1020-Diploma in Mechanical Engineering - 4020340- III Semester - Thermal Engineering I -Unit 1 : Basics of Thermodynamics - (First Half) PPT Prepared By… U. ARAVIND, B.E., M.I.S.T.E., LECTURER / MECHANICAL ENGINEERING LAKSHMI AMMAL POLYTECHNIC COLLEGE K.R. NAGAR, KOVILPATTI - 628 503 THOOTHUKUDI DISTRICT, TAMIL NADU Mobile No: 9943244068 Email Id: aravindulaganathanmech@gmail.com aravindmech@lapc.in

1. 6 April 2022 U.Aravind, Lect/Mech, LAPC 1
Lakshmi Ammal Polytechnic College
Department of Mechanical Engineering
Subject Code : 4020340
Year & Semester : II / III
Scheme : N
By…
U.Aravind,
Lecturer/Mechanical Engineering,
LAPC, Kovilpatti.

2. 6 April 2022 U.Aravind, Lect/Mech, LAPC 2

3. •Understand the basics of systems and
law of thermodynamics and Processes
•Study the different types of Fuels and
their combustion properties
•Study the types, functions and
performance test of IC engines
•Understand the vapour compression
and vapour absorption refrigeration
system
•Study the equipment's used for air
conditioning
6 April 2022 U.Aravind, Lect/Mech, LAPC 3

4. 6 April 2022 U.Aravind, Lect/Mech, LAPC 4
Lakshmi Ammal Polytechnic College
Department of Mechanical Engineering
Subject Code : 4020340
Subject Name : Thermal Engineering I
Year & Semester : II / III
Scheme : N
Unit : I
Topic: Basics of Thermodynamics

5. 6 April 2022 U.Aravind, Lect/Mech, LAPC 5

6. 6 April 2022 U.Aravind, Lect/Mech, LAPC 6

7. 6 April 2022 U.Aravind, Lect/Mech, LAPC 7

8. 1. Mass (m) :
•It is the amount of matter contained
in the object.
•It is denoted by “m”
•Unit: “kg”
6 April 2022 U.Aravind, Lect/Mech, LAPC 8

9. 2. Weight (W) :
•It is the force of gravity acting on
an object.
•It is denoted by “W”
•Unit: “N”
•Weight = mass x acceleration = mg
•g=9.81 m/Sec2
6 April 2022 U.Aravind, Lect/Mech, LAPC 9
W = mg

10. 6 April 2022 U.Aravind, Lect/Mech, LAPC 10

11. 3. Volume (V) :
•It is the space occupied by a object.
•It is denoted by “V”
•Unit: “ m3 ”
6 April 2022 U.Aravind, Lect/Mech, LAPC 11

12. 4. Density (ρ) :
•Mass per unit volume of a object is
called Density.
•It is denoted by “ρ”
•Unit: “kg/m3”
•Density = mass/volume = m/V
6 April 2022 U.Aravind, Lect/Mech, LAPC 12
ρ = m/V

13. 6 April 2022 U.Aravind, Lect/Mech, LAPC 13

14. 5. Specific Weight (w) :
•Weight per unit volume of a object
is called Specific weight.
•It is denoted by “w”
•Unit: “N/m3”
•Specific Weight = Weight/volume
= W/V
6 April 2022 U.Aravind, Lect/Mech, LAPC 14
w = W/V

15. 6. Specific Volume (VS) :
•Volume per unit mass of a object is
called Specific weight.
•It is denoted by “VS”
•Unit: “m3/kg”
•Specific Volume = Volume/mass = V/m
6 April 2022 U.Aravind, Lect/Mech, LAPC 15
VS = V/m = 1/ρ

16. 6 April 2022 U.Aravind, Lect/Mech, LAPC 16

17. 7. Specific Gravity (S):
•It is also called “Relative Density”
•It is the ratio between Density of
Object and Density of Standard
Fluid.
•Specific Gravity =
Density of Object
Density of Standard Fluid
•It is Denoted by “S”
•Unit: “No Unit”
6 April 2022 U.Aravind, Lect/Mech, LAPC 17
1 Litre = 1 x 10-3 m3

