Second Law of
Thermodynamics - Entropy
Chapter 7
So far…
We have studied the second law by
looking at its results
We don’t have a thermodynamic
property that can describe it
In this chapter we will develop a
mathematical definition for entropy
 Entropy is a quantitative measure of
disorder
How does the second law relate
to disorder?
Disorder naturally increases, just like
water flows down hill and gases
expand
Natural processes proceed
spontaneously toward disorder
Disorder
Cleaning
your room
takes a lot of
effort
 Messing it up
is much
easier
Disorder
 Dissolving
sugar in water
increases
disorder
 Sugar
dissolves
spontaneously
 Sugar and
water don’t
separate
spontaneously
Energy
It takes the same amount of energy to
clean your room as to mess it up.
It takes the same amount of energy to
dissolve sugar as to separate it.
We all know that these are not
equivalent, reversible, processes.
Understanding Entropy
 What is it?
 That’s hard to put your hands around
 It’s a measure of disorder
 It’s a measure of how concentrated energy is
 The units of entropy are J/K
 We actually can define entropy about as well
as we can define energy (Not well at all)
 Remember – we define energy as the ability
to do work
Rudolf Clausius
 German Physicist
 Developed the
concept of entropy
http://scienceworld.wolfram.com/
biography/Clausius.html
Clausius
Inequality
Two part system
Reversible Cyclic
Device (heat
engine)
Piston Cylinder
Device
Consider an energy
balance for the entire
system
Ein  Eexit  Ec
QR  Wrev  Wsys   dEc
Wc  Wrev  Wsys
QR  Wc  dEc
If the heat engine is completely
reversible then…
Q
Q
 QR TR

Q T
 QR  TR
Q
T
This is based on Chapter 6,
where we said that heat
transfer rate in a reversible
system is proportional to the
temperature
That means that we can rewrite the energy balance,
and solve for the net work (the work of the combined
system.
 QR   Wc  dEc
Wc  TR
Q
T
 dEc
That means that we can rewrite the energy balance,
and solve for the net work (the work of the combined
system.
Wc  TR
Q
T
 dEc
To find the total net
work we need to
integrate over time
Wc  TR 
Q
T
0
  dEc
Why is this a cyclic
integral?
But we know that it is impossible for a system to
exchange heat with only one reservoir, and produce
work!!
Wc  TR 
Q
T
That means Wc can’t
be positive, but it can
be zero or negative!!
Wc  TR 
Q
T
0
Since TR is always positive…

Q
T
0
This is called the
Clausius inequality
It is equal to 0 for the
reversible case, and is
less than 0 for the
irreversible case
Can we prove that it is 0 for the reversible case?
If the “system” is
reversible, we can run it
backwards.
If it produces negative work
going one way, it will
produce positive work going
the other way
But we know you can’t exchange
heat with only one reservoir, and
produce work!!
Therefore, for a reversible system, the cyclic
integral must be 0!!!
 Q 

0


  T rev
What kind of properties have
cyclic integrals equal to 0?
Consider a piston cylinder
device
When you have gone
through a complete cycle,
the volume is back to where
it started
The cyclic integral is 0
The cyclic integral of
a property is 0!!!
 Qnet 
 S , the entropy


 T int,rev
Remember, only the cyclic integral is
equal to 0
Consider a cycle composed of two
processes
Arbitrary
2
Reversible
1
Specific Volume
Process 1 (from
state 1 to state 2)
is reversible
Process 2 (from
state 2 back to
state 1 is
arbitrary
Arbitrary
2
Reversible
1
Specific Volume

2
1
Since the process
is a cycle, the total
change in entropy
is 0
The change in
entropy from state1
to 2 is
The change in
entropy from state
2 to 1 is
 Qnet 

  S1  S2   0
 T  rev
So, the definition of a change in
entropy becomes….
Arbitrary
2
S2  S1  
2
1
Reversible
1
Specific Volume
 Qnet 


