The document provides information on various thermodynamics concepts: - A pure substance has a constant composition, while a mixture consists of multiple substances. - A system is the quantity of matter under analysis, and can be open, closed, or isolated based on mass and energy transfers. - Thermodynamic properties include intensive properties like temperature and pressure, and extensive properties like volume and energy which depend on system mass. - Processes involve system state changes or energy transfers. Equilibrium occurs when properties are uniform throughout the system.

1. Chapter No 1
Fundamentals of
Thermodynamics
Marks 20

2. Concept of Pure Substance
 A pure substance or chemical substance is a material
that has a constant composition
 All elements are pure substances. Sugar, salt, and
baking soda are pure substances
 Air is a mixture of several gases regarded as Pure
substance.
 Working substance – The working substance is most
producing and absorbing device is gas and vapour or
vapour and liquid.
 e.g. Petrol, ammonia, Freon.

3. Thermodynamic system
‘‘System’’ is defined as the quantity of
matter or region in space upon which the
attention is concentrated for the sake of
analysis.

4. Types of system
1) Closed System - Only transfer of energy but no
mass transfer across the boundary.
2)Open System - In this system, both mass & energy
may cross the boundary.
3)Isolated System - In this system, there is no transfer
of mass & energy across the boundary, e.g. Thermos

5. Thermodynamic Property
 Properties may be of two types
1)Intensive properties- The properties which
are independent on mass of the system are
called as intensive properties. E.g. pressure,
temperature, density etc.
 Extensive properties- The properties of
system which depend on mass of the system
are called as extensive properties. E.g. total
volume, total mass, total energy, enthalpy,
entropy, weight etc.

6. Thermodynamic Process
 A process occurs when the system
undergoes a change in a state or an energy
transfer at a steady state.
 substance which is being heated in a
closed cylinder undergoes a non-flow
process Closed systems undergo non-flow
processes.
 A process may be a flow process in which
mass is entering and leaving through the
boundary of an open system.

7. Thermodynamic Equilibrium
1. Thermal equilibrium. The temperature of the system
does not change with time and has same value at all
points of the system.
2. Mechanical equilibrium. There are no unbalanced
forces within the system or between the
surroundings.
3. Chemical equilibrium. No chemical reaction takes
place in the system and the chemical composition
which is same throughout the system does not vary
with time.

8. POINT FUNCTION
 When two properties locate a point on the graph
(co-ordinate axes) then those properties are
called as point function.
 Examples. Pressure, temperature, volume etc.
dV = V2 − V1 (an exact differential).

9. PATH FUNCTION
 There are certain quantities which cannot be
located on a graph by a point but are given by
the area or so, on that graph.
 In that case, the area on the graph, pertaining to
the particular process, is a function of the path
of the process. Such quantities are called path
functions.
 Examples. Heat, work etc.

10. Energy
 Stored energy- it is energy possessed by system
within its boundaries. Kinetic energy, potential
energy, internal energy etc.
 Transient Energy - It is defined as the energy,
Which is capable of crossing boundaries.
 Example – heat, work.

11. Heat
 Heat - is that form of energy which transferred
from one body to another on account of
temperature difference.
 Work: work is a transient form of energy. When
a force acts upon a body causing the body to
move and to overcome continually a resistance.
Work is said to be done.

12. Difference between Heat & Work
Sr.
no Heat work
1
It occurs due to temperature
difference
It caused due to displacement
or motion
2
It is low grade energy It is high grade energy
3
In stable condition, there is
no restriction for heat
transfer.
In stable condition, there is
no work transfer.
4
It is denoted by Q It is denoted by w

13. Enthalpy
 It is Total heat content of the system and defined
as energy which is algebraic sum of internal
energy and flow work.
H= u + PVs kJ/Kg
Where H= Total Enthalpy
u = Internal Energy
P = Absolute Pressure

14. Entropy
 It means Transformation.
 Entropy is a thermodynamic property of a
working substance which increases with the
addition of heat and decreases with removal of
heat. Entropy is represented by symbol S or ϕ.
 Change in Entropy = dQ / T
 Where, T = Absolute temperature
dQ = Change in heat
ΔS = Change in Entropy
Unit of Entropy KJ/Kg0K