18. • For liquids are compared
with water at 4°C which has a
density of 1.0 kg per litre.
• For gases are commonly
compared with air, which has a
density of 1.29 grams per litre

19. For Brain
• Ex 1 Density of liquid mercury is 13.6 kg
per litre. What is the specific gravity of
liquid mercury?
• Ex 2 Density of gas carbon dioxide is 1.976
grams per litre. What is the specific gravity
of gas carbon dioxide?
13.6
1.53

20. 8. Pressure (p):
•Force acting on Unit area is called
pressure.
•Pressure =
Force
Area
=
F
A
•It is Denoted by “p”
•Unit: kN/m2
6 April 2022 U.Aravind, Lect/Mech, LAPC 20

21. Pressure α
𝟏
𝐀𝐫𝐞𝐚

23. 1 bar = 1 x 100 kN/m2
1 KPa (Kilo Pascal) = 1 x kN/m2
1 MPa (Mega Pascal) = 1 x MN/m2
1 Pa = 1 x N/m2
1 MPa = 1 x 103 kN/m2

24. For Brain
• Ex 1 If 8 bar is equal to how much KN/m2
• Ex 2 If 10 KPa is equal to how much
KN/m2
• Ex 3 If 10 MPa is equal to how much
KN/m2
8 bar = 800 KN/m2
10 KPa = 10 KN/m2
10 MPa = 10 x 103 KN/m2

25. 9. Temperature (t):
•It is the degree of Hotness or
Coldness of a Body
•It is Denoted by “t”
•Unit: °C, K, F
6 April 2022 U.Aravind, Lect/Mech, LAPC 25
K = °C + 273

26. For Brain
• Ex 1 If 0°C bar is equal to how much K
• Ex 2 If 80°C bar is equal to how much K
0°C = 273 K
80°C = 353 K

27. Standard
Temperature and
Pressure
(S.T.P) Condition
Normal
Temperature and
Pressure
(N.T.P) Condition
Standard Temperature
= 15°C = 288 K
Normal Temperature =
0°C = 273 K
Standard Pressure = 1
atm = 101.325 kN/m2
Normal Pressure = 1
atm = 101.325 kN/m2
6 April 2022 U.Aravind, Lect/Mech, LAPC 27

28. Heat (Q)
•It is Flow of energy from high temperature
object to low temperature object
•It is Denoted by “Q”
•Unit: KJ
•Heat Transferred = mass x specific heat x
temperature difference
6 April 2022 U.Aravind, Lect/Mech, LAPC 28
Q = m Cs (T2 – T1)

29. 6 April 2022 U.Aravind, Lect/Mech, LAPC 29
• Specific heat capacity is defined by the amount of heat
needed to raise the temperature of 1 gram of a
substance 1 degree Celsius (°C)

30. 6 April 2022 U.Aravind, Lect/Mech, LAPC 30

31. Specific Heat Capacity at
Constant Volume (Cv)
• Amount of heat needed to raise the 1 kg of a
Gas through 1K at constant volume.
•It is Denoted by “Cv”
•Unit: KJ/kg K
6 April 2022 U.Aravind, Lect/Mech, LAPC 31
Cv = 0.718 kJ/kg K

32. Specific Heat Capacity at
Constant Pressure (CP)
• Amount of heat needed to raise the 1 kg of a
Gas through 1K at constant pressure.
•It is Denoted by “CP”
•Unit: KJ/kg K
6 April 2022 U.Aravind, Lect/Mech, LAPC 32
Cv = 1.005 kJ/kg K

33. Work (w)
• If a Object moves through a distance by the
action of Force is called as Work.
• Unit: KN-m
6 April 2022 U.Aravind, Lect/Mech, LAPC 33
Work = Force x Distance moved

34. 6 April 2022 U.Aravind, Lect/Mech, LAPC 34

35. Power (P)
• It is the rate of doing work.
• Unit: KW or KJ/Sec
6 April 2022 U.Aravind, Lect/Mech, LAPC 35
• Power =
𝐖𝐨𝐫𝐤 𝑫𝒐𝒏𝒆
𝐓𝐢𝐦𝐞 𝐭𝐚𝐤𝐞𝐧

