Chapter 20

Entropy and second law of thermodynamics

1 Content
In this chapter we will introduce the second law of the thermodynamics. The following
topics will be covered.

Second law of thermodynamics

Reversible processes
Entropy
The Carnot engine
Refrigerator
Real engines

2 Second law of thermodynamics

If a closed system is in a configuration that is not the equilibrium configuration, the
most probable consequence will be that the entropy of the system will increase
monotonically.
If an irreversible process occurs in a closed system, the entropy of the system always
increases; it never decreases.
In a process that occurs in a closed system the entropy of the system increases for
irreducible processes and remains constant for reversible process. The entropy never
decreases. The second law of thermodynamics can be written as

S  0

3 Third law of thermodynamics

Nernst proposed what he calls the heat theorem (third law of thermodynamics). The
entropy of any object at 0 K is zero.

lim S  0
T 0

4 Non perfect, but realizable heat engine

An engine is a device for converting heat energy into work. The way a typical engine
operates is to absorb heat from some substance or reservoir at high temperature, to permit
the working substance to do work and thereby be cooled, and then reject some heat to a
reservoir at a lower temperature.

1
 It not only absorbs heat QH from a reservoir at temperature TH, but also rejects heat
QL to some second reservoir at some lower temperature TL. The engine does some work

W = QH – QL.

where W is the work done by the system (gas).

5 Non perfect, but realizable refrigerator

A refrigerator is an engine runs backward. Work is done on the refrigerator, and the
net effect is to cool reservoir and heat a hotter reservoir.

It removes heat QL from a reservoir at lower temperature TL and rejects heat QH to a

reservoir at higher temperature TH. The work W (= QH – QL) is done on the refrigerator to
make it function.

6 Traditional thermodynamic statements of the second law of thermodynamics

2
6.1 Lord Kelvin (William Thomson, 1st Baron Kelvin )

(26 June 1824 – 17 December 1907) was an Irish mathematical physicist and engineer.
At Glasgow University he did important work in the mathematical analysis of electricity
and thermodynamics, and did much to unify the emerging discipline of physics in its
modern form. He is widely known for developing the Kelvin scale of absolute
temperature measurement. He was given the title Baron Kelvin in honor of his
achievements and is therefore often described as Lord Kelvin. The title refers to the River
Kelvin, which flows past his university in Glasgow, Scotland.
He also had a later career as an electric telegraph engineer and inventor, a career that
propelled him into the public eye and ensured his wealth, fame and honor.

6.2 Rudolf Clausius

Rudolf Julius Emanuel Clausius (January 2, 1822 – August 24, 1888), was a German
physicist and mathematician and is considered one of the central founders of the science
of thermodynamics By his restatement of Sadi Carnot's principle known as the Carnot
cycle, he put the theory of heat on a truer and sounder basis. His most important paper,

3
On the mechanical theory of heat, published in 1850, first stated the basic ideas of the
second law of thermodynamics. In 1865 he introduced the concept of entropy.

6.3 Kelvin’s formulation of the second law of thermodynamics ((Kelvin’s

impossible engine))
Any device that converts heat into work by mean of a cyclic process is called a heat
engine. Perfect heat engine: It extracts heat from a reservoir and performs an equivalent
amount of work without producing any other effect on the environment.

(a) It is impossible for any cyclic process to occur whose sole effect is the extraction
of heat from a reservoir and the performance of an equivalent amount of work.
(b) It is impossible by means of any inanimate agency to derive mechanical work
from any portion of matter by cooling it below the lowest temperature of its
surroundings.

6.4 Perfect refrigerator ((Clausius impossible engine))

It removes heat Q from the reservoir at low temperature and transfer it to the reservoir
at high temperature without affecting the environment in any other way.

4
 Perfect refrigerator

(a) It is impossible to construct a perfect refrigerator. Heat cannot be taken in a

certain temperature and converted into work with no other change in the system
or the surrounding. In other words, heat cannot flow by itself from a cold to a hot
place.
(b) It is impossible for a self-acting machine to convoy heat continuously from one
body to another which is at a high temperature.

6.5 Derivation of from Kelvin’s impossible engine to the Clausius impossible

engine

or equivalently

5
 One can show that the Kelvin and the Clausius statements are equivalent to one
another by proving that if the Kelvin statement is not true, neither is the Clausius
statement. Then we show that if the Clausius statement is not true, neither is the Kelvin
statement.
Suppose that the Kelvin’s statement is false and that we have an engine which
removes heat from a reservoir and does work. We now permit the impossible Kelvin
engine to run a conventional engine backboard as a refrigerator, removing heat Q1 from
the cold reservoir and delivering Q2 to the hot reservoir. Since the internal energy of these
engine remains the same, W = Q and W = Q2 – Q1. Therefore Q2 = Q1 + Q. The
combination of the two devices is a self-acting device which removes a quantity of heat
Q2 from a cold reservoir and delivers it to the hot one in violation of the Clausius
statement.

6.6 Derivation of Clausius impossible engine to the Kelvin’s impossible engine

6
 Assuming that the Clausius statement is false, we construct an engine which takes Q2
from a cold reservoir and delivers it to a hot one, and we operate it simultaneously with a
conventional engine which removes Q2 from a hot reservoir, does work W, and delivers
Q1 to the cold reservoir. The combination of these devices removes heat Q2 – Q1 from a
cold reservoir and does work W. The two devices working together therefore violate the
Kelvin’s statement.

7 Carnot engines for an ideal gas

A theoretical engine developed by Sadi Carnot. A heat engine operating in an ideal
reversible cycle (now called a Carnot Cycle) between two reservoirs is the most efficient
engine possible. This sets an upper limit on the efficiencies of all other engines. It is of
interest to exhibit explicitly how such an engine operating quasi-statically between heat
reservoirs can be constructed.
Such an engine is the simplest conceivable engine and is called a “Carnot engine.”
The Carnot engine goes through a cycle consisting of four steps, all performed in a quasi-
static fashion. After four steps, the engine is back in its initial state and the cycle is
completed.

All real engines are less efficient than the Carnot engine because they all operate
irreversibly so as to complete a cycle in a brief time interval.

