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MATERIALS THERMODYNAMICS - With Emphasis on Chemical Approach (With CD-ROM)© World Scientific Publishing Co. Pte. Ltd.http://www.worldscibooks.com/engineering/8274.html

27

Chapter 3

The Second Law of Thermodynamics

3.1 Grades of Energy

Thermodynamics deals with energy. The first law is concerned with the conservation of energy in energy transfer. Energy may be transferred from one place to the other or transformed in one form to another, but the total energy of the universe (the system + the surroundings) is always conserved.

Now a question arises as to in which direction energy flows in a natural process, and if there is any law which governs the direction of the energy flow. Thus, the primary interest in thermodynamics is to predict the direction of chemical or physical processes, in particular the spontaneous, natural direction. Some examples we observe in our everyday life are, � Energy always flows spontaneously from a higher temperature to a lower temperature,

but not the reverse (thermal energy). � Energy always flows naturally from a higher pressure to a lower pressure (mechanical

energy). � Energy always flows from a higher voltage potential to a lower voltage potential

(electrical energy). � Wood burns spontaneously in air if ignited, but the reverse process, i.e., the

spontaneous recombination of the combustion products to wood and oxygen in air, has never been observed in nature (chemical energy).

� Ice at 1 atm. pressure and a temperature above 0°C always melts spontaneously, but water at 1 atm. pressure and a temperature above 0°C never freezes spontaneously in nature (phase transition).

There will be numerous examples of processes which proceed in one direction, but not in the other direction spontaneously. The above examples strongly support the view that energy contained in the source from which the energy is drawn must be different in quality or grade from the energy contained in the sink into which the energy flows.


28 Materials Thermodynamics with Emphasis on Chemical Approach

Thermal energy in water

Suppose we have two buckets of water of the same amounts, but one at temperature of 90°C, and the other at 40°C (Refer to the figure below.) Now thermal energy (heat) (Q) is taken out of the hot water by cooling from 90°C to 80°C (∆T = -10°C). The energy so extracted is transferred to the cold water, and heats the cold water from 40°C to 50°C (∆T = +10°C), assuming that there is no heat loss and the heat capacity of water is constant. The energy Q which was originally in the hot water is now in the cold water. Here a question arises as to whether it would be possible to let the same amount of energy Q return to the hot water spontaneously so that the temperature of the hot water recovers its original temperature, 90°C (∆T = +10°C) – see the following figure. Our everyday experience disproves it. If it were the case, the hot water could get hotter while the cold water becomes colder. Even though we talk about the same amount of energy Q, say, 1,000 joules, the energy in the hot water and that in the cold water are different in terms of ability or usefulness. This example clearly reveals that energy has quantity, and also quality as well.

It would be very useful if we could find a function or property of a system by which the quality (or grade) of energy is quantitatively determined. As it determines the direction of its flow, the energy grade of a system must be a function of the thermodynamic driving forces of the system which provides the impetus for the transfer of the energy from the system. When the thermodynamic driving forces in the system decrease, the energy in the system is degraded. In this regard, the energies contained in the sources in the above examples are at the higher energy grades than those in the sinks. For example, the energy contained in a higher temperature system is at the higher energy grade than that in a lower temperature system.

When two systems exchange thermal energy in the mode of heat, the energy grades of both systems will be changed. Suppose two systems at different temperatures are placed in contact with each other. Then thermal energy will be transferred in the mode of heat from the high temperature system to the low temperature one. The following diagram schematically represents the changes of the energy and temperature of the systems:

90 80 70

T(oC)

50 40 30

T(oC)

Q

90 80 70

T(oC)

50 40 30

T(oC) Q

Yes. No.


The Second Law of Thermodynamics 29

- System A (high temperature) and System B (low temperature) are brought into contact with each other.

- As the thermal energy is transferred, System A experiences decrease in the amount of energy and also its temperature, whereas in System B both the amount of energy and its temperature increase.

- The transfer of thermal energy continues until the temperatures of both systems arrive at the same temperature of Te.

- During the thermal process, the energies of both systems change:

System A: o eA A AU U U→ → (energy loss)

System B: o eB B BU U U→ → (energy gain)

- According to the first law, the energy loss of System A is equal to the energy gain of System B, and thus the total energy is conserved:

o o e eA B A B A BU U U U U U+ = + = +

- The temperatures of both systems change:

System A System B o eA AT T T→ → o

B BT T← ←

(decrease) (increase)

- Note that the energy in System A has experienced decrease in quantity and also in quality (grade) as the temperature of A has decreased, whereas the energy in System B is increased in quantity and also quality.

- The net effect of the above heat transfer process is that the total energy is conserved, but System A has left with a permanent degradation of its initial energy. The energy in System B has been upgraded, but the so-upgraded energy is not useful (or available) for the purpose of returning the so-degraded energy of System A back to the level of its initial quality. The only way to bring the energy grade of System A back to the initial value by means of thermal energy is to transfer energy from a system which is at the

temperature higher thanoAT .

From the above example, it can be concluded that,

� the increase of the energy grade or energy quality of a system by heat transfer is possible, but only at the expense of lowering the energy grade of another system whose energy grade is higher than that of the system, and

A

B

oAU AU

eAU

oBU BU

eBU

oBT

BT

eT

oAT

AT

System A o oA A,U T

System B o oB B,U T



� the so-upgraded energy of the system is of no use for recovering the energy grade of the other so-degraded system, and

� the universe (the sum of both systems in the present case) has suffered from a permanent degradation of the energy.

Through the above discussions, it is now clear that factors which determine the extent of change in the grade of energy of a system is not just the quantity of energy transferred, but also the temperature of the system at which the energy transfer occurs.

Consider the following example in which thermal energy is transferred by heat flow:

A

B

oAU AU

oBU BU

oBT

BT

oAT

AT

A

B

oAU AU

oBU BU

oBT

BT

oAT

AT

Thermal energy (Q) released from System A by cooling can be transferred to System B directly and thus raise the temperature of System B.

Q

Q

Q

Q

Thermal energy (Q) released from System B by cooling cannot be transferred to System A directly and is unavailable for raising the temperature of System B.

Hot T1

Warm T2

q

q

Path 1

Path 2

Cold T3

q

Path 3

Yes No



Thermal energy q is to flow in the mode of heat from the hot body to the cold body. The heat flow may occur either

(1) directly from the hot to cold bodies, i.e., Path 1, or (2) from the hot to warm, and then to cold bodies, i.e., Path 2 + Path 3.

The degradation or decrease in quality of the thermal energy q is greater in Path 1 than in Path 2, as in the latter path further degradation should occur through Path 3 to reach the same final state. Thus,

� Degradation in Path 1 > Degradation in Path 2 � Degradation in Path 1 > Degradation in Path 3

In other words,

Path 1 is more irreversible than either Path 2 or Path 3.

From the view point of the conversion of thermal energy into work (w), the above analysis can be interpreted as follows:

� Once the thermal energy q is transferred to a lower temperature, say, from T1 to T3, it loses the capacity to do work which might have been done if a chance were given during the course of the transfer.

� The loss of capacity to do work is greater in Path 1 than in Path 2, as, after Path 2, there is a chance to do work during further cooling from T2 to T3 (Path 3).

� The loss of capacity to do work and the degradation (or decrease in quality) of thermal energy shares the same origin; that is, the transfer of thermal energy.

Therefore, the quantification of degradation of thermal energy or the loss of capacity to do work must include both the amount of thermal energy transferred in the mode of heat, q, and the temperature T at which the transfer occurs. � The larger the amount of heat flow, the greater the extent of degradation. � The lower the temperature of the body to which heat flows, the higher the degree of

degradation.

Therefore,

Extent of degradation ∝

Thermodynamics may be defined as a physical science concerned with the transfer of heat and the performance of work accompanying various physical and chemical processes. The first law of thermodynamics is the law of energy conservation. Explain, using an example, why the first law is not enough to answer the following comments:

“Some things happen spontaneously; some other things don’t.”

Example 3.1

T

q



Heat reservoir T1

Engine

q1

q2

w

T1 > T2

Heat reservoir T2

Refer to the following two figures: The above example demonstrates that the first law of thermodynamics alone is not enough to completely deal with processes which occur naturally. We now search in several different ways a function or property which can be a barometer of the direction of a natural or spontaneous process. 3.2 Heat Engines and Entropy

A heat engine is a device for converting heat (thermal energy) into work (e.g., steam engine, internal combustion engine). The figure on the right is the schematic representation of a heat engine.

• Each cycle takes thermal energy of heat (q1) from the high temperature heat reservoir, and

• uses some of it to generate work (w), and • rejects the unused portion (q2, thermal energy) to the

low temperature heat reservoir The efficiency of the engine is defined as

1

The amount of thermal energy expended to conduct work

The amount of thermal energy supplied to the engine

w

qε = =

Where “ ” is to ensure ε to be positive.

The prime interest we may have in dealing with a heat engine is to know the maximum amount of work that can be obtained from each cycle of the heat engine. We already know that the maximum work can be obtained when the process is conducted in a reversible manner (refer to Chapter 2.1). It is possible to devise a cyclic process which consists of processes which are all reversible.

System

Surroundings

q

w Tsys

Tsur

Tsys<Tsur

System

Surroundings

q

w Tsys

Tsur

Tsys>Tsur

U U

The process satisfies ∆U = q + w, and occurs spontaneously as indicated by

the arrows.

The process satisfies ∆U = q + w, but does not occur spontaneously as

indicated by the arrows.



Carnot Cycle

The Carnot cycle is the cyclic operation of an idealized (ideal gas) engine which runs in four steps: two reversible isothermal steps and two reversible adiabatic steps. After completing a cycle, the system returns to its initial state so that all of the state functions restore their initial values.

The figure below shows the Carnot engine, in which heat q1 is taken in from the high temperature heat reservoir, work w is done through running a complete cycle of the Carnot engine, and heat q2 is rejected to the low temperature heat reservoir.

It is clear from the figure that work is done by the Carnot engine through Step 1 and Step 2 as the gas expands, and work is done on the engine through Step 3 and Step 4 as the gas is compressed. The net amount of work (w) done by the engine is the balance of the work done by the engine and the work done on the engine.

Now we discuss each step of the Carnot cycle one by one.

Step 1: Reversible isothermal (T1) expansion:

2

1

V

Vw PdV= −∫

1PV nRT=

2

11 1

1V

Vw nRT dV

V= − ∫

11 1

2

Vw nRT ln

V=

Heat Reservoir (high temperature)

T1

Heat Reservoir (low temperature)

T2

Carnot Engine

1q

2q

w

Step 1 Isothermal (T1) expansion

Step 2 Adiabatic expansion

Step 3 Isothermal (T2) compression

Step 4 Adiabatic compression

a

b

c

d

P

V

One cycle of the Carnot engine

a

b P

V

1 1PV

2 2PV

1PV nRT=

a′ b′

Work (w1) = area (abb′a′)

= 2

1

V

VPdV−∫



The engine takes heat q1 from the high temperature (T1) heat reservoir, and the gas in the engine expands isothermally and reversibly. The figure shown above explains how to obtain the amount of work done by the engine in Step 1. Since the gas is at the constant temperature T1, the internal energy (U) stays unchanged:

1 0U∆ =

1 1 1U q w∆ = +

11 1

2

Vw nRT ln

V=

21 1

1

Vq nRT ln

V=

This is the amount of heat taken in from the high temperature heat reservoir to do work w1 by the reversible isothermal expansion. Note that in this isothermal expansion the temperature of the gas is the same as that of the high temperature heat reservoir (T1).