36. Energy
• It is capacity to do work
6 April 2022 U.Aravind, Lect/Mech, LAPC 36
Potential Energy Kinetic Energy
It is the energy
possessed by the object
due to its height
It is the energy
possessed by the object
due to its velocity

37. Law of Conservation of Energy
• Energy can neither be
created nor destroyed
• Energy can be converted
from one form into another
form
6 April 2022 U.Aravind, Lect/Mech, LAPC 37

39. Thermodynamic System:
• It is a definite space or matter where
thermodynamics process takes place.
Surroundings:
• Everything outside the system is called
surrounding.
Boundary:
• It separates system from the surrounding. It
may be real or imaginary surface/layer.
6 April 2022 U.Aravind, Lect/Mech, LAPC 39

40. Universe = System + Surroundings

41. Types of System
•Open System
•Closed System
•Isolated System
6 April 2022 U.Aravind, Lect/Mech, LAPC 41

42. Open System:
• In this system, both mass and energy can exchanged with
surroundings.
• Ex: Gas Turbine and Steam Turbine.
Closed System:
• In this system, energy can exchanged with surroundings
but not mass.
• Ex: Pressure Cooker
Isolated System:
• In this system, both mass and energy cannot exchanged
with surroundings.
• Ex: Thermo Flask
6 April 2022 U.Aravind, Lect/Mech, LAPC 42

43. No
Yes

45. Property:
• Any measurable characteristics of a system is
called property.
Properties are classified into 2 types
1. Intensive or Intrinsic properties
•They do not depend on mass of a system
•Ex: Pressure, Temperature, Density
2. Extensive or Extrinsic properties
•They depend on mass of a system
•Ex: Volume, Weight, Energy
6 April 2022 U.Aravind, Lect/Mech, LAPC
45

46. States:
• It is the condition of the system describes by
its properties.
Thermodynamic Process:
• It changes the state of the system.
1. Heating
2. Cooling
3. Expansion
4. Compression
6 April 2022 U.Aravind, Lect/Mech, LAPC
46

47. Point Function Path Function
It does not depends on the
flow of path of the process
It depends on the flow of path
of the process
Point function is also called as
the state function.
Path function is also called as
the area function.
It is located by a point It is given by area
Ex: Pressure, Volume,
Temperature
Ex: Heat, Work
6 April 2022 U.Aravind, Lect/Mech, LAPC 47

48. Thermal Equilibrium
• If temperature is same at all points in a system, then system
is called as Thermal Equilibrium.
Thermodynamics Equilibrium
• When a system is in thermal, mechanical and chemical
equilibrium, then it is called as Thermodynamics
Equilibrium.
6 April 2022 U.Aravind, Lect/Mech, LAPC 48

49. Laws of Thermodynamics
• Zeroth law of
Thermodynamics
• First law of
Thermodynamics
• Second law of
Thermodynamics
6 April 2022 U.Aravind, Lect/Mech, LAPC 49

50. Zeroth law of Thermodynamics :
• It states that “When two systems are in
thermal equilibrium with third system,
then they must be in thermal equilibrium
with each other”.
6 April 2022 U.Aravind, Lect/Mech, LAPC
50

53. First law of Thermodynamics :
• It states that “The amount of heat energy
supplied to the system is equal to the sum
of change in Internal energy of the system
and Work done by the system”.
6 April 2022 U.Aravind, Lect/Mech, LAPC
53
Q = ∆U + W

54. Second law of Thermodynamics :
I. Kelvin-Planck Statement
It is impossible to build a heat engine which
will give 100% efficiency. That is always some
heat rejection is there.
6 April 2022 U.Aravind, Lect/Mech, LAPC
54

55. Second law of Thermodynamics :
I. Kelvin-Planck Statement
6 April 2022 U.Aravind, Lect/Mech, LAPC
55

56. Second law of Thermodynamics :
II. Clausius Statement
It states that, Heat will not flow from cold
body to hot body without the aid of external
work.
6 April 2022 U.Aravind, Lect/Mech, LAPC
56

57. • Heat Received /Supplied /
added to the System
Q +Ve
• Heat Rejected /Lost from the
system
Q -Ve
• Increase in Internal Energy ∆U +Ve
• Decrease in Internal Energy ∆U -Ve
• Work Output/Work performed
by the System
W +Ve
• Work Input/Work done on the
System
W -Ve
6 April 2022 U.Aravind, Lect/Mech, LAPC 57

