Derivation of Entropy from Second Law
Consider the following combined system consisting of a
Carnot heat engine and a piston cylinder system
R
Qrev
Qrev
Wrev
E
WC =Wsys +Wrev
Q

Combined
system
Wsys
System
First Law applied to the combined system
dEC  Qrev  WC
WC  Qrev  dEC
For a Carnot Engine (reversible cycle)
Qrev TR

Q
T
substituting
WC 

Q
T
Qrev  Q 
TR
T
 TR  dEC
130
Let the system undergo a cycle while the Carnot cycle
undergoes one or more cycles. Integrate over entire cycle
(recall system energy is a state property)
 WC  
Q
T
 TR   dEC
The net work for one cycle is
 Q 
WC  TR   
 T  boundary
The integral must be evaluated at the system boundary
The combined system (cycle) draws heat from a single
reservoir while involving work WC
Based on K-P statement the combined system cannot
produce net work output  WC  0
 Q 
0


 T  boundary
This is the Clausius Inequality which is valid for all
thermodynamic cycles, reversible or irreversible.
131
For a reversible cycle (no irreversibilities in the system)
 Q 
 Int  0
 T  rev

Since the cycle integral of (Q/T) is 0, the quantity Q/T
is a state property, it does not depend on the path (similar
 dE  0 ).
We call this new property entropy, S,
 Q 
dS   
 T  int
rev
units : kJ/K
For a process where the system goes from state 1 to 2
 Q 
S 2  S1  12  
 T  int
rev
Specific entropy s= S/M
 Q / M 
2  q 
s2  s1  12 

 int 1   int
T

 rev
 T  rev
132
The entropy change between two specified states is the
same whether the process is reversible or irreversible
T
Irreversible
process
2
Internally
reversible process
1
S
To calculate S2-S1 evaluate  Q / T for the reversible path
Consider a cycle made up of an irreversible process
followed by a reversible process
T
Irreversible
process
1
2
Internally
Reversible process
S
Applying Clausius Inequality to the cycle
Q
2  Q 
1  Q 



1   2   int  0
T
 T b
 T  rev
133
 Q 
   S1  S 2   0
 T b
 Q 
S 2  S1  12  
 T b
2
1
Remove inequality sign to get the entropy balance
equation for a closed system
 Q 
S 2  S1  12    S gen
 T b
Change in entropy
of the system
Entropy transfer
to the system by
heat transfer
Entropy generated in
the system due to
irreversibilities
 0 irreversib le process

