Elements of Mechanical Engineering
Chapter 3 Properties of Gases
Prepared by :- Mr. Mitesh D. Gohil
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Outline
3.1 Gas Laws like Boyleβs Law, Charleβs law, Gay lussac law
3.2 Combined gas law or Ideal gas equation
3.3 Basic gas processes
3.4 General Procedure for Applying the first law of thermodynamics to
different processes
3.5 Constant Volume Process / Isochoric process in Close System
3.6 Constant Pressure Process / Isobaric Process in Close System
3.7 Constant Temperature Process / Isothermal Process / Hyperbolic Process
3.8 Reversible adiabatic Process
3.9 Polytropic Process
3.10 The combination of Polytropic law (πππ = ππ¨π§π¬π. ) and Equation of State
(PV = mRT)
Ideal Gas:-
Ideal gas is βwhich would behave in an ideal manner at all pressure
and temperature such a gas will follow all gas laws and
characteristics gas equation.
Real Gas:-
The behavior of all real gas at high pressure and low temperature is
different and in that condition real gas will not follow gas laws.
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3.1 Gas Laws
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P β1
V(T= Constant)
PV = Constant
P1V1 = P2V2 = Constant β¦(1)
Boyleβs law
The absolute pressure of a given mass of an ideal gas is inversely
proportional to its volume, if the temperature of the gas is kept constant.
3.1 Gas Laws⦠continue
Q. State the following : Boyles Law, Charleβs Law GTU : April 2010
Fig. P-V diagram for constant
temperature process from state 1 to
2 in close system
5
Charlesβs Law
Volume of a given mass of an ideal gas is directly proportional to its absolute
temperature, if the pressure of gas is kept constant.
V β T (P= Constant)
V
T= Constant
V1T1=V2T2= Constant β¦(2)
Gay lussac law
The absolute pressure of a given mass of an ideal gas is proportional to
temperature, if the volume of gas is kept constant.
P β T (V= Constant)
P
T= Constant
P1T1=P2T2= Constant β¦(3)
3.1 Gas Laws⦠continue
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Avogadroβs law Avogadroβs law states that equal volume of different ideal gases at the
same pressure and temperature contain the same number of molecules.
OR
Avogadroβs law states that the volume of a g mol of all gases at the
pressure of 1 atm. and temperature of 0Β° C is the same, and equal to 22.4
liters.
1 kg mole contains 6.023 Γ 1023 molecules.
A mole (n) of a substance has numerically equal to the molecular weight
(M) of the substance.
Example:- 32 gm π2 = 1 g mole
28 kg π2= 1 kg mole
Mole of gas, n
n =m
M,kg moles β¦(4)
Where , m = mass of substance, kg
M= Molecular weight, kg/kg mole
3.1 Gas Laws⦠continue
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3.2 Combine gas law
Combine gas law / Characteristic gas equation / Equation of state / Ideal gas
equation
For an ideal gas, the equation of state represents the relationship between pressure
(p), volume (V) and Temperature (T).
π π, π, π = 0
It is possible to change the state 1 to
state 2 by infinite number of paths.
However we will follow path 1-3-2.
Because in that path, from 1-3 there is
constant volume process, in that we can
apply gay lussac law and from 3-2 there
is constant pressure process, in that we
can apply charlβs law.
Q. State the following : Characteristic gas equation. GTU : April 2010
Q. Derive characteristic equation of a perfect gas. GTU : Dec 2008
Fig. P-V diagram for any process from state
1 to 2 in close system
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P1V1T1
=P2V2T2
= constant(β΅ V3 = V1 & P3 = P2) β¦(7)
Put the value of T3 from Eq. (5) to (6)
V3P3T1P1
=V2T2
P1V3T1
=P3V2T2
For Process 1-3
P1T1=P3T3
(Gay lussac law) β¦(5)
For Process 3-2
V3T3=V2T2
(Charlesβs law) β¦(6)
Fig. P-V diagram for any process
from state 1 to 2 in close system
3.2 Combine gas law⦠continue
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So, for any fix mass of gas (close system), the change of state is connected by equation
ππ
π= ππππ π‘πππ‘
ππ
π= ππ
ππ = ππ π β¦(8)
ππ£ = π π β΅ π£ =π
πβ¦(9)
Where , R= Characteristic gas constant
π£ = ππππππππ ππππ’ππ
Equation (8) is known as Combine gas law / Characteristic gas equation /
Equation of state / Ideal gas equation for an ideal gas.
Q. Derive Expression pV/T = constant with help of Boyleβs law and Charlesβs law. GTU : May 2013
3.2 Combine gas law⦠continue
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R depends on the kind of gas.
