University of Babylon /College Of Engineering
Electrochemical Engineering Dept.
Second Stage /Thermodynamics
Entropy
  1


QC
QH
QH
TH
QC
QH
1

TC
TH

QC
TC
TH
TC
If used numerical value for heats , Q H is positive and Q C negative , lead to :

Q
Q
QH
Q
 C  H  C 0
TH
TC
TH
TC
Thus for a complete cycle of a Carnot engine
Q
T
0
As known for any cyclic process    0 , and integration for complete cycle

dQrev
0
T
thus it can be concluded that
and its equal to :
dS 
dQ
or dQrev  TdS
T
S System  
Q
T1
S Surroundings 
Q
T2
Q
is represented a state function called Entropy S.
T
University of Babylon /College Of Engineering
Electrochemical Engineering Dept.
Second Stage /Thermodynamics
STotal  
Q Q

is a positive value , and for irreversible process the entropy
T1 T2
change is a positive , all spontaneous process [∆S ≥ 0 ]
STotal  
Q Q Q(T1  T2)


T1 T2
T1T2
Since T1 > T2 , the total entropy changes result from this irreversible process is
positive , ∆S Total becomes smaller as the difference between T 1 and T2 get
smaller.
Since entropy is a state function , the entropy change of any system and its
surroundings considered together is positive and approach zero.
Where ∆S Total = ∆S System + ∆S Surroundings ≥ 0
When a process is reversible and adiabatic dQ
is constant .
Rev.
= 0 , thus the entropy of system
During irreversible adiabatic process and the process is said to be isotropic
dQrev  TdS
 dQ
rev
S2

 T dS
S1
S2
QRe v 
 TdS
S1
The integration represent an area under curve , in this case entropy is the lost work.
University of Babylon /College Of Engineering
Electrochemical Engineering Dept.
Second Stage /Thermodynamics
Heat Engine
Heat engine involve a working fluid to and from which heat is transferred
 Receive heat Q from high temperature source.
 Concert part of this heat to work.
 Reject the remain with heat Q to low temperature sink.
For Carnot engine and by equating the two below equations

T1  T2
W
and  
T1
Q
 T 
Q(T1  T2 )
W T1  T2

W 
 Q1  2 
Q
T1
T1
T1 

This is the maximum work can be got ,
During the first case , the work lost is W  (STotal )(T2 )
Generally T2 is represented surroundings temperature .
Entropy changes of an ideal gas
For ideal gas goes from initial conditions (P 1V1T1) to final conditions (P 2V2T2)
d U= d Q – d W
(first law )
For reversible process , the first law equation becomes :
d U= d Q Rev – P d V
H = U + PV ( enthalpy definition )
dH=dU+PdV+VdP
d H = d Q Rev – P d V + P d V + V d P
d Q Rev = d H – V d P
RT
dQRe v  C P dT 
dP
P
dQRe v
dT
dP
 CP
R
( divided by T)
T
T
P
University of Babylon /College Of Engineering
Electrochemical Engineering Dept.
Second Stage /Thermodynamics
dT
dP
R
T
P
dS  C P
Integration from initial state T 1 and P1 to final state T 2 and P2
T2
 CP
S  S 2  S1 
T1
P
dT
 R ln 2
T
P1
Calculate entropy change at constant pressure process
Q
C dT
S 
 P
T
T
But
CP
 A  BT  CT 2  DT 2
R
S  R[ A  BT  CT 2  DT 2 ]
 R[ A ln
dT
T
T2
C
D 1 1
 B(T2  T1 )  (T22  T12 )  (  )
T1
2
2 T2 T1
By use mean heat capacity definition
T2
C Pms 
C
P
dT / T
T1
ln(T2 / T1 )
The subscript ms denote a mean value specific to entropy calculation

CPms
D 
 A  BTlm  TamTlm C 

R
(T1T2 ) 2 

Where Tam is arithmetic mean temperature , and T lm is the logarithmic mean
temperature
Tlm 
T2  T1
ln(T2 / T1 )