The
Ideal Gas
1
Ideal gas equation of state
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Property tables provide very accurate information about the
properties.
It is desirable to have simple relations among the properties that are
sufficiently general and accurate.
Any equation that relates P-v-T are called Equation of state.
Property relations that involve other properties of a substance at
equilibrium states are also referred to as equations of state.
The simplest and best-known equation of state for substances in
gas phase is the ideal-gas equation of state.
2
Ideal gas equation of state
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The vapor phase of a substance is called gas when it is above the
critical temperature.
Vapor implies a gas that is not far from a state of condensation.
In 1802, Charles and Lussac experimentally determined the
following (Ideal gas equation of state):
Pv  RT
R is the Gas constant
R= Ru / M
Ru is universal gas constant
Ru = 8.314 KJ/Kmol . K
M is Molar Mass of the gas (molecular weight)
M is the mass of one mole (kmole) of substance in grams (kilograms)
The values of R and M are given in Table A-1 for several substance.
3
P-v-T relation for ideal gases
PV  mRT (v V / m )
PV  NRuT (mR  NMR  NRu )
Pv  RT
Pv  RuT
(v V / N )
The properties of an ideal gas for a fixed mass at two different states
are related to each other by
PV
m
RT
m1  m2
PV
PV
PV
PV
1 1
2 2
1 1

and
 2 2
RT1 RT2
T1
T2
4
Ideal gas
An ideal gas is an imaginary substance that obeys the P-v-T relation.
The aforementioned relation approximates the behavior of real gas at
low densities.
At low pressure and high temperature, the density of a gas decreases,
and the gas behaves like an ideal gas.
In the range of practical interest, many familiar gases such as air,
nitrogen, oxygen, hydrogen, helium, Aragon, neon, krypton, and
even heavier gases such as carbon dioxide can be treated as ideal
gases with negligible error (often less than 1 %).
Dense gases such as water vapor in steam power plants and
refrigerant vapor in refrigerators, however, should not be treated as
ideal gases.
Instead, the property tables should be used for these substances.
5
Is steam an Ideal Gas? It depends !!!!!
Figure on the
right shows the
percentage of
error involved in
assuming steam
to be an ideal
gas, and the
region where
steam can be
treated as an
ideal gas.
6
Is Water vapor an ideal gas?
At pressure below 10 KPa, water
vapor can be treated as an ideal
gas, regardless of its
temperature, with negligible
error (less than 0.1%).
At higher pressures, however, the
ideal-gas assumption yields
unacceptable errors.
In air-conditioning applications ,
where the pressure of the water
vapor is very low (ideal gas
relations can be used)
In steam power plant applications,
they should not be used
7
Compressibility Factor Z
Real gases deviate from ideal-gas behavior significantly at
state near the saturation region and the critical point.
This deviation at a given temperature and pressure can
accurately accounted for by introducing a correction factor
called the compressibility factor
Pv  RT
Pv
1
RT
Pv
Z
RT
v actual
Z
v ideal
For real gases Z can be grater than
or less than unity.
Real Gas
Z>1
Ideal Gas
Z=1
Z<1
8
Compressibility Factor Z for Nitrogen
9
Compressibility Factor Z for H2
Similar charts have been
prepared for different gases
and they are found to be
qualitatively similar.
Can quantitative
similarity be achieved?
10
Generalized Compressibility chart
This can be done if the coordinates are modified such that PR =
P/Pcr and TR = T/Tcr where PR and TR are the reduced pressure and
reduced temperature.
The Z factor is the
pressure and
Z
the same reduced
P
RT
same for all gases at
temperature.
This is called the
principle of
corresponding states.
Re duced pressure, PR
11
Observations
A- The deviation of a gas
from unity is greatest in
the vicinity of the critical
point.
B- It can be noticed that at
many states the
compressibility factor Z is
approximately unity :
1.
At very low pressure
(PR<<1),
2.
At high temperatures
(TR>2) except when
PR >>1.
12
Pseudo reduced specific volume
3- PR and vR
Z
2- TR and vR
P
RT
When P and v or T and v are
given, generalized chart
can still be used
vR is pseudo reduced specific
volume defined as:
vactual
vR 
RTcr / Pcr
Pseudoreduced
volume lines
Re duced pressure, PR
Lines of constant vR are available on the
compressibility chart (Figure A-30 pp. 867-869)
13
Example 2-11

Determine the specific volume of
refiregerant-134 at 1 MPa and 50 C,
using:
Thermodynamic tables
B. The ideal gas low
C. The generalized compressibility chart.
A.

Find the percentage error in the values
obtained in B and C compared to the
value obtained in part A.
14
Equation of state
Equation of State predicts the P-v-T behavior of a gas quite accurately
Pv  RT
15
Specific Heats
16
Importance
We know it takes more energy to warm up
some materials than others
•For example, it takes
about 10 times as much
energy to warm up 1 kg of
water, as it does to warm
up the same mass of iron.
• it is desirable to have a
property that will enable
us to compare the energy
storage capabilities for
different substances.
17
Specific Heat (heat capacity)
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Definition: It is the energy required to raise the
temperature of a unit mass by one degree
Units
 kJ/(kg 0C)
or kJ/(kg K)
 cal/(g 0C) or cal/(g K)
 Btu/(lbm 0F) or Btu/(lbm R)