Example, For air, R= 287 Nm
kg K
Significance of R
R represents the magnitude of work that can be done by 1 kg of gas when subjected to
1 degree change in temperature.
Putting the value of m in terms of mole (n) in equation (8)
Where, R = Universal gas constant= 8.31 kJ/ (kg mol) (K)
R does not depends on the kind of gas and it is same for all gases.
Real gases are not exactly obey this equation of state for all p, V, T value.
Real gases at very low pressure and very high temperature will not obey this equation
of state.
PV = nMRT β΅ n =m
Mβ¦(10)
PV = n RT (Putting, MR = R ) β¦(11)
3.2 Combine gas law⦠continue
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Application of first law of thermodynamics as applied to closed system
1. Constant Volume Process / Isochoric process,
V = Constant
2. Constant Pressure Process / Isobaric Process,
P = Constant
3. Constant Temperature Process / Isothermal Process / Hyperbolic Process,
T = constant
4. Reversible adiabatic Process,
Q = 0, or PVΞ³ = Constant
Where Ξ³ = Adiabatic index
5. Polytropic Process,
PVn = Constant
Where n = Polytropic index
Q. What are basic gas processes ? GTU : May 20123.3 Basic gas processes
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Fig. P-V Diagram for different basic gas processes
3.3 Basic gas processes β¦ continue
Q. How the basic processes are shown graphically on P-v diagram ? GTU : May 2012
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3.4
1) Work done, W= P dV
2) Heat Supply, Q
Ξ΄Q = m Cp or v dT
Q = m Cp or v βT
3) Change in Internal Energy, βU
dU = m CvdT
βU = mCvβT
4) Application of 1st law of thermodynamic,
Ξ΄Q β Ξ΄W = dU
Q βW = βU
Above equation could be derived from constant volume process and true for all processes.
Where, πΆπ = Specific heat at constant pressure
πΆπ£ = Specific heat at constant volume
β¦(13)
β¦(14)
β¦(15)
β¦(16)
General Procedure for applying the 1st law of thermodynamics to closed
system
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5) Change in Enthalpy, βH
h = u + pv
dh = du + p dv + v dp (differentiated above equation with respect to h)
6) Equation of state,
PV = mRT
7) Relation between Cp and Cv,
Cp β Cv = R
Equation Cp β Cv = R, could be derived from constant pressure process and true
for all processes.
Cp
Cv= Ξ³
Where, H = Enthalpy (J)
h = Specific Enthalpy (J/kg)
u = Internal energy (J)
Where, πΆπ = Specific heat at constant pressure
πΆπ£ = Specific heat at constant volume
πΎ ππ πππππππ‘ππ πππππ₯
β¦(17)
β¦(18)
β¦(19) β¦(20)
3.4General Procedure for applying the 1st law of thermodynamics to closed
system β¦ continue
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3.5 Constant Volume Process / Isochoric process in Close System
State 1 :- P1, V1, T1
State 2 :- P2, V2, T2
V1 = V2
V = Constant
dV = 0
1) Work done, W= P dV = 0
2) Heat Supply, Q
Ξ΄Q = m Cv dT
Q = m Cv βT
Q = m Cv (T2 β T1)
β¦(21)
β¦(22)Fig. P-V diagram for constant
volume process from state 1 to 2 in
close system
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3) Application of 1st law of thermodynamic,
Ξ΄Q β Ξ΄W = dU
Q βW = βU
m Cv T2 β T1 β 0 = βU
βU = m Cv T2 β T1
Above equation (23) is derived from constant volume process and true
for all processes.
For infinitesimal process, change in internal energy, ππ’ = π πΆπ£ ππ
For unit mass, Cv =du
dT v
Another word,
du β dT
So, Internal energy of substance is function of temperature only.
β¦(23)
3.5 Constant Volume Process / Isochoric process in Close Systemβ¦cont.
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3.6 Constant Pressure Process / Isobaric Process,
P = Constant
Fig. P-V diagram for constant
pressure process from state 1 to 2
in close system
State 1 :- π1, π1, π1
State 2 :- π2, π2, π2
π1 = π2 = π
1) Work done,
W =
1
2
π ππ
= π
1
2
ππ (β΅P= constant)
= π π 12
= π π2 β π1
= π2π2 β π1π1 (β΅ π1= π2 = π)
= ππ π2 βππ π1 (β΅ PV = mRT)
= ππ π2 β π1 β¦(24)
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π = ππ π2 β π1
3.6 Constant Pressure Process / Isobaric Process, β¦Continue
2) Heat Supply, Q
πΏπ = π πΆπ ππ
π = π πΆπ (π2 β π1)
3) Change in Internal Energy, βπ
ππ = π πΆπ£ππ
βU = mCv(T2 β T1)
4) Application of 1st law of thermodynamic,
πΏπ β πΏπ = ππ
π βπ = βπ
π πΆπ π2 β π1 β ππ π2 β π1 = mCv(T2 β T1)
πΆπ β Cv = π
Above equation (27) is called Meyerβs Equation
..(25)
(From eq.(14))
..(26)
..(27)
Q. With usual notation prove that Cp β Cv = R GTU : June 2009
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For another relationship
π βπ = βπ
π = βπ +π
= π2 β π1 + π2π2 β π1π1= π2 + π2π2) β (π1+π1π1= π»2 βπ»1
Heat transfer in a constant pressure process is equal to change in enthalpy of gas.
