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Chapter 6.
The second law of Thermodynamics
1
Ranque and his tube: No moving parts
compressed air
2 kmoles, 4 atm., 294K
COLD
1 kmole, 1 atm.,
255K
HOT
hollow tube
1 kmole, 1 atm.,
333K
Is this a crazy idea or what?
2
In earlier chapters we alluded to aspects of thermodynamics which
require further discussion or analysis.
1. Many processes do not occur in nature even though they
are not forbidden by the first law. Is there some other law stating
that these processes cannot occur?
2. Is there some general way to decide if a process is
reversible or irreversible?
3. Can we write the first law solely in terms of state variables?
We examine these questions in this chapter.
We can convert đW to a state variable by multiplying it by the
integrating factor (1/P): đWr  dV
where V is a state variable
P
Similarly if we multiply đQ by an integrating factor which turns out
to be (1/T) we obtain an exact differential. We write đQr
S=entropy (reversible processes)
T
 dS
3
The r in the above equations emphasizes that we are dealing with
reversible processes.
Hence dU= đQ- đW becomes dU  TdS  PdV
The boxed equation is the first law in terms of state variables.
It connects any two neighboring equilibrium states.
Of course we must still prove that dS is an exact differential.
The boxed formula is probably the most important
equation in thermodynamics.
4
Irreversible processes.
A reversible process is one whose direction can be reversed
by an infinitesimal change in some property. It is a quasi-static
process in which no dissipative forces are present. All reversible
processes must be quasi-static, but not all quasi-static processes are
reversible (e.g., slowly letting the air out of a tire). In a reversible
process everything (including the surroundings) can be restored to
its initial state.
All natural processes are, in actual fact, irreversible.
How can we determine if a process is reversible or irreversible?
On the other hand any process involving dissipative work,
such as rubbing two solids together, is irreversible.
5
Dissipative work is done on the system. As an example consider a
system consisting of a stirrer and fluid. Work done by rotating the
stirrer is converted to “heat” energy. To reverse the process, the same
amount of heat would need to be extracted from the fluid to perform
the original amount of work, with no other changes. This does not
violate the first law. A new law, the second law of thermodynamics,
will be introduced which forbids this reverse process.
Consider a gas in one portion of a chamber which is then
allowed to undergo a free expansion into an evacuated section of the
chamber. The gas will not compress itself back to the original
volume. Again any process which attempts to restore the gas to its
original volume, with no other changes, would violate the second
law.
Consider heat flowing from a high-T body to a lower-T body
in the absence of other effects. The reverse process does not take
place. It is impossible to create a device to simply reverse the
process.
6
Again the second law would be violated.
The impossibility of the above reverse processes occurring is
contained in the second law of thermodynamics.
Clausius statement: It is impossible to construct a device that
operates in a cycle and whose sole effect is to transfer
heat from a cooler body to a hotter body.
Kelvin-Planck statement: It is impossible to construct a device
that operates in a cycle and produces no other effect than
the performance of work and the exchange of heat with a
single reservoir.
Carnot’s theorem
No engine operating between two reservoirs can be more
efficient than a Carnot engine operating between the same two
reservoirs.
Remember that a Carnot engine is an ideal reversible engine
and hence can be run as a refrigerator.
7
High T
R2=Carnot refrigerator (so reversible)
I= Engine (not an ideal reversible one)
T2
Q2
Q2
I
W
R2
Q2  W
Q2  W
T1
Low T
In the drawing all quantities are
magnitudes. The diagram is of a
self-contained device operating
between two reservoirs. One part of
the device is a Carnot refrigerator
and the other part is an engine. The
system is adjusted so that the work
output from I is used to drive R2.
Let us assume that the efficiency of I is greater than R2. We would
then have W  W
and so Q  Q
Q2
Q2
2
2
The heat extracted from the low temperature reservoir is
( Q2  W )  ( Q2  W )  Q2  Q2
By our assumption, this is positive.
8
The heat deposited to the high-T reservoir is:
Q2  Q2
Therefore the net result of this cyclic system is to extract some heat
from the low temperature reservoir and deposit it into the high-T
reservoir, without any work being done by the surroundings. This
violates the Clausius statement of the second law. Hence our
original assumption is false and the efficiency of I must be less than,
or at most equal to, the efficiency of the Carnot cycle.
(I )  (R )
9
Suppose now that I is replaced by a Carnot engine R1. There is no
reason that we cannot do this. The analysis proceeds as before and,
based on the above analysis, ( R1)  ( R 2)(1)
The nice thing about our Carnot devices is that they are reversible.
Hence we can reverse the engines so that R1 is now the refrigerator
and R2 is the heat engine. Again the analysis proceeds as before and
we obtain
( R 2)  ( R1)( 2)