((Nicolas Léonard Sadi Carnot (1796–1832)))

His famous work on the motive power of heat (Réflexions sur la puissance motrice du
feu, 1824) is concerned with the relation between heat and mechanical energy. Carnot
devised an ideal engine in which a gas is allowed to expand to do work, absorbing heat in
the process, and is expanded again without transfer of heat but with a temperature drop.
The gas is then compressed, heat being given off, and finally it is returned to its original
condition by another compression, accompanied by a rise in temperature. This series of
operations, known as Carnot's cycle, shows that even under ideal conditions a heat engine
cannot convert into mechanical energy all the heat energy supplied to it; some of the heat
energy must be rejected. This is an illustration of the second law of thermodynamics.
Carnot's work anticipated that of Joule, Kelvin, and others.

((Note)) How Kelvin and Clausius discovered Carnot’s ideas?

7
Carnotcycle the classical blog on thermodynamics

http://carnotcycle.wordpress.com/2012/08/04/how-kelvin-and-clausius-discovered-
carnots-ideas/

7.1 Overview of the processes in a Carnot Cycle.

7.2 Isothermal process

A(a, 1) to B (b, 2) is an isothermal expansion. The gas is placed in contact with the high
temperature reservoir TH. The gas absorbs heat QH. The gas does work WAB in raising the
piston.

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 V2
dV V
W12    PdV   RTH V   RTH ln( 2 )
V1
V1
E12  0  Q12  W12
V
QH  Q12  W12  RTH ln( 2 )
V1

where W12 is the work done on the system (gas).

7.3 Adiabatic process

B (b, 2) to C (c, 3) is an adiabatic expansion. The base of the cylinder is replaced by a

thermally nonconducting wall. No heat enters or leaves the system. The temperature falls
from TH to TL. The gas does work WBC.

V3 1 
dV  V R
W23    PdV    P2V2    P2V2 (

) |VV32   (T  T )
V2
V 1   1 H L
 CV (TH  TL )
Q23  0
E23  W23  CV (TH  TL )

where

R
  1
CV

W23 is the work done on the system during this process.

7.4 Isothermal process

9
 The gas is placed in contact with the cold temperature reservoir. C (c, 3) to D (d, 4) is
an isothermal compression. The gas expels energy QL. W34 is work done on the system
(gas).

V4
dV V V V
W34    PdV   RTL    RTL ln( 4 )  RTL ln( 3 )  RTL ln( 2 )
V3
V V3 V4 V1
E 34  0  Q34  W34
V2
Q34  Q L  W34   RTL ln( )
V1

((Note-2))

In the following way, we show that

V3 V2
 .
V4 V1

From the relations,

 1  1
THV2  TLV3
 1  1
THV1  TLV4

we have

V3 V2
 .
V4 V1

((Note-2)) Another proof

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From the relations

P1V1  P2V2
 
P2V2  P3V3
P3V3  P4V4
 
P4V4  P1V1

In multiplying on both sides, we have

   
( P1V1 )( P2V2 )( P3V3 )( P4V4 )  ( P2V2 )( P3V3 )( P4V4 )( P1V1 )

or

   
V1V2 V3V4  V2V3 V4V1

or

1  1  1  1 
V1 V3  V2 V4

or

V3 V2

V4 V1

7.5 Adiabatic process

D (d, 4) to A (a, 1) is an adiabatic compression. The gas is again placed against a
thermally nonconducting wall. So no heat is exchanged with the surroundings. The
temperature of the gas increases from TL to TH. W41 is the work done on the system (gas).

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 V1
dV
W41    PdV    P4V4


V4
V
 CV (TH  TL )
Q41  0
E 41  W41  CV (TH  TL )

((Mathematica))
Green lines: isothermal process
Red lines: adiabatic process ( = 5/3 here)
P
6

V
50 100 150 200

8. Carnot cycle. PV diagram

12
 V
QH  RTH ln( 2 )
V1
V3 V
QL   RTL ln( )  RTL ln( 2 )
V4 V1
W  W12  W23  W34  W41
V2 V
  RTH ln( )  CV (TH  TL )  RTL ln( 2 )  CV (TH  TL )
V1 V1
V2
  R(TH  TL ) ln( )  QL  QH
V1

W is the total work done on the system during the process. For convenience we redefine
W by -W. The new definition of W is the work done by the system.

W  QH  QL

QL QH
 (Carnot cycle)
TL TH

The efficiency R is defined as

W Q  QL T
R   H 1 L
QH QH TH

It is dependent only on the temperatures TH and TL. Although proved for a perfect gas
engine, it must be true for reversible engines. We mean that for a reversible engine the
following two processes are possible.

((Reversible engine))

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What is the definition of the reversible engine?

In reversible engine, all the processes are reversible.

9 Reversible cycle for an ideal gas

An arbitrary reversible cycle, plotted on a PV diagram consists of a family of

isotherm lines and adiabatic lines. We can approximate an arbitrary cycle as closely as
we wish by connecting the isotherms by short, suitably chosen, adiabatic lines. In this
way, we form an assembly of long; thin Carnot cycles.
The traversing the individual Carnot cycles in sequence is exactly equivalent to
traversing the jagged series of isotherms and adiabatics that approximate the actual cycle.

In Carnot cycle,

we put

Q1 = QH, Q2 = -QL

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We extend the Eq. given by

Q1 Q2
 0
T1 T2

by writing the isotherm-adiabatic sequence of lines

Qi
T i
0
i

In the limit of infinitesimal temperature differences between the isotherms, we have

dQr
 T
0 (reversible cycle).

15
16
 P

30

25

20

15

10

V
50 100 150 200 250 300

Fig. Combination of the isothermals and adiabatics.

10 Entropy in the reversible process

Entropy is a measure of disorder of a state. Entropy can be defined using macroscopic
concepts of heat and temperature

dQr
dS  (reversible process)
T

Entropy can also be defined in terms of the number of microstates, W, in a macrostate

whose entropy is S,

S = kBlnW

where W is the number of micro-states. This will be discussed in the next chapter
(Chapter 20S).
The entropy of the universe increases in all real processes. This is another statement
of the second law of thermodynamics. The change in entropy in an arbitrary reversible
process is
f
dQr
S  
i
T

For any reversible cycle, in general,

dQr
 T
0

17
This is called Clausius’ theorem. The importance of Clausius’ theorem is that it permits
us to define a new physical quantity called the entropy, or more precisely, the entropy
difference.

The integral symbol indicates the integral is over a closed path.

We can move around on a PV diagram all over the plane, and go from one condition to
another. In other words, we could say that the gas is in a certain condition a, and then it
goes over to some other condition b.