 T  rev
Or…
 Qnet 
dS  

 T  rev
6-2
The Entropy Change Between
Two Specific States
The entropy change
between two
specific states is the
same whether the
process is
reversible or
irreversible
Entropy is a “state
function”
S2  S1  
2
1
 Qnet 


 T  rev
Consider an adiabatic reversible process…
Q  0
So… S=0
A reversible, adiabatic process is
isentropic!!
But what if the process isn’t
reversible?
S 2  S1  
2
1
 Qnet 


 T irreversible
S  S 2  S1  
2
1
Sgen depends on
the process
 Qnet 

  S gen
 T 
Sgen is always
positive or 0
What kind of things cause a
system to be irreversible?
Friction
Unrestrained gas expansion
Mixing of two fluids
Etc
These factors plus heat transfer all
cause entropy to be generated
Consider an isolated system
Because it is isolated,
there is no heat
transfer or work
0
S 2  S1  
2
1
 Qnet 


 T 
Entropy always increases in an
isolated system
The equality holds if
all the processes
inside the system
are reversible
S  0
Now consider the universe
It can be considered to be an isolated
system!!
Suniverse  S system  S surroundings  0
Or…
S gen  S system  S surroundings  0
This is the second law of thermodynamics,
expressed mathematically!!
You can consider any system to be
isolated, if you draw your system
boundaries out far enough
Sgen is
always
greater than
0 for real
systems
Stotal  S gen  S system  S surroundings  0
Hopefully you have a better feel
for entropy now
How do you calculate entropy
changes for real systems?
 If it’s reversible adiabatic, S=0
 If it’s isothermal heat transfer
S 2  S1  
2
1
Qrev
 Qnet 

 
T
 T  rev
 If it’s not an ideal gas, just look it up on
the property tables
Entropy is a property
The
entropy of
a pure
substance
is
determined
from the
tables, just
like any
other
property
Some Remarks about Entropy
 Processes can occur in a certain direction only, not in
any direction such that
S gen  0
 Entropy is a non-conserved property, and there is no
such thing as the conservation of entropy principle.
The entropy of the universe is continuously
increasing.
 The performance of engineering systems is degraded
by the presence of irreversibilities, and entropy
generation is a measure of the magnitudes of the
irreversibilities present during that process.
Isentropic Processes
The entropy of a fixed mass can be
changed by
 Heat transfer
 Irreversibilities
If the entropy does not change, it is
isentropic
Engineering Devices
Many engineering
devices are
essentially
adiabatic
Most devices perform best when
irreversibilities are eliminated
Isentropic model serves as an
idealization of a real process
These devices work best when they
are isentropic
Assuming a device is isentropic
simplifies calculations
The assumption
that a process is
isentropic, gives us
a connection
between the inlet
and outlet
conditions – just
like assuming
constant volume, or
constant pressure
Property Diagrams Involving
Entropy
S 
Qint,rev
T
Qint,rev  TdS
2
Qint,rev   TdS
1
This area has no meaning for irreversible processes
Consider some special cases of
the Ts diagram
Isothermal Process
1
2
Isentropic Process
1
Q=0
Q
2
Entropy
Entropy
T-s Diagram for the Carnot
Cycle
Temperature
1
Isentropic
compression
Isothermal
expansion
W=Qin-Qout
Qin
2
Isentropic
expansion
3
4
Isothermal Qout
compression
Entropy
Mollier Diagrams
(h-s Diagrams)
For adiabatic,
steady flow
devices, h is a
measure of work,
and s is a
measure of
irreversibilities
Mollier Diagram for Steam, see
Appendix A-10
Summary
 We developed a definition for entropy
based on the Clausius inequality
 Qnet 
 S , the entropy


 T int,rev
Summary
 We showed that the change in
entropy is related to a reversible
process
S2  S1  
2
1
 Qnet 


 T  rev
 But that entropy is a state function
Summary
 A reversible, adiabatic process is also
isentropic
 The opposite statement is not true
 Many engineering devices are
adiabatic and work best when they
are reversible
 An isentropic model is often used to
determine the best results we can obtain
from an engineering device