Simple mathematical definition:
E  mCT
18
Constant volume and Constant
pressure specific heats, Cv and Cp
Cv
can be viewed as energy
required to raise the
temperature of a unit mass by
one degree as the volume is
maintained constant.
Cp
can be viewed as energy
required to raise the
temperature of a unit mass by
one degree as the pressure is
maintained constant.
Cp > Cv
19
Mathematical form of Cv
Consider constant volume system. Heat
it from T1 to T2.
E=U+KE +PE
E  U
dE  dU But dE mCvdT
du  CvdT
 u 
Cv   
 T  v
20
Mathematical form of Cp
We’ll worry about the math later, but…
 h 
Cp  

 T  p
h includes the internal energy and the work required to
expand the system boundaries
h  u  Pv
21
Observations
Cp is always bigger than Cv. This is
because it takes more energy to warm up
a constant pressure system due to the
system boundaries expansion.
That is you need to provide the energy to
increase the internal energy
do the work required to move the system
boundary
22
Observations (continued)
 Both are expressed in terms of u or h, and T, which
are properties and thus Cv and Cp are also
properties.
 Because they are properties, they are independent of
the type of process!!
 u 
Cv   
 T  v
 h 
Cp  

 T  p
23
Cv dependence on T for an ideal gas.
Recall that :
 u 
Cv   
 T  v
Joule found experimentally that
the internal energy of an ideal gas
is a function of temperature only
u  u (T )
Hence, Cv is at most function of T for an
ideal gas.
The partial derivative becomes ordinary
derivative for an ideal gas.
.
du
Cv 
dT
24
Cp dependence on T for an ideal gas.
Recall that
 h 
Cp  

 T  p
h  u  Pv
h  u( T )  RT  h( T )
Hence, Cp is at most function of T for
an ideal gas.
The partial derivative becomes
ordinary derivative for an ideal gas.
dh
Cp 
dT
25
Internal energy and Enthalpy as
functions of Cv and Cp for an ideal gas.
du  Cv (T )dT
u  u2  u1   Cv T dT
2
1
dh  C p (T ) dT
h  h2  h1   C p T dT
T2
T1
26
To carry out these integrations, we need
Cv and Cp as functions of T.
 Analytical expressions are available in
Table A-2c.
 In this table, Cp is given as
Cp = a + bT + cT2 + dT3
 The constants a, b, c, and d are
tabulated for various gases.
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27
Method 1
2
2
h   C p dT   (a  bT  cT  dT )dT
1
2
3
1

b T T
h  aT 
2
2
2
2
1
  cT
3
2
T
3
3
1
  d T
T
4
4
2
4
1
This is inconvenient!! Only do it if you really need to be very
accurate!!
Isn’t there an easier way?
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
Method 2
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These integrations of Eqs. 2-34
and 2-35 were tabulated in Table
A-17 page 849.
u  uo  
T
T 0 0
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
C v dT
h  ho  
T
T0  0
C p dT
T0=0 K was chosen to be an arbitrary reference.
This choice has no effect on u and h
The u and h data are given in KJ/kg for air.
Other gases in KJ/Kmol.
29
Method 3
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The variation of specific
heat for gases with complex
molecules are higher and
increase with temperature.
The variation of specific
heats is smooth and can be
approximated as linear over
small temperature interval
(a few hundred degrees or
less)
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Method 3
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Assume Cp and Cv is
constant over a short
temperature range (a few
hundred degrees or less).
The constant specific
heats are evaluated at the
average temperature
(T1+T2)/2.
u2-u1=Cv,av(T2-T1)
h2-h1=Cp,av(T2-T1)
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Three Ways to Calculate u
32
Example

Air at 300 K and 200 kpa is heated at
constant pressure to 600 K. Determine
the change in the internal energy of air
per unit mass using:

Data from air tables (Table A-17)
 The functional form of the specific heat
(Table A-2c)
 The average specific heat value (Table A2b).
33
Important Observation
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The previous relations are not
restricted to any kind of process.
The presence of constant volume
specific heat in an equation should
not lead to the concept that this
equation is valid only for constant
volume process.
The constant volume or constant
pressure part of the name defines
only how they are measured for each
substance (see figure).
Once we have Cv or Cp as function
of T, we can perform the integration
for any process.
34
Specific heat relations of an Ideal Gas
Cp is modeled in the Appendix A-2(c) as a
function of temperature – so you could calculate
dh, but what if you want to calculate du?
You’d need Cv. There is no corresponding Cv
table !!
Recall that
h=u+RT
dh = du + RdT
CpdT=CvdT+ RdT
Cp=Cv + R
35
Specific Heat Ratio
Cp
k
Cv
k also varies with temperature, but this
variation is very mild.
k = 1.4 for diatomic gases (like air)
k = 1.667 for noble gases
36
Specific heats of Solids and
Liquids
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The specific volume of
incompressible
substances remain
constant during a process.

The Cv and Cp values of
incompressible
substances are identical
and are denoted by C.
Cp = C v = C
37
Internal energy of Solids and Liquids
du  CV dT  CdT
u  CT  C(T2  T1 )
38
Enthalpy of Solids
h  u  Pv
dh  du  Pdv  vdP
But dv is 0 if the system
is incompressible
0
h  u  vP  CT  vP
h  u  Cavg T
Small
for
solids
39
Enthalpy of Liquids
h  u  vP  CT  vP
We have two cases:
Constant pressure process,
P  0
h  u  Cavg T
Constant temperature process,
h  vP
T  0
40