We will require that the transition, made from a to b reversible. Now we go around
the path 1 (a→K1→b→K3 →a). Then we have

dQr dQr

aK 1b
T
 
bK 3 a
T
0 (1)

Next we go around the path 2 (a→K2→b→K3 →a),

dQr dQr

aK 2 b
T
 
bK 3 a
T
0 (2)

Subtracting Eq.(1) from Eq.(2),

dQr dQr

aK 1b
T
 
aK 2 b
T

which does not depend on the path taken.

We define the entropy to go from a to b by a reversible process

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 b
dQr
Sb  S a  
a
T

The entropy of a system is a function of the thermodynamic co-ordinates whose change is

equal to the integral of dQr/T between the terminal states, integrated along reversible path
connecting two states.

11 T-S diagram for the Carnot cycle

For each infinitesimal amount of heat that enters a system during an infinitesimal portion
of a reversible process,

dQr  TdS .

In the case of a reversible adiabatic process, we have

dQr
dS  0
T

If T is not zero, dS = 0 and S is constant. Therefore, during a reversible adiabatic process,

the entropy of a system remains constant. In other words, the system undergoes an
isentropic process.
If two equilibrium states are infinitesimally near, then we have

dQ  TdS
dQ dS
T
dT dT

At constant volume,

 dQ   dS 
   CV  T  
 dT V  dT V

and at constant pressure,

 dQ   dS 
   CP  T  
 dT  P  dT  P

The work done by the system is the area of the rectangle:

dE  TdS  PdV

 dE  0   TdS   PdV

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Then we have

 TdS   PdV  W
where W is the work done on the system and –W is the work done by the system.

((Note))

In a Carnot cycle,

Q1 Q2
  S
T1 T2

Then the work done by the system (-W) is given by

Q1 T
Area= S (T1  T2 )  (T1  T2 )  Q1 (1  2 )
T1 T1

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 Area T
The efficiency  R  1 2
Q1 T1

12 Reversible Engine
We show that no engine can do more work than a reversible one. Suppose that A is a
reversible engine (a Carnot cycle), and that B is also a reversible engine.

A: reversible engine (Carnot cycle) B: reversible engine

We consider the combined engine of A and B. Since A (Carnot cycle) is the reversible
engine, we have

Note that A is a reversible engine. This system is equivalent to the engine of A+B.
Net effect is to extract a net heat W1-W from the reservoir at TL and convert it into work.
This process is prohibited because of Kelvin’s impossible engine: it is impossible for any
cyclic process to occur whose sole effect is the extraction of heat from a reservoir and the
performance of an equivalent amount of work.
Thus one can get

W1-W≤0. (1)

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Since B is the reversible engine, we have

This process is prohibited because of the Kelvin’s impossible engine. Hence

W-W1≤0. (2)

From Eqs.(1) and (2), one can get W = W1. So if both engines are reversible, they must
both do the same amount of work.

Here is the Carnot’s brilliant conclusion.

If one engine is reversible, it makes no difference how it is designed.

13 Carnot inequality (irreversible engine)

For a reversible engine, we have

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 Q1 Q2
 0
T1 T2

where T1>T2, Q1>0, and Q2<0.

The efficiency R is defined as

WR Q1  Q2 Q Q T
R    1 2  1 2  1 2
Q1 Q1 Q1 Q1 T1

For an irreversible engine

The combination of A and B gives

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Then we have

W1  WR

for a fixed Q1 from the high temperature reservoir. The efficiency 1 is given by

W1 WR T
1    R  1 2
Q1 Q1 T1

Since WI  Q1  Q21 (Q21<0), we get

Q1  Q21 T Q21 T
1 2 or  2
Q1 T1 Q1 T1

Finally we obtain the Carnot inequality.

Q1 Q21
  0,
T1 T2

or

Q1 Q2
  0, for irrevercible cycle (for convenience Q21 = Q2).
T1 T2

For any irreversible cycle, in general,

24
 dQ
 T
0

The integral symbol indicates the integral is over a closed path.

14 Entropy in the irreversible process (I)

We consider the cycle which include both reversible and irreversible process. One has
the Clausius inequality.

dQ
 T
0

We go around the path. Then we have

dQirr dQr
a b
 T
 
ba
T
0
Irreversible Re versible

which leads to

dQirr dQr
a b
 T
 
a b
T
 Sb  S a  S
Irreversible Re versible

Second law of thermodynamics

In any process in which a thermally isolated system (Q = 0) goes from one
macrostate to another macrostate, the entropy tends to increase.

dQirr
 T
 S ,

25
In a small irreversible change, we hae

dQirr
 dS
T

When the system is isolated or the system is adiabatic (no heat exchange between the
system and surroundings),

dS  0

This means that the entropy of an isolated system either remains constant or increases.

((Note)) Clausius impossible engine

We consider the Clausius impossible refrigerator. The change of entropy can be
evaluated as

S 
Q ( Q )

T  TH   0
Q L
TH TL TH TL

This is inconsistent with the second law of thermodynamics (S>0).

Fig. Clausius impossible engine.

15 Entropy in the irreversible process (II)

In order to define the entropy change S for an irreversible process that takes one
from an initial state i to a final state f of a system, we find a reversible process that
connects states i and f. We then calculate
f f
dQ dQ
S  S f  Si   r   irr  0
i
T i
T

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In order to find the entropy change for an irreversible path between two equilibrium
states, find a reversible process connecting the same states, and calculating the entropy
change.

((Example-1)) Contact of two systems with different temperatures

Suppose that waters (mass m) are in two containers separately. The temperatures of these
containers are T1 and T2. The heat capacity of water per unit mass is C. We consider that
these two containers are in contact. The final temperature is Tf = (T1+T2)/2 in the thermal
equilibrium. The change of entropy for the container 1 is

Tf
mCdT T
S1  
T1
T
 mC ln( f )
T1

The change of entropy for the container 2 is

Tf
mCdT Tf
S 2   T
T2
 mC ln(
T2
)

The resultant change of entropy is

Tf T
S  S1  S 2  mC[ln( )  ln( f )]
T1 T2

or

27
 2
Tf
S  S1  S 2  mC ln( )0
T1T2

where

2
T T  Tf 
T f  1 2  T1T2 or    1
2  T1T2 

We note that when T2  T1 , we have

mC (T2  T f )  mC (T f  T1 )

or

T1  T2
Tf  .
2

((Example-2))
Pure heat transfer, not involving any work, is irreversible in energy transfer, if it takes
between two systems having different temperatures.

