
P
v

P
w
e
i
v
v
P
P
Tv
iew
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T
Thermodynamics Tables
 The relationship among thermodynamic properties are too complex to be
expressed in simple equations and thus, properties are measured and/or
calculated and then presented in a tabulated form.
Table A-1
Saturated liquidvapor region
Table A4 T entry
Table A5 P entry
w Su Tab
at pe le
er r
(o hea A6
r v te
ap d
or
)
two properties will fix the
state.
 In two phase regions, any two
properties (except P and T)
will fix the state. P and T are
dependent on each other.
Table A7
Compresse Liquid
 In single-phase regions, any
Table A8
Saturated ice-vapor
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Enthalpy − A Combination Property
 In the analysis of certain types of processes, particularly in power
generation and refrigeration, we frequently encounter the combination of
internal energy U, and pressure-volume product PV. That is
H  U  PV
h  u  pυ
 Before 1930, h was referred to as heat content or total heat.
 After 1930, it is referred to as enthalpy (from the Greek word enthalpien
which means to heat)
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Saturated Liquid and Saturated Vapor States
Table A-4
 Saturated liquid-vapor
mixture falls under the P-v
(or T-v) dome.
 Its properties can be obtained
from Water Tables A-4 and
A-5
P
=
co
ns
t.
T
vf
vg
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Saturated Liquid and Saturated Vapor States
Table A-5
 In Table A-5 (page 832),
Pressure is listed in the left
column as the independent
variable.
 Use whichever table is
convenient.
v fg  vg  v f
h fg  hg  h f
 Enthalpy of vaporization or
latent heat
 the amount of energy needed to
vaporize a unite mass of
saturated liquid at a given
temperature or pressure
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Example 3-1:
Saturated Liquid and Saturated Vapor
 A rigid tank contains 50 kg of saturated liquid water at 90oC. Determine
the pressure in the tank and the volume of the tank. (Table A-4)
 Always first you must know the state!!!
(Answers: 70.14 kPa, 0.0518 m3)
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Example 3-2:
Saturated Liquid and Saturated Vapor
 A piston-cylinder device contains 0.06 m3 of saturated water vapor at 350-
kPa pressure. Determine the temperature of the vapor and the mass of the
vapor inside the cylinder. (Table A-5)
(Answers: 138.86oC, 0.114 kg)
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Example 3-3:
Saturated Liquid and Saturated Vapor
 A mass of 200 g of saturated liquid water is completely vaporized at a
constant pressure of 100 kPa. Determine (a) the volume change and (b) the
amount of energy added to the water.
(Answers: 0.3368 m3, 451.6 kJ)
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Saturated Liquid-Vapor Mixture
 In the saturated liquid-vapor mixture, the mass fraction of vapor is called
the QUALITY (x) and is expresses as
m
m
x

m m m
g
f
g
g
total
 Derivation:
V  V f  Vg
mv  m f v f  mg v g
 ( m  m g )v f  m g v g
 v  (1  x)v f  xvg
v  v f  x (v g  v f )
 v  v f  xv fg
where v fg  v g  v f
Gas
mg vg
P or T
Liquid
mf vf
f
 f    g
g
28
Average Properties

In the saturated mixture region, the
average value of any intensive
property y is given as:
y  y f  x( yg  y f )
 y f  x y fg

where f stands for saturated liquid and
g for saturated vapor. For example:
v  vf  xvfg
h  hf  xhfg
u  uf  xufg
s  sf  xsfg
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Average Properties
In the saturated mixture region, the average value of any intensive property y is
given as:
y  y f  x( yg  y f )
01
 y f  x y fg = yg
where f stands for saturated liquid and g for saturated vapor. For example:
When x = 0 we have all liquid, and y = yf
When x = 1 we have all gas, and y = yf + yfg = yg
The average properties of the mixtures are always between the values of the
saturated liquid and the saturated vapor properties. That is
y f  yavg  y g
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X=0
X=1
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Saturated Liquid-Vapor Mixture
T  Tsat at given P
P=
co
ns
t
.
T
T
Tsat
P  Psat at given T
vf
v
vg v
v f  v  v g at given P or T
u f  u  u f at given P or T
P
Psat
T=
c
P
vg
h f  h  h f at given P or T