Comparing equations (1) and (2) we conclude that
(R1)  (R2)
All Carnot engines (operating between the same
temperatures) have the same efficiency.
In the above analysis no particular working substance was assumed.
The efficiency of the Carnot cycle is independent of the working
substance.
10
Breaking news!
Two devils have invaded our laboratory from their den at the center
of the Earth.
The red devil is capable of violating the Clausius Statement of
the 2nd Law. That is, she can take energy from a low
T reservoir and deposit it into a reservoir at higher T, without
any other changes, in some cyclic process.
(She argues that it’s OK because the 1st Law is not violated.)
The blue devil is capable of violating the Kelvin-Planck Statement
of the 2nd Law by extracting energy from a single reservoir and
performing some work in a cyclic process without depositing any
energy into a reservoir at lower T. (He argues that it’s OK because
the 1st Law is not violated.)
11
We wish to demonstrate that, if the Clausius Statement of the 2nd
Law is violated, then the Kelvin-Planck statement is also violated.
We begin with an ordinary engine (blue). It takes some energy from
the high T reservoir, does some work and then dumps some waste
energy into the low T reservoir.
But the dreaded red devil (Violator of the
Clausius statement) sneaks into the system.
High T
The devil takes the energy that the engine
T2
deposited into the low T reservoir and
Q1
puts it back into the high T reservoir.
Q2
W
This complete system then takes
energy from the high T reservoir
Q1
Q1 engine
and performs some work with
no change in the low T reservoir.
Low T
T1
This violates the Kelvin-Planck
statement of the 2nd Law!
12
In other words, the Kelvin-Planck statement says that it is impossible
to construct any cyclical system (no matter how complex!) which
takes energy from a single reservoir and perform useful work. But
this statement is violated if we permit a violation of the Clausius
statement (with the aid of the red devil).
A violation of one statement of the 2nd Law leads to a violation of the
other statement of the 2nd Law. Hence they are equivalent.
13
We wish to demonstrate that, if the Kelvin-Plank Statement of
the 2nd Law is violated, then the Clausius statement is also
A device (R) would like to extract some
violated.
energy from the low T reservoir and
deposit the same amount of energy into the
High T
T2
high T reservoir. This does not violate the 1st
Law. We know that this is impossible as it
Q2  Q1 Q1 would cause Herr Clausius to be very
Q2
unhappy. Enter the blue devil (which
R
W
Q1 violates the Kelvin-Planck Statement.) He
extracts some energy from the high T
reservoir and uses it to perform an equal
Low T
T1
amount of work without depositing energy
into the low T reservoir.
He uses this work to drive the refrigerator. The result of this system is
to take a certain amount of energy from the low T reservoir and
deposit the same amount into the high T reservoir, without any
influence from, or change of, the environment external to the system.14
In other words, the Clausius Statement says that it is impossible to
construct any cyclical system (no matter how complex!) which
takes energy from a low T reservoir and transport it to a high T
reservoir with no change in the system or envionment. But this
statement is violated if we permit a violation of the Kelvin-Planck
statement (with the aid of the blue devil).
Again a violation of one statement of the 2nd Law leads to a
violation of the other statement of the 2nd Law. Hence they are
equivalent.
We are now going to study entropy in some detail. In preparation
for this discussion we prove a theorem given on the next slide.
15
To emphasize that our development is completely
general we introduce a generalized force Y (which
could be P) and a generalized displacement X
(which could be V) and we consider some
reversible process if as shown on the diagram.
isotherm
adiabat
reversible process
Y
i
a
b
f
adiabat
X
We now demonstrate the following: Any arbitrary
reversible process, in which the temperature may
change in any manner, can be replaced by two
reversible adiabatic processes connected by a
reversible isothermal process in such a way that
the heat exchanged over the isothermal process
equals that exchanged over the original arbitrary
process.
In the above diagram, adiabats are drawn passing
through i and f.
17
Curve ab represents an isothermal process. It is
selected so that the area under iabf is equal to
the area under the original curve. This ensures
that the work done in the two processes is equal.
Since the two paths connect the same states, the
change in internal energy must be the same
also. Therefore the heat exchanged must also
be the same. However, for the curve iabf, the
heat exchange takes place only during the
isothermal part. This completes the proof.
18
Now we consider some reversible cycle as
shown below.
adiabats
T2
P
.i
.f
reversible
cycle
T1
V
We draw a large number of adiabats, dividing
the cycle into a large number of narrow strips,
one of which is shown on the diagram. The
isotherms (T1 and T2) are drawn as discussed
above. We now have a Carnot cycle.
19
Heat energy Q2 enters the system during the
isothermal process at T2
Heat energy Q1 leaves the system during the
isothermal process at T1
We have previously shown, for a Carnot cycle,
Q1
T1