58. Solved Problems

59. 1. A closed system receives 100 kJ of heat and
performs 135 kJ of work. Calculate the change in
Internal Energy.
Given Data:
•Heat Received, Q = 100 kJ (+Ve)
•Work Performed, W = 135 kJ (+Ve)
To Find: Change in Internal Energy
Solution:-
By First law of Thermodynamics, Q = ∆U + W
∴ ∆U = Q – W
= 100 – 135 = -35 kJ
-ve Sign indicate decreases in Internal Energy. 59
Change in Internal Energy, ∆U = -35 kJ

60. 2. An input of 1 kw-hr work into a system(closed) increases
the internal energy of the system by 2000 kJ. Determine
how much heat is lost from the system.
Given Data:
•Work Input, W = -1 kw-hr (-Ve)
= -1k
J
sec
x 3600 sec = -3600 kJ
•Increased Internal Energy, ∆U = 2000 kJ (+Ve)
To Find: Heat loss
Solution:-
By First law of Thermodynamics, Q = ∆U + W
∴ Q = 2000 + (-3600) = -1600 kJ
-ve Sign indicate Heat loss from the system. 60
Heat Loss from the system, Q = -1600 kJ
1 Watt = 1 Joule per second (1W = 1 J/s)
1 Hour = 3600 Seconds
Convert kw-hr into kJ

61. 3. In a non-flow process there is a heat loss of 1055 kJ
and an Internal energy increase of 210 kJ. How much
work is done?
Given Data:
•Heat Loss, Q = -1055 kJ (-Ve)
•Increased Internal Energy, ∆U = 210 kJ (+Ve)
To Find: Work done,
Solution:-
By First law of Thermodynamics, Q = ∆U + W
∴ W = Q – ∆U
= -1055 kJ – 210 kJ = -1265 kJ
-ve Sign indicate Work Input.
61
Work Done, W= -1265 kJ

62. Home Work
Problems

63. 1. A closed system receives 125KJ of heat
while it performs 160KJ of work.
Determine the change in Internal Energy.
2. An input of 2Kw-hr work into a closed
system increases the internal energy of
the system by 3000KJ. Determine how
much heat is lost from the system.
63

64. Thermodynamic
Processes of Perfect
Gases
4/6/2022 U.Aravind, Lect/Mech, LAPC 64

65. Gas Vapour
A gas is a substance whose
evaporation from liquid state
is Complete
A vapour is a substance whose
evaporation from liquid state
is incomplete
It cannot be liquefied by
applying pressure
It can be liquefied by applying
pressure
Ex: Oxygen, Nitrogen,
Hydrogen
Ex: Ammonia, Wet Steam,
Freon
4/6/2022 U.Aravind, Lect/Mech, LAPC 65

66. Laws of Perfect Gases
4/6/2022 U.Aravind, Lect/Mech, LAPC 66

67. 1. Boyle’s Law
•It states that, “Volume is inversely
proportional to pressure at
constant temperature”
•V = C x
𝟏
𝐏
P1 V1 =P2V2
4/6/2022 U.Aravind, Lect/Mech, LAPC 67
V α
𝟏
𝐏 C is Constant
PV = C

68. 4/6/2022 U.Aravind, Lect/Mech, LAPC 68

69. 4/6/2022 U.Aravind, Lect/Mech, LAPC 69

70. 4/6/2022 U.Aravind, Lect/Mech, LAPC 70

71. 2. Charles Law
•It states that, “Volume is directly
proportional to temperature at
constant pressure”
•V = CT
V1
T1
=
V𝟐
T𝟐
4/6/2022 U.Aravind, Lect/Mech, LAPC 71
V α T
C is Constant
𝐕
𝐓
= C

72. 4/6/2022 U.Aravind, Lect/Mech, LAPC 72

73. 4/6/2022 U.Aravind, Lect/Mech, LAPC 73

74. 3. Joule’s Law
•It states that, “Internal Energy is
directly proportional to
temperature”
•∆U α ∆T
•∆U = C.∆T
•∆U = mCv ∆T
∆U = mCv (T2 − T1)
4/6/2022 U.Aravind, Lect/Mech, LAPC 74
U α T
C is Constant
Put C = mCv
∆T = T2 − T1

