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Chapter 5
WORKING WITH
IDEAL GAS
In thinking about the world we are fixed with the same kind of
problems as a cartographer who tries to cover the curved face of
the earth with a sequence of plane maps. We can only expect an
approximate representation of reality from such a procedure and all
rational knowledge is therefore necessarily limited.
−Fritjof Capra (The Tao of Physics)
In this chapter, we will learn to apply the first law of thermodynamics
to closed, simple compressible systems containing ideal gas, which is indeed
an imaginary gas. However, the behaviour of real gases at low pressures
are often approximated to the behaviour of an ideal gas, the properties of
which are related to each other in a very special way. We will also learn
about these relationships in this chapter.
36 Chapter 5
5.1 Definition of an Ideal Gas
The volume occupied by the molecules of an ideal gas is assumed to be
negligibly small compared to the total gas volume. Also, molecules of an
ideal gas are assumed to have no influence on each other. Hence, an ideal
gas satisfies the following two conditions:
•An ideal gas at an equilibrium state can be described by the ideal gas
equation of state,
PV =nRT (5.1)
where Pis the absolute pressure,Vis the volume, nis the
amount of gas present, Ris the universal gas constant,andTis
the absolute temperature. Absolute pressure is the actual pres-
sure at a point. When we measure the pressure at a point using
most pressure-measuring devices which read zero in the local atmo-
sphere, what we get is the gauge pressure. Absolute pressure is
then obtained by adding the local atmospheric pressure to the gauge
pressure. For a description on absolute temperature, see Section 5.2.
For information on the universal gas constant R, see Section 5.3.
•The specific internal energy of an ideal gas is a function of tempera-
ture only. That is to say
u=u(T)for an ideal gas. (5.2)
In other words, the internal energy of an ideal gas experiences abso-
lutely no change when the volume occupied by an ideal gas or the
pressureexertedonanidealgasisvariedsoastochangethein-
termolecular distances, while maintaining the temperature constant.
It should not be surprising, because we know that intermolecular
phenomena are totally absent in an ideal gas, and thus altering the
distances among the molecules of an ideal gas shall have absolutely
no influence on the internal energy content of an ideal gas. How-
ever, changing the temperature of an ideal gas alters the behaviour
of the individual gas molecules, and therefore the internal energy is
changed.
The behaviour of real gases at low pressures can be approximated by
(5.1) and (5.2). It is therefore, a real gas at low pressures may be considered
an ideal gas.
Working with Ideal Gas 37
5.2 Absolute Temperature Scale
The scales of temperature in the SI units are the Celsius scale (◦C),
which is also known as centigrade scale, and the Kelvin scale (K). Temper-
atures in these scales are related to each other by
TK=TC+ 273.15
where TKis the temperature in the Kelvin scale and TCis its equivalent
in the Celsius scale.
Kelvin scale is an absolute temperature scale, which implies the
lowest attainable temperature in the scale is zero. There are no negative
temperatures on an absolute temperature scale. The concept of absolute
temperature is one of the most confusing point in the subject of thermody-
namics for a beginner of this subject. The concept of absolute temperature
will become clear when one goes on to learn the second law of thermo-
dynamics. For now, let us accept the concept of an absolute temperature
scale without further elaboration and proceed.
5.3 Different Forms of
the Ideal Gas Equation of State
•When using the ideal gas equation of state in the form given by
(5.1), we take Pin kPa, Vin m3,andTin K. The amount of gas
present, denoted by n, is taken in kmol. And, Rtakes the value 8.314
kJ/kmol ·K for any gas, and therefore it is known as the universal
gas constant.
•The ideal gas equation of state is also used in the form
PV =mRT (5.3)
where Pis in kPa, Vis in m3,andTis in K. The mass of gas present,
denoted by m, is taken in kg. Therefore Rtakes the unit kJ/kg ·K.
When Ris in kJ/kg ·K, it is known as the specific gas constant.
38 Chapter 5
•We could also write the ideal gas equation of state in the form
Pv =RT (5.4)
where Pis in kPa, and Tis in K. If v=V/n is the molar volume
in m3/kmol, then Ris the universal gas constant, taking the value
8.314 kJ/kmol ·K for all gases. If v=V/m is the specific volume in
m3/kg, then Ris the specific gas constant in kJ/kg ·K.
Combining (5.1) and (5.3), we relate the universal gas constant and the
specific gas constant by
n×universal gas constant =m×specific gas constant
which gives
specific gas constant =universal gas constant
M
where M=m/n is known as the molar mass, and it takes the unit
kg/kmol. The universal gas constant takes the value 8.314 kJ/kmol ·Kfor
all gases, and the value of molar mass is specific to the gas. Therefore, the
numerical value of the specific gas constant, which depends on the molar
mass of the gas, is specific to the gas and its value changes from one gas
to the other. Please note it is common to use the same symbol Rfor both
the universal gas constant and the specific gas constant, since the units
will determine which one is in use.
Keep in mind that the numerical value of molar mass is the same as that
of the relative molecular mass, whereas molar mass takes the unit g/mol
or kg/kmol and the relative molecular mass is dimensionless.
5.4 Internal Energy and Cv
The change in the specific internal energy of any substance with
respect to temperature while the volume is maintained constant can be
Working with Ideal Gas 39
expressed as
Cv= lim
δT→0
u(v, T +δT )−u(v, T )
δT
= lim
δT→0
u(T+δT)−u(T)
δT at constant volume
=du
dT v
(5.5)
which is the definition of Cvknown as the specific heat at constant
volume. It is one of the most extensively used specific heats in thermo-
dynamics.
Since uof an ideal gas is a function of Tonly, uis independent of its
volume. Therefore, the definition of Cvgiven by (5.5) simplifies to
Cv= lim
δT→0
u(T+δT)−u(T)
δT =du
dT for an ideal gas. (5.6)
A visual representation of the specific heat at constant volume is given
in Figure 5.1, which shows a typical plot of uversus Tfor an ideal gas.
We know that the slope of the tangent to the uversus Tcurve at point A,
shown by the dashed line in Figure 5.1, represents an infinitesimal change
in uwith Tat point A. This slope quantifies the specific heat at constant
volume at the temperature represented by point A. Since the slope may
vary from one temperature to another, it is obvious that Cvmay also vary
with temperature. Since uis a function of Talone for an ideal gas, Cvof
an ideal gas is a function only of T.
p
A
u
T
H
HY uversus Tcurve
tangent at A
Z
Z}
r
Figure 5.1 An example of uversus Tcurve.