Q2
T2
Now we revert to a notation in which heat entering a
system is positive and heat leaving the system is
negative. Hence Q1  Q1 and we have
Q1
T1

Q2
T2
0
{Remember that “heat energy” means an energy transfer by the
20
heating process.}
Continuing with the other Carnot cycles we have
Q1
T1

Q2
T2

Q3
T3

Q4
T4
   0
The Q’s represent infinitesimal heat transfers.
Taking the limit, the sum becomes an integral and we have

đQ
R
T
0
Clausius Theorem: The integral of đQ/T around any
reversible cycle equals zero.
21
We can use the Clausius Theorem to show the following:
Consider 2 points (i, f) on any reversible cycle giving
0 

đQ
R
T
f
i
path 1  
đQ
T
i
f
đQ
T
path 2
 path 1   path 2
f
i
f
đQ
T
i
đQ
T
đQ/T is independent of the path taken. It depends only
on the state of the system at the initial and final points.
Hence we can introduce a state function S, called the
entropy, by
dS 
đQR
T
Reversible process!
22
The main points so far in this chapter are:
First Law: dU  TdS  PdV
Second Law:
Clausius statement: It is impossible to construct a device that
operates in a cycle and whose sole effect is to transfer
heat from a cooler body to a hotter body.
Kelvin-Planck statement: It is impossible to construct a device
that operates in a cycle and produces no other effect than
the performance of work and the exchange of heat with a
single reservoir.
T1
A Carnot engine is the most efficient engine.   1 
T2
Clausius Theorem

đQ
R
T
0
dS 
đQR
T
23
In preparation for the next point we review the efficiency.
Q1
T1
W Q 2  Q1


1
1
Carnot !!
Q2
Q2
Q2
T2
Q1
Remember that Q1 is negative. Hence
is negative.
Q2
Q1
Hence the smaller (more negative)
is, the smaller the
Q2
efficiency.
Q1
1
Q1


For example, if
,
 1 the efficiency is zero, while if
Q2
2
Q2
the efficiency is 0.5
24
Now consider an irreversible cycle between the same two reservoirs.
Since I < R this gives
Q1 I
Q2 I