dE1  dW1  dQ1  dQ1  T1dS1

dE2  dW2  dQ2  dQ2  T2 dS 2

Here the total internal energy is constant.

dE1  dE2  0  dQ1  dQ2

The newly created entropy,

28
 dQ1 dQ2 dQ1 dQ1 T T
dStotal  dS1  dS 2      ( 2 1 )dQ1
T1 T2 T1 T2 T1T2

The heat fows from high temperature to low temperature. If dQ1>0, then T2>T1. So that

dStotal  0

16. Adiabatic free expansion (irreversible process)

An adiabatic free expansion of an ideal gas i.e. where a greater volume suddenly
becomes available to the gas is an irreversible process which proceeds through a chaotic
non-equilibrium path. Nonetheless we can characterize the beginning and end points and
the net values of relevant changes in energy. Since the gas expands against a vacuum it
does no work and thus

Wi  f  0 .
since there is no motion of the boundary (nothing to push against; there is no movable
piston). Combining this with our requirement that the process is adiabatic, we have

Ei  f  Qi  f  Wi  f  0  0  0

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If we are dealing with an ideal gas, then the absence of a change in the internal energy
implies that the temperature is the same before and after the expansion even though no
temperature is defined during the irreversible process: Tf = Ti.
In order to calculate the entropy of this process, we need to find an equivalent
reversible path that shares the same initial and final state. A simple choice is an
isothermal, reversible expansion in which the gas pushes slowly against a piston. Using
the equation of state for an ideal gas this implies that

PiVi  Pf V f

The initial and final states a (Pi, Vi) and b (Pf, Vf) are shown on the P-V diagram. Even
though the initial and final states are well defined, we do not have intermediate
equilibrium states that take us from the state a (Pi, Vi) and the state b (Pf, Vf).

Fig. Note that the irreversible process (green line) cannot be described in such a line in
the P-V phase diagram. The isothermal process is denoted by the blue line.

We thus replace the free expansion with an isothermal expansion that connects states i
and f. Then the entropy can be calculated as follows.

E  0
RT
dQr   dW  PdV  dV (reversible process)
V
Vf V
dQ
f
dV V
S   r  R   R ln( f )
V
T V
V Vi
i i

Since Vf>Vi, S is positive. This indicates that both the entropy and the disorder of the
gas increase as a result of the irreversible adiabatic expansion.

30
17. The entropy for the adiabatic free expansion (microscopic staes)
Entropy can be treated from a microscopic viewpoint through statistical analysis of
molecular motions. We consider a microscopic model to examine the free expansion of
an ideal gas. The gas molecules are represented as particles moving randomly. Suppose
that the gas is initially confined to the volume Vi. When the membrane is removed, the
molecules eventually are distributed throughout the greater volume Vf of the entire
container. For a given uniform distribution of gas in the volume, there are a large number
of equivalent microstates, and the entropy of the gas can be related to the number of
microstates corresponding to a given macrostate.

Fig. The volume of the system in the initial state is Vi (the macrostate). The volume of
cell (the microstate) is Vm. The number of cells (sites) is given by the ratio Vi/Vm.

We count the number of microstates by considering the variety of molecular locations

available to the molecules. We assume that each molecule occupies some microscopic
volume Vm . The total number of possible locations of a single molecule in a macroscopic
initial volume Vi is the ratio

Vi
wi  ,
Vm

which is a very large number. The number wi represents the number of the microstates,
or the number of available sites. We assume that the probability of a molecule occupying
any of these sites are equal.

31
 Neglecting the very small probability of having two molecules occupy the same site,
each molecule may go into any of the wi sites, and so the number of ways of locating N
molecules in the volume becomes

N
V 
  i  .
N
Wi  wi
 Vm 

Similarly, when the volume is increased to Vf, the number of ways of locating N
molecules increases to
N
N Vf 
Wf  wf    .
 Vm 

Then the change of entropy is obtained as

S  S f  Si
 k B ln W f  k B ln Wi
N N
V  V 
 k B ln f   k B ln i 
 Vm   Vm 
 Nk B [ln(V f )  ln(Vm )]  Nk B [ln(Vi )  ln(Vm )]
 Nk B [ln(V f )  ln(Vi )]
Vf
 Nk B ln( )
Vi

When N  N A , we have

Vf V
S  N Ak B ln( )  R ln( f )
Vi Vi

We note that the entropy S is related to the number of microstates for a given macrostate
as

S  k B ln W .

The more microstates there are that correspond to a given macrostate, the greater the
entropy of that macrostate. There are many more microstates associated with disordered
macrostates than with ordered macrostates. Therefore,it is concluded that the entropy is a
measure of disorder. Although our discussion used the specific example of the free

32
expansion of an ideal gas, a more rigorous development of the statistical interpretation of
entropy would lead us to the same conclusion.

18. Example:calculation of the entropy in the reversible process

Fig. Four paths (in the P-V phase diagram) used for the calculation of the
change in entropy. A1  A3 (isothermal, denoted by red). A1  A4  A3
(partly adiabatic, denoted by blue). A1  A0  A3 (denoted by green).
A1  A2  A3 (denoted by purple).

As is shown, the change in the entropy is defined for the reversible process. Here we
calculate the entropy along the four paths, where the path starts from the point A1 and
reaches the point A3 in the P-V diagram as shown above. It is shown that the change of
entropy is independent of the path chosen.

(a) Path-1 (isothermal process): A1 → A3 (denoted by red line)

PV  RTi , E  0 , Q   W  PV

Then the entropy change is

V2 V
PdV 2
dV V
S  V T  R V V  R ln( V12 ) . (1)
1 1

33
(b) Path-2 (adiabatic process): A1 → A4 → A3 (denoted by blue line)

Path: A1 → A4: (adiabatic process)

Q  0 , S1  0 ,

which means that the change of entropy is zero.