40 Chapter 5
The unit for Cvused in this textbook is kJ/kg ·Kifuin (5.6) is taken
as the specific internal energy. The unit for Cvused is kJ/kmol ·Kifu
in (5.6) is taken as the molar internal energy. Since a unit change in the
Kelvin scale (K) is the same as a unit change in the Celsius scale (◦C),
the numerical value of Cvin kJ/kg ·K is identical to that in kJ/kg ·◦C.
Similarly, the numerical value of Cvin kJ/kmol ·K is the same as that in
kJ/kmol ·◦C.
Equation (5.6) for an ideal gas can be rearranged to give
du =CvdT (5.7)
which is also applicable for any substance undergoing a constant-volume
process, see (5.5). Integrating (5.7) between the initial and final equilibrium
states of a process, we get
uf
uo
du =Tf
To
CvdT
where uoand ufare the respective specific (or molar) internal energies at
the initial and the final equilibrium states of the process, and Toand Tfare
the respective temperatures at the initial and the final equilibrium states of
the process.
Since uis a property, the above reduces to
Δu=uf−uo=Tf
To
CvdT (5.8)
which can be used to evaluate Δufor an ideal gas undergoing any process,
constant-volume or not, or for any substance undergoing a constant-volume
process.
If Cvof (5.8) is in kJ/kg ·K, then
ΔU=mΔu=mTf
To
CvdT
where mis the mass of the substance in kg. If Cvof (5.8) is in kJ/kmol ·K,
then
ΔU=nΔu=nTf
To
CvdT
where nis the amount of the substance in kmol.
Working with Ideal Gas 41
5.5 Enthalpy and Cp
The change in the specific enthalpy of any substance with respect to
temperature while the pressure is maintained constant is expressed as
Cp= lim
δT→0
h(P, T +δT )−h(P, T )
δT
= lim
δT→0
h(T+δT)−h(T)
δT at constant pressure
=dh
dT P
(5.9)
which is the definition of Cpknown as the specific heat at constant
pressure. It is another extensively used specific heat in thermodynamics.
The specific enthalpy hgiven by (4.2) becomes
h=u+RT for an ideal gas, (5.10)
when the Pv term of (4.2) is eliminated using (5.4). Since uat an equi-
librium state of an ideal gas is a function of temperature alone and Ris a
constant, it is obvious from (5.10) that hof an ideal gas is also a function
of temperature alone. That is to say
h=h(T)for an ideal gas.
Since hof an ideal gas is a function of Tonly, it is independent of its
pressure. Therefore, Cpof (5.9) simplifies to
Cp= lim
δT→0
h(T+δT)−h(T)
δT =dh
dT for an ideal gas. (5.11)
Like Cv,Cpmay also vary with temperature. The variation of Cpwith
temperature would, of course, be different for different substances. Since
his a function of Talone for an ideal gas, Cpof an ideal gas is a function
of Tonly. The units of Cpis the same as that of Cv(discussed in Section
5.4.)
Equation (5.11) for an ideal gas can be rearranged to give
dh =CpdT (5.12)
42 Chapter 5
which is also applicable for any substance undergoing a constant-pressure
process, see (5.9). Integrating (5.12) between the initial and final equilib-
rium states of a process, we get
hf
ho
dh =Tf
To
CpdT
where hoand hfare the respective specific (or molar) enthalpies at the
initial and the final equilibrium states of the process, and Toand Tfare the
respective temperatures at the initial and the final equilibrium states of the
process.
Since his a property, the above reduces to
Δh=hf−ho=Tf
To
CpdT (5.13)
which can be used to evaluate Δhfor an ideal gas undergoing any pro-
cess, constant-pressure or not, or for any substance undergoing a constant-
pressure process.
If Cpof (5.13) is in kJ/kg ·K, then
ΔH=mΔh=mTf
To
CpdT
where mis the mass of the substance in kg. If Cpof (5.13) is in
kJ/kmol ·K, then
ΔH=nΔh=nTf
To
CpdT
where nis the amount of the substance in kmol.
5.6 Relating Ideal Gas Specific Heats
The specific heats of an ideal gas are related to each other in a very
simple manner. Let us now see how to obtain this relationship. Take the
expression for the specific enthalpy of an ideal gas, h=u+RT , given by
Working with Ideal Gas 43
(5.10), and differentiate it with respect to temperature T.Sincebothh
and uare functions of temperature alone for an ideal gas, we get
dh
dT =du
dT +Rfor an ideal gas.
Substituting Cv=du/dT of (5.6) and Cp=dh/dT of (5.11) in the
above expression, we get
Cp=Cv+Rfor an ideal gas. (5.14)
If the specific heats are in kJ/kmol ·KthenRis the universal gas con-
stant. If the specific heats are in kJ/kg ·KthenRis the specific gas
constant.
The ratio between the specific heats is known as the specific heat
ratio γ, and is defined as
γ≡Cp
Cv
(5.15)
The specific heat ratio is also known as the isentropic exponent,
and is sometimes denoted by k.
Combining (5.14) and (5.15) so as to eliminate Cp,weget
Cv=R
γ−1for an ideal gas. (5.16)
Combining (5.15) and (5.16) so as to eliminate Cv,weget
Cp=γR
γ−1for an ideal gas. (5.17)
5.7 Data on Ideal Gas Specific Heats
Ideal gas specific heats are sensitive to changes in temperature. How-
ever, the effect of temperature is negligibly small on the specific heats of
monoatomic gases. Therefore, for monoatomic gases, such as argon, he-
lium, etc., Cv= 12.5 kJ/kmol ·KandCp= 20.8 kJ/kmol ·Kcouldbe
used independent of temperature. Specific heats change rather slowly with
44 Chapter 5
temperature for diatomic gases, such as O2,H2,andN2. For polyatomic
gases, such as CO2,CH4,C2H6, etc, specific heats vary significantly with
temperature.
Data obtained for Cpof a gas at different temperatures while maintain-
ing low pressures, can be fitted using algebraic equations such as the one
given below as a function of temperature:
Cp=a+bT +cT2+dT3
where the constants a,b,cand dtake different values for different gases
as shown in Table 5.1. Algebraic expressions for Cpare available in the
literature∗for a large number of gases that are assumed to behave as ideal
gas. Substituting the algebraic equation describing Cpin (5.14), we can
get the algebraic equation for Cvas a function of T.