Q1
Q2

T1
T2
Q1 I
T1
or

Q2 I
T2
or, for an infinitesimal heat transfer,
As before
I

đQ
T
0
đQ1I
TI

đQ2 I
T2
0
0
In this expression đQ/T is not an entropy. (!!!!!!!!!)
Combining our result with the expression for a reversible
process we have
đQ
Clausius Inequality.
T  0

This is sometimes taken as a statement of the second law.
Now we have a way to decide whether or not a process is reversible!25
Now we wish to examine, again, the change in entropy of an
irreversible process.
Consider going between two equilibrium states if by some
irreversible process. We complete the cycle by fi by some
reversible process. Because part of the cycle is irreversible


f
đQ I
T
i
i

f
đQ I
T

đQ
T

f
i
f

i
i
đQ R
T
đQ R
T
f
đQI
T

i
f
đQR
T
0
0
f
 S S   dS  i
f
đQ I
T
i
The change in entropy is greater than the integral going from
the initial state to the final state in the irreversible process!
26
Combining with the reversible case:
dS 
For isentropic processes: dS = 0 and so
For an adiabatic process đQ = 0 and so
đQ
T
đQ
T
0
dS  0
Now consider an isolated system. Then đQ =0 and so dS  0
The entropy of an isolated system increases in any
irreversible process and is unaltered in any reversible
process.
It is the system’s entropy that governs the direction of
spontaneous change.
27
Let us again consider an isolated system. Suppose we start at
some initial state which has an entropy Si .Now we begin
some process. As the process proceeds, the entropy can only
increase or, at best, remain unaltered. If now, at some stage,
we reverse the direction of the process, the entropy must
continue to increase, or remain the same. If the entropy
continues to increase, the system can never regain the initial
state with entropy Si . Hence it is obvious that , for a process
to be reversible, the entropy cannot change.
We now can state when a process is irreversible:
In an isolated system, an irreversible process is one in
which the entropy increases.
Now suppose the isolated system is in equilibrium, so no changes
are taking place in the system. Then dS system  0
If dS system  0 , this implies a change towards equilibrium. Since
the entropy of the system can only increase, this means that the
equilibrium state is the state of maximum possible entropy. This
is true regardless of the type of process taking place in the system.
Thus, at equilibrium, the entropy of an isolated system has its
maximum value with respect to all possible variations and the
condition for this maximum is that dS system  0
An example of an isolated system is a universe. We must then have
Suniverse  Ssystem  Ssurroundings  0
NOTE: One of these two terms can be negative!!!
Suniverse  0
29
This principle of increasing entropy gives a
direction for the sequence of natural events.
The laws of mechanics are second order
equations in time and are unaltered by the
transformation t  -t.
According to
mechanics all physical processes can run
backwards. This is not so. The process can
only go in a direction which increases the
entropy. Entropy provides the “arrow of time”
for physical processes.
As will be discussed later, increasing entropy
is associated with increasing disorder.
30
Entropy and available energy.
What is the significance of the statement that entropy is
created in an irreversible process? Of course, it tells us whether or
not a process is possible. For example a chemist might wish to know
if two substances will react under certain conditions. If the reaction
would decrease the entropy of the universe, it is not possible.
Something about the energy is irretrievably lost during an
irreversible process, even though the energy is conserved. This is
often stated as the “availability of the energy to do useful work”.
Consider a simple situation:- the transfer of Q from a high-T
reservoir T2 to a low-T reservoir T1 through a conductor.
This reduces the capacity of Q to be converted to do useful work.
Let us consider this in more detail.
Suppose an even lower temperature reservoir T0 exists.
Consider a Carnot engine operating between T2 and T0
and extracts energy Q from the high-T reservoir.
T0
W
1
Q
T2
31
T2
Q
same
Q
T1
Q
C
C
W
W
T0
Increasing T of
reservoirs
32
Now suppose that the heat Q had been transferred to the reservoir
at T1
and we now operate the same Carnot engine between
T1 and T0 extracting energy Q from the T1 reservoir, then
T0
W
1
Q
T1
The difference W  W  is the loss in work that Q can do as a
result of it being transferred from the reservoir at T2 to the
reservoir at the lower temperature T1 .
 T0 T0 
Q Q
W  W  Q    T0     T0 S
W  W
 T1 T2 
 T1 T2 
S is the increase in entropy which has resulted from the energy
transfer.
Although we have chosen to discuss a simple case, the result is
true in general. Although the energy is not lost, it has been degraded
in that it is less available do useful work.
33
To summarize:
All natural processes are irreversible
Every irreversible process makes energy less available to
do external work.
All natural processes continually make energy less available
to do useful work.
34
EXAMPLE: We consider an irreversible engine.
T2
Q2
I
W
Because the process is irreversible, we cannot
draw the cycle. We consider one complete cycle.
S ( I )  0
Since S is a state variable:
Q2
Q1
S (reservoirs )  

