1. Equivalence of the statements
&
Causes of irreversibility
Prepared By
Janak Kanojia
Mtech
PDPU
(17MSE006)

2.  We know the two statements of second law of thermodynamics i.e. Kelvin-Planck & Clausius
statements.
 These statements may appear to be unconnected to each other at first glance.
 But it can be easily shown that they are virtually two parallel statements of second law.
 Which is exactly what it is proved in equivalence of statements.
“By showing that violation of one statements implies the violation of the second
and
vice-versa”
Equivalence of Kelvin-Planck & Clausius
statements

3.  Hence there will be two possibilities for the systems i.e.
a. Violation of Clausius statement Violates Kelvin-Planck’s statement.
b. Violation of Kelvin-Planck’s statement Violates Clausius statement.
Leads to
Leads to
Hot Reservoir
Cold Reservoir
H.P H.E
Hot Reservoir
Cold Reservoir
H.E H.P
a)Violation of Clausius statement b)Violation of Kelvin-Planck statement
Q1
Q1
Q2
Q1
Wnet =Q1-Q2
W=Q1
Q1
Q2
Q1+Q2

4. Cold Reservoir
Q1
Combined System
Wnet =Q1-Q2
Q1-Q2
Q1in=Q1out
H.
P
H.
E
Q1
Violation of Clausius statement
• Lets assume H.P violates Clausius statement requiring no work input to transfer heat from cold to hot reservoir.
• Now we take a H.E which follows Kelvin-Planck statement, working between two thermal reservoirs such that it
draws amount of heat Q1 equal to discharge of H.P, producing net work and rejecting Q2 heat to cold reservoir.
• As Q1in=Q1out for hot reservoir may then be eliminated so we see that the H.P and H.E acting together constitute
a heat engine operating in cycle producing work while exchanging heat only with one body at fixed temperature.
This violates the Kelvin-Planck Statement.
Violation of Clausius statement
Violation of Kelvin-Planck
statement

5. Violation of Kelvin-Planck statement
• Lets assume H.E violates Kelvin-Planck statement i.e. it works with 100% efficiency there is no heat loss.
• Now we take a H.P which follows Clausius statement, which extracts heat from cold reservoir and transfer to hot
reservoir at higher temperature with expenditure of work equal to what is provided by H.E.
• Here H.E and H.P together constitute a heat pump working in cycle whose sole effect is to transfer heat from
lower temperature to higher temperature body violating Clausius statement.
Cold Reservoir
Q1
Combined
System
Hot Reservoir
Q2
Q2
Q net= (Q1+Q2)in – (Q1)out
Violation of Kelvin-Planck
statement
Violation of Clausius statement

6. Causes of Irreversibility
• A reversible process is one which can be turned back such that both the system and the
surroundings return to their original states, with no other change anywhere else in the
universe.
• Hence for a process to be reversible it is carried out infinitely slowly with infinitesimal
gradient, so that every state passed through by system is in equilibrium.
 So what causes irreversibility ?
a. Lack of equilibrium during the process
b. Involvement of dissipative effects
A
B
V
P

7. • For a thermodynamics equilibrium, the process has to have following characteristics.
a. Mechanical equilibrium
b. Chemical equilibrium
c. Thermal equilibrium
• Any lack of equilibrium causes spontaneous change which is irreversibility.
Example: Consider an insulated container divided into chamber A and B by diaphragm chamber A contains mass
of gas and chamber B is evacuated, hence there will be flow of gas from A-B until pressure in both chamber is
equal.
By second law of thermodynamic it can be proved that this process is irreversible. Let us install an engine which
develops work at expense of expansion of gas passing through the engine for reversing of the system let a heat
source, consuming work develop by engine heat the gas in chamber B so that pressure in A is returned to its initial
value and chamber B is evacuated.
Cause due to lack of equilibrium
A BA B
Diaphragm E
W
Heat
source
Q=W

8.  The net result is a cycle, in which we observe
that net work output W is accomplished by
exchanging heat with a single reservoir
only violating kelvin-Planck statement.
 Hence the free expansion is irreversible.
E
W
Heat
source
Q=W

9. Irreversibility due to dissipative effects
The irreversibility of a process may be due to dissipative effects in which the work is done without
producing any equivalent increase in kinetic or potential energy of any system.
The transformation of work always involves dissipation of energy in the form of :
 Mechanical friction
 Magnetic hysteresis
 Electrical resistance
 Viscosity or fluid viscosity
 Inelasticity