S gen  0 reversible process
 0 impossible process

Note: Moran and Shapiro uses symbol  instead of Sgen
134
For an isolated system (adiabatic closed system)
Sisol 
Since Sgen  0 
2
1
Q
T
 S gen
Sisol  0
This is the Increase in Entropy Principle which simply
stated says that for an isolated system the entropy always
increases or remains the same
Isolated
system
surroundings
W,Q
system
W,Q
S total  Ssystem  Ssurr  0
135
The increase of entropy principle does not say that the
entropy of a system cannot decrease
Isolated
system
Ssurr= +3 kJ/kg
Ssys=
-2 kJ/kg
Q
A process that is both adiabatic and reversible is referred
to as isentropic, and for a closed system
S isentropic  1
2
Q
T
 S gen  0
entropy is constant, S1=S2.
Microscopic Point of View:
Entropy is the measure of molecular disorder or
randomness. As a system becomes more disordered, the
position of the molecules becomes less predictable and
the entropy increases. Entropy is the lowest in a solid
because molecules are held in place and simply vibrate
and highest in a gas where the molecules are free to move
136
in any direction. Entropy of a pure crystalline substance
at absolute zero temperature is zero since the state of each
molecule is known  Third Law of Thermodynamics
Disorganized energy does not have potential to do useful
work,
M
M
Organize the KE of the molecules (on average moving in
one direction) by expanding through a nozzle increases
the ability to do useful work. Some of the IE of the
molecules is converted in to KE resulting in a temperature
drop and corresponding drop in entropy.
The paddle-wheel work done on a
gas increases the level of disorder
(entropy) of the gas and thus the
energy is degraded during this process
Heat is a form of “disorganized energy”
and some entropy will flow with it
T increases
M
Q
137
Evaluation of Entropy Change in a Closed System
System
Q
W
Apply First Law neglecting KE and PE effects, no shaft
work
In differential form du  q  w  q  Pdv
q  du  Pdv
For an internally reversible process ds 
q
T
so
Tds  du  Pdv
recall h  u  Pv  du  d (h  Pv)  dh  Pdv  vdP
substitute du into the above equation
Tds  (dh  Pdv  vdP)  Pdv
Tds  dh  vdP
These are the Gibb’s equations (or simply Tds equations)
138
For a system undergoing a process from state 1 to state 2
du 2 P
 1 dv
T
T
dh
v
s 2  s1  12 ds  12  12 dP
T
T
s 2  s1  12 ds  12
Note the terms on the RHS are state properties, so the
terms on the LHS, i.e., entropy, must also be a state
property
Although we derived the Tds equations for a reversible
process since entropy is a state property we can integrate
these equations to get the change in entropy between any
two states for any process
c ( P, T )dT 2 P
s2  s1 
 1 dv
T
T
c ( P, T )dT 2 v
s2  s1  12 P
 1 dP
T
T
2 V
1
The above integration is not straight forward, so entropy
is tabulated along with u and h in the steam tables and
plotted on T-s and h-s (Mollier) diagrams.
139
Entropy tabulated in steam tables is relative to an
arbitrary reference state
The value of entropy at a given state is determined just
like for the other state properties u and h
T
Liquid
P,Ts
Vapor
P,Ts
Liqud-vapor
S=(1-x)sf + xsg
s
140
Recall for an internally reversible process dS 
Therefore,
Qint  TdS
Q
T
rev
Heat added in process from 12 is Qint  12 TdS
rev
= Area under curve
For an isentropic process S is constant
T
2
1
S
No heat transfer  area under the curve is zero
141
Entropy Change for an Ideal Gas
For an ideal gas du  cV (T )dT , dh  cP (T )dT , Pv  RT
c (T )dT
du 2 P
dv
 1 dv  12 V
 R 12
T
T
T
v
v 
c (T )dT
s 2 (T2 , v2 )  s1 (T1 , v1 )  12 V
 R ln 2 
T
 v1 
s 2  s1  12
and
c (T )dT
dh 2 v
dP
 1 dP  12 P
 R 12
T
T
T
P
P 
c (T )dT
s 2 (T2 , P2 )  s1 (T1 , P1 )  12 P
 R ln 2 
T
 P1 
s 2  s1  12
The value of the specific entropy is set to zero at 0K and 1
atm, e.g., s(0K, 1 atm)= 0
The specific entropy at temperature T and 1 atm is
s (T ,1 atm)  s (0K,1 atm)  0T
c P (T )dT
 1 atm 
 R ln

T
1
atm


s o  s (T ,1atm)  0Tc p (T )
dT
T
142
Values of s o are tabulated as a function of T for air in
Table A-22
so
2
1 c p (T )
dT
dT 1
dT
 02c p (T )
 0 c p (T )
 s2o  s1o
T
T
T
Therefore, s 2  s1
 s
o
2
s
o
1
  R ln P 
2
 P1 
If the specific heats cP and cV are taken as constant
s 2  s1  cV 12
v 
dT
 R ln 2 
T
 v1 
 T2 
 v2 
s 2  s1  cV ln   R ln 
 T1 
 v1 
s 2  s1  c P 12
P 
dT
 R ln 2 
T
 P1 
 T2 
 P2 


s 2  s1  c P ln   R ln 
 T1 
 P1 
143
144