T2
T1
S(universe)  S(system)  S(reservoirs)
Must have S(universe)  0
Q1
T1

Q2
T2

I 
Since
Q1
T
 1
T2
Q2
Q1
T1
W
Q2
Q1
0

T1
Q2  Q1
Q2

T1 
 I  1     R
T2 


Q2
T2
1
Q1
T1

T2
Q2
Q1
Q2
I  R
35
EXAMPLE: A 0.5kg sample of water (the system) at 90oC cools
to room temperature (surroundings) at 293K. Calculate the entropy
J
changes.
cP  4180
kg  K
Since the entropy is a state variable, the change in entropy is the same
as if the water were cooled reversibly from T2  363 K to T1  293 K
P is constant, so
S(system)  
 T1 
dT
 mc P 
 mc P ln 
T
T
 T2 
T2
đQ

J   293 
 ln

S (system)  (0.5kg) 4180
kg  K   363 

T1
S (system)  448
J
K
The heat enters the room (surroundings), which is at constant T and so
its entropy increases (the entropy change must be positive).
The energy by the heating process that leaves the water is
T1
Q(system)   đ Q  mc P  dT  mc P (T1  T2 )
T2
36
The energy that enters the surroundings is mc P (T2  T1 )
 T2  T1 
J
  499
S(surroundin gs)  mc P 
K
 T1 
J
S(surroundin gs)  499
K
S(universe)  S(system)  S(surroundin gs)
J
J
J
S(universe)  448  499  51
K
K
K
J
S (universe)  51  0
K
37
EXAMPLE: Two reservoirs are separated by a diathermic
wall. We consider an interval of time over which some heat is
exchanged, the amount being sufficiently small so that the
temperatures of the reservoirs do not
adiabatic walls
change appreciably.
TH  500 K
TL  200 K
diathermic wall
Q  4  10 5 J
(S )universe  (S )system  (S )Hres  (S )Lres
(S )universe
QH QL
4  105 J 4  105 J




TH
TL
500 K
200 K
(S )universe
J
 1200
K
Can the heat flow be in the opposite direction? Why not?
38
EXAMPLE (Problem 6.5):A thermally insulated 50 ohm resistor
carries a current of 1A for 1 sec. The initial temperature of the
resistor is 100C, its mass is 5 grams and the specific heat capacity
at constant pressure is 850 J/(kgK).
(a) What is the entropy change of the resistor?
(b) What is the entropy change of the universe?
2
Q
i
Rdt
đ


(a) P  i 2 R đQ  i 2 Rdt 
  mc P mc P 
dT
 dT  P
i2 Rdt
dT 
mcP
Tf
S  mcP 
Ti
(1A)2 (50)(1s)
T 
 11.8K
J
3
(5  10 kg)(850
)
kg  K
Ti  283 K Tf  294.8K
mc P dT
đQ
dS 
but đ Q  mc P dT ds 
T
T
 Tf
dT
 mcP ln
T
 Ti


 