Path: A4 → A3 (constant volume)

PV  RTi , E  Q  CV T ,

Then the change of entropy is

Ti
dt T V  V 
S 2  CV   CV ln( i )  CV (  1) ln 2   R ln 2  .
T4
T T4  V1   V1 

where

 1
 1  1 Ti  V2 
TiV1  T4V2 , or  
T4  V 1 

Then we have

V 
S  S1  S 2  R ln 2  . (2)
 V1 

(c) Path-3: A1 → A0 → A3 (denoted by green line)

Path: A1 →A0

Q  CV T ,

T0
dT T 
S1  CV   CV ln 0  .
Ti
T  Ti 

Path: A0 →A3

Q  C P T ,

34
 Ti
dT T 
S 2  C P   C P ln i  .
T0
T  T0 

Then we have

S  S1  S 2
T  T 
 C P ln i   CV ln 0 
 T0   Ti 
T  T 
 ( R  CV ) ln i   CV ln 0 
 T0   Ti 
T 
 R ln i 
 T0 

or

V2
S  R ln( ), (3)
V1

where

P2V1  RT0 . P2V2  RTi ,

or

Ti V2
 .
T0 V1

(d) Path-4: A1 → A2 → A3 (denoted by purple line)

Path: A1 →A2

Q  C P T ,

T2
dT T 
S1  C P   C P ln 2  .
Ti
T  Ti 

Path: A2 →A3

Q  CV T ,

35
 Ti
dT T 
S1  CV   CV ln i  .
T2
T  T2 

Then we have

S  S1  S 2
T  T 
 C P ln 2   CV ln i 
 Ti   T2 
T  T 
 ( R  CV ) ln 2   CV ln i 
 Ti   T2 
T 
 R ln 2 
 Ti 

or

V2
S  R ln( ), (4)
V1

since

P3V2  RT2 . P2V1  RTi

or

T2 V2
 .
Ti V1

((Conclusion))
For any reversible cycle, in general, we have

dQr
 T
 S  0

19 Entropy for the ideal gas during the reversible process

We now consider the reversible process from the initial state (Pi, Vi, Ti) to the final
state (Pf, Vf, Tf).

PiVi  RTi
Pf V f  RT f

36
 If a system absorbs an infinitesimal amount of heat dQ during a reversible process,
the entropy change of the system is equal to

dQ
dS 
T

Consider one of the expression for dQ of an ideal gas.

dE  dQ  PdV
dE  CV dT
dQ  CV dT  PdV

Dividing by T

dQ dT P
dS   CV  dV
T T T

Since

PV  RT , for ideal gas,

we have

Tf Vf
dT R
S   CV   dV
Ti
T Vi V
Tf Vf
 CV ln  R ln
Ti Vi
Tf V
 CV ln  (C p  CV ) ln f
Ti Vi
Tf V
 CV [ln  (  1) ln f ]
Ti Vi

The change in entropy depends only on the properties of the initial and final states. It
does not depend on how the system changes from the initial to the final state.

In general, for n mole (n = N/NA)

S  nCV ln T  nR ln V  const
N N
 CV ln T  R ln V  const
NA NA

37
where

3R 3 N A k B
CV  
2 2
R  N Ak B

Then we have

3
S  Nk B ( ln T  ln V )  const
2
 Nk B ln(T 3 / 2V )  const
1
 1
 Nk B ln(T V )  const
1
 Nk B ln(TV  1 )  const
 1

where

CP 5 / 2 5
  
CV 3 / 2 3
2
 1 
3

In the adiabatic process (S = constant), we have

TV  1 =const

or

PV  = const

The physical meaning of this equation for S will be discussed in Chapter 20S.

We also note that

S P 1
  Nk B ,
V T V

leading to the Boyle's law.

20 Condition for equilibrium

38
During the infinitesimal irreversible process,

T (e)  T
dQirr dQirr
  dS
T (e) T

where dQirr is the heat moving from the surrounding (temperature T(e)) to the system
(temperature T). For convenience here we use dQirr = dQ.

The first law of the thermodynamics can be written in the usual form

dE  dQ  dW

And the inequality becomes

dQ  dE  dW  TdS
dE  PdV  TdS  0

This inequality holds during any infinitesimal portion and, therefore, during all
infinitesimal portions of irreducible process.

During the irreversible process by imposing the condition that two of the thermodynamic
co-ordinates remain constant, then the inequality can be reduced to a simpler form.

(a) If V and E are constant.

dS>0.

which means that the entropy of a system at constant E and V increases during an
irreversible process, approaching a maximum at the final state of equilibrium.

39
(b) If T and V are constant, the inequality reduces to

d ( E  TS )  0
dF  0

expressing the result that the Helmholtz function (F = E - ST) of a system at constant T
and V decreases during an irreversible process and becomes a minimum at the final
equilibrium.

(c) If T and P are constant, the inequality reduces to

d ( E  PV  TS )  0
dG  0

expressing the result that the Gibbs function (G = F+PV) of a system at constant T and P
decreases during an irreversible process and becomes a minimum at the final equilibrium
state.

21 Intensive and extensive parameters (definitions)

Intensive variable
P (pressure)
T (temperature)
 (chemical potential)

Extensive variable
V (volume)
C (heat capacity)
E (Internal energy)
Q (heat)
S (entropy)
F (free energy)
N (number of particles)

40
 The macroscopic parameter specifying the macro-state of a homogeneous system can
be classified into two types.
Let y denotes such a parameter. Consider that the system is divided into two parts by
introducing a partition, and denoted by y1 and y2, the values of this parameter for the two
subsystems.
Then two cases can arise
(1) One has y1 + y2 = y, in which the parameter y is said to be extensive.
(2) One has y1 = y2 = y, in which the parameter y is said to be intensive.

In simple terms, one can say that an extensive parameter get doubled if the size of the
system is doubled, while an intensive parameter remains unchanged.
The mass (M) and the volume (V) of a system are extensive parameters. The density 
of a system is an intensive parameter:   M / V .
Indeed, it is clear that the ratio of any two extensive parameters is an intensive
parameter.
The mean pressure (P) of a system is an intensive parameter, since both parts of a
system, after subdivision, will have the same pressure as before. Similarly, the
temperature T of a system is an intensive parameter. The internal energy E of a system is
an extensive quantity. The total energy of the system is the same after subdivision as it
was before: E  E1  E2 . The entropy S is an extensive quantity, because the heat Q is an
extensive quantity. The heat capacity C is an extensive quantity.

 E 
T   T (intensive), E (extensive), S (extensive).
 S V
 E 
P    P (intensive), E (extensive), V (extensive).
 V  S

41
  F 
P    P (intensive), F (extensive), V (extensive).
 V T
 F 
S    S (extensive), F (extensive), T (intensive).
 T V

22. Selected problems

22.1

Problem 20-22*** (SP-20) (10-th edition)

An insulated Thermos contains 130 g of water at 80.0°C. You put in a 12.0 g ice at
0°C to form a system of ice +original water. (a) What is the equilibrium temperature of
the system? What are the entropy changes of the water that was originally the ice cube (b)
as it melts and (c) as it warms to the equilibrium temperature? (d) What is the entropy
change of the original water as it cools to the equilibrium temperature? (e) What is the net
entropy change of the ice + original water system as it reaches the equilibrium
temperature.