Another word,
πβ β ππ
So, enthalpy of substance is function of temperature only.
3.6 Constant Pressure Process / Isobaric Process, β¦Continue
m Cp dT = dH (from equation (25))
dh = Cp dT
Cp =dh
dTp
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3.7 Constant Temperature Process / Isothermal Process
T = constant
Fig. P-V diagram for constant
temperature process from state 1
to 2 in close system
State 1 :- π1, π1, π1
State 2 :- π2, π2, π2
π1 = π2 = π
PV=mRT (from the equation of state)
PV=constant
(β΅ In close system mR = const. & T = const.)
PV=C
β΄ π1 π1 = π2π2 = πΆ
So, this process follow the Boyleβs law
Q. Explain Isothermal Process. For Isothermal process, Find expression of work done,
Change in internal Energy, Change in Enthalpy and Heat transfer.
GTU : Dec 2012
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3.7 Constant Temperature Process / Isothermal Processβ¦.Continue
1) Work done, W
W =
1
2
π ππ (from eq. (13)
=
1
2πΆ
πππ (β΅PV= C)
= πΆ lnπ 12
= πΆ lnπ2 β lnπ1
= πΆ lnπ2π1
= π1π1 lnπ2π1
(β΅ π1π1 = π2π2 = πΆ)
= ππ π1 lnπ2π1
(β΅ PV = mRT)
W = ππ π1 lnπ1π2
(β΅ π1π1 = π2π2) β¦(28)
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2) Change in Internal Energy, βπΌ
dU = m CvdT
βU = mCvβT (From equation (15))
βU = 0 (βT = 0, Isothermal Process) β¦(29)
3) Application of 1st law of thermodynamic,
πΏπ β πΏπ = ππ
π βπ = βπ (From equation (16))
π = βπ +π
π = ππ π1 lnπ1π2
(From (28)&(29) β¦(30)
4) Change in Enthalpy, βπ―
h = u + pv (From equation (17))
H=U+PV
π»2 β π»1 = (π2 β π1) + (π2π2 β π1π1)
π»2 β π»1 = 0 (β΅βπ = 0 & π1π1 = π2π2) β¦(31)
3.7 Constant Temperature Process / Isothermal Processβ¦.Continue
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3.8 Reversible adiabatic Process,
π = 0, & πππΎ = Const.
Q. Prove that π·π½πΈ = ππππππππ, for an adiabatic process.
( GTU : Dec 2010, June 2011, Dec 2011)
In Adiabatic process we know, π = 0 ππ πΏπ = 0
Now From State of equation,
ππ = ππ π
πππ
ππ+ π
ππ
ππ= ππ
(differentiated above equation
with respect to T)
πππ + πππ = ππ ππ β¦(32)
From the 1st law of thermodynamic,
πΏπ β πΏπ = ππ
ππ + πΏπ = 0 (β΅πΏπ = 0)
ππΆπ£ππ + πππ = 0 (from eq.(13) &(15)) β¦(33)
ππ =βπππ
ππΆπ£
Where Ξ³ = Adiabatic index
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3.8 Reversible adiabatic Process, β¦.Continue
πππ + πππ = ππ βπππ
ππΆπ£
πππ + πππ =π
πΆπ£βπππ
πππ + πππ =πΆπ β πΆπ£
πΆπ£βπππ (β΅π = πΆπ β πΆπ£)
πππ + πππ = πΎ β 1 βπππ (β΅πΆπ
πΆπ£= πΎ)
πΎ πππ + πππ = 0
πΎππ
π+ππ
π= 0
πΎ lnπ + lnπ = lnπΆ(Integrating the above
equation)
ln ππΎ + lnπ = lnπΆ
ln πππΎ = lnπΆ
πππΎ = πΆ β¦(34)
Substituting the value of dT in eq. (32) from eq. (33)
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1) Work done, W
π =
1
2
π ππ (Form eq.(13))
=
1
2
πΆπβπΎ ππ (β΅ πππΎ = πΆ)
= πΆπ1βπΎ
1 β πΎ1
2
=πΆ
1 β πΎπ21βπΎ β π1
1βπΎ
=1
1 β πΎπΆπ2
1βπΎ β πΆπ11βπΎ
=1
1 β πΎπ2π2 β π1π1 (β΅ π2π2
πΎ = π1π1πΎ = πΆ)
=1
1 β πΎππ π2 βππ π1 β΅ ππ = ππ π
=ππ π2 β π11 β πΎ
β¦(35)W
3.8 Reversible adiabatic Process, β¦.Continue
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2) Application of 1st law of thermodynamic,
πΏπ β πΏπ = ππ
π βπ = βπ
π = ββπ (β΅Q = 0)
π = mCv(π1 β π2) (β΅ βU = mCv π2 β π1 ) β¦(36)
3.8 Reversible adiabatic Process, β¦.Continue
Q. Define adiabatic process. Derive the relation between P, V, T for this process. Also
derive the expression for work done and change in internal energy for this process.