Gas ab×102c×105d×109
Hydrogen 29.11 -0.1916 0.4003 -0.8704
Oxygen 25.48 1.520 -0.7155 1.312
Air 28.11 0.1967 0.4802 - 1.966
Carbon dioxide 22.26 5.981 -3.501 7.469
Water vapour 32.24 0.1923 1.055 -3.595
Table 5.1 Values of the constants in Cp=a+bT +cT2+dT3for a few
selected gases valid in the temperature range 273 K to 1800 K, where Cpis in
kJ/kmol ·KandTis in K.
Table 5.2 shows the values of ideal gas specific heats of few selected
gases at 300 K. Specific heats are given both in kJ/kg ·K and in kJ/kmol ·K.
Multiplying the specific heat given in kJ/kg ·K by the molar mass, we get
the specific heat given in kJ/kmol ·K. The last column of Table 5.2 shows
the values of the specific heat ratio γ.
This table also shows that the specific heats of monoatomic gases are
almost the same, when taken in the unit of kJ/kmol ·K, and the specific
heats of diatomic gases are also nearly the same. It is common to assume
that air also falls into the group of diatomic gases. For polyatomic gases,
∗See, for example, Table A-2 of ¸Cengel, Y.A. & Boles, M.A. 1998 Thermody-
namics: an engineering approach, 3rd Edition, McGraw-Hill International Editions.
Working with Ideal Gas 45
specific heats differ from gas to gas. Note that specific heats of ideal gases
are also known as zero-pressure specific heats, and are sometimes
denoted by Cvo and Cpo, since all real gases approach ideal-gas behaviour
at low (or near zero) pressures.
GAS CpCvCpCvγ
molar mass (in kJ/kg ·K) (in kJ/kmol ·K)
(in kg/kgmol)
Argon
39.950 0.5203 0.3122 20.786 12.472 1.666
Helium
4.003 5.1930 3.1159 20.788 12.473 1.666
Hydrogen
2.016 14.3230 10.1987 28.875 20.561 1.404
Nitrogen
28.013 1.0400 0.7432 29.134 20.819 1.399
Air
28.97 1.005 0.718 29.105 20.793 1.400
Carbon monoxide
28.010 1.0410 0.7442 29.158 20.845 1.399
Carbon dioxide
44.010 0.8457 0.6568 37.219 28.906 1.288
Ethane
30.070 1.7668 1.4903 53.128 44.813 1.186
Table 5.2 Ideal gas specific heats at 300 K for a few selected gases.
It should be borne in mind that for real gases, the specific heats depend
not only on the temperature but also on the pressure or volume, and for in-
compressible substances, such as liquids and solids, both constant-pressure
and constant-volume specific heats are approximately the same, and thus
Cp≈Cv.
46 Chapter 5
5.8 Evaluation of ΔUfor an Ideal Gas
In this section we will learn to evaluate ΔU, the internal energy differ-
ence, of an ideal gas in three different ways.
5.8.1 Using an Expression for Cv
Suppose you are asked to evaluate the specific internal energy change
of nitrogen when its temperature is increased from 300 K to 600 K at low
pressures given the expression
Cp/R =a+bT +cT 2+dT 3+eT 4
where a= 3.675, b=−1.208×10−3,c= 2.324×10−6,d=−0.632×10−9,
e=−0.226×10−12,Cptakes the unit of R,andTis in K.
Since nitrogen is held at low pressure, it is assumed to behave as an
ideal gas. Substituting the algebraic equation describing Cpin (5.14), we
get
Cv=(a+bT +cT2+dT3+eT4)R−R
=(a−1) R+bRT +cRT2+dRT3+eRT4
Substituting it in (5.8) and integrating the resulting expression from 300
K to 600 K, we get
Δu=(a−1) R(600 −300)
+bR 6002−3002
2+cR 6003−3003
3
+dR 6004−3004
4+eR 6005−3005
5
Using the given numerical values of a,b,c,dand e,andR= 8.314
kJ/kmol ·K in the above, we get Δu= 6345.5kJ/kmol.
The accuracy of the result obtained depends, of course, on the ac-
curacy of the algebraic equation used. This method is, however, a very
inconvenient method for hand calculations.
Working with Ideal Gas 47
5.8.2 Using the Ideal-Gas Property Table
Alternatively, we can use the ideal-gas property tables, similar to
Table 5.3, to evaluate Δuof nitrogen as
Δu=uat 600 K −uat 300 K
= 12574 kJ/kmol −6229 kJ/kmol = 6345 kJ/kmol
In the above calculations, Δuis evaluated using the uvalues obtained
from ideal-gas property tables that are compiled by integrating (5.7) with
the help of computers using very accurate algebraic equations describing
Cvas a function of temperature. This method therefore gives accurate
values of Δu. Also, it is very easy to use.
T h u
0 0 0
220 6,391 4,562
230 6,683 4,770
240 6,975 4,979
250 7,266 5,188
300 8,723 6,229
- - -
- - -
600 17,563 12,574
- - -
- - -
Table 5.3 Ideal gas enthalpy and internal energy for N2,whereTis in K,
and hand uare in kJ/kmol ·K.
An important aspect of the ideal gas property tables, such as Table 5.3
and others†, is that the entries do not include pressure as an entry. It is
†See, for example, the following tables:
Table A-17 to Table A-25 of ¸Cengel, Y.A. & Boles, M.A. 1998 Thermodynamics:
an engineering approach, 3rd Edition, McGraw-Hill International Editions
Table A-5M to Table A-11M of Wark, K. 1989 Thermodynamics, 5th Edition,
McGraw-Hill International Editions.
48 Chapter 5
because data on ideal gas properties of gases are meaningful only at low
pressures since ideal gas behaviour is imitated by real gases only at low
pressures.