  (5  10 3 kg) 850 J  ln 294 .8 

kg  K   283 


39
J
S(resistor )  0.174
K
(b) Surroundings isolated, so
S(surroundin gs)  0
J
S(universe)  0.174
K
40
Remember:
đQ
dS 
T
(reversible process)
đ Q into system, the entropy increases
đ Q out of system, the entropy decreases
For any closed cycle (because U and S are state variables):
U  S  0

R
đQ
T
0
41
EXAMPLE:
Consider one cycle of this reversible
process. The heat capacities are:
P(Pa)
1370 .85 C
274.85 C
P2
b
c
(a) Q=?
J
CV  8
K
J
CP  10
K
b
J
 đ Q  ca CVdT  8 K (274K ) Qab  2192J
P1
a
0.85 C
9  10 3
V1
V (m3 )
J
Q

C
dT

10
(1096K ) Q bc  10960J
d
 đ b P
K

d
548.85 C
J
Q

C
dT

8
( 822K ) Q cd  6576J
V

3  đ
K
20  10
c
a
J
V2
Q

C
dT

10
( 548K ) Q da  5480J
 đ d P
K
For the cycle U  0
W  1096 J
(b) P  ? W=area enclosed W  (V2  V1 )( P2  P1 )
1096 J
P 
11  10 3 m3
Qcycle  1096 J
( P2  P1 ) 
W
(V2  V1 )
P2  P1  9.96  104 Pa
42
(c)

 Tf 
S  
S  C ln 

T
T
 Ti 
 Tb 
J  548 
J

 CV ln   8 ln
S ab  5.545
K  274 
K
 Ta 
đ Q  C dT
đQ  ?
T
Sab
 Tc 
J  1644 



Sbc  CP ln   10 ln
K  548 
 Tb 
Scd
Sda

đQ
T
 Td
 CV ln
 Tc

J  822 
  8 ln

K  1644 

 Ta 
J  274 

 CP ln   10 ln
K  822 
 Td 
0
J
Sbc  11.10
K
Scd
S da
J
 5.545
K
J
 11.10
K
because S is a state variable (like U)
43
EXAMPLE (Problem 6.1): A Carnot engine operates on 1 kg of
methane which we will consider to be an ideal gas. The value of
the ratio of specific heats is 1.35. If the ratio of the maximum to the
minimum volume is 4 and the efficiency is 0.25, find the entropy
increase of the methane during the isothermal expansion.
1 kg CH 4
b
T2

P
c
 1
adiabat
a
1kg
16kg
kmole
n  0.0635 kmole
  1.35 Vd  4Vb
adiabat

n
Q1
T
 1  1  0.25
Q2
T2
  0.25
Q1
T
3
 1 
Q2
T2
4
T2 
T1
isothermal expansion:

d
4
T1......(1)
3
V 
U  0 so Q2  W   PdV  nRT2 ln  c 
 Vb 
b
c
V
 Vc 
Q2
S(isothermal expansion) 
 nR ln .....(2)
T2
 Vb 
adiabatic expansion
Using (1)
 Vc

 Vd



 1
TV  1  constant
T1 3


T2 4
 Vc

 Vd
Vc Vc Vd  3 

  
Vb Vd Vb  4 
1
0.35
T2 Vc
 1
  3
   
 4
(4)
 T1Vd
1
 1
 1
 3
  
4
1
0.35
Vc
 1.76
Vb
From (2)
J
S(iso exp)  (0.0625kmole )( 8.314  10
) ln( 1.76)
kmole  K
3
J
S(isothermal expansion )  294
K
for CH 4
Energy enters the engine during this part of the cycle so the
entropy increases.
45
EXAMPLE (Problem 5.11): A reversible engine is connected to 3
reservoirs as shown below. During some integral number of complete
cycles the engine absorbs 1200J from the reservoir at 400K and
performs 200J of mechanical work.
(a) Find the quantities of energy exchanged with the other reservoirs,
and state whether the reservoir gives up or receives energy
(b) Find the change in entropy of each reservoir.
(a) U  Q  W
U  0 after n cycles
Q  W Q1  (Q2  Q3 )  W
W=200J
Q1  1200 J
R
Q2
T1  400 K
T2  300 K
Q3
1200 J  200 J  (Q2  Q3 )
T3  200 K
(Q2  Q3 )  1000 J .....(1)
For n complete reversible cycles, the entropy change of the engine=0
(see slide 20)
Q1 Q2 Q3