Step 2: Reversible adiabatic expansion:

The engine at this step is now insulated so that no heat is allowed to be exchanged with the surroundings. The gas expands from V2 to V3 and the pressure changes from P2 to P3, as depicted in the figure below.

Since the expansion is conducted under the adiabatic condition, the temperature of the gas decreases. The expansion continues until the temperature has become the same as the temperature of the low temperature heat reservoir (T2).

For the adiabatic change of the gas, the relationship of PV = nRT does not hold, but

instead the relationship of γPV c= (constant) applies (see Example 2.11).

3

22

V

Vw PdV= −∫

PV cγ =

3

22

1V

Vw c dV

V γ= − ∫

1 13 2

2 1

V Vw c

−γ −γ−= −

− γ

2 32 3c P V PVγ γ= =

b

c

P

V b′ c′

3 3PV

2 2PV

Work (w2) = area (bcc′b′)

= 2

1

V

VPdV−∫

PV cγ =



This is the work done by the engine through the reversible adiabatic expansion. From the ideal gas law, we know the relationships of P2V2 = nRT1 and P3V3 = nRT2, and thus the last equation becomes

2 12

( )

1

T Tw nR

−= −

− γ

P

V

C

Cγ =

P VC C R− =

2 2 1( )Vw nC T T= −

0q = (adiabatic)

U q w∆ = +

2 2 1( )VU nC T T∆ = −

As can be seen above, by the reversible adiabatic expansion, the engine does work w2

against the external pressure, and accordingly the internal energy decreases by ∆U2 and the temperature of the gas decreases from T1 to T2.

In order to bring the engine back to the initial starting point, the gas which has expanded through the above two steps must now be compressed to the initial volume.

Step 3: Reversible isothermal compression:

3 3 2 22 1

PV P Vw

−= −

− γ

c

d

P

V c′ d′

3 3PV

4 4P V 2PV nRT=

Work (w3) = area (dcc′d′)

= 4

3

V

VPdV−∫

4

33

V

Vw PdV= −∫

2PV nRT=

4

33 2

1V

Vw nRT dV

V= − ∫

33 2

4

Vw nRT ln

V=



This is the amount of work done on the engine through the compression of the gas. Since the gas is at the constant temperature T2, the internal energy (U) stays unchanged:

3 0U∆ =

3 2 3U q w∆ = +

33 2

4

Vw nRT ln

V=

42 2

3

Vq nRT ln

V=

This is the amount of heat (q2 < 0) rejected by the gas to the low temperature heat reservoir. Note that in this isothermal compression the temperature of the gas is the same as that of the low temperature heat reservoir (T2).

Step 4: Reversible adiabatic compression:

Now the engine is insulated again so that no heat exchange with the surroundings is allowed. The gas is compressed from V4 to V1 and the pressure changes from P4 to P1. The process is graphically represented in the figure below.

Since the compression is conducted under the reversible adiabatic condition, the temperature of the gas rises. The compression continues until the temperature of the gas reaches that of the high temperature heat reservoir (T2) which is also the temperature of the gas at the beginning of the cycle. As seen in Step 2, for the reversible adiabatic change of the gas, the relationship of PV = nRT does not hold, but instead the

relationship of γPV c= (constant) applies.

In a similar way to Step 2,

4 1 2( )Vw nC T T= −

4 1 4( )VU nC T T∆ = −

As seen above, by the reversible adiabatic compression, the work w4 is done on the

a

d

P

V a′ d′

4 4P V

1 1PV

Work (w4) = area (add′a′)

= 1

4

V

VPdV−∫

PV cγ =



engine against the internal pressure, and accordingly the internal energy increases by ∆U4 and the temperature of the gas rises from T2 to T1.

The complete Carnot cycle is summarized in the following figure:

In a complete cycle, the gas follows the path “abcd”. During the cycle,

� the amount of heat taken in from the high temperature heat reservoir: 21 1

1

Vq nRT ln

V=

� the amount of heat rejected to the low temperature heat reservoir: 42 2

3

Vq nRT ln

V=

� the amount of work done by the engine:

1 2 3 4w w w w w= + + +

11 1

2

Vw nRT ln

V= 2 2 1( )Vw nC T T= −

33 2

4

Vw nRT ln

V= 4 1 2( )Vw nC T T= −

P1,V1

P2,V2

P3,V3

P4,V4

Step 1 Isothermal expansion

q1, w1, T1

w2,

T1→ T2

q2, w3, T2

w4,

T2 → T1

V →

P ↑


Step 3 Isothermal

compression

Step 4 Adiabatic

compression

q1,

w1 T1

w2 T1→ T2

Insulation

w3 T2

q2

w4 T2→T1

Insulation

T2

Carnot cycle

a

b

c

d



311 2

2 4

VVw nRT ln nRT ln

V V= +

21 1

1

Vq nRT ln

V=

42 2

3

Vq nRT ln

V=

1 2( )w q q= − +

For further consideration on the last equation, let us employ the equations for adiabatic conditions given in Step 2 and Step 4.

For Step 2,

2 32 3P V PVγ γ=

2 2 1P V nRT= , 3 3 2PV nRT=

1 11 22 3TV T Vγ− γ−=

For Step 4, in a similar way,

1 12 14 1T V TVγ− γ−=

Combination of the last two equations yields,

32

1 4

VV

V V=

This last result is useful to bring a very important relationship when it is combined with q1 and q2 in the Carnot cycle.

21 1

1

Vq nRT ln

V= 4

2 23

Vq nRT ln

V=

32

1 4

VV

V V=

1 2

1 2

0q q

T T+ =



This equation is resulted from the reversible cyclic heat engine which is a closed system, in which heat q1 is taken in at T1 and heat q2 is removed at T2, both of which are under the isothermal conditions. Since the sum of changes of any state property must be zero in a cyclic process, this last equation suggests that there exists a state property which is related to q/T. The change of this property during the expansion is q1/T1 and its change during the compression is q2/T2, and the sum of these two changes for a complete cycle is zero, which is the necessary and sufficient condition for a state property. This property is termed entropy. The concept of entropy will be further refined after discussing the efficiency of the Carnot engine.

1

The amount of thermal energy expended to conduct work

The amount of thermal energy supplied to the engine

w

qε = =

1 2( )w q q= − +

2

1

1q

qε = + 2

1

1T

Tε = −

This is a remarkable result: The efficiency depends only on the temperatures of the reservoirs, and is independent of the nature of the engine, working substance, or the type of work performed.

The above two equations indicate that only a fraction of the heat taken in from the high temperature reservoir is converted to work. This fraction is

1 2

1

q q

q

+ or 1 2

1

T T

T

−

It is obvious from the above relationship that complete conversion of heat into work is possible only if q2 is equal to zero which means that T2 is equal to 0K, and thus it indicates that the complete conversion of heat into work is impossible in practice.

Refer to the diagram of the Carnot cycle in the text. An engine operates between 1,200°C (T1) and 300°C (T2), and Step 1 (isothermal expansion) involves an expansion where the pressure of the gas drops from 6×105 N m-2 to 4×104 N m-2. The working substance is one mole of an ideal gas.

(1) Calculate the efficiency of the heat engine. (2) Calculate heat absorbed in Step 1. (3) Calculate the amount of heat rejected in Step 3 (isothermal compression).

Example 3.2

1 2

1 2

0q q

T T+ =



(1) 2

1

1T

Tε = − 0 611.ε = or 61.1%

(2) 21 1

1

Vq nRT ln

V=

PV nRT=

11 1

2

Pq nRT ln

P=

P1 = 6×105 N m-2

P2 = 4×104 N m-2 n = 1mole R = 8.314J mol-1K-1 T1 = 1473K

1 33 164q , J=

(3) 1 2

1 2

0q q

T T+ =

1 33 164q , J=

T1 = 1473K T2 = 573K

1 12 900q , J= −

Entropy

Let us return to the important relationship obtained for a reversible cyclic process:

1 2

1 2

0q q

T T+ =

Since the heat exchanges are conducted reversibly,

1 1473T K=

2 573T K=



1 2

1 2

0,rev ,revq q

T T+ =

If heat exchanges are conducted in an infinitesimal amount,

1 2

1 2

0,rev ,revq q

T T

δ δ+ =

Since the operation is a cyclic process,

0i ,rev

i

q

T

δ=∫�

This last equation applies not only to the Carnot cycle, but also to any cycle which is reversible (later in Example 3.5 we will see that this equation is equally valid for any cyclic process which includes irreversible steps).

A reversible cycle can be approximated by a series of Carnot cycles as seen on the right in the following figure. In the figure, a number of curved lines are drawn for the isothermal and adiabatic processes. The entropy change of each step of individual Carnot cycles is cancelled by the neighboring Carnot cycle, except all outmost steps indicated by thick lines in the figure. The sum of the entropy changes of these surviving steps must also be zero. The approximation can be close enough to the real arbitrary cycle by making the individual Carnot cycles infinitesimally small.

So far we have dealt with a cycle which is combination of reversible steps only. However, it is reminded that the equation in the above is equally valid for a cycle which involves irreversible steps. This is discussed in Example 3.5.

A real cycle

The cycle approximated by a series of Carnot cycles

(thick zigzag curves)

Each of these is a Carnot cycle.

V →

P ↑

V →

P ↑



Now it is in order to examine whether or not each term in the above equation, i.e.,

i ,rev iq T represents the change of a state property of the system. If it is really the case, its value for all processes which share the same initial and final states must be the same, irrespective of whether the process is reversible or irreversible. The following figure shows two reversible cycles, i.e., cycles “1a2r1” and “1b2r1”.

For the cycle “1a2r1”,

2 1

1 20i ,rev

i a r

q q q

T T T

δ δ δ = = + ∫ ∫ ∫�

For the cycle “1b2r1”,

2 1

1 20i ,rev

i b r

q q q

T T T

δ δ δ = = + ∫ ∫ ∫�

Combination of the above two equations yields,

2 2

1 1a b

q q

T T

δ δ = ∫ ∫

In the above equation, both integral terms indicate the change of the property, q/T, when the state is changed from State 1 to State 2 in the figure. The term on the left side is the change along the path “1a2” and the term on the right side is along the path “1b2”, and both are the same. This result ensures that the quantity of q/T for a reversible path is independent of the choice of the path as long as the initial and final states are not altered. Thus it is a function of the state. In other words it is the change of a property of the system. This property is termed entropy with the symbol of S.

revqdS

T

δ≡

Since the entropy is a state function, its change for a path must be the same irrespective of the path being reversible or irreversible. That is, the above relationship holds for both reversible and any irreversible processes, even though the equation includes qrev which is the heat flow for a reversible process (refer to Example 3.6 and page 77).

If a system undergoes a change in state, say, from State 1 to State 2, the entropy change by the change of the state is given by integrating the above equation:

•

•

1

2

a

b

r

V

P



This equation can be integrated by expressing q as a function of T.

1. Entropy change of the system (sysS∆ ):

(1) Step 1: Isothermal expansion : 11

1sys,

qS

T∆ =

(2) Step 2: Adiabatic expansion: 2 0sys,S∆ = , as q = 0.

(3) Step 3: Isothermal compression: 23

2sys,

qS

T∆ =

(4) Step 4: Adiabatic compression: 4 0sys,S∆ = , as q = 0.