Another aspect of the individual entries of internal energy and enthalpy
in the ideal gas property tables is that individual entries of uand hdepend
on the respective reference states chosen as described below. Let us set
the reference temperature as Tref , and the corresponding uand hvalues
are, say, uref and href , respectively. Integrating (5.7) and (5.12) between
Tref and T,weget
u=uref +T
Tref
CvdT
h=href +T
Tref
CpdT
The reference state chosen may differ from one thermodynamic property
table to the other. For example, the reference state chosen in Table 5.3 is
0K,andthehand uvalues at this reference state are set to zero. The
individual entries of uand hat a chosen Tmay differ from one table to
another, since the individual entry depends on the reference state. However,
the change in internal energy or enthalpy calculated using individual entries
from a particular table is independent of the reference state, as shown below
for the case of Δuas
Δu=u2−u1
=uref +T2
Tref
CvdT −uref +T1
Tref
CvdT
=T2
T1
CvdT
5.8.3 Using an Average Value for Cv
Yet another, but an approximate method for the evaluation of Δuis to
take Cvas a constant about an average value, (Cv)av , over the temperature
range of our interest, which reduces (5.8) to the following approximate
Working with Ideal Gas 49
relationship:
Δu≈(Cv)av (Tf−To)(5.18)
Similarly, if Cpis assumed to be a constant about an average value,
(Cp)av, over the temperature range of our interest, then (5.13) reduces to
the following approximate relationship:
Δh≈(Cp)av (Tf−To)(5.19)
Suppose (Cv)at 300 K and (Cv)at 600 K are given as 0.743 kJ/kg ·Kand
0.778 kJ/kg ·K, respectively. Since these values are somewhat close, we
can evaluate the algebraic mean of the Cvvalues given at 300 K and 600
K, which gives (Cv)av = 0.761 kJ/kg ·K. Using this value in (5.18), we
get
Δu=0.761 ×(600 −300) kJ/kg
= (228.3kJ/kg)×(28.013 kg/kmol)
= 6395 kJ/kmol
Since Δuis evaluated using the average value for Cv, the result is
only an approximate result. Yet, this method is widely used owing to its
simplicity, particularly when hand calculations are carried out. However,
the above method may give inaccurate results if the temperature interval
considered is very large.
5.9 Worked Examples
The following tools are useful in working out thermodynamic problems
on closed simple compressible systems containing ideal gas:
Tool 1: The first law of thermodynamics applied to closed simple com-
pressible systems, given by Qin +Win =ΔU, which is also the
principle of conservation of energy.
Tool 2: Evaluation of ΔUusing data on specific heats, or data on internal
energy itself.
Tool 3: The ideal gas equation of state that interrelates pressure, tem-
perature, volume and mass (or amount of substance) of an ideal gas.
50 Chapter 5
Example 5.1
A closed rigid container, shown in Figure 5.2,
with negligible heat capacity has a volume of 1 m3. It holds air at 300 kPa
and 303 K. Heat is supplied until the tem-
perature of air reaches 500 K. Take Cvfor
air as 0.718 kJ/kg ·K. The molar mass of
air is 29 kg/kmol. Determine the amount
of heat supplied to air assuming air behaves
as an ideal gas.
Figure 5.2
air
container
Solution to Example 5.1
The amount of heat supplied to the air, denoted by Qin, should be found.
Air in the closed container is taken as a closed system. The first law applied to
the closed system, that is air, gives Qin +Win =ΔU. The container is rigid
and therefore the shape and size of its boundary cannot be changed. Thus, no
work is done at the boundary of the system in changing its shape or size and
thereby compressing or expanding the air contained in the container. There is
also no other forms of work involved, and therefore Win = 0. Thus, the first
law gives
Qin =ΔU(5.20)
If we can evaluate ΔUby some means, then Qin will be known from (5.20).
Since air is assumed to behave as an ideal gas, (5.8) is used to evaluate Δuas
Δu=500 K
303 K
(0.718 kJ/kg ·K)dT
=(0.718 kJ/kg ·K)×(500 K−303 K) = 141.5kJ/kg
Therefore,
ΔU= 141.5×mkJ (5.21)
where mis the mass of air in the closed tank in kg, and the value of mis not
known.
Thereisonlyonewaytocalculatem, which is to use the ideal gas equation of
state in the form PV =mRT , in which Ris the specific gas constant calculated
as
R=8.314 kJ/kmol ·K
29 kg/kmol =0.287 kJ/kg ·K
Working with Ideal Gas 51
We know the values of P,Vand Tat the initial state of the system, and
thereforeweworkoutthevalueofmas
m=PV
RT=(300 kPa)(1m3)
(0.287 kJ/kg ·K) (303 K)=3.45 kg (5.22)
Substituting mof (5.22) in (5.21), we get
ΔU= 141.5×3.45 = 488 kJ (5.23)
Combining (5.20) and (5.23) so as to eliminate ΔU,wegetQin = 488 kJ
which is the amount of heat supplied to the air to increase its temperature from
303 K to 500 K.
Example 5.2
If the closed rigid container of Example 5.1
weighs 10 kg and is made of steel having the specific heat 0.5 kJ/kg ·◦C,
determine the amount of heat supplied to increase the temperature of air
to 500 K.
Solution to Example 5.2
The closed rigid container of Example 5.1 is made of a material with negli-
gible heat capacity, and therefore the container absorbs a negligible amount of
heat as we heat the air to reach 500 K. Thus, we ignored the amount of heat
supplied to the container, and evaluated only the amount of heat supplied to
the air to increase its temperature to 500 K.
In this example, the container is said to be made of steel. So, as we heat
the air to 500 K, the container also is heated to 500 K absorbing part of the
heat supplied. We, therefore, have to take the closed system to include both the
air and the container. Let us denote the amount of heat supplied to the closed
system as Qin, and we need to evaluate the value of Qin.
Since there are no work modes involved in the process, Win = 0. Thus, the
first law applied to the closed system gives
Qin =ΔU(5.24)
52 Chapter 5
where ΔUis made up of the following components:
ΔU=ΔUair +ΔUcontainer (5.25)
Since air in this example undergoes the same property changes as the air in
Example 5.1, ΔUair is evaluated following the same steps as in Example 5.1.
Therefore, we get
ΔUair = 488 kJ
ΔUcontainer is evaluated as
ΔUcontainer =mass ×specific heat ×temperature increase
=(10kg)×(0.5kJ/kg ·◦C)×(500 K−303 K)
= 985 kJ
Substituting the above in (5.25), we get
ΔU= 488 kJ + 985 kJ = 1473 kJ
which is used in (5.24) to obtain
Qin = 1473 kJ
which is the amount of heat supplied to increase the temperature of the air
from 303 K to 500 K when the air is contained in a steel container of 10 kg
mass and 0.5 kJ/kg ·◦C specific heat.
Let us closely examine the equation
ΔUcontainer =mass ×specific heat ×temperature increase
which was used to determine the internal energy increase of the container
in Example 5.2. The term specific heat used here can be either Cvor
Cp,sinceCp≈Cvfor incompressible substances, such as liquids and solids.
Since specific heats are sometimes referred to as the heat capacities, the
statement “container with negligible heat capacity” in Example 5.1 sim-
ply means the specific heat of the material of construction of the container
is negligibly small. So ΔUcontainer is negligibly small for a container with
negligible heat capacity.