15. Laws of thermodynamics
 Zeroth law of thermodynamics - If two systems are each is
thermal equilibrium with a third system, then the two systems are
also in thermal equilibrium with one another.
Or
 If System A and System C and System B and System C are
thermal equilibrium with each other than system A, B are also in
thermally equilibrium with each other

16. Law of conservation of energy
Developed the concept of energy and
hypothesis that it can be neither created nor
destroyed; this came to be known as the law
of the conservation of energy. The first law
of thermodynamics is merely one statement
of this general law/principle with particular
reference to heat energy and mechanical
energy i.e., work.

17. First Law of Thermodynamics
 When a system undergoes a thermodynamic cycle then
the net heat supplied to the system from the surroundings
is equal to net work done by the system on its
surroundings.
OR
 “Heat and work are mutually convertible but since
energy can neither be created nor destroyed, the total
energy associated with an energy conversion remains
constant”.
OR
 No machine can produce energy without corresponding
expenditure of energy, i.e., it is impossible to construct a
perpetual motion machine of first kind”.

18. Limitations of first law of
thermodynamics
 The first law states that the work transfer is equal to heat transfer and does
not place any restriction on the direction of flow but the reversal of process
not violet the first law.
 - According to this statement the first law the potential energy can be
converted into kinetic energy and kinetic energy can be converted into
potential energy but in natural practice this do not happen
 - E.g. A car driver up hill , the level of fuel in the tank drop. If this process
reverse i.e. car is coasted down the hill, the fuel consume cannot be
reproduced.
 - Hence from above example the reversal of process is not true without the
aid of external work.

19. Application of First Law of
Thermodynamics
 Steady Flow Energy equation- (S.F.E.E.)- in many practical
problems, the rate at which the fluid flows through a machine or
piece of apparatus constant. This is steady flow.
 Assumptions-
1. The mass flow through the system remains constant.
2. Heat transfer rate is constant.
3. Work transfer rate constant.

21. Application of First Law of
Thermodynamics
 Boiler- which generates steam on heating
water.
 Heat is supplied to system Q1-2 is positive.
 According to steady flow equation
 There is no work-done W 1-2 = 0
 Assuming change in KE & PE Negligible &
m = 1kg
 h1 + gz1 + C1
2/2 + Q 1-2 = h2 + gz2 + C2
2/2 +
W 1-2
 h1 + 0 + 0 + Q 1-2 = h2 + 0+ 0 + 0
 Q 1-2 = h2 - h1

22.  Condenser - Heat is rejected to system Q1-2
is nequative.
 There is no work-done W 1-2
 According to steady flow equation
 Assuming change in KE & PE Negligible &
m = 1 kg
h1 + gz1 + C12/2 + Q 1-2 = h2 + gz2 +
C22/2 + W 1-2
h1 + 0 + 0 -Q 1-2 = h2 + 0+ 0 + 0
Q 1-2 = h1 - h2

23.  Turbine = Turbine converts energy of the working
substance in to work.
 Heat is rejected to system Q1-2 , is zero
 There is work-done W 1-2 positive.
 According to steady flow equation
 Assuming change in KE & PE Negligible & m = 1 kg
 h1 + gz1 + C12/2 + Q 1-2 = h2 + gz2 + C22/2 + W 1-2
h1 + 0 + 0 + 0 = h2 + 0+ 0 + W 1-2
W 1-2 = h1 - h2

24.  Compressor- compressor is device, which
compresses air and supplies the same at high
pressure, work is supplied. i.e. work is done
on the system
Heat is rejected to system Q1-2 , is zero
There is work-done W 1-2 nequative.
According to steady flow equation
h1 + gz1 + C1
2/2 + Q 1-2 = h2 + gz2 + C2
2/2 - W
1-2
h1 + 0 + 0 + 0 = h2 + 0+ 0 - W 1-2
W 1-2 = h2 - h1

