1.  WHY 2ND LAW OF THERMODYNAMICS:-
 The 1st law of thermodynamics states that a
certain energy flow takes place when a system
under goes a process or change of state.
δW=δQ
 But , it does not give any information on
whether that process or change of state is
possible or not.

2.  ACCORDING TO 1ST LAW OF THERMODYNAMICS:-
 Work is completely converted in to heat or
heat is completely converted into work.
 (δW = δQ and δQ=δW)
 Potential energy can be transformed into
kinetic energy or kinetic energy can be
transformed into Potential energy.
 (PE→KE and KE→PE)
 Heat flows from hot to cold or from cold to
hot.
 (Th → Tl and Tl → Th)
 Gas expands from high pressure to low
pressure or from low pressure to high

3.  LIMITATION OF 1ST LAW OF THERMODYNAMICS:-
 Heat is not completely converted into
work.(Q ˃ W)
 kinetic energy can not be transformed into
Potential energy.(KE ≠> PE)
 Heat flow from cold to hot is not possible.(Tl
≠> Th)
 Gas expands from low pressure to high
pressure is not possible.

4.  LIMITATION OF 1ST LAW OF THERMODYNAMICS:-
 1st law does not help to predict whether
the certain process is possible or not.
 A process can be proceed in particular
direction only , but 1st law does not gives
information about direction.
 1st law not provides sufficient condition for
certain process to take place.

5.  2ND LAW OF THERMODYNAMICS:-
Heat can not flow itself from cold body to hot
body.
 The 2nd law of thermodynamics is also
used to determine the theoretical limits for
the performance of mostly used in
engineering systems like heat engines and
refrigerators.

6.  DIRECTION OF CHANGE:-
 The second law of thermodynamics asserts that
processes occur in a certain direction and that the
energy has quality as well as quantity.
 In above example, two bodies are different
temperature are brought into contact, heat energy
flows from the body at high temperature to that at low
temperature. Heat energy never flow from lower
temperature level to higher temperature without
applying external work.
Hot
Container
Cold
Surroundings
Possible
Impossible

7.  KELVIN PLANK STATEMENT:-
 It is impossible to construct a heat engine
that operates on cycle to receive heat from
single reservoir and produce equivalent
amount of work.
 It implies that it is impossible to build a heat
engine that has 100% thermal efficiency .

8.  CLAUSIUS STATEMENT:-
 It is impossible to construct a device as
heat pump that operates in a cycle and
produces no effect other then the transfer
of heat from a lower temp body to higher
temp body.
 Heat can not itself flow from colder body to
hater body.

9.  EQUIVALENCE OF THE TWO STATEMENTS:-
 It can be shown that the violation of one
statement leads to a violation of the other
statement, i.e. they are equivalent.
1. Violation of Kelvin-plank statement
leading to violation of Clausius statement
.
2. Violation of Clausius statement leading to
violation of Kelvin-plank statement .

10.  VIOLATION OF KELVIN-PLANK STATEMENT LEADING
TO VIOLATION OF CLAUSIUS STATEMENT:-

11.  VIOLATION OF CLAUSIUS STATEMENT LEADING TO
VIOLATION OF KELVIN-PLANK STATEMENT :-

12.  ENTROPY:-
 Is a thermodynamic function of the state of the
system
 Can be interpreted as the amount of order or
disorder of a system
 As with internal energy, it is the change in entropy is
important and not its absolute value.
 Change in Entropy:-
The change in entropy S of a system when an
amount of thermal energy Q is added to a system by a
reversible process at constant absolute temperature
T is given by:
S = Q / T

13.  EXAMPLE:-
1. A heat engine removes 100 J each cycle from a heat
reservoir at 400 K and exhausts 85 J of thermal
energy to a reservoir at 300 K. Compute the change
in entropy for each reservoir.
 Since the hot reservoir loses heat, we have that:
S = Q / T = -100 J / 400 K = -0.25 JK-1
For the cold reservoir we have:
 S = Q / T = 85 J / 300 K = 0.283 JK-1
Therefore:
The increase in entropy of the cold reservoir is greater
than the decrease for the hot reservoir.

14.  ENTROPY OF IDEAL GAS:-
 An expression for the entropy change of an ideal
gas can be obtained from Eq.
 ds =
𝑑𝑢
𝑇
+
𝑃 𝑑𝑣
𝑇
----- Eq. 1
Or
 ds =
𝑑ℎ
𝑇
-
𝑣 𝑑𝑃
𝑇
----- Eq. 2
 by employing the property relations for ideal
gases. By substituting du = cv dT and P =
𝑹𝑻
𝑽
into
Eq. 1, the differential entropy change of an ideal gas
becomes
 ds = cv
𝑑𝑇
𝑇
+ R
𝑑𝑣
𝑣

15.  The entropy change for a process is obtained by
integrating this relation between the end states:
ΔS = 𝑐 𝑣
𝑑𝑇
𝑇
+ 𝑅 ln
𝑣2
𝑣1
------ Eq. 3
 A second relation for the entropy change of an ideal
gas is obtained in a similar manner by substituting dh
= cpdT and v =
𝑅𝑇
𝑃
into Eq. 2 and integrating. The result
is
ΔS = 𝑐 𝑝
𝑑𝑇
𝑇
+ 𝑅 ln
𝑝1
𝑝2
------ Eq. 4

16.  From Eq.(3) and (4), we see that for constant volume
process,
ΔS = 𝑐 𝑣
𝑑𝑇
𝑇
= 𝑐 𝑣 ln
T2
T1
For constant pressure process, we have
ΔS = 𝑐 𝑝
𝑑𝑇
𝑇
= 𝑐 𝑝 ln
T2
T1
For isothermal process,
ΔS = R ln
𝑣2
𝑣1
= R ln
𝑝1
𝑝2

17.  THE CARNOT PRINCIPLE:-
 The Carnot cycle is a thermodynamic process
that describes how a fluid is used to convert
thermal energy into work.
 Characteristics:-
 High Efficiency
 Multi-Source Engine
 Better reliability and easier maintenance
 Reversible
 Safe, discrete and oxygen-free
 Modularity and flexibility

18.  P-V DIAGRAM:-
 A pressure volume diagram is used to describe
corresponding changes
in volume and pressure in a system.
 The PV diagram, called an indicator diagram,
was developed by James Watt and his
employee John Southern (1758–1815) to
improve the efficiency of engines.

19.  This is the PV diagram…..
 isothermal segments (AB and CD) occur when
there is perfect thermal contact between the
working fluid and one of the reservoirs, so that
whatever heat is needed to maintain constant
temperature will flow into or out of the working
fluid, from or to the reservoir.

20.  adiabatic segments (BC and DA) occur when
there is perfect thermal insulation between the
working fluid and the rest of the universe,
including both reservoirs, thereby preventing
the flow of any heat into or out of the working
fluid.

21.  THERMAL EFFICIENCY:-
 The thermal efficiency of any engine working
between the temperatures of T1 and T2 is i.e.
increase the temperature difference under
which the engine works.