Working with Ideal Gas 53
Example 5.3
Helium gas is contained in a rigid vessel of 0.5
m3at 500 kPa. The gas is agitated violently by a stirrer that transfers 250
kJ of work to the gas. Assuming that the system undergoes an adiabatic
process, determine the final pressure of the gas. Assume that Cvfor helium
is a constant at 12.46 kJ/kmol ·K.
Solution to Example 5.3
The final pressure of helium, denoted by Pf, could be calculated only in
one way, that is, by using the ideal gas equation of state at the final state as
PfVf=nRT
f. Since the vessel is rigid, the volume of the vessel remains a
constant. Therefore, Vf=0.5m
3. Substituting the known values in the above
equation, we get
Pf×0.5=8.314 nT
f(5.26)
where Pfis in kPa, nis in kmol and Tfis in K. Calculation of Pffrom (5.26)
is possible only if we know the values of nand Tf.
Let us calculate the value of nusing the ideal gas equation of state at the
initial state as PoVo=nRT
o, in which Po= 500 kPa, Vo=0.5m
3,andTois
not known. Therefore, we get
500 ×0.5=8.314 nT
o(5.27)
where nis in kmol as told already, and Tois in K.
There are two independent equations (5.26) and (5.27), and there are four
unknowns which are Pf,To,Tfand n. We require two more independent
equations to solve for the unknowns.
We have been told at the beginning of this section that we can use three
tools. And, we have so far used only one which is the ideal gas equation of state.
Let us examine the problem statement to see if the first law of thermodynamics
canbeusedinthisexample.
It is said that the system undergoes an adiabatic process, and therefore Qin
= 0. It is also said that the system has rigid boundary and that 250 kJ of stirring
work is done on the system. Therefore, Win = 250 kJ. The first law applied to
closed system, Qin +Win =ΔU, thus gives
ΔU= 250 kJ (5.28)
Now, there are three independent equations (5.26), (5.27) and (5.28), and
five unknowns Pf,To,Tf,nand ΔU. We still require two more independent
equations to solve for the unknowns.
54 Chapter 5
Out of the three tools, we have so far used two which are the ideal gas
equation of state and the first law of thermodynamics. The remaining tool is
the evaluation of ΔUusing data on specific heats, or data on internal energy
itself.
Since the value of Cvis given in the problem statement and since helium
behaves as an ideal gas, we evaluate ΔUusing (5.8) as
ΔU=(nkmol)TfK
ToK
(12.46 kJ/kmol ·K)dT
=12.46 n(Tf−To)kJ (5.29)
Eliminating ΔUfrom (5.28) and (5.29), we get
12.46 n(Tf−To) = 250 (5.30)
Now, there are three independent equations (5.26), (5.27) and (5.30), and
four unknowns Pf,To,Tfand n. We require one more independent equation
to solve for the unknowns.
We have used all the three tools available, and there are no other means
to get that missing equation from the data available in the problem statement.
Therefore, we have to work with what is available to determine the value of Pf.
There is a way to do so. Let us eliminate Tfand Tofrom (5.30) using (5.26)
and (5.27), respectively, as below.
12.46 n0.5Pf
8.314 n−500 ×0.5
8.314 n= 250
which gives Pf= 833.6 kPa.
Example 5.4
An unidentified gas of mass 6 g is contained in
a piston-cylinder device of negligible heat capacity. The pressure of the gas
is 1.7 bar, its volume 0.002 m3, and its temperature 27◦C. The gas expands
receiving 445 J of heat. The temperature of the gas becomes 115◦C, its
pressure reduced, and its volume increased. The work done by the gas on
its surroundings is 100 J. Determine the values of Cv,Cp,andthemolar
mass of the gas, assuming ideal gas behaviour.
Working with Ideal Gas 55
Solution to Example 5.4
Since ideal gas behaviour is assumed, we use (5.8) to determine the value
of Cv. Taking Cvas a constant, we get from (5.8)
Δu=Cv(Tf−To)(5.31)
Since To=27
◦CandTf= 115◦C, we get (Tf−To)=88
◦C, which is
equivalent to 88 K since a unit change in ◦C is the same as a unit change is K.
Therefore, (5.31) gives
Cv=Δu/(88 K)(5.32)
Now we need to calculate the internal energy change. The gas receives 445
J of heat and do 100 J of work. Taking the gas as the closed system, the first
law of thermodynamics is applied as
ΔU=Qin +Win = 445 J+(−100 J) = 345 J
The mass of the gas is 6 g, and thus
Δu=ΔU/m = 345 J/6g=57.5J/g (5.33)
Eliminating Δufrom (5.32) and (5.33), we get
Cv=57.5J/g
88 K=0.653 J/g ·K=0.653 kJ/kg ·K(5.34)
It is possible to determine the value of Cpusing Cp=Cv+Rgiven by (5.14)
for an ideal gas, provided we know the values of Cvand R. We, of course, know
the value of Cvin kJ/kg ·K from (5.34), and therefore we need the value of R
in kJ/kg ·K.
Since the values of pressure, volume, mass and temperature at the initial
state of the gas is given in the problem, we find Rusing the ideal gas equation
of state given by (5.3) as
R=(1.7×100 kPa)(0.002 m3)
(0.006 kg) (273 + 27) K=0.189 kJ/kg ·K(5.35)
which is the specific gas constant.
Substituting the value of Cvfrom (5.34) and Rfrom (5.35) in (5.14), we
get the value of Cpas
Cp=Cv+R
=0.653 kJ/kg ·K+0.189 kJ/kg ·K=0.842 kJ/kg ·K
56 Chapter 5
The molar mass could be found as
M=universal gas constant
specific gas constant
=8.314 kJ/kmol ·K
0.189 kJ/kg ·K=43.99 kg
kmol ≈44 kg
kmol
Going through Table 5.2, we see that the numerical values of Cv,Cpand
Mevaluatedabovematchwiththoseofcarbondioxide.
Example 5.5
A thermally insulated rigid box of negligible
heat capacity contains two compartments of equal volume. Initially, one
compartment contains air at 5 bar and 25◦C, and the other is evacuated.
The dividing partition of negligible mass, is then raptured. Calculate the
final air temperature and pressure. It is a common assumption that air at
low or moderate pressures behaves like an ideal gas. The value Cvfor air
at room temperature is 718 J/kg K.