75. 4. Regnault’s Law
•It states that, the specific heat at
Constant pressure (Cp) and specific
heat at Constant volume (Cv) do not
change with the change in pressure
and temperature of the gas.
4/6/2022 U.Aravind, Lect/Mech, LAPC 75

76. 5. Avogadro's Law
•It states that, “Equal volume of all
gases at the same temperature and
pressure contains the same number
of molecules”
•V = kT
•
𝐕
𝐧
= k
𝐕𝟏
𝐧𝟏
=
𝐕𝟐
𝐧𝟐
4/6/2022 U.Aravind, Lect/Mech, LAPC 76
V α n k is Constant

77. 4/6/2022 U.Aravind, Lect/Mech, LAPC 77

78. Characteristic Gas Equation
Consider Specific Volume VSof a Gas
~By Boyle’s Law: VSα
1
P
~By Charle’s Law: VSαT
Combine the both, we get
VSα
1
P
x T
VS= R x
1
P
x T
∴ PVS= RT
P
V
m
= RT
~ This is the Characteristics Gas Equation
4/6/2022 U.Aravind, Lect/Mech, LAPC 78
R is Gas
Constant
Specific Volume
=
𝐕𝐨𝐥𝐮𝐦𝐞
𝐌𝐚𝐬𝐬
=
𝐕
𝐦
PV = mRT

79. General Gas Equation
From Characteristic Gas Equation For a Gas at State 1
~P1V1 = mRT1
From Characteristic Gas Equation For a Gas at State 2
~P2V2 = mRT2
Combine the both, we get
P1V1
P2V2
=
mRT1
mRT2
P1V1
P2V2
=
T1
T2
This is the General Gas Equation
4/6/2022 U.Aravind, Lect/Mech, LAPC 79
𝐏𝟏𝐕𝟏
𝐓𝟏
=
𝐏𝟐𝐕𝟐
𝐓𝟐

80. Relation between two specific heats
(𝐂𝐩, 𝐂𝐕) & Gas Constant (R)
By first law of Thermodynamics
~𝐐 = ∆U + W
In Constant Pressure Process, we get
~𝐐 = m𝐂𝐩(𝐓𝟐-𝐓𝟏)
~∆U = m𝐂𝐕(𝐓𝟐-𝐓𝟏)
~W = m𝐑(𝐓𝟐-𝐓𝟏)
Putting equations 2,3,4 in
m𝐂𝐩(𝐓𝟐-𝐓𝟏) = m𝐂𝐕(𝐓𝟐-𝐓𝟏) + m𝐑(𝐓𝟐-𝐓𝟏)
m(𝐓𝟐-𝐓𝟏)𝐂𝐩 = m(𝐓𝟐-𝐓𝟏) [𝐂𝐕 + R]
𝐂𝐏 = 𝐂𝐕 + R
4/6/2022 U.Aravind, Lect/Mech, LAPC 80
1
2
3
4
𝐂𝐏 − 𝐂𝐕 = R
R = 0.287 kJ/kg.k

81. Ratio of Specific Heats (γ)
•γ =
𝐒𝐩𝐞𝐜𝐢𝐟𝐢𝐜 𝐡𝐞𝐚𝐭 𝐚𝐭 𝐜𝐨𝐧𝐬𝐭𝐚𝐧𝐭 𝐩𝐫𝐞𝐬𝐬𝐮𝐫𝐞
𝐒𝐩𝐞𝐜𝐢𝐟𝐢𝐜 𝐡𝐞𝐚𝐭 𝐚𝐭 𝐜𝐨𝐧𝐬𝐭𝐚𝐧𝐭 𝐯𝐨𝐥𝐮𝐦𝐞
Universal Gas Constant (𝐑𝐔)
•Universal gas constant = Molecular weight
of the gas x Characteristic gas constant
• RU= M x R
4/6/2022 U.Aravind, Lect/Mech, LAPC 81
γ =
𝐂𝐏
𝐂𝐕
KJ/Kg.mole.K
𝐑𝐔= 8.314 kJ/kg.mole.K