0
T1 T2 T3
Q2 Q3
Q
1200 J

 1 
T2 T3
T1
400 K
46
Q2 Q3
J

 3 .....(2)
T2 T3
K
From (1) Q3  Q2  1000 J
Placing in (2)
Q2 Q2  1000 J
J

 3
T2
T3
K
T3
J
Q2  Q2  1000 J  3 T3
T2
K
2
Q2  Q2  1000 J  600 J
3
From (1)
Q2  1200 J
Q3  Q2  1000 J
(b) For the reservoirs S   Q
Q1
1200 J
S1  

T
400 K
S3  
Q3
200 J

T
200 K
T
S1  3
J
K
S2  1 KJ
Q3  200 J
(Out of engine)
(Into engine)
where Q is for the engine
Q2
 1200 J
S2  

T
300 K
S2  4 KJ
ΔS1+ ΔS2 + ΔS3=0 (Of course)
It is the requirement that the entropy change be zero that heat is
transferred into the reservoir which has the middle temperature.
47
EXAMPLE: A mass of liquid at a temperature T1 is mixed with an
equal mass of the same liquid at a temperature T2. The system is T2
thermally isolated. Calculate the entropy change of the universe in
terms of cp and the temperatures and show that it is necessarily
positive. (Assume reversible process connecting states.)
Q(1)  Q(2)  0 mcP (T f  T1 )  mcP (T f  T2 )  0
1
T f  T1  T2  T f 2T f  T1  T2
T f  (T1  T2 ).....(1)
2
đ Q1 đQ 2  mc P dT   mc P dT 
dS  dS 1  dS 2 


 

T
T
 T 1  T  2
T
T


 T f 
dT 
 dT

  T f 



S  mcP  


mc
ln

ln


P 



T
T 


T
 T2 
  T1 

T

f
f
1
2
2




 T f  T f 
T
f
 
S  mcP ln     mcP ln 
 T T  
 T1  T2 
 1 2  
 T  T 
2 

S  2mcP ln  1


 2 T1T2 
(universe)
48
One way to show that this is an increase in entropy is to show that
T1  T2
1
2 T1T2
Let
T2
x
T1
f ( x) 
df
1
1 x

 3/2
dx 2 x 4x
d2 f
1
3(1  x)
  3/2 
2
dx
2x
8 x5 / 2
1 x
2 x
df
0
dx
at x  1
x 1
d2 f
1
 0
2
4
dx
Hence the minimum value of x=1 (temperatures the same) and
so the minimum change in entropy is zero
A second way is as follows:
1  x 1  x  2 x  2 x (1  x )2  2 x
f 


1
2 x
2 x
2 x
49
EXAMPLE: The value for c P for some substances
can be reasonably represented by cP  a  bT
(a) Find the heat absorbed and the increase in entropy of a mass m of
the substance when its temperature is increased at constant
pressure from T1 to T2.
(b) From a plot for the specific heat for Cu (see Figure 4.1) find the
change in the molal specific entropy for Cu, when the temperature
is increased at constant pressure, from 500 K to 1200 K.