Thus,

1 21 2 3 4

1 2

0sys sys, sys, sys, sys,q q

S S S S ST T

∆ = ∆ + ∆ + ∆ + ∆ = + = (see page 38)

2. Entropy change of the surroundings (surS∆ )

(1) Step 1: Isothermal expansion of the system: 11

1sur ,

qS

T

−∆ = ( heat loss)

(2) Step 2: Adiabatic expansion of the system: 2 0sur ,S∆ = , as q = 0.

(3) Step 3: Isothermal compression of the system: 23

2sur ,

qS

T

−∆ = (heat gain)

(4) Step 4: Adiabatic compression of the system ( 4sur ,S∆ ): 4 0sur ,S∆ = , as q = 0.

Thus,

1 21 2 3 4

1 2

0sur sur , sur , sur , sur ,q q

S S S S ST T

∆ = ∆ + ∆ + ∆ + ∆ = − + =

3. Entropy change of the universe (uniS∆ )

0uni sys surS S S∆ = ∆ + ∆ =

In summary, for a reversible cyclic process the entropies of the system, the surroundings and the universe all remain unaltered.

For a system which undergoes the process of the Carnot cycle, calculate the entropy change of the system, the surroundings and the universe (the system + the surroundings).

Example 3.3

2

1revq

ST

δ∆ = ∫



It can be seen that the increase in the entropy of the system is exactly the same in numerical value as the decrease in the entropy of the surroundings, and thus, the total entropy (the entropy of the universe) is unaltered. In the reversible process, thus, we may say that:

Entropy of the amount of 1

1

q

Thas been transferred from the surroundings to the system.

In the previous example (Example 3.3), when we consider the step of isothermal expansion only, what are the entropy changes of the system, surroundings and the universe?

Example 3.4

1

1sys

qS

T∆ =

1

1sur

qS

T∆ = −

0uni sys surS S S∆ = ∆ + ∆ =

System Isothermal expansion at T1

Surroundings Heat reservoir at T1

1q

Universe

An irreversible step is inserted into the Carnot cycle as shown in the figure below. The irreversible step added is an adiabatic free expansion. Determine the entropy changes of the system (the revised cycle), the surroundings and the universe. Assume that the working substance is n moles of ideal gas. Determine also the efficiency ε.

Example 3.5





a (P1,V1)

P

V

Step A Adiabatic free expansion

b′ (PA,VA) b (P2,V2)

c (P3,V3)

d (P4,V4)



Step 1: Reversible isothermal expansion (a→b′)

(1) Heat : A1 1

1

Vq nRT ln

V= (> 0)

(2) Work : 11 1

A

Vw nRT ln

V= (< 0)

(3) Entropy change : 1 A1

1 1

q VS nR ln

T V∆ = =

Step A: Adiabatic free expansion (b′→b)

(1) Heat : A 0q = (adiabatic)

(2) Work : A 0w = (free expansion)

(3) Entropy change - The internal energy does not change for an adiabatic free expansion

( 0 0 0U q w∆ = + = + = ).

- Thus there will be no change in temperature (T1) - However, the entropy change is not equal to zero. We may find the

entropy change by reversing the process since the entropy change for Step b′→b must be the same in numerical value as the entropy change for the Step b→b′, but in opposite sign.

- The entropy is a state function, and thus we may choose the isothermal compression for Step b→b′ (Temperature: constant at T1).

- Ab b'

2

VS nR ln

V→∆ = , and thus 2A b b'

A

VS S nR ln

V→∆ = ∆ =

- Refer to the further discussion given in Example 3.13.

Step 2: Reversible adiabatic expansion (b→c)

(1) Heat : 0q =

(2) Work : 2 2 1( )Vw nC T T= − (< 0)

(3) Entropy change : 2 0S∆ = since 0q = . (Note that T = constant at T1 for

adiabatic free expansion, but T1→T2 for adiabatic reversible expansion.)

Step 3: Reversible isothermal compression (c→d)

(1) Heat : 42 2

3

Vq nRT ln

V= (< 0)

(2) Work : 33 2

4

Vw nRT ln

V= (> 0)

(3) Entropy change : 2 43

2 3

q VS nR ln

T V∆ = = (< 0)

Step 4: Reversible adiabatic compression (d→a)

(1) Heat : 0q =



(2) Work : 4 1 2( )Vw nC T T= − (> 0)

(3) Entropy change : 4 0S∆ =

Entropy change of the system (sysS∆ : Total entropy change for the modified cycle)

1 A 2 3 4sysS S S S S S∆ = ∆ + ∆ + ∆ + ∆ + ∆

A 2 4

1 A 3

0 0sysV V V

S nR ln nR ln nR lnV V V

∆ = + + + +

2 4

1 3sys

V VS nRln

V V

∆ =

From the two adiabatic steps,

32

1 4

VV

V V= (Refer to earlier discussion)

1 0sysS nRln∆ = =

Recall that the modified cycle includes an irreversible step, but the entropy change of the system for the cycle is zero. This is another evidence that entropy is a state function.

Next, let us examine the entropy change of the surroundings.

Step 1: Reversible isothermal expansion

(1) Heat : A1 1 1

1,sur

Vq q nRT ln

V= − = −

(2) Entropy change : 1 A1

1 1

,sur,sur

q VS nR ln

T V∆ = = −

Step A: Adiabatic free expansion

(1) Heat : A 0,surq =

(2) Entropy change : A 0,surS∆ =

Step 2: Reversible adiabatic expansion

(1) Heat : 2 0,surq =

(2) Entropy change : 2 0,surS∆ =

Step 3: Reversible isothermal compression

(1) Heat : 43 3 2

3,sur

Vq q nRT ln

V= − = −



(2) Entropy change : 3 43

2 3

,sur,sur

q VS nR ln

T V∆ = = −

Step 4: Reversible adiabatic compression

(1) Heat : 4 0,surq =

(2) Entropy change : 4 0,surS∆ =

Entropy change of the surroundings (surS∆ )

1 A, 2 3 4sur ,sur sur ,sur ,sur ,surS S S S S S∆ = ∆ + ∆ + ∆ + ∆ + ∆

A 4

1 3sur

V VS nR ln nR ln

V V∆ = − −

From the two adiabatic steps,

32

1 4

VV

V V=

2

Asur

VS nR ln

V∆ = (> 0)

Entropy change of the universe (uniS∆ )

uni sys surS S S∆ = ∆ + ∆

2

Auni

VS nR ln

V∆ = (> 0)

The above analysis of a cyclic process may be summarized as follows:

� The entropy change of the system (the engine) in a cyclic process is zero (no entropy change) irrespective of the cycle being reversible or irreversible.

� The entropy change of the surroundings in a cyclic process is zero if the entire process is reversible, but greater than zero if the process includes any irreversible step.

� The entropy change of the universe in a cyclic process is zero if the process is reversible, but greater than zero if the process includes any irreversible step. Thus the entropy of the universe increases each time the cycle is completed.

The efficiency of the revised cycle:

(1) Work done by the modified cycle (wsys)



11

Asys

Vw nRT ln

V= + 2 1( )VnC T T− + 3

24

VnRT ln

V + 1 2( )VnC T T−

11

A

VnRT ln

V= + 3

24

VnRT ln

V (< 0)

(2) Heat taken in from the surroundings: (qsys)

A1 1

1sys

Vq q nRT ln

V= = (> 0)

(3) Efficiency of the modified cycle: (ε)

sys

sys

w

q

−ε =

11

Asys

Vw nRT ln

V= + ( 1 1

2 2A A

V VnRT ln nRT ln

V V− ) + 3

24

VnRT ln

V

and we know 32

1 4

VV

V V=

Then 11 2

A

( )sysV

w nR ln T TV

= − + 22

A

VnRT ln

V

2

1

1T

Tε = − −

22

A

A1

1

VT ln

VV

T lnV

When all steps are reversible

2

1

1revT

Tε = −

22

A

A1

1

rev

VT ln

VV

T lnV

ε = ε −

V2 > VA and VA > V1. Thus the second term on the right side is positive.

ε < εrev The above result demonstrates that the efficiency of a cyclic engine which involves an irreversible step is lower than that of the Carnot cycle in which all steps are reversible.



That is, the efficiency of the Carnot engine is the maximum that a cyclic engine can obtain, and all practical engines which involve irreversible step(s) have efficiencies lower than that of the Carnot engine.

From the results in Example 3.5,

A

1sys,irr

VS nR ln

V∆ = + 2

A

VnR ln

V = 2

1

VnR ln

V (= sys,revS∆ )

A

1sur ,irr

VS nRln

V∆ = − (cf, 2

1sur ,rev

VS nRln

V∆ = − )

2

Auni

VS nR ln

V∆ = (> 0)

Note from the above result that the entropy change (increase) of the system is the same for both the reversible and irreversible processes, whereas the entropy change (decrease) of the surroundings for the irreversible process is less than that for the reversible process. The net result is that the total entropy, i.e., the entropy of the universe, increases due to the irreversible step involved. The entropy change of the system can be viewed as follows:

( sysS∆ ) ( A

1

VnR ln

V) ( 2

A

VnR ln

V)

The sum is always the same as the entropy change for the reversible process.

Let us consider a partial process of the Carnot cycle as shown in the figure below. The process starts at State “a” and ends at State “c” via States “b′” and “b”. Determine the entropy changes of the system, the surroundings, and the universe.

Example 3.6





a (P1,V1)

P

V

Step A Adiabatic free expansion

b′ (PA,VA) b (P2,V2)

c (P3,V3)

d (P4,V4)

Entropy change of the system

Entropy transferred from the surroundings

Entropy produced (generated) inside the system = +



Suppose that the following two figures represent Path A and Path B, respectively”

Path A Path B

Note that Path B is the reverse of the compression paths in the Carnot cycle. Note also that both paths share the same starting point “a” and the finishing point “c”, and the processes are fully reversible in both paths. However, the amount of thermal energy taken in by the reversible Path A (qrev = q1) is not the same as that taken in by the reversible Path B (qrev = q2). When the thermal energy supplied is divided by the temperature at which it is delivered, it becomes the entropy change of the process:

1A

1

qS

T∆ = and 2

B2

qS

T∆ =

But we know that A BS S∆ = ∆ as the entropy is a state function. Therefore,

1 2

1 2

q q

T T=

This result assures that even though the thermal energy (heat, q) is not a state property (different for different paths), the entropy (q/T) is indeed a state property (independent of the path taken).

When the Carnot engine completes its cycle, the amount of thermal energy entered into the system during the expansion is not the same as that left during the compression. The former is larger than the latter. We know that the difference has been converted into work. That is, the amount of thermal energy of the system (q) alone is not conserved. However, the quantity of q/T is conserved. As a proof of this, calculate the change of entropy of the system for following two reversible paths: 1) Path A: An isothermal expansion followed by an adiabatic expansion 2) Path B: An adiabatic expansion followed by an isothermal expansion

Example 3.7

Isothermal (T1) expansion

Adiabatic expansion

Isothermal (T2) compression

Adiabatic expansion

a (P1,V1)

b (P2,V2)

c (P3,V3)

d (P4,V4)

P

V

q1


Adiabatic expansion


Adiabatic expansion

a (P1,V1)

b (P2,V2)

c (P3,V3)

d (P4,V4)

P

V

q2



Just as for a Carnot engine running in the forward direction we have an engine running in the reverse direction.

For a complete cycle

The coefficient of performance of a heat pump (η) is defined as

w

q1=η

A Carnot engine can be run in reverse and used to transfer energy in the mode of heat from a low temperature reservoir to a high temperature reservoir. This type of a device is called a heat pump, if it is used as a heat source. It is called a refrigerator, if it is used to remove heat. Prove that, either for heat pump or for refrigerator, work must be done on the engine.