82. Internal Energy (U)
•It is the Energy stored by a gas which is the total of
kinetic energy and potential energy due to the
motion of molecules and atoms.
•It Changes with change in temperature.
∆U = m𝐂𝐕(𝐓𝟐 − 𝐓𝟏)
4/6/2022 U.Aravind, Lect/Mech, LAPC 82

83. Enthalpy (h)
•It is the amount of heat content used or
released in a system at constant pressure.
•Sum of Internal Energy and Pressure-
Volume product is called enthalpy.
h = U +PV
•In differential form, ∆h = ∆U + PdV
•Change in enthalpy = Mass x Specific heat
at constant pressure x Temperature
Difference
4/6/2022 U.Aravind, Lect/Mech, LAPC 83
∆h = m𝐂𝐏(𝐓𝟐− 𝐓𝟏)

84. Entropy (S)
•It is the measure of amount of energy in
the system which is unavailable to do
work.
•Entropy increases with heat addition and
decreases with heat rejection.
•Change in entropy
dS =
𝐝𝐐
𝐓
dS = 𝐓𝟏
𝐓𝟐 𝐝𝐐
𝐓
4/6/2022 U.Aravind, Lect/Mech, LAPC 84

85. Formula’s and Values

86. 1. Characteristic gas equation : PV = mRT
2. General gas equation:
𝐏𝟏𝐕𝟏
𝐓𝟏
=
𝐏𝟐𝐕𝟐
𝐓𝟐
= ---- = C
3. 𝐂𝐏 − 𝐂𝐕 = R
4. 𝐑𝐔 = M R
5. Change in Internal Energy, ∆U = m𝐂𝐕(𝐓𝟐-𝐓𝟏)
6. Change in Enthalpy, ∆h = m𝐂𝐏(𝐓𝟐− 𝐓𝟏)
7. Change in Entropy, dS = 𝐓𝟏
𝐓𝟐 𝐝𝐐
𝐓
8. R = 0.287 kJ/Kg.k
9. 𝐑𝐔 = 8.314 kJ/Kg.K
10.𝐂𝐏 = 1.005 kJ/Kg.K
11.𝐂𝐕 = 0.718 kJ/Kg.K
4/6/2022 U.Aravind, Lect/Mech, LAPC 86

87. Solved Problems

88. 1. A gas at a pressure of 3 bar and temperature 27°C occupied a
volume of 1.5𝐦𝟑. If the gas constant is 287 J/Kg.K. Determine the
mass of the gas.
Given Data:
• Pressure, P = 3 bar
= 3 x 100 kN/m2
= 300 kN/m2
• Temperature, T= 27°C
= 27 + 273 = 300 K
• Volume, V = 1.5m3
• Gas Constant, R = 287 J/kg.K =
287
1000
= 0.287 kJ/kg.K
To Find: Mass, m
Solution:-
From Characteristic gas equation : PV = mRT ∴ m =
PV
RT
m =
300 x 1.5
0.287 x 300
=5.226 kg
88
1 bar = 1 x 100 kN/m2
K = °C + 273
m = 5.226 Kg

89. 2. Find the volume of 2 kg of air at STP.
Given Data:
• Mass of the air, m = 2 kg
• Temperature, TSTP = 288 K
• Pressure, PSTP = 101.325 kN/m2
• Gas Constant, R = 0.287 kJ/kg.K
To Find: Volume, 𝑉𝑆𝑇𝑃
Solution:-
From Characteristic gas equation : PV = mRT
PSTPVSTP = mR TSTP
∴ VSTP =
mRTSTP
PSTP
VSTP =
2 x 0.287 x 288
101.325
=1.6315
89
At STP
𝐕𝐒𝐓𝐏 = 1.6315𝐦𝟑