Solution to Example 5.5
As shown in Figure 5.3, compartment A contains air initially, and compart-
ment B contains nothing, that is vacuum. When the partition is raptured, air in
compartment A rushes into compartment B. Because air rushes to occupy the
entire box, the pressure and temperature of air occupying the entire box have
highly nonuniform distributions. If we allow a considerable length of time to
pass, then it is possible for air occupying the entire box to reach an equilibrium
state, having a final temperature Tfand a final pressure Pfuniformly distributed
everywhere within the box. We need to find the numerical values of Tfand Pf.
To be able to work out this problem, it is important to choose the system
with care. Let us choose the closed system to include air in compartment A,
compartment B, partition, and the rigid vessel. The boundary of the system
chosen is represented by the dashed line in Figure 5.3.
Working with Ideal Gas 57
initial state, final state,
AB
air at
PA=5 bar
TA=298 K
vacuum
Pf=?
Tf=?
(VA=V
B)
with partition without partition
Figure 5.3 The initial and final states of the system of Example 5.5.
Since no heat or no work crosses the boundary, the first law applied to the
chosen closed system yields
ΔU=0 (5.36)
which means there is no change in the total internal energy of the closed system.
The total internal energy change of the chosen closed system is made up of the
following components:
ΔU=ΔUair in A +ΔUvessel +ΔUpartition
where compartment B makes no contribution towards the total internal energy
change of the system since it contains nothing. The vessel has negligible heat
capacity and the partition has negligible mass, and therefore their contributions
to the total internal energy change of the system are negligible. Thus, we get
ΔU=ΔUair in A =0 (5.37)
which means the internal energy of air that was initially occupying compartment
A remains a constant.
Since the mass of this air remains the same, we conclude that the specific
internal energy of air that was occupying compartment A remains a constant,
which means
Δuair in A =0 (5.38)
Air is assumed to behave as an ideal gas, and Cvof air is taken a constant.
Therefore, with the help of (5.8), we expand (5.38) to give
Cv[Tf−TA]=0 (5.39)
58 Chapter 5
where TAis the temperature of air occupying compartment A at the initial state,
and Tfis the temperature of air occupying the entire box at the final state.
Equation (5.39) gives
Tf=TA= 298 K=25
◦C(5.40)
That is to say the temperature of air has not changed even though the volume
occupied by air has doubled. This should not come as a surprise since the specific
internal energy of air, taken as an ideal gas, is a function of temperature alone.
There was no change in the specific internal energy of air according to (5.38),
and therefore there is no change in its temperature, either.
Now, we have to find the final pressure Pf. Notethatwehavealreadyused
two of the three tools mentioned at the beginning of this section. We are left
with only one tool which is the ideal gas equation of state. Application of the
ideal gas equation of state to air at its initial and final states gives the following
respective equations:
PAVA=nRT
Aand (5.41)
Pf(VA+VB)=nRT
f(5.42)
where PA,VAand TAare the respective pressure, volume and temperature of
air at its initial state, and Pf,(VA+VB)and Tfare the respective pressure,
volume and temperature of air at its final state. The amount of air occupying
compartment A at the initial state is the same as that occupying the entire box
at the final state, and is denoted by n.
Eliminating the term (nR) from (5.41) and (5.42), we get
Pf(VA+VB)
Tf
=PAVA
TA
(5.43)
Compartments A and B are of equal volume, and therefore VA=VB.We
also know from (5.40) that Tf=TA. Using the above facts and the information
that the pressure of air in compartment A at the initial state, PA,is5barin
(5.43), we get
Pf=2.5bar
Example 5.6
A thermally insulated, rigid vessel with negligi-
ble heat capacity is divided into two compartments A and B by a non heat
Working with Ideal Gas 59
conducting partition of negligible mass such that VA=4VB.Eachcon-
tains nitrogen which behaves ideally. Compartment A is at 1 bar and 300
K, and compartment B is at 4 bar and 600 K. Determine the equilibrium
temperature and the equilibrium pressure reached after the removal of the
partition. The value Cvfor nitrogen may be assumed a constant at 0.743
kJ/kg ·K.
Solution to Example 5.6
This problem is similar to that of Example 5.5, except for the fact that
compartment B is not vacuum but has nitrogen in it. The system boundary
shown by the dashed line of Figure 5.4 clearly defines the system, which consists
of nitrogen in both compartments A and B, the partition and the rigid vessel.
Since neither heat nor work crosses the system boundary, the first law applied
to closed systems yields
ΔU=0 (5.44)
which means that there is no change in the total internal energy of the system.
initial state,
final state,
AB
1bar
300 K
4bar
600 K
final pressure Pf=?
final temperature Tf=?
(VA=4V
B)
with partition
without partition
Figure 5.4 The initial and final states of the system of Example 5.6.
60 Chapter 5
Ignoring the internal energy changes of the partition and the vessel, the total
internal energy change of the system is written as
ΔU=ΔUN2in A +ΔUN2in B =mAΔuN2in A +mBΔuN2in B
where mAand mBare the respective masses of nitrogen in compartments A
and B at the initial state.
Since nitrogen contained in both the compartments A and B behaves as
an ideal gas, and since nitrogen has constant Cv, we may expand the above
expression as
ΔU=mACv(Tf−TA)+mBCv(Tf−TB)(5.45)
where Tfis the temperature of nitrogen at the final state, TAis the temperature
of nitrogen in compartment A at the initial state, and TBis the temperature of
nitrogen in compartment B at the initial state.
Combining (5.44) and (5.45), we get
mACv(Tf−TA)+mBCv(Tf−TB)=0
Substituting the respective temperatures of nitrogen in compartments A and
B at the initial state in the above equation and eliminating Cv,weget
mA(Tf−300 K)+mB(Tf−600 K)=0 (5.46)
The respective masses of nitrogen in A and in B at the initial state are
evaluated using the ideal gas equation of the form PV =mRT as
mA=(1 bar)VA
R(300 K)and mB=(4 bar)VB
R(600 K)(5.47)
Substituting mAand mBof (5.47) in (5.46), we get
(1 bar)VA
R(300 K)(Tf−300 K)+(4 bar)VB
R(600 K)(Tf−600 K)=0 (5.48)
Since VA=4VB, (5.48) gives Tf= 400 K.
Now, we have to determine Pf, the pressure at the final state. Note that
we have already used all the three tools mentioned at the very beginning of
this section. And, therefore, it may seem impossible to find the value of Pf.
However, there is a fundamental law known as the principle of conservation of
mass, which we have not used.
