1. Entropy
Two reversible adiabatic can not intersect each other
CA
B
C
isotherm
RA
AB and CB are two reversible adiabatic and they intersect at B.
AC is a reversible isotherm. Through AC we can transfer heat to the cycle ABC which is
A close cyclic process. From the cycle ABC there is no heat loss but work out put. So
This violates 2nd law of thermodynamics. Hence two RA can not intersect each other.
p
v
i
f
a b
RA
Rev isotherm
Any rev path
Through i, f there is a
Rev process. We construct
two rev adiabatic through
i and f, then a rev isotherm
ab such that:
Area under iabf = area under if,
First law for if : , πππ = π’ π β π’π + π€ππ
Process iabf: πππππ = π’ π β π’π + π€ππππ
π€ππ = π€ππππ, π π πππ = πππππ
p
v
πππππ = πππ + π ππ + π ππ = π ππ =πππ
Thus any rev path can be substituted by a
reversible Zig-zag path, between the same end
states, consisting of a rev adiabatic + rev
isotherm+ another rev adiabatic.
2. Clausius theorem
p
v
Any rev cyclic process
RA
Rev isotherm
T1
T2
Lets us consider a smooth close curve which represents a rev cycle. Then we take two
Elemental parts from this cycle as shown in the picture (upper part and lower part). The
upper part cab represented by an isotherm+ two RA and similarly the lower part of the
curve. The upper isotherm is at a temp say T1 and the lower isotherm is at T2. Now this
elementary process can be thought of as a Carnot engine which receives heat dq1 at T1
and rejects heat dq2 at T2. So for this elementary cycle
ππ1
π1
=
ππ2
π2
ππ ππ π€π πππ π€ππ‘β π‘βπ
Signs of heat transfer then we get
ππ1
π1
+
ππ2
π2
= 0. If we add all the elementary cycles such
that we cover the entire cyclic process then we get :
dq1
dq2
ππ1
π1
+
dq2
T2
+
dq3
T3
+ β― .
dqn
Tn
= 0 or
ππ
π
= 0 , hence
ππ
π
ππ π πππππππ‘π¦ ππππ π’, π, π£ πππ β π€βππβ
is called entropy , ds.
R1
R2
a
b
π
π ππ
π π 1
+ π
π ππ
π π 2
= 0, ο π
π ππ
π π 1
= π
π ππ
π π 2
Hence entropy is a property, independent of path
So, ππ =
ππ πππ£
π
or, π
π ππ
π
= π
π
ππ = π π βπ π
Clausius theorem
3. ππ =
ππ πππ£
π
; If ππ πππ£ = 0, then the process is called rev adiabatic and then ds=0
Hence s=Const;
T
S
A B
CD
ππ
ππ2
RA
AB is a general process
Either rev or irreversible,
All other processes are rev.
Consider an elemental cycle, as shown in the picture. ππ = heat supplied at π1,
ππ2 = βπππ‘ ππππππ‘ππ ππ‘ π2
1 β
ππ2
ππ πππ¦
β€ 1 β
ππ2
ππ πππ£
ππ2
ππ πππ¦
β₯
ππ2
ππ πππ£
ππ πππ¦
ππ2
β€
ππ πππ£
ππ2
ππ πππ¦
ππ2
β€
π
π2
; π ππππ
ππ πππ£
ππ2
=
π
π2
Or,
Or,
Or,
Or,
ππ πππ¦
π
β€
ππ2
π2
ο
ππ πππ¦
π
β€ ππ , πππ πππ¦ ππππππ π
For a reversible process:
ππ
π
= ππ
For any other process : ππ β₯
ππ
π
For any cyclic process:
ππ
π
β€ ππ
ππ
π
β€ 0
πΌπ
ππ
π
= 0, cycle is reversible,
ππ
π
< 0, cycle is irreversible and feasible
ππ
π
> 0, cycle is impossible
Clausius Inequality
4. Entropy principle
For any process undergone by the system, ππ β₯
ππ
π
, if the process is reversible then ππ =
ππ
π
For isolated system, dq=0, so, ππ β₯ 0, if isolated process is reversible, π‘βππ ππ = 0, π = π
For irreversible process, ππ > 0, π π πππ‘ππππ¦ πππ€ππ¦π πππππππ ππ
Causes of entropy increase: for a closed system entropy increases due to
(a) External interaction, (b) internal irreversibility
ππ = ππ ππ₯π‘ + ππ πππ‘
=
ππ
π
+ ππ πππ‘; so, ππ >
ππ
π
; ππππ¦ πππ πππ£πππ ππππ ππππππ π ππ πππ‘ = 0
ππ =
ππ
π
+ ππ πππο π 2 β π 1 = 1
2 ππ
π
+ π πππ1β2
ππ
ππ‘
=
π ππ£
π
+ π πππ
Entropy change Entropy transfer Entropy generation
dq
T
π πππ
π 2 β π 1
For an open system ππ
π
π πππ
π π
ππ π π +
π ππ£
π
+ π πππ = π π π π +
ππ ππ£
ππ‘
, for steady state
ππ ππ£
ππ‘
= 0
ππ π π +
π ππ£
π
+ π πππ = π π π π + π2 π 2 β π1 π 1
Integral form of the equation
Change of entropy
within the CV
system