((Solution))
Lf = 333 x 103 J/kg = 333 J/g
Lv = 2256 x 103 J/kg = 2256 J/g
Cwater = 4180 J/kg.K = 4.180 J/g.K
Cice= 2220 J/kg.K = 2.22 J/g.K
M = 0.130 kg water at 353 K
m = 0.012.kg ice cube at 273 K.

(a) Equilibrium temperature

Q1  m[ LF  Cwater (Te  273)]

Q2  MCwater (353  Te )

From the condition that Q1  Q2 , we have

Te = 339.5 K.

(b)

LF
m  14.637 J / K
273

(c)

42
 Te
mCwater ln( )  10.936 J / K
273

Then
LF T
Sice  m[  Cwater ln( e )]  25.57 J / K
273 273

(d)

Te
S water  MCwater ln( )]  21.18 J / K
353

(e)

S net  Sice  S water  4.395 J / K

________________________________________________________________________
22.2

Problem 20-20*** (SP-20) (10-th edition)

Expand 1.00 mol of an monatomic gas initially at 5.00 kPa and 600 K from initial
volume V1 = 1.00 m3 to final volume Vf = 2.00 m3. At any instant during the expansion,
the pressure P and volume V of the gas are related by

P  5.00 exp[(Vi  V ) / a ]

With P in kPa, Vi and Vf in m3, and a = 1.00 m3. What are the final (a) pressure and (b)
temperature of the gas? (c) How much work is done by the gas during the expansion? (b)
What is S for the expansion? (Hint: use two simple reversible processes to find S).

((Solution))
R = 8.314472 J/mol K
n =1 mol
monatomic gas
CV = 3R/2
P(V) =5 exp[(Vi-V)/a] kPa
a = 1.0 m3.
Ti = 600 K
Pi = 5 kPa

43
(a)

Pf = P(Vf = 2)= 5/e = 1.8394 kPa

(b)

Pf V f  RT f
Pf V f 1.84  103  2.0
Tf    442.6 K
R 8.314472

(c) The work done on the system is

Vf Vf
Vi  V
W    PdV  5  exp[ ]dV
Vi Vi
a
V V Vf
 5a[exp( i }]Vi
a
V  Vf
 5a[1  exp( i )]  3.16 J
a

The work done by the system is 3.16 J

(d)

1 P dT dV
dS  dE   CV R
T T T V

Then we have

3 T V
S R ln( f )  R ln( f )  0.237 R  1.97 J / K
2 Ti Vi

44
________________________________________________________________________
22.3

Problem 20-35*** (SP-20) (10-th edition)

The cicle in Fig. represents the operation of a gasoline internal combustion engine.
Volume V3 = 4.00 V1. Assuming the gasoline-air intake mixture is an ideal gas with  =
1.30. What are the ratios (a) T2/T1, (b) T3/T1, (c) T4/T1, (d) P3/P1, and (e) P4/P1? (f) What
is the engine efficiency?

((Solution))
 = 1.30
State-1 (P1, V1, T1)
State-2 (P2 = 3P1, V2 = V1, T2)
State-3 (P3, V3 = 4V1, T3)
State-4 (P4, V4 = 4V1, T4)

(a)

P2V2 P1V1

T2 T1 T2 3P1V1
,  3
3P1V1 P1V1 T1 P1V1

T2 T1

(b) The path 2-3 is adiabatic; TV-1 = constant

 1  1
T2V2  T3V3 T3 V
 1
1 1
 1
 1  1   1  0.3  0.66
T2V1  T3 (4V1 ) 1 T2 (4V1 ) 4 4

or

45
 T3  0.66T2  0.66  3T1  1.98T1

(c) The path 4-1 is adiabatic; TV-1 = constant

 1  1
T4V4  T1V1
T4 V 1 T4  0.66T1
 ( 1 ) 1  ( ) 1  0.66
T1 V4 4

(d)
P3V3 P1V1

T3 T1 P3 1
, 
P3 V1 T3 1 1 P1 2
  2 
P1 V3 T1 4 2

(e)
P4V4 P1V1

T4 T1 P4
,  0.165
P4 V1 T4 1 P1
   0.66  0.165
P1 V4 T1 4

(f)

The path 2-3 is adiabatic.

1
W23  ( P V  P V )  3.4nRT1
 1 3 3 2 2
The path 4-1 is adiabatic.
1
W41  ( PV  P V )  1.13nRT1
 1 1 1 4 4
The path 1-2 is isobaric.
n 1
Q12  nCV (T2  T1 )  ( RT2  RT1 )  ( P V  PV )  6.67 RT1
 1  1 2 2 1 1
W12  0

The path 3-4 is isobaric.

W34  0

The total work W

W  W12  W23  W34  W41  2.27nRT1

46
The engine efficiency  is

W
  0.34
Q12

23. Hint of SP-20 and HW20

23.1

Problem 20-11** (SP-20) (10-th edition)

In an experiment, 200 g of aluminum (with a specific heat of 900 J/kg K) at 100°C is

mixed with 50.0 g of water at 20°C, with the mixture thermally isolated. (a) What is the
equilibrium temperature? What are the entropy changes of (b) the aluminum, (c) the
water, and (d) the aluminum-water system?

mAl=0.2 kg, CAl = 900 J/kg K, TiAl = 100 °C = 373 K

mwater = 0.05 kg, Cwater=4180 J/kg K, Tiwater = 20 °C = 293 K

Hint:
330 1
S Al  mAl C Al  dT
Ts T

________________________________________________________________________
23.2

Problem 20-18** (SP-20) (10-th edition)

A 2.0 mol sample of an ideal monatomic gas undergoes the reversible process shown
in Fig. The scale of the vertical axis is set by Ts = 400 K and the scale of the horizontal
axis is set by Ss = 20.0 J/K. (a) How much energy is absorbed as heat by the gas? (b)
What is the change in the internal energy of the gas? (c) How much work is done by the
gas?