90. 3. A cylinder contains 3 kg of oxygen at gas at 5 bar pressure and temperature
27°C. Determine the volume of the cylinder. 𝐑𝐔 = 8.314 kJ/kg.mole.K
Given Data:
• Mass, m = 3 kg
• Pressure, P = 5 bar = 5 x 100 kN/m2
= 500 kN/m2
• Temperature, T= 27°C = 27 + 273 = 300 K
• Universal Gas Constant, RU = 8.314 kJ/kg.mole.K
To Find: Volume, V
Solution:-
From Characteristic gas equation : PV = mRT
We need to Find R
We Know RU = M x R
R =
RU
M
=
8.314
32
= 0.2598
∴ V =
mRT
P
V =
3 x 0.2598 x 300
500
=0.4676 𝐦𝟑
90
V = 0.4676 m𝟑
M - Molecular Weight
Molecular Weight of Oxygen is 32

91. 4. A gas whose original pressure, volume and temperature were 120 kN/𝐦𝟐
,
0.125 𝐦𝟑 , 30°C respectively is compressed such that its pressure and
temperature are raised to 600 kN/𝐦𝟐
and 70°C respectively. Determine the
new Volume of the gas.
Given Data:
• Initial Pressure, P1 = 120 kN/m2
• Initial Volume, V1 = 0.125 m3
• Initial Temperature, T1 = 30°C = 27 + 273 = 303 K
• Final Pressure, P2 = 600 kN/m2
• Final Temperature, T2 = 70°C = 27 + 273 = 343 K
To Find: Final Volume, V2
Solution:-
From General gas equation :
𝐏𝟏𝐕𝟏
𝐓𝟏
=
𝐏𝟐𝐕𝟐
𝐓𝟐
V2 =
P1V1
T1
x
T2
P2
=
120 x 0.125 x343
303 x 600
= 0.0283 𝐦𝟑
91
𝐕𝟐 = 0.0283 m𝟑

92. 5. A gas at a temperature of 20°C and pressure of 2bar occupies a volume of
0.1𝐦𝟑
. If the gas expanded to a pressure of 0.2bar and volume of 0.5𝐦𝟑
. What
would be the final temperature of the gas?
Given Data:
• Initial Temperature, T1 = 20°C = 27 + 273 = 293 K
• Initial Pressure, P1 = 2bar = 2 x 100 = 200 kN/m2
• Initial Volume, V1 = 0.1 m3
• Final Pressure, P2 = 0.2 bar = 0.2x 100 = 20 kN/m2
• Final Volume, V2 = 0.5 m3
To Find: Final Temperature, T2
Solution:-
From General gas equation :
𝐏𝟏𝐕𝟏
𝐓𝟏
=
𝐏𝟐𝐕𝟐
𝐓𝟐
𝐓𝟐 =
𝐏𝟐𝐕𝟐
𝐏𝟏𝐕𝟏
x 𝐓𝟏
92
𝐓𝟐 = 146.5 K

93. 6. A Gas having molecular weight 28 occupies a volume of 0.15𝐦𝟑
at
pressure of 2 bar and a temperature of 20°C. Find the mass and
volume of the gas at 0°C and 1 bar pressure. Find also the density of
the gas at 0°C and 1 bar pressure. Take 𝐑𝐔 = 8.314 kJ/kg.mole.K.
Given Data:
• Molecular Weight, M = 28
• Initial Volume, V1 = 0.15 m3
• Initial Pressure, P1 = 2bar = 2 x 100 = 200 kN/m2
• Initial Temperature, T1 = 20°C = 20 + 273 = 293 K
• Final Temperature, T2 = 0°C = 0 + 273 = 293 K
• Final Pressure, P2 = 1 bar = 1 x 100 = 100 kN/m2
• Universal Gas Constant, RU = 8.314 kJ/kg.mole.K
To Find:
• Mass, m
• Final Volume, V2
• Density 93

94. Solution:-
I. To Find Mass, m
From Characteristic gas equation :𝐏𝟏𝐕𝟏 = 𝐦𝐑𝐓𝟏
We need to Find R
We Know RU = M x R
R =
RU
M
=
8.314
28
= 0.2969
∴ m =
𝐏𝟏𝐕𝟏
𝐑𝐓𝟏
m =
200 x 0.15
0.2969 x 293
= 0.3448
94
m = 0.3448 kg