on
st .
v
saturated mixture
v
Quality is an intensive property
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Example 3- 4:
Saturated Liquid-vapor mixture (continued)
 A rigid tank contains 10 kg of water at 90oC. If 8 kg of water is in the
liquid form and the rest is in the vapor form, determine (a) the pressure in
the tank and (b) the volume of the tank.
(Answers: 70.14 kPa, 4.73 m3)
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Example 3-5:
Saturated Liquid-vapor mixture (continued)
 An 80-L vessel contains 4 kg of refrigerant 134a at a pressure of 160 kPa.
Determine a) the temperature of the refrigerant, b) the quality, c) the
enthalpy of the refrigerant, and d) the volume occupied by the vapor phase.
(Answers: -15.62oC, 0.158, 62.7 kJ/kg, 0.0777 m3)
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Superheated Vapor Table A-6
 In the region to the right of the saturated vapor line, a substance exists as
superheated vapor (single phase).
P
T=
co
ns
t.
vg
v
v
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Superheated Vapor
.
T
T  Tsat at given P
P=
co
ns
t
T
Tsat
P  Psat at given T
vf
v
vg v
v  v g at given P or T
u  u g at given P or T
P
Psat
T=
c
P
h  h g at given P or T



on
st .
superheated vapor
vg
v
v
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Example 3-6
 Determine the internal energy of water at 200 kPa and 300°C.
P=
co
ns
t
.
T
T
Tsat
vf
vg v
v
 Answer u= 2808.8 kJ/kg
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Compressed liquid Table A-7 only for water
 In the region to the left of the saturated liquid line, a substance exists as
compressed liquid.
T
v
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Compressed Liquid
P=
co
ns
t
.
T
T
Tsat
T  Tsat at given P
P  Psat at given T
vf
vg v
v
v  v f at given P or T
u  u f at given P or T
h  h f at given P or T



compressed liquid
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A general approximation
In the absence of compressed liquid data (available only for water), a general
approximation is to treat compressed liquid as saturated liquid at the given
temperature. Such that
Acceptable
P
Wrong
P=
5
Mp
T
a.

5 Mpa
T =2
264
80
T=
8
v vf
Approximate value
Precise value
v
y y

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v vf
Wrong value
Precise value
0
v
f @T
If the compression is moderate, the properties do not vary significantly with
pressure. But they do vary with temperature
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Linear Interpolation
A
B
100
5
130
y
200
10
130  100 y  5

200  100 10  5
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Linear Interpolation (Continued)
Now
X1=
X =
X2=
T
140
143
145
Psat
y1= 0.3615
y=?
y2= 0.4154
y  y1
x  x1

y 2  y1 x 2  x1
x  x1
y  y1 
 ( y2  y1 )
x2  x1
Psat  0.3615 
Psat  0.394
143  140
145  140
 ( 0.4154  0.3615)
kPa
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Example 3-7
Superheated Vapor

Determine the temperature of water at a state of P = 0.5 MPa and h = 2890 kJ/kg.
 First go to saturated liquid vapor table get hf and hg at 0.5 MPa




T C
200
T
250
h kJ/kg
2855.8
2890
2961.0
(Answers: 216.4 oC)
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Example on Compressed Liquid
Example 3-8:
 Determine the internal energy of
compressed liquid water at 80oC
and 5 MPa using (a) data from the
compressed liquid table and (b)
saturated liquid data. What is the
error involved in the second case?
(Answers: 333.82 kJ/kg, 334.97
kJ/kg, 0.34%)
pa
5M
263.99
80
80
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Example 3-8 more
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The Use of Steam Table to Determine
Properties
Example 3-9:
 Determine the missing properties and the phase descriptions in the
following table for water.
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Reference State and Reference Values
 The values of u, h, and s cannot be measured directly, and they are
calculated from measurable properties using the relations between
thermodynamic properties. However, those relations give the changes in
properties, not the values of properties at specified state.
 For water, the state saturated liquid at 0.01oC is taken as the reference
state, and the internal energy and entropy are assigned zero values at that
sate.
 For refrigerant 134a, the state saturated liquid at -40oC is taken as the
reference state, and the enthalpy and entropy are assigned zero values at
that state.
 In thermodynamics we are concerned with the changes in properties, and
the reference chosen has no consequences in the calculations. i.e. we
always need the change in u, h, and s in the real life ….and not the absolute
value even if the reference temp. is different we do not care….!
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