22

2

qS

T∆ = 1

11

qS

T∆ =

021 =∆+∆=∆ SSScycle

From the first law ∆U = (q1 + q2) + w = 0

1 2

1 2

0q q

T T+ =

Example 3.8

12

2

1T

w qT

= − −

02 <q

012

1 >−T

T

w > 0 Work is done on the engine.

Heat reservoir T2

Engine

Heat reservoir T1

q2

q1

w

T2 > T1



Note that the coefficient of performance of a heat pump of a refrigerator, unlike the efficiency of a heat engine, can be greater than unity. Exercises

3.1 The following diagram shows the operation cycle of a Carnot refrigerator. The refrigerator operates between 25°C (T2) and -10°C (T1) and step 2 involves heat absorption of 500 J.

(1) Calculate the coefficient of the refrigerator. (2) Calculate the total work done per cycle.

1 2

1 2

0q q

T T+ = from the earlier discussion.

1

2 1

T

T Tη =

−

P1,V1

P2,V2

P3,V3

P4,V4

Step 4 Isothermal

compression

q2, w1, T2

w2, T2 → T1

w4, T1 → T2

V →

P ↑


Step 3 Adiabatic

compression

Step 2 Isothermal expansion



3.3 Energy Dispersion and Entropy

Transfer of thermal energy induces degradation or decrease in quality of energy, which in turn results in the loss of capacity to do work. Natural or spontaneous processes always occur in the direction by which thermal energy is degraded. Entropy is not energy, but a kind of the measure of the amount of energy in a physical system not available to do work. However, the definition of entropy, ∆S = qrev/T, which has been introduced earlier is in fact the definition of the change of entropy, not entropy itself. In order to define entropy itself explicitly, we need to consider the transfer of thermal energy in a microscopic point of view. Macrostates and Microstates

Discussions so far have been based on macroscopic matter, and made no reference to the microscopic nature of matter. As we know that atoms or molecules of matter provide storage to the energy transferred to the matter, a question arises as to what happens in the matter in the molecular point of view when thermal energy has been transferred to or from the matter. To answer this question, we first define two important terms: macrostates and microstates.

Dictionaries define, “macro-” as the prefix of the word “macroscopic” which designates a size scale very much larger than that of atoms and molecules, and “micro-” as the prefix of the word “microscopic” which designates a size scale comparable to the subatomic particles, atoms, and molecules.

But a macrostate and a microstate in thermodynamics do not just designate something big and small in size. In thermodynamics, a microstate isn't just about a smaller amount of matter or a smaller size of matter, but it implies a detailed look at the energy that molecules or other particles in a system have.

Let us start our discussion with an example of a simple system. Suppose that there is a system consisting of three molecules and three units of energy to share among them. The macrostate of the system is the one having the physical size of three molecules with the energy of three units. In classical thermodynamics, the macrostate is simply called the state. The macrostate or state does not care as to how the energy is distributed among the molecules. As long as the total number of molecules and the total units of energy are kept unaltered, the system is at the same macrostate. As for the microstate, however, we need to look at the energy distribution in detail. What we need to consider is the number of ways the total energy can be distributed among the molecules (or atoms). For the present example, there are several different ways of distributing these three units of energy to the three molecules. One way is that the energy is shared equally and each molecule has one unit of energy. The other way is that one molecule has all three units of energy and the other two have none. There is still another way: two units to one molecule, one unit to the other one molecule, and none to the last molecule. The following table shows all possible combinations.



The system of (n = 3, ε = 3)

These ten possible combinations are referred to as microstates of the system. In other words, the system has ten different microstates and thus it takes one of these ten microstates at any one instant in time. The specific phrase, “one of these …at any instant in time” is important. The system cannot take two or more microstates at the same time, but only one microstate at a time. The system will be in one of the microstates at one instant, and in the next instant the system can immediately move to another microstate. The system has 10 different choices for its microstate, but no options for its macrostate; i.e., it is under only one macrostate. The more number of microstates a system has, the more number of choices the system can take for the next instant.

Let us consider more about the above system. First we define a “configuration” as the collection of those microstates that possess identical distributions of energy among the accessible energy levels without distinguishing individual molecules. According to this definition of configuration, the system above has three different configurations. If all microstates are equally probable, the probability of any one configuration is proportional to the number of microstates: that is, the system will be at Configuration I for 10% of the time, at Configuration II for 30% of the time, at Configuration III for 60% of the time. As the number of molecules and the number of energy units increases, the number of accessible microstates will grow explosively; if a system has the size of 1000 molecules with the total energy of 1000 units, the number of available microstates will be around 10600- a number that greatly exceeds the number of atoms in the observable universe.

A more rigorous definition of the microstate is; “A microstate is one of the large numbers of different accessible arrangements of the molecules' motional energy (the translational, rotational, and vibrational modes of molecular motion) for a particular macrostate.”

One microstate is something like an instantaneous snapshot of the energy of all the individual molecules in the macrostate. In the next instant the system immediately changes to another microstate. A single microstate of a system has all the energies of all

Energy level

4 3 2 1 0

Microstates 1 2 3 4 5 6 7 8 9 10 Configurations I II III

No. of macrostates= 1

No. of microstates = 10

A B C

A

B C

B

A C

C

A B

A

B

C

A

C

B

B

A

C

B

C

A

C

B

A

C

A

B



the molecules on specific energy levels at one instant. All of the energy of a system can only be in one microstate at any time.

Energy Dispersion and Entropy

A macrostate is the thermodynamic state of a system that is exactly characterized by the system's properties such as P, V, T, and number of moles of each constituent. Thus, if its observable properties (P, V, T, …) do not change, a macrostate does not change over time, but the microstate does.

The energy of a system becomes more dispersed when the number of microstates available to occupy, i.e., the number of accessible microstates, becomes larger - there are more choices the system can take for distributing its energy at one instant.

The bottom line is that the change in the number of accessible microstates is the determiner of the spontaneous direction of a natural process.

Entropy is now defined to be the measure of the spontaneous dispersal of energy (in the previous sections it was called degradation of energy) for a system. As the number of microstates that are accessible for a system indicates all the different ways that the energy can be arranged in that system, the larger the number of accessible microstates, the greater the entropy of the system at a given temperature.

The correlation of the entropy with the number of microstates is shown by the Boltzmann equation which is one of the most celebrated equations in statistical thermodynamics:

S k lnW=

where W is the number of microstates, and k is Boltzmann's constant (1.38x10-23 J K-1).

Then the change of entropy between two different states (the initial and final states) will be,

final

initial

WS k ln

W

∆ =

How is this equation related to the following expression of the entropy change developed from the macroscopic approach of classical thermodynamics?

T

qS rev=∆

These two expressions, the latter from macroscopic approach and the former from microscopic considerations, are indeed related to each other, and the relationship is clarified in Examples 3.14 and 3.17.

Why does S depend not on W, but on the logarithm of W? Suppose we have two systems: System 1 (entropy of S1 and the number of microstates of W1) and System 2 (entropy of S2 and the number of microstates of W2). If we now redefine these two



systems as a single system, then the entropy of the new system will be the sum of the two: S = S1 + S2. But the number of microstates will be the product W1×W2 because for each microstate of System 1, System 2 can be in any of W2 numbers of microstates.

Now the entropy expressed by the logarithm of W warrants the additive nature of entropy. Provided that we are able to calculate the number of microstates for a given macrostate, the Boltzmann equation gives us answers to the relation between molecular motion and entropy of a system, i.e., the relationship between

Suppose that we have two hypothetical metal blocks: Block 1 having mass of three atoms and energy of 6 units, and Block 2 having mass of three atoms and energy of 2 units. There will be a number of ways of distributing the energy to the atoms in each block. The following diagram shows the number of ways of distributing the energy between the atoms, i.e., the number of microstates each block has:

* Config.: configuration number, Micro.: Number of microstates

S1, W1 S2, W2 S1, W1 S2, W2

S, W=W1W2 1 1S k lnW= 2 2S k lnW=

System 1 System 2 New System

1 2S k lnW k lnW= +

1 2S S S= +

1 1S k lnW= 2 2S k lnW=

( )1 2S k ln W W=

S k lnW=

1 2W WW=

1 2( )S k ln W W=

1 2S k lnW k lnW= +

Molecules (or atoms or ions) constantly energetically speeding, colliding with each other, moving distances in space (or, just vibrating rapidly in solids)

What we define as its entropy of the system

Block 1 (n = 3, ε = 6)

Conf.

Micro.

En

erg

y

•••

••

•

• •• •

• ••

•

•

•

• •

• •

•

I II IV III V VI VII

3 6 3 6 3 6 1

W = 28

0 1 2 3

4

5

6

Block 2 (n = 3, ε = 2)

Conf.

Micro.

En

erg

y

••

• •• •

I II

3 3

W = 6

0 1 2 3

4

5

6

No energy transfer

××××



Block 1 has 7 different arrangements (configurations) of atoms to the accessible energy levels, while Block 2 has 2 different configurations. Each of the configurations has its own number of microstates. For example, Configuration I of Block 1 has three different ways of arranging three atoms in those two energy levels. Let us name the atoms A, B and C, then the diagram on the right shows three different arrangements. In this way we can find that Block 1 has 28 microstates (W = 28) in total, and Block 2 has 6 (W = 6). If the combination of Block 1 and Block 2 is considered to form one system, but without allowing energy transfer between them, then the number of microstates of the system will be W = 168 (= 28×6).

Now suppose that one unit of energy is transferred from Block 1 to Block 2 so that the energy of Block 1 is reduced from 6 to 5 units, while the energy of Block 2 is increased from 2 to 3 units. Microstates of the two blocks in the new system are shown in the diagram below. The number of microstates of the combined system (Block 1 + Block 2) is now 210 (= 21×10), which is larger than that of the previous case where it was 168. Therefore, transfer of one unit of energy from Block 1 to Block 2 results in increase of the number of microstates, and thus the transfer must be spontaneous or natural.

Further transfer of energy by one unit equalizes the energy of both blocks. The result is shown in the following figure:

En

erg

y

0 1

2

3

4

5

6 A

B C A

B

C A B

C

Block 1 (n = 3, ε = 6→5)

Conf.

Micro.

En

erg

y

••

•

• •• •

• •• •

•

•

•

•

I II IV III V

3 6 3 6 3

W = 21

0 1 2 3

4

5

6

Block 2 (n = 3, ε = 2→3) εεεε

Conf.

Micro.

En

erg

y

••

•

••• •

I II

3 6

W = 10

0 1 2 3

4

5

6

• •

III

1

Block 1 (n = 3, ε = 5→4)

Conf.

Micro.

En

erg

y

••

•

• •• •

••

•

• •

I II IV III

3 6 3 3

W = 15

0 1 2 3

4

5

6

Block 2 (n = 3, ε = 3→4) εεεε

Conf.

Micro.

En

erg

y

••

•

• •• •

••

•

• •

I II IV III

3 6 3 3

W = 15

0 1 2 3

4

5

6



The number of microstates of the combined system is 225 (= 15×15), and thus this transfer of energy is also spontaneous.

Further transfer of energy will not occur spontaneously as it will decrease the number of microstates because it is the reverse of the above transfer sequence.

The above results are graphically shown in the following figure:

Note that the number of microstates reaches the maximum when both blocks share the same energy in the present case (i.e., the same temperature). No further transfer of energy will occur spontaneously. Note also that the energy disperses to become uniform, through which the number of microstates reaches the maximum.