T2

(a) đ Q  mc P dT Q  m c P dT  m (a  bT)dT
T1

b 2
2 
Q  ma(T2  T1 )  (T2  T1 )
2

T
2
đ Q mc P dT m(a  bT)dT
(a  bT )
ds 


S  m 
dT
T
T
T
T
T1


 T2 
S  m a ln    b(T2  T1 )
 T1 


50
J
kmole  K
J
3
c

29
.
75

10
at 1000K P
kmole  K
3
c

25
.
5

10
(b) From the graph, at 400K P
This gives b  7.08
s  2.27  104
J
kmole  K 2
a  2.27  104
J
kmole  K
 1200 
J
J
  7.08
ln
(1200  500 )K
2
kmole  K  500 
kmole  K
s  2.48  104
J
kmole  K
51
Example: A system is taken reversibly around the cycle
a b c  d  a
(a)
(b)
(c)
(d)
Calculate the heat transferred in each process.
Does the cycle operate as an engine or a refrigerator?
Find the efficiency of this cycle operating as an engine.
What is the coefficient of performance of this cycle
operating as a refrigerator.
500
b
c
a
d
R
4
3R
4
T (K )
200
J
S 
K
52
đ Q  TdS
đ Q  dU  đ W
(a)
 3R R 
 R 3R 
Qab  Qcd  0 Qbc  (500)
   250 R Qda  (200) 
  100 R
 4 4
4 4 
Qab  Qcd  0 Qbc  250R Qda  100R
(b) Q is positive for the cycle and there is no change in internal
energy so the work is positive and hence it operates as a
heat engine.
W Q bc  Q da 250R  100R 150R
(c)  



Q bc
Q bc
250R
250R
(d) COP 
heat out 100 R

W
150 R
  0.60
COP  0.67
53
(Treat as an ideal gas)
compressed air
2 kmoles, 4 atm., 294K
COLD
1 kmole, 1 atm.,
255K
HOT
hollow tube
1 kmole, 1 atm.,
333K
J
cP (air )  2.9110
kmole  K
dU  đQ  đ W C V dT  TdS  PdV
4
PdV  VdP  nRdT
CV dT  TdS  VdP  nRdT
nRT
TdS  CV dT  nRdT  VdP  ncP dT 
dP
P
 Tf 
 Pf
dT
dP
dS  nc P
 nR
S  nc P ln   nR ln
T
P
 Ti 
 Pi



54
Consider 1kmole, 4 atm, 294K
1kmole, 1atm, 255K
kJ
S  7.38
K
Consider 1kmole, 4 atm, 294K
1kmole, 1atm, 333K
kJ
S  15.15
K
The total change in entropy is then,
kJ
S  22.5
0
K
The scheme does not violate any law of thermodynamics!
55
Example: Construct a reversible process to show explicitly that the
entropy increases during the free expansion of an ideal gas.
For the free expansion of an ideal gas we have:
W  0 Q  0  U  0  T  0
To calculate the change in entropy we will consider a
reversible isothermal process. (Quasistatic expansion to final V)
dS 
đ QR
T
dU  PdV PdV nRdV



T
T
V
Vf
dV
S  nR 
 nR ln 
V
 Vi
Vi
Vf
S  0

 and

Vf  Vi
entropy increases
Ways to increase entropy: (1) add energy (2) increase volume
56
Recapitulation.
In this chapter we have introduced the extremely important concept
of entropy. Entropy and energy are very different.
Although there is a conservation law for energy, there is no
conservation law for entropy. (S is conserved only in a reversible
process.)
For example when hot and cold water are mixed, the energy
flow out of the hot water equals the energy flow into the cold
water. The total energy has not changed. However the entropy
increase of the cold water is greater than the entropy decrease of
the hot water. Hence the entropy of the system after the transfer
is greater than the entropy before the transfer. Where did the
additional entropy come from? Some entropy is created in the
mixing process and once created it can never be destroyed.
1st Law: Energy can neither be created nor destroyed.
2nd Law: Entropy can be created but it can never be destroyed.
57
Where do we go from here?
We will apply the 2nd Law to several situations.
We will also discuss the TdS equations. These useful equations
for entropy are written in terms of the measurable quantities
c P , c V ,  and  .
58