47
Hint

Ss = 20 J/K, Ts = 400 K.
3
CV  R , R = 8.314472 J/mol K, n = 2.0
2

(a)
Q   dQ   TdS
(b)
E  nCV T
________________________________________________________________________
23.3

Problem 20-16** (SP-20) (10-th edition)

An 8.0 g ice at -10°C is put into a Thermos flask containing 100 cm3 of water at 20°C.
By how much has the entropy of the cube-water system changed when equilibrium is
reached? The specific heat of ice is 2220 J/kg K.

Cice = 2220 J/kg K, Cwater = 4180 J/kg K

LF = 333 x 103 J/kg
mice = 0.008 kg (ice cube), Mwater= 100 cm3 = 0.10 kg (water)

Change of entropy:

1
273 m L Ts 1
Sice  miceCice  dT  ice F  miceCwater  dT
263 T 273 273 T

48
________________________________________________________________________
23.4

Problem 20-30** (HW-20) (10-th edition)

A 500 W Carnot engine operates between constant-temperature reservoirs at 100°C

and 60.0°C. What is the rate at which energy is (a) taken in by the engine as heat and (b)
exhausted by the engine as heat?

W = QH – QL=500 W (we assume that W is defined as the work done by the gas).
TH = 373 K, TL = 333 K

Carnot cycle:

QH QL

TH TL

________________________________________________________________________
23.5

Problem 20-34** (SP-20) (10-th edition)

An ideal gas (1.0 mol) is the working substance in an engine that operates on the
cycle shown in Fig. Processes BC and DA are reversible and adiabatic. (a) Is the gas
monatomic, diatomic, or polyatomic? (b) What is the engine efficiency?

49
Hint:

n = 1 mol.
PA = P0, VA = V0
PB = P0, PB = 2V0
PC = P0/32 PC = 16V0
PC = P0/32 PC = 8V0

(a) The path BC is adiabatic.

 
PBVB  PCVC
(b)
QAB  CP (TB  TA )
1
WBC  (P V  P V )
 1 C C B B

The engine efficiency  is defined as

Wtot

QAB

________________________________________________________________________
23.6

Problem 20-62 (HW-20) (10-th edition)

Suppose 2.00 mol of a diatomic gas is taken reversibly around the cycle shown in the
T-S diagram of Fig., where S1 = 6.00 J/K and S2 = 8.00 J/K. The molecules do not rotate
or oscillate. What is the energy transferred as heat Q for (a) path 1 → 2, (b) path 2 → 3,
and (c) the full cycle? (d) What is the work W for the isothermal process? The volume V1
in state 1 is 0.200 m3. What is the volume in (e) state 2 and (f) state 3? What is the

50
change Eint for (g) path 1 → 2, (h) path 2 → 3, and (i) the full cycle? (Hint: (h) can be
done with one or two lines of calculation using Section 19-8 or with a page of calculation
using section 19-11.) (j) What is the work W for the adiabatic process?

Hint:

diatomic gas, n = 2
5 7 C
CV  R , CP  R,   P  1.40
2 2 CV
S1 = 6.0 J/K, S2 = 8.0 J/K
T1 = 350 K, T3 = 300 K
V1 = 0.2 m3.

(a) Q12  T1S  T1 ( S 2  S1 )

(b) Q23  0
(d) E12 = 0

24. Link

Carnot engine (Wikipedia)

http://en.wikipedia.org/wiki/Carnot_heat_engine

Entropy (Wikipedia)
http://en.wikipedia.org/wiki/Entropy

_____________________________________________________________________
APPENDIX-I

51
Refregerator

Performance of Refrigerator
Coefficeint of performance (COP) K

QL
K
|W |

((Note))
Efficiencies of real engine 

|W |

QH

________________________________________________________________________
APPENDIX-II
Empirical temperature and absolute temperature
Our fundamental theorem shows us that the ratio QL/QH has the same value for all
reversible engines that operate between the same empirical temperature tH and tL; that is,
this ratio is independent of the special properties of the engine, provided it is reversible. It
depends only on the empirical temperatures tH and tL. We may therefore write:

QL
 F (tH , tL ) ,
QH

where F (tH , tL ) is a universal function of the two temperatures tH and tL.

We shall now prove that the function F (tH , tL ) has the following property:

52
 F (tH , t0 )
F (tH , tL )  ,
F (tL , t0 )

where t0 is arbitrary.

Fig.1 t0<tL. A series connection of the reversible cyclic engines A1 and A2. A1 and A2
work between the temperatures tH and tL, and tL and t0, respectively. If A1 absorbs
an amount of heat QH at tH and gives up an amount of heat QL at tL during a cycle.

53
 If A2 absorbs an amount of heat QL at tL and gives up an amount of heat Q0 at t0
(Tomonaga).

Fig.2 t0>tL.A1 and A2 are two reversible cyclic engines which work between the
temperatures tH and tL, and tL and t0, respectively. If A1 absorbs an amount of heat
QH at tH and gives up an amount of heat QL at tL during a cycle. If A2 absorbs an
amount of heat QL at tL and gives up an amount of heat Q0 at t0 (Fermi).

We consider a process consisting of series connection of the reversible engine A1 and the
reversible engine A2 as shown in Fig.1 and Fig.2. For the engine A2, we have the relation

Q0
 F (tL , t0 ) .
QL

Similarly for the engine A1 (tH>tL), we have

QL
 F (t H , t L ) .
QH

Then we get

Q0 Q0 QL
F (tH , t0 )    F (tL , t0 ) F (tH , tL ) ,
QH QL QH

or

54
 F (tH , t0 )
F (tH , tL )  .
F (tL , t0 )

Here we note that t0 is arbitrary. Note that t0<tL (in Fig.1) and t0>tL (Fig.2). We choose
t0  t0 . We may keep it constant in all our equations. It follows that the function f(t) can
be defined as

1
f (t )  .
F (t , t0 )

So we have

f (tL )
F (t H , t L )  .
f (tH )

We place

 (t )  f (t ) ,

where  is an arbitrary constant. Using this equation, we get

QL  (t L )
 .
QH  (t H )

 is regarded as a new temperature. Note that (t) increases with increasing an empirical
temperature. (t) expresses the relation between the empirical temperature and the new
temperature (the absolute temperature). Note that (t) is not uniquely determined. (t) is
indeterminate to the extent of an arbitrary multiplicative constant factor .

QL TL
 (Carnot cycle)
QH TH

The efficiency is given by

TL
 1 .
TH

The efficiency becomes 1 at TL = 0 K.