As for the view of degradation of energy, let us look at the energy transfer again. For Block 1, it initially had 6 units of energy, but, after losing one unit of energy to Block 2, it is left with 5 units of energy. However, these 5 units of energy are not the same as before not only in the quantity of energy, but also in its quality. These remaining 5 units of the energy have been degraded, since these energies are not available for conducting all the work the initial 6 units of energy could do.

The energy in Block 2 has been upgraded by receiving one unit of energy from Block 1, but these upgraded energies are not available or useful for the energy of Block 1 to recover its initial grade. It is not possible for Block 2 to return one unit of energy back to Block 1 on its own right. If it could, the number of microstates would have decreased.

When we calculate the entropy of these three hypothetical systems,

� Before energy transfer : 168S k ln= � After energy transfer by one unit : 210S k ln= � After energy transfer by two units : 225S k ln= (uniform energy distribution)

Microstates 168 210 225 210 168

Block 1 ε 6 5 4 3 2

W 28 20 15 10 6

Block 2 ε 2 3 4 5 6

W 6 10 15 21 28

50

100

150

200

250 No

. of m

icrostates



It is clear from the above that the total energy disperses toward uniform distribution, i.e., toward the maximum number of microstates, or toward the maximum entropy.

In the previous discussion, we have seen that a particular configuration can have several different arrangements of molecules (or atoms) to the accessible energy levels (microstates). Here we discuss in a systematic way as to how to determine the number of microstates for a given configuration.

Consider the number of ways of distributing total n particles in a number of different energy levels in such a way that

n0 particles in the ε0 energy level, n1 particles in the ε1 energy level, n2 particles in the ε2 energy level, ………………………………… ni particles in the εi energy level.

Then the total number of ways to arrange n particles is n!, but there are,

n0! ways of arranging n0 particles in the ε0 energy level: not distinguishable, n1! ways of arranging n1 particles in the ε1 energy level: not distinguishable, n2! ways of arranging n2 particles in the ε2 energy level: not distinguishable, ………………………………………………………………………………. ni! ways of arranging ni particles in the εi energy level: not distinguishable.

Therefore, not all n! ways are distinguishable, since the particles at a same energy level are merely changing their sequential positions at the same level. Thus the number of distinguishable ways of arrangements (the number of microstates), W, is given by

!!.....!!

!

210 innnn

nW =

If E units of energy are distributed among n distinguishable particles, the total number of accessible microstates W is given by

( 1)

( 1)

n E !W

n ! E !

+ −=−

n = 3 E = 5

Five units of energy are distributed among three distinguishable particles. Calculate the total number of accessible microstates in the system.

Example 3.9



(3 5 1)

(3 1) 5

!W

! !

+ −=−

W =21

!!.....!!

!

210 innnn

nW =

600,831!1!2!4!3!2

!12 ==W

Let us define the notation {a, b, c,..} in which the numbers from the left, a, b, c, …, represent the occupancy of the energy levels from the lowest one upwards. Calculate the number of microstates for a system consisting of five energy levels in a state with occupation of the energy levels {2,3,4,2,1}.

Example 3.10

n = 2+3+4+2+1 = 12

Calculate the entropy of the system which consists of 12 particles in a state with occupation of the energy levels {2,3,4,2,1}.

Example 3.11

S k lnW=

W = 831,600 k = 1.38066 × 10-23 J K-1

23 1

22 -1

1 38066 10 (831 600)J K

1 88 10 J K

S . ln ,

.

− −

−

= ×

= ×

A system consists of 10 particles distributed over four energy levels {5,3,2,0}. If a single particle is excited by one energy level, what would be the new distribution of particles in the energy levels which maximizes the entropy of the system?

Example 3.12



Exciting a particle from the configuration {5,3,2,0} leads to the following three possibilities: {4,4,2,0}, {5,2,3,0} and {5,3,1,1}. Then,

{4,4,2,0} → 150,3!0!2!4!4

!10 ==W

{5,2,3,0}→ 520,2!0!3!2!5

!10 ==W

{5,3,1,1}→ 040,5!1!1!3!5

!10 ==W

When the volume of a system is increased without change in energy, its energy levels become closer together, i.e., more energy levels become accessible to molecules which are within the original energy range. Thus, while the original molecular motional energy is still the same in the larger volume (isothermal), the system has more energy levels, meaning that the number of accessible microstates increases.

This can be conceptually proved by employing quantum mechanics. The energy (translational energy) for a particle in a cubic box is given by

23

2 2 2 2( )

2

x y zh n n n

mV

+ +ε =

where m is the mass of the particle, h is the Plank constant, nx, ny and nz are the quantum numbers which can have positive integer values, and V is the volume of the box.

Upon expansion of the box (i.e., increase in V), the energy (ε) of each quantum level, i.e., a given (nx, ny, nz), decreases as can be seen in the above equation. Since the total energy of the gas does not change upon expansion into vacuum, the gap between the energy levels narrows as the volume expands, and thus more accessible energy levels are provided, leading to increase in the number of accessible microstates.

The following figure schematically represents the above discussion:

The configuration which maximizes the entropy is thus {5,3,1,1}: spreading out to the higher energy level.

In Example 3.5, an adiabatic free expansion step was inserted to know the effect of an irreversible step on the entropy change of the system. Although there was no heat transfer and no work done during the expansion, there was a change of entropy of the system. The entropy change was found in the Example by reversing the process; i.e., by compressing the expanded gas back to its initial volume. Explain the entropy change along with an adiabatic free expansion in view of the change in the number of microstates.

Example 3.13



3.4 Equilibrium Criterion and Entropy

Equilbirum Criterion

When a system is left to itself, it would either remain unchanged in its initial state, or move spontaneously to some other state. If the former is the case, the initial state is indeed the equilibrium state. If the latter is the case, however, the system is initially in a non-equilibrium state, and the system will spontaneously move to the equilibrium state.

From the discussions done so far, we may be able to draw the following summary:

� All real processes involve some degree of irreversibility and thus all real processes lead to an increase in the total entropy: Stot = Ssys + Ssur, and ∆Stot > 0.

� The total entropy does not change in the reversible process (∆Stot = 0). � The reversible process is the succession of equilibrium states. � At equilibrium, the total entropy or the number of accessible microstates is maximum. From the molecular point of view, the equilibrium state is the state at which thermal energy is dispersed to the maximum: i.e., a state at which the number of accessible

Stot

Process path

The total entropy is the maximum at equilibrium.

Gas Vacuum Expanded gas

Before expansion After expansion

Energy levels Accessible

energy levels



microstates is the maximum. Therefore, the following is the condition to be satisfied for a state to be at equilibrium:

0=dW

where W is the total number of microstates (accessible) which was given earlier as,

!!.....!!

!

210 innnn

nW =

Suppose that we have n particles which are distinguishable from each other, but do not interact with each other. All n particles are contained in a box. We are interested in the total number of possible distributions of the particles in the box whether each of them is on the left or right half. The number of distributions may be called the number of microstates. If the total number of particles is four, then the statistics will be like what we can see in the following:

Note that the curve is symmetric about the configuration of equal distribution (2 and 2).

Suppose that we now increase the number of particles to distribute, and the following is the selected statistics with 10 particles. Note that the smallest number of microstates is still unity as with less number of particles, but the largest number has grown to 252, and the plot shows much sharper peak at the configuration having equal number of particles in both halves (i.e., uniform distribution).

When we have 100 particles, the number of microstates of the most probable configuration (equal distribution of 50:50) will explosively grow as given below:

30100!10

50!50!W = =

Recall that the number of microstates defined by above equation is in fact the total number of microstates for a given configuration, not for all configurations of the a system. Under what conditions can we assume the above equation to represent approximately the total number of microstates of the whole system?

Example 3.14

Left half Right half

Particles No. of microstates

Probability Number of microstates

Left Right 0 4 1 1/16 1 3 4 4/16 2 2 6 6/16 3 1 4 4/16 4 0 1 1/16

Total 16 16/16(=1)

1 2 3 4 5 6



For 1,000 particles, the number of microstates of the most probable configuration becomes to be

3001000!10

500!500!W = =

As the number of particles is increased, the number of microstates of the most probable configuration becomes dominating and virtually represents the total number of microstates. That is,

W(most probable configuration) ≈ W(total)

The systematic way of finding the maximum number of microstates for a configuration, which must be the most probable configuration, is to take derivative of the equation for the number of microstates and put it equal to zero:

!!.....!!

!

210 innnn

nW =

Ln = Number of particles at the left half

Rn = Number of particles at the right half

R Ln n n= −

!

! ( )!L L

nW

n n n=

−

Taking logarithm for convenience,

! ! ( )!L LlnW ln n ln n ln n n= − − −

Applying the Stirling’s approximation,

!ln n n ln n n= − for large n

Particles No. of microstates

Probability Number of microstates

Left Right 0 10 1 1/814 1 9 10 10/814 2 8 45 45/814 3 7 120 120/814 4 6 210 210/814 5 5 252 252/814 6 4 210 210/814 7 3 120 120/814 8 2 45 45/814 9 1 10 10/814 10 0 1 1/814

Total 814 1

50 100 150 200 250 300



( ) ( )L L L LlnW n ln n n ln n n n ln n n= − − − −

Taking derivative with respect to nL,

0L

L

d lnW n nln

dn n

− = =

12Ln n=

12Rn n=

Now the procedure seen in the above example is extended to be applicable to more general cases:

!!.....!!

!

210 innnn

nW =

Taking the natural logarithm,

( )0 1 2! ! ! !lnW lnn ln n lnn ln n= − + + +⋯

or

! !ii

lnW ln n ln n= −∑

Applying the Stirling’s approximation !ln x x ln x x= −

( ) ( )i i ii

lnW n ln n n n ln n n= − − −∑

Taking derivative,

( )i i i i ii

d lnW n d ln n ln n dn dn= − + −∑

ii i i i

ii

dnd lnW n ln n dn dn

n

= − + −

∑

( )i ii

d lnW ln n dn= −∑

At the maximum of W, 0ln =Wd



( ) 0i i

i

ln n dn =∑

This equation represents the condition for the maximum number of microstates of the

most probable configuration, and in fact the total number of microstates of the system,

provided that n is sufficiently large. However, there are two constraints which must be

met;

(1) The total energy (U) must be constant:

∑ =εi

ii Un or in the differential form ∑ =εi

ii dn 0

(2) The total number of atoms (or molecules) must be constant:

∑ =i

i nn or in the differential form ∑ =i

idn 0

Applying so-called the undetermined multipliers,

where α and β are constant.

or Combining the last two equations,

( ) 0i ii

ln n dn =∑

∑ =εβi

ii dn 0

∑ =αi

idn 0

( ) 0i i ii

ln n dn+ α + βε =∑

0i iln n + α + βε =

ieeniβε−α−=

∑ ∑ βε−α−==i i

iieenn

∑ βε−

βε−=

i

i

i

i

e

e

n

n

Summation of these three equations yields,

dni ≠ 0 Therefore,

Summation yields



This equation gives the number of particles (ni) which will be found at the energy level of εi, or the probability (ni / n) that a particular particle will be found at the energy level of εi for the maximum number of microstates (W), or for the maximum entropy (S = k ln W), or at equilibrium (d ln W = 0). Partition Function

The constant β introduced in the above equation is related to temperature and given as*,

where k is Boltzmann’s constant: 23 -11 38 10 J Kk . −= ×

Substitution yields,

The partition function defined above is the sum of all the energy states allowed. The partition function is a function of temperature T and microstate energies εi’s, and the microstate energies are determined by other thermodynamic variables such as the number of particles and the volume, as well as microscopic quantities like the mass of the

constituent particles. The partition function is a particularly useful function which allows us to calculate all the thermodynamic properties of the system, once it is known. Partition function encodes how the probabilities are partitioned among the different microstates, based on their individual energies. If all states are equally probable (equal energies) the partition function is representative of the total number of possible states.

kT

1=β

∑ε−

ε−

=

i

kT

kTi

i

i

e

e

n

n

∑ε

−=

i

kTi

eZ

Z

e

n

n kTi

iε−

=

Definition of a function termed Partition function

* Readers who are interested in more discussion on this are suggested to refer to books on statistical thermodynamics.