GTU : Jan 2011
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3.9 Polytropic Process,
PVn = Constant
Where n = Polytropic index
1) Work done, W
W =
1
2
P dV (Form eq.(13))
= 12CVβn dV (β΅ PVn = C)
= CV1βn
1 β n1
2
=C
1 β nV21βn β V1
1βn
=1
1 β nCV2
1βn β CV11βn
=1
1 β nP2V2 β P1V1 (β΅ P2V2
n = P1V1n = C)
=1
1 β nmRT2 βmRT1 β΅ PV = mRT
W =mR T2 β T11 β n
β¦(37)
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2) Change in Internal Energy, βπ
dU = m CvdT (from eq.(15))
βU = mCv T2 β T1 β¦(38)
3) Application of 1st law of thermodynamic,
Ξ΄Q β Ξ΄W = dUQ = W+ βU
=mR T2 β T11 β n
+mCv T2 β T1 (From eq. 37 &(38))
=m Cp β Cv T2 β T1
1 β n+mCv T2 β T1
(β΅R = Cp β Cv)
= m Cv
CpCvβ 1
1 β n+ 1 T2 β T1
= m CvπΎ β 1
1 β n+ 1 T2 β T1
(β΅Cp
Cv= Ξ³)
Q = m CvπΎ β n
1 β nT2 β T1
β¦(39)
3.9 Polytropic Process,β¦.Continue
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π = π πΆπ π2 β π1 (Putting, πΆπ£πΎβπ
1βπ= πΆπ)
Where, πΆπ=Polytropic specific heat
3.9 Polytropic Process,β¦.Continue
From eq. (39)
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3.10 Combination of Polytropic law (πππ = ππ¨π§π¬π. ) and Equation of State (PV = mRT)
Polytropic law (πππ = πΆπππ π‘. )
P1V1n = P2V2
n
β΄P1P2
=V2V1
n
β¦(40)
Polytropic law (πππ = πΆπππ π‘. )
P1V1n = P2V2
n
P1V1 V1nβ1
= P2V2 V2nβ1
mRT1 V1nβ1
= mRT2 V2nβ1 β΅ PV = mRT
T1T2
=V2V1
nβ1
β¦(41)
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From the eq. (40)
From the eq. (41)
P1P2
1n
=V2V1
β¦(40)(a)
T1T2
1nβ1
=V2V1
β¦(41)(b)
From the eq. (40)(a) & (41)(b)
T1T2
1nβ1
=P1P2
1n
T1T2
=P1P2
nβ1n β¦(42)
3.10 Combination of Polytropic law (πππ = ππ¨π§π¬π. ) and Equation of State (PV = mRT)β¦
Q. State the following : Boyles Law, Charleβs Law GTU : April 2010
Q. Derive characteristic equation of a perfect gas. GTU : Dec 2008
Q. State the following : Characteristic gas equation. GTU : April 2010
Q. Derive Expression pV/T = constant with help of Boyleβs law and Charlesβs law. GTU : May 2013
Q. What are basic gas processes ? GTU : May 2012
Q. With usual notation prove that Cp β Cv = R GTU : June 2009
Q. Explain Isothermal Process. For Isothermal process, Find expression of work done,
Change in internal Energy, Change in Enthalpy and Heat transfer.
GTU : Dec 2012
Q. Prove that π·π½πΈ = ππππππππ, for an adiabatic process. GTU : Dec 2010,
June 2011, Dec
2011
Q. Define adiabatic process. Derive the relation between P, V, T for this process. Also
derive the expression for work done and change in internal energy for this process.
GTU : Jan 2011
Q. How the basic processes are shown graphically on P-v diagram ? GTU : May 2012
Questions of GTU Exam
32
Thank Youβ¦
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