((Comment by S.Tomonaga)) S. Tomonaga, What is physics? (Iwanami, 1979).

"I do not know who gave a proof for the universal function. When I borrowed a book
of Clausius from Riken, this proof was written on the book by pencil, next to the proof

55
given by Clausius. This book was bought from Carl Runge in Germany by Riken. The
proof was written in German. So I think that the proof might be given by Runge."

((Comment by M.S.)) I found similar proof in the book of Enrico Fermi

(Thermodynamics, 1936).

REFERENCES
E. Fermi, Thermodynamics (Dover Publication, 1936). p.39-41
S. Tomonaga, What is physics? Iwanami (1979, in Japanese).

________________________________________________________________________
Sin-Itiro Tomonaga
March 31, 1906 – July 8, 1979) was a Japanese physicist, influential in the
development of quantum electrodynamics, work for which he was jointly awarded the
Nobel Prize in Physics in 1965 along with Richard Feynman and Julian Schwinger.
http://en.wikipedia.org/wiki/Sin-Itiro_Tomonaga

Enrico Fermi;
29 September 1901– 28 November 1954) was an Italian-American physicist, best
known for his work on Chicago Pile-1 (the first nuclear reactor), and for his contributions
to the development of quantum theory, nuclear and particle physics, and statistical
mechanics. He is one of the men referred to as the "father of the atomic bomb".[4] Fermi
held several patents related to the use of nuclear power, and was awarded the 1938 Nobel
Prize in Physics for his work on induced radioactivity by neutron bombardment and the
discovery of transuranic elements. He was widely regarded as one of the very few
physicists to excel both theoretically and experimentally.
http://en.wikipedia.org/wiki/Enrico_Fermi

APPENDIX-III
1. Born diagram
(N. Hashitsume, Introduction to Thermal and Statistical Mechanics, Iwanami, 1980,
in Japanese)

In thermodynamics, we often use the following four thermodynamic potentials, E, F,

G, and H. The diagram (called the Born's diagram) was introduced by Born (Max). In
order to memorize this diagram, we give interpretation for the letters. The sun (S;
entropy) pours lights on the trees (T; temperature). The water falls from the peak (P;
pressure) of mountain into the valley (V; volume). We draw a square with four vertices
noted by S, T, P, and V. . The light propagates from the point S to the point T. The water
flows from the point P to the point V. These two arrows are denoted by the vectors given
by ST (the direction of light flow) and PV (the direction of water flow). These vectors
are perpendicular to each other. The four sides of the square are denoted by E, F, G, and
H in a clockwise direction. Note that the side H (H: heaven) is between two vertices S
(sun) and P (peak).

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 S

H E

P V

G F

T
Fig. Born diagram. S: entropy. T: temperature. P: pressure. V: volume. H: heaven
(between S and P). E→F→G→H (clockwise). The water flow from P (peak) to V
(valley). The sun light from S (sun) to T (tree).

(i) The natural variables of the internal energy E is S and V.

dE  TdS  PdV

The sign before T is determined as plus from the direction of the vector ST ( ST : the
direction of light). The sign before P is determined as minus from the direction of the
vector VP   PV ( PV ; the direction of water flow).

(ii) The natural variables of the Helmholtz energy F is V and T.

dF   SdT  PdV

The sign before S is determined as minus from the direction of the vector TS   ST
( ST : the direction of light). The sign before P is determined as minus from the direction
of the vector VP   PV ( PV ; the direction of water flow).

(iii) The natural variables of the Gibbs energy G is P and T.

dG  VdP  SdT

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The sign before S is determined as minus from the direction of the vector TS   ST
( ST : the direction of light). The sign before V is determined as plus from the direction of
the vector PV ( PV ; the direction of water flow).

(iv) The natural variables of the enthalpy H is S and P.

dH  TdS  VdP .

The sign before T is determined as minus from the direction of the vector ST ( ST : the
direction of light). The sign before V is determined as plus from the direction of the
vector PV ( PV ; the direction of water flow).

Maxwell's relation

(i) The internal energy E  E ( S ,V )

For an infinitesimal reversible process

dE  TdS  PdV

showing that

 E   E 
T   and P   
 S V  V  S

The Maxwell’s relation;

 T   P 
    
 V  S  S V

(ii) The enthalpy H  H ( S , P ) is defined as

H  E  PV

For an infinitesimal reversible process

dH  dE  PdV  VdP
 TdS  PdV  PdV  VdP
 TdS  VdP

showing that

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  H   H 
T   and V   .
 S  P  P  S

The Maxwell’s relation:

 T   V 
   
 P  S  S  P
(iii) The Helmholtz free energy F  F (T ,V ) is defined as

F  E  ST or E  F  ST

For an infinitesimal reversible process

dF  dE  SdT  TdS
 TdS  PdV  SdT  TdS
  PdV  SdT

showing that

 F   F 
P    and S   
 V T  T V

The Maxwell’s relation:

 S   P 
   
 V T  T V

(iv) The Gibbs free energy G  G (T , P ) is defined as

G  H  ST  ( E  PV )  ST  F  PV

Then we have

dG  VdP  SdT

showing that

 G   G 
S    and V  
 T  P  P T

(1) The Maxwell’s equation:

59
  S   V 
    
 P T  T  P

 T   P 
Here we consider the Maxwell's relation     
 V  S  S V

 P 
For   , in the Born diagram, we draw the lines along the vectors PS and SV . The
 S V
resulting vector is PV  PS  SV (the direction of water flow)

T P

V E S

 T 
For   , in the Born diagram, we draw the lines along the vectors TV and VS . The
 V  S
resulting vector is TS  TV  VS   ST (anti-parallel to the propagating direction of
 P 
light). Then we have the negative sign in front of   such that
 S V

 T   P 
    
 V  S  S V

 S   P 
(ii) Maxwell's relation    
 V  T  T V

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  S   P 
Here we consider the Maxwell's relation    
 V T  T V

 S 
For   , in the Born diagram, we draw the lines along the vectors SV and VT . The
 V T
resulting vector is ST  SV  VT (the direction of sun light)

P S

T F V

 P 
For   , in the Born diagram, we draw the lines along the vectors PT and TV . The
 T V
resulting vector is PV  PT  TV (the direction of water flow). Then we have the
 P 
positive sign in front of   such that
 T V

 S   P 
   
 V T  T V

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P S

T F V

62