25. Second Law of Thermodynamics
 Kelvin Planck statement: - “It is impossible to
construct a heat engine working in a cyclic process,
whose sole effect is to convert the heat energy supplied
into equivalent amount of work. It means that heat
engine cannot fully convert the heat supplied into
mechanical work.” (Applicable for heat engine)
 Clausius statement: “It is impossible for machine (heat
pump or refrigerator) to transfer heat from a body at low
temperature to a body at high temperature to a body at
high temperature, without aid of external source.”
(Applicable for heat Pump & Refrigerator)

26. Equivalence of Kelvin Planck
statement
 Violation of kelvin Plank Statement –
 Consider the heat-engine having 100% Efficiency i.e.
violation of KP Statement. Such a heat engine will
convert the heat energy supplied Q2 into equivalent
amount of work (We). (Q2= We).
 This work produced can be utilized to drive a heat
pump, which receive amount of Q1 from cold body
and rejects an amount of heat (Q1+Q2) to hot body as
shown in fig.

27. Violation of kelvin Plank
Statement
 If the combination of heat
engine and heat pump is
considered as single
system, the result will be a
device , which delivers heat
as Q1 from cold body to
hot body without having
any external work , thus
violation of claussious
statement.
 Hence violation of Kelvin
plank statement results in
violation of classious
statement.

28. violation of classious statement
 Consider the heat pump having which violate
classious statement.as it transfer heat from
cold body to hot body without external work
 Now let heat engine absorbs an Q2 and
produces work, We = Q2 - Q1 , Where Q1 is
the heat rejected to cold body and Q2 heat
received from hot body.
 If the combination of heat engine and heat
pump is considered as single system, the
result will be a device whose aim is to deliver
the work, We = Q2 - Q1
 It means that total amount of heat received
from heat source is converted into work, thus
violating KP Statement.
 Hence violation of classious statement results
in violation of Kelvin plank statement.

29. Perpetual motion machine
 A machine which
violates the first law of
thermodynamics is
called the perpetual
motion machine of
the first kind
(PMM1). Such a
machine creates its
own energy from
nothing and does not
exist.
 Without violating the first
law, a machine can be
imagined which would
continuously absorb heat
from a single thermal
reservoir and would convert
this heat completely into
work. The efficiency of
such a machine would be
100 per cent. This machine
is called the perpetual
motion machine of the
second kind (PMM2).

30. Application of second law of
thermo dynamics
 There are three important applications of Statement
of Second Law of Thermodynamics namely:
a) Heat Engine
b) Heat Pump and
c) Refrigerator

31. Heat engine
As shown in the figure below having source of temperature (T1) and
Sink at temperature (T2). The amount of heat taken from source is Q1.
Out of this amount of heat work done by the engine is W and remaining
part of heat rejected to the sink.
1. Heat is added. This is at a relatively high temperature, so the heat can
be called Q1.
2. Some of the energy from that input heat is used to perform work (W).
3. The rest of the heat is removed at a relatively cold temperature (Q2).

32. Heat engine
 An efficiency of the heat
engine can be calculated
as: (efficiency = COP)
 COP = Work done (W) /
Input Heat (Q1)
= We / Q2
= (Q2 – Q1) / Q2
 In terms of temperature
COP= (T2-T1)/ T2

33. Heat Pump
 is a thermodynamic device which transfer heat from
low temperature body and gives out the same to high
temperature body.
COP = Work done (Q2)
Input Heat (Q2 – Q1)
= Q2 / (Q2 – Q1)
In terms of temperature COP= T2 / (T2-T1)

34. Refrigerator
 It is a thermodynamic device which absorbs heat
from low temperature body and gives out the same to
high temperature body.
 COP = Work done (Q1) / Input Heat (Q2 – Q1)
 = Q1 / (Q2 – Q1)
 In terms of temperature COP= T1 / (T2-T1)

35. Differentiate between heat pump
and refrigerator
Sr no Heat Pump Refrigerator
1 Function is to supply more and
more amount of heat to hot body
from cold body
Function is to absorb more and more
amount of heat from cold body and give
to hot body and maintain temperature of
cold body.
2 COP = Q2 / (Q2 – Q1) COP = Q1 / (Q2 – Q1)
3 COP Is more than COP of
Refrigerator for same working
temperature.
COP Is less than COP of pump for same
working temperature.
4 COP= T2 / (T2-T1) COP= T1 / (T2-T1)