Suppose that we have a hypothetical system in which n atoms are distributed in two energy levels. Using the Boltzmann distribution, discuss the dependence of the entropy change on 1/T, i.e.,

revqS

T∆ =

Example 3.15

Boltzmann distribution



According to the Boltzmann distribution,

∑ε−

ε−

=

i

kT

kTi

i

i

e

e

n

n

For two energy levels 1 and 2

1 2

i

kTi

kT kT

n e

ne e

ε−

ε ε− −=

+

1 1

2 2

0 and for level 1

and for level 2

n

n

ε =ε = ε

1 1

1 kT

n

ne

ε−=

+

2

1

kT

kT

n e

ne

ε−

ε−=

+

Then the distribution of atoms in level 1 and level 2 can be represented by the ratio of the number of atoms in level 2 to that in level 1, which is given by

2

1

kTne

n

ε−=

It is seen that the ratio is dependent on the temperature. If the temperature approaches 0K, the ratio becomes zero, meaning that n2 = 0 and thus all atoms are in the ground state (level 1). On the other hand, if the temperature is increased very high, the ratio approaches unity, which means the atoms are evenly distributed between the two levels. As the temperature is increased, therefore, more particles are accessed to the higher energy level, and thus the number of accessible microstates increases, and then the entropy of the system increases. It can be said that the entropy is a kind of function of the above ratio:

2

1

nS f

n

=

Suppose we have two systems; one at a cold temperature Tc, and the other at a higher temperature Th, both of which have two energy levels and n atoms. Then, the ratios (λ’s) are given by



2

1

c

c

c

kTT

T

ne

n

ε− λ = =

2

1

h

h

h

kTT

T

ne

n

ε− λ = =

When the two systems are placed in contact with each other, the temperature will be equalized to the equilibrium temperature (Te).

2h c

eT T

T+=

Then,

2

1

e

e

e

kTT

T

ne

n

ε− λ = =

Let us assume that Th is higher than Tc by α times, that is, h cT T= α .

c

c

kTT e

ε−λ =

1

c

h

kTT e

ε − α λ =

21 c

e

kTT e

ε − +α λ =

The number of microstates of the high temperature system will decrease by cooling from Th to Te, and on the other hand the number of microstates of the cold temperature system will increase by heating from Tc to Te. The ratio of the microstates at Te to that at Th of the high temperature system will be given by,

1(1 )e c

h

T kT

T

e

−α ε α +α

λ=

λ

According to the Boltzmann equation for entropy, the entropy change due to the change in state is proportional to the logarithm of the ratio of the number of microstates of the initial and final states. Thus the entropy change by changing the temperature from Th to Te will be proportional to the logarithm of the above equation:

1

(1 )e

h

T

T c

lnkT

λ − α ε= λ α + α

The above equation clearly predicts that the entropy change is inversely proportional to the temperature for a given temperature difference (i.e., for a given α). The above equation also predicts decrease of the entropy upon cooling, since the value of α is greater than unity. In a similar way, for heating of the low temperature system to the equilibrium temperature,



1

1e

c

T

T c

lnkT

λ α − ε = λ + α

Note in the above equation that it predicts that the entropy is also inversely proportional to the temperature for a given temperature difference and it increases upon heating. The change of the total entropy may also be predicted by adding the above two equations:

e

h

T

T

lnλ

+λ

e

c

T

T

lnλλ

= 1

(1 ) ckT

− α ε α + α

1

1 ckT

α − ε + + α =

( )( )

21

1 ckT

α − ε α + α

> 0

The above proves that heat transfer always results in increase of the total entropy of the universe (for the present case, the sum of the high temperature and low temperature systems).

The macrostate (or simply state) of a system can be fixed by fixing some of the variables like U, V, n and T. Not all of them are independent. When we fix V and n, for instance, U is dependent on T. Express internal energy in terms of partition function.

Example 3.16

∑ ε= iinU

∑ε

−ε=

i

kTi

i

eZ

nU

Z

e

n

n kTi

iε−

=

o oi

kTi

i

N Nu U e

n Z

ε−= = ε∑

∑ε−

=i

kTi

eZ

∑ε−ε=

i

kTii

ekTdT

dZ2

Multiplying Z

kTNo2

to both sides

2o

i

o kTi

i

N kT NdZe

Z dT Z

ε−= ε∑

2oN kT dZ

uZ dT

=

Internal energy

Partition function

Taking derivative

By comparison of the two equations

For internal energy per mole (u) of particles



Recall

i ilnW n ln n n ln n= −∑

and

=

ε−

Z

enn

kT

i

i

Combining the two equations,

i i

kT kT

i

n nlnW n ln n e ln e

Z Z

ε ε− − = −

∑

i i i

kT kT kTi

i i

n n nlnW n ln n e ln n ln Z e e

Z Z ZkT

ε ε ε− − −= − + + ε∑ ∑ ∑

∑−

=i

kTi

eZε

o o εi

kTi

i

N Nu U e

n Z

ε−= = ∑

o

nulnW n ln n n ln n n ln Z

N kT= − + +

2o

d ln Zu N kT

dT=

2o

V

d ln Zu N kT

dT =

1dx d ln x

x=

As the volume is constant,

Express entropy in terms of partition function. Make use of the results obtained in the previous example.

Example 3.17



o

nulnW n ln Z

N kT= +

2o

V

d ln Zu N kT

dT =

V

d ln ZlnW n ln Z nT

dT = +

S k lnW=

V

d ln ZS nk ln Z nkT

dT = +

Per mole of particles,

o oV

d ln Zs N k ln Z N kT

dT = +

From the previous example,

o

nulnW n ln Z

N kT= +

Classical thermodynamics defines the change of entropy a system by

T

dqdS rev=

Statistical thermodynamics, on the other hand, defines entropy by

S k lnW=

or the change of entropy by

dS k d lnW= .

Discuss how they are related to each other.

Example 3.18



For constant temperature,

o

nd lnW du

N kT=

oNu U

n=

dUd lnW

kT=

revdqdU = at constant volume

revdqd lnW

kT=

revdqkd lnW

T=

3.5 Entropy Changes and Entropy Productions

If we are faced with the problem of deciding whether a given process will proceed spontaneously, we might intuitively think whether there is enough energy available. We might suppose that it is perhaps the energy of the system that tends to become minimum and then would be tempted to suggest that,

- if the reaction is exothermic (∆H < 0), it takes place spontaneously, and - if the reaction is endothermic (∆H > 0), it does not take place spontaneously.

However, there are numerous reactions that are endothermic (heat absorbing), and occur spontaneously. For example, the phase transformation of white tin (Sn) to gray tin is exothermic:

Sn(white, 298K) = Sn(gray, 298K) ∆H = -2,100 J mol-1

but white tin is more stable at 298K. Thus energy alone then is not sufficient. Another example is dissolution of solid Na2NO3 in water. It dissolves, even though the process is endothermic (∆H > 0).

T

dqdS rev=

From classical thermodynamics

dS kd lnW= From statistical thermodynamics



Moreover, if the energy of the system decreases during a spontaneous change, its surroundings must experience increase in energy by the same amount in order to satisfy the first law of thermodynamics. This implies that the increase in energy of the surroundings is also spontaneous. Therefore energy alone does not provide a decisive clue as to whether or not a process of interest is spontaneous.

Entropy Changes

In order to clarify how a system changes of its state, we need to know not only the amount of energy in the system, but also the direction of distribution of the energy. We already discussed that it is entropy that satisfies the requirements, and the entropy change was defined as,

revqS

T∆ =

Application of this equation is now discussed using a simplifed process.

Let us consider a cylinder which contains water and water vapor at temperature T. The cylinder is in thermal contact with a heat reservoir, and they are in thermal equilibrium with each other at temperature T. The system (cylinder) and the surroundings (heat reservoir) are schematically shown below: Now thermodynamic states of the above setting can be summarized as follows:

� The temperature inside the cylinder is the same as that of the heat reservoir at T. � The pressure inside the cylinder is the saturation vapor pressure of water at T (Pi). � The external pressure (Po) is kept the same as the internal pressure Pi, i.e., Po = Pi. � The whole setting is in equilibrium and the piston does not move in either direction.

Next, the external pressure is suddenly decreased by ∆P, and thus the piston moves out due to the pressure imbalance (Refer to the following figure). After one mole of water has vaporized, the external pressure is restored to the saturation vapor pressure of water (Pi) at T.

water vapor

water

Heat Reservoir

Frictionless piston

Pi Po

T

T T



During the above process, the following events will occur:

� The volume inside the cylinder expands and the internal pressure decreases. � Water vaporizes and heat flows from the reservoir as the vaporization is endothermic. � The piston (frictionless) moves outwards and conducts work against the new external

pressure (Pext = Po - ∆P). � After one mole of water has vaporized, the system (cylinder: water + water vapor) and

the surroundings restores the new equilibrium at temperature T and pressure Pi. � But the amount of water vapor in the system at the new equilibrium state has been

increased by one mole. Otherwise the system at the new equilibrium is the same as before (the same T and P).

In the process described above, the system (cylinder) conducted work against the external pressure of Pext = Po - ∆P, and hence the amount of work done by the system is

( )VPPw o ∆−−=

where V = molar volume of water vapor. The negative sign in the equation (i.e., the negative amount of work) warrants that the work was done by the system and thus the system has lost energy.

The amount of heat (thermal energy) transferred from the heat reservoir to the cylinder can be found from the first law:

wqU +=∆ Thus,

( )VPPUq o ∆−+∆=

As the reservoir (surroundings) has lost this amount of thermal energy as heat to the system, the entropy change of the surroundings (which is so large that its temperature virtually does not change by losing the heat of q) is given by,

T

qSsur −=∆ or

T

dqdSsur −= for an infinitesimal change,

Note that the entropy change of the reservoir (∆Ssur) is not constant, but varies with the degree of irreversibility; that is, it changes with change in ∆P:

( )osur

U P P VqS

T T

∆ + − ∆∆ = − = −

The larger ∆P is, the less amount of work the cylinder (the system) performs, and the smaller the amount of thermal energy the surroundings loses to the system, and thus the smaller the entropy change (decrease as q < 0) of the surroundings becomes.

When the above process is conducted in a reversible manner by infinitesimally small change in P, i.e., ∆P → 0, then the above equation will become,



rev osur

q U P VS

T T

∆ +∆ = − = −

The entropy change of the surroundings as a function of the degree of irreversibility can be graphically represented as follows:

We are now ready to determine the entropy change of the system by applying our knowledge in two basic points:

(1) Entropy is a state function. (2) When a process undergoes in a reversible manner, the sum of the entropies of the

system and of the surroundings is equal to zero:

0sys surS S∆ + ∆ =

Reminder: A reversible process is the process in which the system and surroundings can be restored to the initial state from the final state without producing any changes in the thermodynamics properties of the universe (the system + the surroundings).