Working with Ideal Gas 61
Application of the principle of conservation of mass to the given closed sys-
tem yields
total mass of nitrogen in the vessel at the final state
= total mass of nitrogen in the vessel at the initial state
which gives
mfinal =mA+mB(5.49)
where mfinal is the total mass of nitrogen in the vessel at the final state, and it
is related to Pfby the ideal gas equation of state according to
mfinal =Pf(VA+VB)
RT
f
(5.50)
Substituting mA,mBand mfinal from (5.47) and (5.50), respectively, in
(5.49), we get
Pf(VA+VB)
RT
f
=(1 bar)VA
R(300 K)+(4 bar)VB
R(600 K)(5.51)
Since VA=4VBand Tf= 400 K, (5.51) gives Pf=1.6bar.
Example 5.7
A thermally insulated rigid box of negligible
heat capacity is divided into two compartments having a volume of 0.5
m3each by a piston that is kept from moving by a pin. The piston does
not permit the gases to leak into each other. Initially, one compartment
contains nitrogen at 5 bar and the other compartment contains hydrogen
at 2 bar. The entire system is at 300 K. The mass of the piston is 2 kg and
it is made up of copper of specific heat 0.386 kJ/kg ·◦C. The pin holding
the piston is removed and the piston is allowed to reach an equilibrium
position. The piston is assumed to move without friction. Determine the
pressures, temperatures and volumes of the gases at the final equilibrium
state. Assume ideal gas behaviour. Take Cvas 0.743 kJ/kg ·K for nitrogen,
and Cvas 10.199 kJ/kg ·K for hydrogen. Molar mass of nitrogen is 28
kg/kmol and that of hydrogen is 2 kg/kmol.
62 Chapter 5
Solution to Example 5.7
The system boundary shown by the dashed line of Figure 5.5 clearly defines
the system, which consists of nitrogen in compartment A, hydrogen in compart-
ment B, the piston and the rigid vessel.
initial state final state
A
nitrogen
PA=5 bar
TA=300 K
VA=0.5 m3
B
hydrogen
PB=2 bar
TB=300 K
VB=0.5 m3
(pistonisheldbyapin) (pistonisfreetomove)
Figure 5.5 The initial and final states of the system of Example 5.7.
l
AB
nitrogen
Pfbar
TfK
VAf m3
hydrogen
Pfbar
TfK
VBf m3
pin
h
q
q
qq
qqqq
q
qpq
qqqq
q
qqppqq
q
qq
"
"
b
b
Once the pin holding the piston is removed, the piston is free to move until
the net force acting on the piston becomes zero. The force balance on the piston
in the absence of friction is written as follows:
pressure in compartment A x cross-sectional area of the piston =
pressure in compartment B x cross-sectional area of the piston,
which gives that, at the final equilibrium state, the pressures in the two
compartment on either side of the piston equal each other. Let us denote this
pressure as Pf(see Figure 5.5).
Since the piston is made up of copper, which is a good heat conductor, the
temperatures of the gases as well as the piston are all the same; 300 K at the
initial state and Tfat the final state (see Figure 5.5). The respective volumes of
nitrogen and hydrogen at the final state are given by VAf and VBf (see Figure
5.5). We are to determine the numerical values of Pf,Tf,VAf and VBf.
Since neither heat nor work crosses the system boundary, the first law applied
to closed systems yields ΔU=0, which means that there is no change in the
total internal energy of the system. Ignoring the internal energy change of the
Working with Ideal Gas 63
vessel of negligible heat capacity, the total internal energy change of the system
is written as
ΔU=ΔUA+ΔUB+ΔUp=0 (5.52)
where the subscript Adenotes the nitrogen in compartment A, Bdenotes the
hydrogen in compartment B, and pdenotes the piston.
Expanding (5.52) in terms of masses and specific heats, we get
mA(Cv)A(Tf−300)+mB(Cv)B(Tf−300)+mp(specific heat)p(Tf−300) = 0
(5.53)
in which the initial temperatures of hydrogen, nitrogen and piston are all 300 K,
and the final temperatures are all TfK. When rearranged, (5.53) gives
[mA(Cv)A+mB(Cv)B+mp(specific heat)p][Tf−300 K]=0 (5.54)
Since [mA(Cv)A+mB(Cv)B+mp(specific heat)p] of (5.54) takes a finite
value, we get
Tf= 300 K(5.55)
Now, we go on to determine Pf, which is the final pressure in both the
compartments, using the ideal gas equation of state and the principle of con-
servation of mass as follows. The masses of gases in compartments Aand B
remain constant during the process. Therefore, applying the ideal gas equation
of state to the gases in compartments Aand Bat the initial and final states
yields the following equations, with reference to Figure 5.5:
PfVAf
Tf
=(5 bar)(0.5m3)
300 K(5.56)
and PfVBf
Tf
=(2 bar)(0.5m3)
300 K(5.57)
Substituting the value of Tffrom (5.55) in (5.56) and (5.57), we get
PfVAf =(5bar)(0.5m3)and (5.58)
PfVBf =(2bar)(0.5m3) (5.59)
which has three unknowns Pf,VAf and VBf.
There are no other equations except (5.58) and (5.59) that are in terms of
the unknowns. We therefore need one more independent equation in terms of one
or more of the three variables Pf,VAf and VBf , to solve for the unknowns. We
have indeed used up all three of our tools along with the principle of conservation
64 Chapter 5
of mass. The situation seems hopeless except for the fact that the total volume
of the system remains a constant. Therefore, we get
VAf +VBf =1m3(5.60)
Substituting VAf from (5.58) and VBf from (5.59) in (5.60), we get
(5 bar)(0.5m3)
Pf
+(2 bar)(0.5m3)
Pf
=1m3(5.61)
and thus
Pf=3.5bar (5.62)
Substituting the value of Pffrom (5.62) in (5.58), we get
VAf =(5 bar)(0.5m3)
3.5bar =0.714 m3(5.63)
Substituting the value of VAf from (5.63) in (5.60), we get VBf = 0.286 m3.
In working out thermodynamic problems on closed simple compressible
systems with ideal gases, we started out with the following three tools:
Tool 1: The first law of thermodynamics applied to closed simple com-
pressible systems, given by Qin +Win =ΔU, which is also the
principle of conservation of energy.
Tool 2: Evaluation of ΔUusing data on specific heats, or data on internal
energy itself.
Tool 3: The ideal gas equation of state that interrelates pressure, tem-
perature, volume and mass (or amount of matter) of an ideal gas.
However, while solving the problems we realized that the following two
tools must be added to the list:
Tool 4: Principle of conservation of mass.