95. II. To Find Final Volume, 𝐕𝟐
From Characteristic gas equation :𝐏𝟐𝐕𝟐 = 𝐦𝐑𝐓𝟐
∴ 𝐕𝟐 =
𝐦𝐑𝐓𝟐
𝐏𝟐
=
0.3448 x 0.2969 x 273
100
= 0.27947 𝐦𝟑
II. To Find D𝐞𝐧𝐬𝐢𝐭𝐲
Density =
𝐌𝐚𝐬𝐬
𝐕𝐨𝐥𝐮𝐦𝐞
=
𝟎.𝟑𝟒𝟒𝟖
𝟎.𝟐𝟕𝟗𝟒𝟕
= 1.233
95
𝐕𝟐 = 0.27947 m𝟑
Density = 1.233 kg/m𝟑

96. 6. A mass of air has an initial pressure if 1.3 MN/𝐦𝟐, volume
0.014𝐦𝟑
and temperature 135°C. It is expanded until its final
pressure is 275 KN/𝐦𝟐 and its volume becomes 0.056𝐦𝟑 .
Determine i) mass of the air ii) final temperature of the air.
Take R = 0.287 kJ/kg.K
Given Data:
• Initial Pressure, P1 = 1.3MN/m2
= 1.3 x 103 = 1300 kN/m2
• Initial Volume, V1 = 0.014 m3
• Initial Temperature, T1 = 135°C = 135 + 273 = 408 K
• Final Pressure, P2 = 275 kN/m2
• Final Volume, V2 = 0.056 m3
• Gas Constant, R = 0.287 kJ/kg.K
To Find:
• Mass, m
• Final Temperature, T2
96

97. Solution:-
I. To Find Mass, m
From Characteristic gas equation :𝐏𝟏𝐕𝟏 = 𝐦𝐑𝐓𝟏
∴ m =
𝐏𝟏𝐕𝟏
𝐑𝐓𝟏
m =
1300 x 0.014
0.287 x 408
= 0.1554
II. To Find Final Temperature, 𝐓𝟐
From Characteristic gas equation :𝐏𝟐𝐕𝟐 = 𝐦𝐑𝐓𝟐
∴ 𝐓𝟐 =
𝐏𝟐𝐕𝟐
𝐦𝐑
=
275 x 0.056
0.1554 x 0.287
= 345.29
97
m = 0.1554 kg
𝐓𝟐 = 345.29 K

98. Review Questions

99. Part – A(1 mark Questions)
1.State Boyle’s law.
2.Define Enthalpy.
3.Write down the characteristic gas
equation.
4.What is the value of universal gas
constant?
5.Write down the relationship between Cp,
Cv and R.
6.Write down the general gas equation.

100. Part – B (2 mark Questions)
1. State i)Boyle’s law ii) Joule’s law
2. State the law of perfect gases.
3. State i) Charle’s law ii) Regnault's law iii) Avogadro's law
4. Define Enthalpy.
5. Define Change in Entropy.
6. Write the general equation for Change in Entropy.
7. What is the difference between vapour and gases? Give two
examples for each.
8. Derive the characteristic gas equation.
9. Derive the relationship between Cp, Cv & R
10. Derive general gas equation.
11. What is the Universal gas constant? What is its relation
between characteristic gas constant?
6 April 2022 U.Aravind, Lect/Mech, LAPC 100

101. Part – C (7 & 8 mark Questions)
1. A gas at a pressure of 4.5bar and temperature 30°C occupies a
volume of 3.1m3 . If the mass is 1.25kg. Determine Gas
Constant.
2. A gas occupies 3m3 at a temperature of 150°C. The pressure of
the gas is 7bar. The gas expands in such a manner that the
volume becomes 9m3 and temperature is 10°C. What is the
pressure of the gas?
3. A gas at a pressure of 20bar and temperature 300°C occupies a
volume of 0.06m3
. It is expanded to a pressure of 120kN/m2
and temperature 30°C. Determine the final volume.
4. A gas at a temperature of 20°C and pressure of 1.5bar occupies
a volume of 0.105m3. If the gas is compressed to a pressure of
7.5bar and volume of 0.04m3
, what will be the final
temperature of the gas? Also determine the mass of the gas.
Take R = 0.29kJ/kg.K
6 April 2022 U.Aravind, Lect/Mech, LAPC 101