We already know the change of the entropy of the surroundings when the process is reversible, and thus the entropy change of the system must be,

∆Ssur

Decrease in the degree of irreversibility 0

∆S

sur

When the process is conducted reversibly.

water vapor

water

Heat Reservoir

Frictionless piston

Pi Po - dP

T

T T

↑↑↑↑↑↑↑↑

q



revsys

qS

T∆ = or rev

sysdq

dST

=

To help better understand the above equations as to why qrev is used even for an irreversible process, we now analyze in depth the transfer of thermal energy in the mode of heat and the conversion of thermal energy into the mode of work. The figure given below explains the transfer of thermal energy in an arbitrary irreversible process:

If the process is conducted in a reversible manner, the transfer and conversion of thermal energy can be represented by the following figure:

Note that, in the case of the reversible process both thermal energy taken in and that expended to conduct work are larger than those in the case of an irreversible process, but the change in the internal energy, ∆U, is the same as it is a state property.

When the above two processes are compared,

Ubefore

Thermal energy expended by conducting work w (or qw,irr)

Thermal energy taken in from the surroundings (q)

Uafter

System U

Surroundings

qrev wrev

∆U = Uafter - Ubefore

Ubefore

Thermal energy expended by conducting work wrev

Thermal energy taken in from the surroundings (qrev)

Uafter

Thermal energy which could have done work, had the process been reversible, but unused and left in the system (qw,irr)

Thermal energy which would have been taken in from the surroundings, had the process been reversible (qw,irr)

Uafter, rev

Uafter, irr

System U

Surroundings

q w

∆U = Uafter - Ubefore

Reversible process

Irreversible process qw,irr q

qrev

These are valid for any process, irrespective of the process being reversible or irreversible, since the entropy is a state function.



Two important points can be noted from the above:

(1) Uafter, irr = Uafter,rev since the internal energy is a state property.

(2)

q + qw,irr = qrev

The change of thermal energy which the system experiences in an irreversible process is the sum of the energy transferred from the surroundings and the thermal energy unexpended because of less amount of work done due to the process being irreversible, and the sum is always the same as the thermal energy transferred in the reversible process.

The following figure shows schematically the entropy changes of the system, the surroundings and the universe as a function of the degree of irreversibility:

It is emphasized again that although the thermal energy appeared in the expression for the entropy change of the system is that for a reversible process, the application of the equation

revsys

qS

T∆ =

surq

ST

∆ = −

revsur

qS

T∆ = −

0sys surS S∆ + ∆ =

The thermal energy transferred from the surroundings

The thermal energy unexpended, but left in the system

The thermal energy transferred from the surroundings

Irreversible process Reversible process

∆Ssys

∆Suiv = ∆Ssys + ∆Ssur

Decrease in degree of irreversibility 0

∆S

Reversible path

• rev

sysq

ST

∆ =



is not limited to the reversible process only, but equally valid for any irreversible processes. The total entropy change is then,

sursystot SSS ∆+∆=∆

T

qS rev

sys =∆ T

qSsur

−=∆

−+=∆T

q

T

qS rev

tot

T

qqS rev

tot−=∆

As rev maxq q= , and thus

revq q≥

0totS∆ ≥

Thus, the entropy of the universe (the system + the surroundings: totS∆ ) increases in irreversible (or natural, or spontaneous) processes. Only if the process is reversible, in other words, if the process is a succession of equilibrium states, the entropy of the universe remains unchanged. In other words the entropy is merely transferred from the system to the surroundings, or vice versa in the reversible process.

The entropy change accompanied by the change in state can be found by integrating the differential expression of the entropy change between the two limiting states:

T

qdSd rev=

Integration

∫=∆2

1

State

State

rev

T

qdS

Suppose we are interested in finding the entropy change due to the change in temperature from T1 to T2 at a constant pressure. Then,

2

1

State revState

dqS

T∆ = ∫



State 1 = T1, State 2 = T2 At constant P,

dHdqrev = dTCdH P=

The above equation allows us to find the entropy change of a material due to the change of temperature. Consider one mole of a perfect gas.

rev Pdq CdS dT

T T= =

2

1

T PT

CS dT

T∆ = ∫

Entropy is defined as for a system, where the equality sign holds for a reversible process and the inequality sign for an irreversible process. We know that δq is not a state function, i.e., it depends on the path the process takes. Using the case of an ideal gas, prove that entropy is a state function.

T

qS

δd ≥

δ δdU q w= + First law of thermodynamics

dU = CV dT δq = dqrev for a reversible process

δw = - PdV for PV work

VPdqdTdC revV −=

V

VdR

T

TdC

T

qdV

rev +=

PV = RT for one mole of a perfect gas

dS = dqrev / T by definition

Example 3.19



From Example 3.19,

The entropy change depends on the volume or pressure change. Refer to Example 3.13.

As entropy is a state function, we are free to choose a path from the initial to final states. The path along which a process takes place reversibly would be most convenient for

2 2 22 1 1

1 1

δstate revVstate

q T VS S C ln R ln

T T V

− = = +

∫

① This side depends only on the initial and final states. (T1,V1 and T2,V2)

② Therefore S is a state function for a perfect gas. Refer to the explanation given in page 42 for more general discussions.

2 2

1 1V

T VS C ln R ln

T V

∆ = +

T1 = T2 = constant

2

1

VS Rln

V

∆ =

When a system undergoes a process at a constant pressure, does the entropy change depend on the temperature?

If one mole of a perfect gas undergoes an isothermal process, is the change of entropy independent of pressure?

Example 3.20

Example 3.21

1

2

PS Rln

P

∆ =

P1V1 = P2V2



thermodynamic calculations, because the heat absorbed or released can be directly related to the entropy change:

This equation tells us that the entropy of a substance held at constant pressure increases when the temperature increases. Whenever a system undergoes a change in state and the amount of work done by the system is less than the maximum possible amount of work, then there is a net increase in entropy.

Suppose that thermal energy q is transferred as heat spontaneously from a system at a fixed temperature T1 to a system at a fixed temperature T2 without performing any work. Prove that the total entropy change of the process, ∆Stot, is positive.

T

qdSd rev= for a reversible process

TT

CS P dd =

PP T

qC

=dδ

System 1 T1

System 2 T2

0d

d⟩=

T

C

T

S P

q

at constant pressure.

w = 0

11 T

qS

−=∆

22 T

qS =∆

−=∆+∆=∆

1221

11

TTqSSS tto

021

21 ⟩

−=∆TT

TTqS tto

Prove the following statement: “An isothermal change in phase (phase transformation) of a substance produced by input of energy as heat always leads to an increase in the entropy of the substance.”

Example 3.22

Example 3.23



We may apply a number of equations developed so far to answer the above questions:

� When heat is transferred without changing temperature,

T

qS rev−=∆

� When heat is transferred with accompanying the change in temperature,

∫=∆2

1

T

T

rev

T

dqS

� When the pressure is constant,

rev Pdq dH C dT= = ,

and thus

∫=∆2

1

T

T

P dTT

CS

T

qS rev=∆

0⟩∆

=∆t

t

T

HS

qrev = ∆Ht (heat of transformation) > 0 T = Tt (transformation temperature)

Develop equations for entropy changes for the following three cases: (1) Thermal energy q is transferred from the heat reservoir I at temperature TI to the

heat reservoir II at temperature TII. Assume the heat transfer causes no temperature change in either reservoir.

(2) Thermal energy q is transferred from the heat reservoir I to the finite mass II (nII moles). Assume there is no temperature change in the heat reservoir I, but the temperature of the finite mass II rises from IIT to TII′.

(3) Thermal energy q is transferred from the finite mass I (nI moles) to another finite mass II (nII moles). Assume the temperature of both masses changes: Mass I from IT to TI′, and Mass II from IIT to TII′.

Example 3.24



Since entropy is a state function, the change in entropy is independent of the path the process takes. In other words, we can take any path that is convenient, as long as the path connects the two states. Two convenient paths will be,

(1) Reversible, isothermal process (T1, P1 → P2) followed by reversible, isobaric process (P2, T1 → T2)

(2) Reversible, isobaric process (P1, T1 → T2) followed by reversible, isothermal process (T2, P1 → P2).

Let us take the first path. The following is the graphical expression of the first path:

Processes

Reservoir I → Reservoir II

Reservoir I → Finite mass II

Finite mass I → Finite mass II

Entropy change

in I (J K-1) II T

qS −=∆

II T

qS −=∆ I

I

(I)I I

T P

T

CS n dT

T∆ = ∫

Entropy change

in II (J K-1) IIII T

qS =∆ II

II

(II)II II

T P

T

CS n dT

T∆ = ∫ II

II

(II)II II

T P

T

CS n dT

T∆ = ∫

Remarks

T

p PT

qdH C dT H C dT

n= → ∆ = =∫ : Find T.

1 1( )TP PT

C CdS dT S J mol K dT

T T− −= → ∆ = ∫

1( )T PT

CS J K n dT ,

T−∆ = ∫

TI

TII

TI

TII

TII

TI

TII

TII

TI

q q q

Develop a general equation for the entropy change of an ideal gas when a system undergoes the change in state from State 1 (P1, T1) to State 2 (P2, T2).

Example 3.25

′

′ ′

′

′ ′

′ ′

′

′



Step 1: Reversible, isothermal process

T

dqdS rev=

revrev dwdqdU +== 0

,revrev dwdq −= PdVdwrev −=

RTPV = for an ideal gas

V

dVRTdwrev −=

V

dVRTdqrev =

V

dVRdS =1

∫=∆2

11

V

V V

dVRS

21

1

VS R ln

V

∆ =

RTVPVP == 2211 and 2

1

1

2

P

P

V

V =

21

1

PS Rln

P

∆ = −

Actual path

P1

P2

T2 T1

Reversib

le iso

therm

al

Reversible, isobaric



Step 2: Reversible, isobaric process

T

dqdS rev=

dHdqrev = when P is constant.

dTCdH P=

dTCdq Prev =

dTT

CdS P=2

∫=∆2

12

T

T

P dTT

CS

The entropy change of the whole process from State 1 to State 2 is then,

21 SSS ∆+∆=∆

2

1

2

1

T PT

C PS dT Rln

T P

∆ = −

∫

Lost Work and Entropy Production

When we look at the equation for work done by the system,

( )VPPw ∆−−= o

it is clear that the system will do more work as ∆P becomes smaller, i.e., the pressure difference between the system and the surroundings becomes smaller. The system will do the maximum work when ∆P → 0, i.e., when the process is reversible.

omax revw w PV= = −

wqU +=∆



omax revq q U PV= = ∆ +

From the First Law,

wqU +=∆

revrev wqU +=∆

actact wqU +=∆

actrevrevact wwqq −=− Subscript “act” = “actual”

( )actrevactrev wwqq −−=

About ( )actrev ww −

The work lost due to the process being irreversible is permanent; that is, the amount of the lost work, wlos, is permanently lost so that it can never be recovered.

� losw < 0 for irreversible processes: the higher the degree of irreversibility, the larger the absolute amount of the lost work (|losw |).

� 0=losw for the reversible process, i.e., no lost work.

Let us consider a system which gains heat q and does work w. The diagram below is the schematic representations of the First Law of Thermodynamics to show the position of the lost work:

U q w∆ = + or w q U= − + ∆

The work that could have been done had the process been reversible.