Tool 5: Any other constraint stated in the problem, such as “the total
volume remains constant”.
Working with Ideal Gas 65
Example 5.8
Show that we require more information about
the problem to solve for the unknowns of Example 5.7 if the heat con-
ducting copper piston is replaced by a piston that is a perfect heat insulator.
Solution to Example 5.8
The problem of this example differs from that of Example 5.7 in the aspect
that the piston of this example does not allow heat to pass from one compartment
to the other. Therefore, the final temperatures in the two compartments cannot
be considered the same, as was in the Solution to Example 5.7.
The unknowns of the problem are Pf, which denotes the equal final pressures
of gases in both the compartments, TAf and TBf , which are the unequal final
temperatures of gases in both the compartments, and VAf and VBf , which are
the unequal final volumes of the gases in both the compartments. To solve for
the five unknowns, Pf,TAf ,TBf ,VAf and VBf , we require five independent
equations in terms of one or more of the unknowns alone. Let us see how to get
the equations required.
Using Tool 1 and Tool 2 we get
mA(Cv)A(TAf −TAo)+mB(Cv)B(TBf −TBo)=0 (5.64)
where mA,(Cv)A,TAo and TAf are the respective mass, Cv, initial temperature
and final temperature of nitrogen in one compartment, and mB,(Cv)B,TBo
and TBf are the respective mass, Cv, initial temperature and final temperature
of hydrogen in the other compartment.
Substituting whatever is known from the problem statement in (5.64), we
get
mA(0.743 kJ/kg ·K)(TAf −300 K)+mB(10.199 kJ/kg ·K)(TBf −300 K)=0
(5.65)
The masses of the gases are found using Tool 3 as follows:
mA=(5 bar ×100 kPa/bar)(0.5m3)
(8.314/28 kJ/kg ·K) (300 K)=2.81 kg (5.66)
mB=(2 bar ×100 kPa/bar)(0.5m3)
(8.314/2kJ/kg ·K) (300 K)=0.08 kg (5.67)
Substituting mAof (5.66) and mBof (5.67) in (5.65), we get
2.81 ×0.743 ×(TAf −300) + 0.08 ×10.199 ×(TBf −300) = 0 (5.68)
66 Chapter 5
which is one of the five equations in terms of the unknowns alone.
Since the masses of gases in compartments Aand Bstay the same during
the process, use of Tool 4 gives
mAf =mAo and mBf =mBo (5.69)
Using Tool 3, (5.69) can be expanded to
PfVAf
TAf
=(500 kPa)(0.5m3)
300 Kand (5.70)
PfVBf
TBf
=(200 kPa)(0.5m3)
300 K(5.71)
which are two more independent equations in terms of the unknowns alone.
Tool 5 gives
VAf +VBf =1m3(5.72)
which is yet another independent equation in terms of the unknowns alone.
We now have four independent equations (5.68), (5.70), (5.71) and (5.72)
to solve for the five unknowns Pf,TAf ,TBf,VAf and VBf . From the problem
statement, it is impossible to get one more independent equation in terms of
one or more of the unknowns. Therefore, more information about the problem
should be provided in order to solve for the unknowns, if the heat conducting
copper piston of Example 5.7 is replaced by a piston that is a perfect heat
insulator.
5.10 Summary
•The ideal gas equation of state is given by
PV =nRT (5.1)
or by
PV =mRT (5.3)
where Pis the absolute pressure in kPa, Vis the total volume in m3,and
Tis the absolute temperature in K. In (5.1), nis the amount of gas in
Working with Ideal Gas 67
kmol, and therefore Ris the universal gas constant which takes the value
8.314 kJ/kmol ·K for all gases. In (5.3), mis the mass of gas in kg, and
therefore Ris the specific gas constant in kJ/kg ·K, the numerical value
of which depends on the gas concerned.
•Specific gas constant is related to the universal gas constant by
specific gas constant =8.314 kJ/kmol ·K
M (in kg/kmol)
where M=m/n, is the molar mass.
•The ideal gas equation of state is also given by
Pv=RT (5.4)
where Pis the absolute pressure in kPa and Tis the absolute temperature
in K. If vis the molar volume in m3/kmol, then Ris the universal gas
constant. If vis the specific volume in m3/kg, then Ris the specific gas
constant.
•For an ideal gas, properties u,h,Cvand Cpdepend only on temperature.
•For an ideal gas, specific (or molar) enthalpy is given by
h=u+RT (5.10)
•Specific heat at constant volume for an ideal gas is defined as
Cv=du
dT (5.6)
•Specific (or molar) internal energy change of an ideal gas undergoing any
process can be evaluated using
Δu=Tf
To
CvdT (5.8)
Note that (5.8) could also be used to evaluate Δufor substances other
than ideal gas provided the process is executed at constant volume.
•To evaluate Δuusing (5.8), Cvmust be known either as a function of
Tor be approximated to a constant. It is also common to obtain Δuas
(uf-uo), where ufand uoare obtained from ideal-gas (or zero-pressure)
property tables.
68 Chapter 5
•Specific heat at constant pressure for an ideal gas is defined as
Cp=dh
dT (5.11)
•Specific (or molar) enthalpy change of an ideal gas undergoing any process
can be evaluated using
Δh=Tf
To
CpdT (5.13)
Note that (5.13) could also be used to evaluate Δhfor substances other
than ideal gas provided the process is executed at constant pressure.
•To evaluate Δhusing (5.13), Cpmust be known either as a function of
Tor be approximated to a constant. It is also common to obtain Δhas
(hf-ho), where hfand hoare obtained from ideal-gas (or zero-pressure)
property tables.
•Specific heats of an ideal gas is related to each other by
Cp=Cv+R(5.14)
•Specific heat ratio (or the isentropic exponent) is defined as
γ=Cp/Cv(5.15)
•Specific heats of an ideal gas are given in terms of γand Ras follows:
Cv=R
γ−1(5.16)
Cp=γR
γ−1(5.17)
•In working out thermodynamic problems on closed simple compressible
systems with ideal gases, the following tools are used:
Tool 1: Principle of conservation of mass.
Tool 2: The first law of thermodynamics applied to closed simple com-
pressible systems, given by Qin +Win =ΔU, which is also the
principle of conservation of energy.
Tool 3: The ideal gas equation of state that interrelates pressure, tem-
perature, volume and mass (or amount of matter) of an ideal gas.
Tool 4: Evaluation of ΔUusing data on specific heats, or data on in-
ternal energy itself.
Tool 5: Any other constraint stated in the problem.