The work that was actually done.

( )actrev ww − = _

The difference is the amount of work lost by the system due to the process being irreversible, and thus is called lost work (wlos). In other words, the system does less amount of work in an irreversible process than in the reversible process by the amount of the lost work (los rev actw w w= − ).



More about the lost work ( )los rev actw w w= −

( )actrevactrev wwqq −−=

( )actrevlos www −=

When heat q is supplied to the system, it adds to the molecular motional energy (i.e., kinetic energy: translational, rotational and vibrational) of the system. Then a portion of the motional energy will be expended in doing work w, which is the energy transfer due to organized motion of particles (atoms, molecules, etc.). The system can do the maximum amount of work when the process goes in the reversible manner (wrev). If the process is irreversible, the amount of work done by the system (wact) is less than wrev by the amount of wlos as seen in the above figure. This difference is termed the lost work, as the system has lost the chance to do this amount of work due to the process being irreversible. In other words, the motional energy of the system equivalent to losw has lost its chance to

conduct work and thus remained in the system as part of the motional energy.

rev rev act actU q w q w∆ = + = +

( )rev act rev actw w q q− = − −

The amount of work lost due to irreversibility ( losw )

The amount of heat which has lost the chance to do work of losw , and thus

remains in the system as motional energy.

•

q

w

∆U

∆U

•

revq actq

revw

actw

Reversible process

Irreversible process

Amount of work lost due to irreversibility (wlos)

Amount of heat (thermal energy) which could have been converted to work, had the process been reversible, but left unused in the system because the process is irreversible (qw,irr)

wlos = qw,irr



losrev wqq −= where q = qact

We know

T

qS rev

sys =∆ , T

qSsur −=∆

sursystot SSS ∆+∆=∆

T

qqS rev

tot−=∆

T

w

T

qS los

sys −=∆ T

wS los

tot −=∆

Since 0≤T

wlos

T

qSsys ≥∆ 0≥∆ totS

For infinitesimal change

T

qdSsys

δ≥ 0≥totdS

Equality : when the process is reversible. Inequality : when the process is irreversible.

It is now clearly seen that the total entropy change (∆Stot), which is the sum of the entropy change of the system (∆Ssys) and the entropy change of the surroundings (∆Ssur), is positive, except a special case which is the process being reversible. Thus, entropy does not play the zero sum game, but the sum of the entropy, i.e., the total entropy increases through irreversible or spontaneous or natural processes. The increase in entropy due to irreversibility is called entropy production or entropy generation.

0≥−=∆T

wS los

tot Entropy production

losrev wqq −=



T

qqS rev

tot−

=∆

Put irr revq q q= −

0w,irrtot

qS

T∆ = ≥

In the above, ( )w,irr revq q q= − is the thermal energy that could have been converted to

work (- losw ), had the process been reversible, but could not, because the process was

irreversible. Recall that work w is the energy transfer due to organized motion of molecules (or atoms, etc), while heat q is the transfer of thermal energy; i.e., the transfer of energy via thermal motion (chaotic, random, disorganized motion of molecules or atoms). Thus the above results may be viewed as mechanical energy (- losw ) having been

degraded to thermal energy (irrq ) due to irreversibility.

More about ( ) w,irrlossys

qwq qS

T T T T∆ = + − = +

q: Thermal energy transferred from the surroundings to the system which has been degraded by doing so.

qw,irr: A portion of the motional (thermal) energy of the system which would have been expended by doing work of wlos had the process been reversible, but has not been expended due to the process being irreversible, and left in the system.

Thus q/T is the entropy change of the system due to thermal energy transferred from the surroundings. As the thermal energy of the surroundings was reduced by q, the entropy of the surroundings has been decreased by the same amount, –q/T.

Thus qw,irr /T is the entropy change of the system due to the thermal energy degraded inside the system.

Entropy transferred from the surroundings:

q

T

Entropy produced in the system:

w,irrq

T

Entropy change of the system:

sysS∆

= +



In the reversible, isothermal expansion, the pressure outside cylinder which resists the expansion is always balanced against the pressure inside the cylinder. The graphical representations of three processes are given below:

(1) (2) (3)

A cylinder (system) contains an ideal gas at P1 and V1. Calculate the entropy change of the system for the following three different processes: (1) The gas is expanded reversibly and isothermally until the volume is doubled (V2

= 2V1). (2) The gas is expanded isothermally against the pressure which is suddenly

decreased to P2 (=½P1) (3) The gas is expanded adiabatically against no pressure (vacuum) until the volume

is doubled.

Example 3.26

State 1

State 2

q, w sys

sur

S

S

∆

∆

∆Ssys

∆Ssur

revq

T−

Decrease in degree of irreversibility 0 ∆

S

• revq

T

Entropy transferred from the surroundings

Entropy produced inside the system,

w,irrq

T

q

T

0

P2

P1

0

P2

P1

0

P2

P1

V2 V1 V2 V1 V2 V1

Work done Reversibly

Work done

Lost work

No work

∫−=2

1

V

VPdVw ( )122 VVPw −−= 0=w

Lost work



� Process 1 and Process 2: Isothermal process → T: constant

� Process 3: Adiabatic process Adiabatic : q = 0 Vacuum : w = 0

But, the internal energy (U) of the ideal gas is a function of temperature alone: (∆U = 0) →(Tintial = Tfinal) →T: constant

� For all three processes, therefore, even though the paths they take are different, the final states are all identical, and thus the entropy change of the system must be the same.

� We may find the solution of the problem by taking Process 1,

∫−=2

1

V

Vrev PdVw

12 2VV = , PV = RT

∫−=1

1

2V

Vrev dV

V

RTw

2revw RT ln= −

revrev wqU +=∆

But 0=∆U (ideal gas)

2revq RT ln=

T

qS rev=∆

2S R ln∆ = In the above analysis, a question arises: it has been seen that the entropy change of the system is given by

T

qS rev=∆

For the case 3) above is an adiabatic expansion, that is, q = 0. Why the entropy change is not equal to zero, but has a finite value of 2lnRS =∆ ? The answer to this question was already given when the Carnot cycle was discussed and also the energy level change along with the volume change was discussed in Example 3.13. The following shows a practical application of what was discussed earlier. The free expansion of a gas into vacuum under an adiabatic condition is certainly a spontaneous process so that the entropy must increase. Remember that qrev in the above formula is not the actual amount of heat transferred (in the present case, q = 0), but the

∆U = q + w = 0



heat to be transferred if the process were reversible. To answer the above question, first, let us reverse the process so that the expanded gas is compressed reversibly to its original volume (V2(=2V1) → V1). The work to be done to the system to compress the gas reversibly is,

PdVdwrev −=

,RTPV = T = constant (ideal gas)

V

dVRTdwrev −=

Integration

22

11

2

lnRTV

VlnRT

V

dVRTw

V

Vrev =

−=−= ∫

0=+=∆ revrev wqU for ideal gas

Ideal gas → U = f(T) only, but T = constant. Hence 0=∆U

Therefore revrev wq −=

2revq RT ln= −

T

qS rev=∆

2S R ln∆ = − for compression

2S R ln∆ = for expansion

Exercises

3.2 Calculate the work done by the reversible, isothermal expansion of 3 moles of an ideal gas from 100 liters initial volume to 300 liters final volume.

3.3 Tin (Sn) transforms from gray to white tin at 286K. The heat of transformation (∆Ht) has been measured as 2.1 kJ mol-1.

Sn(gray, 268 K) → Sn(white, 268 K)

(1) Calculate the entropy change of the system (Tin). (2) Calculate the entropy change of the surroundings. (3) Calculate the total entropy change of the universe (system + surroundings).



3.3 Suppose that the following reaction takes place at 298 K:

Fe2O3 + 2Al = 2Fe + Al2O3

The temperatures of both the system and the surroundings are maintained at 298 K. Calculate the entropy change of the system associated with the reaction. The following data are given:

3.4 One mole of metal block at 1000K is placed in a hot reservoir at 1200K. The metal

block eventually attains the temperature of the reservoir. Calculate the total entropy change of both the system (the metal block) and the surroundings (the reservoir). The heat capacity of the metal is given as

CP = 23 + 6.3x10-3T, J mol-1K-1

3.5 Liquid metal can be supercooled to temperatures considerably below their normal solidification temperatures. Solidification of such liquids takes place spontaneously, i.e., irreversibly. Now one mole of silver supercooled to 940°C is allowed to solidify at the same temperature. Calculate the entropy change of the system (silver). The following data are given:

CP(l) = 30.5 J mol-1K-1 CP(s) = 21.3 + 8.54×10-3T + 1.51×105T-2, J mol-1K-1

ofH∆ = 11,090 J mol-1 (Heat of fusion at Tm = 961°C)

Calculate the entropy change of the surroundings. Does the process proceed spontaneously?

3.6 Two blocks of the same metal with equal mass, but at different temperatures, one at 100°C and the other at 200°C, are brought into contact until they come to the same temperature. Assuming these two blocks are isolated from the surroundings, calculate the total entropy change. The heat capacity of the metal is 24 J K-1mol-1.

3.6 Statements of The Second Law of Thermodynamics

The science of thermodynamics deals with energy and energy transfer. The first law is a statement of the conservation of energy in energy transfer. The nature of energy and energy transfer cannot be fully explained by the first law alone. All of the discussions given in this chapter are in fact on some other natural phenomena, and the governing law behind the discussions is termed the Second Law of Thermodynamics. The following is the collection of statements which cannot be explained with the first law:

� Whenever energy is transferred, the grade of energy cannot be conserved; some energy must be permanently reduced to a lower grade.

Al Fe Fe2O3 Al2O3

o298S (J mol-1K-1) 28.3 27.2 87.5 51.1



� It is impossible to construct a device that will operate in a cycle and produce no effect other than the raising of a weight and the exchange of heat with a single reservoir. (Kelvin-Planck)

� It is impossible to construct a device that operates in a cycle and produces no effect other than transfer of heat from a cooler body to a hotter body. (Clausius)

� Heat cannot pass spontaneously (unaided) from a region of lower temperature to a region of higher temperature. (Clausius)

� It is impossible to construct a heat engine which produces no other effects than the extraction of heat from a single source and the production of an equivalent amount of work.

� The maximum efficiency of a heat engine depends only on the temperatures between which it operates and is independent of the nature of the cycle process.

� It is impossible for a device operating in a cyclic manner to completely covert heat into work.

� It is impossible by a cyclic process to take heat from a reservoir and to convert into work without simultaneously transferring heat from a hot to a cold reservoir. (Kelvin)

� Heat absorbed at any one temperature cannot be completely transformed into work without leaving some changes in the system or its surroundings.

� No process is possible in which the sole result is the absorption of heat from a reservoir and its complete conversion into work.

Various statements in the above deals with a common phenomenon in nature and affirm the existence of a state function termed “entropy.”

� Entropy, like energy, is a fundamental thermodynamic property.

� Entropy is an index of the capacity to do work.

� Change in entropy measures the degree of irreversibility of a process, or the capacity for spontaneous change of a process.

� Every physical or chemical process in nature takes place in such a way as to increase the sum of entropies of all the bodies taking any part in the process.

To sum up, the second law of thermodynamics can be formulated in terms of entropy as:

“The entropy of an isolated system must increase or in the limit remains constant.”

The mathematical expression of the above is,

0S∆ ≥ for an isolated system.

The whole of this chapter have dealt with interpretations and applications of the above inequality equation in a number of different perspectives.

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