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11. Entropy and the second law (Hiroshi Matsuoka) 11.1 Reversible and irreversible processes

In thermal physics, all the macroscopic processes are divided into two classes: reversible processes and irreversible processes. A reversible process of a macroscopic system is a process for which we can find some way of restoring the initial states for both the system and its surroundings. As we will see, almost all real macroscopic processes found in both nature and the lab are irreversible, though we can regard some of them as being approximately reversible. To distinguish these two types of processes on the macroscopic level, we will use entropy and the second law of thermodynamics. Irreversible processes spread energy or microscopic particles

Typically, irreversible processes involve some kind of dispersion or spreading of energy or microscopic particles in space. This spreading of energy or particles is traditionally referred to as a “dissipative” process. The spreading of energy is due to heat transfer driven by a spatial variation of temperature: energy or heat flows from a higher temperature region to its lower temperature surrounding region. The spreading of particles is driven by a spatial variation of density: particles flow from a higher density region to its lower density surrounding region.

Examples of irreversible processes include:

• An adiabatic free expansion of a low-density gas, in which the gas originally confined inside a half of an insulated box expands freely into a vacuum in the other half of the box.

• Friction between two pieces of solid, in which a piece of solid is rubbed against another,

which results in heating up or increasing the temperatures of the surfaces of both pieces and spreading energy inside them.

• Well-known Joule’s experiment, in which a paddle turning in water does some work on the

water and the energy introduced as the work then spreads inside the water to increase the temperature of the water.

• Heat conduction, in which some energy as heat flows from a hot to a cold objects spreading

energy over a larger volume. • Diffusion of particles such as dye molecules in an ink droplet spreading in water.

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Quasi-static processes and purely mechanical processes are reversible processes The most important example of reversible processes is clearly a quasi-static process, whose

effects such as the amounts of heat and work exchanged between a system and the outside during the process can be analyzed in terms of changes in its state variables such as its internal energy, entropy, and enthalpy (more on this, see Ch.9). A purely mechanical process where the temperature of a system involved or the heat flowing into the system does not affect the process is reversible. For example, a planet moving around the sun and a pendulum supported by a frictionless pivot and moving in a vacuum at very low temperatures are practically going through reversible processes.

In thermal physics, the terms quasi-static and reversible are practically equivalent and therefore in the rest of this chapter, we use these two terms interchangeably (more about a logical difference and the practical equivalence between these terms, see Sec.11.8.3). How can we characterize irreversible processes?

To characterize irreversible processes on the macroscopic level or in thermodynamics, we will use entropy and the second law of thermodynamics. To characterize irreversible processes on the microscopic level, on the other hand, turns out to be more difficult than we expect. The problem is that on the microscopic level, dynamical equations such as Newton’s equation of motion or the Schrödinger equation for a wave function are all time reversible and it is still not fully understood how we can explain macroscopic irreversibility using the reversible microscopic dynamical equations. Traditional statements of the second law by Clausius and Kelvin

Originally, the second law was stated in terms of a specific irreversible process. The second law basically declares that there exist irreversible processes in nature so that we can pick up any irreversible process and by declaring that the chosen process is irreversible, we can come up with a statement for the second law. For example, we can state the second law by declaring: “an adiabatic free expansion of a low-density gas is irreversible.”

Historically, Clausius chose heat conduction and stated the second law by asserting that “the heat flow from a hot to a cold systems is irreversible,” while Kelvin chose the full conversion of work into heat and stated the second law by asserting that “a complete conversion of work into heat with no other effect is irreversible,” which implies that it is impossible to construct a perpetual motion machine of the second kind that converts input heat fully into work.

It is a bit confusing that we have so many ways of stating the second law, but we can show all these statements are in fact logically equivalent. Naturally, we suspect that there must be a more general statement for the second law, which we call the “modern” statement.

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The modern statement of the second law The modern or general statement of the second law that does not use a particular irreversible

process employs the entropy of a system to classify all the adiabatic processes that the system can undergo into two groups: reversible or quasi-static processes and irreversible processes. Specifically, the modern statement asserts: “in an adiabatic process, the entropy of a macroscopic system either remains constant or increases, and the entropy increases if and only if the process is irreversible.”

Keep in mind that this statement of the second law is concerned only with adiabatic processes, not with general processes. Sometimes, the second law is misused simply because this restriction to adiabatic processes is overlooked. As we will see below, this restriction to adiabatic processes does not severely limit the application of the second law, since we can place s system of interest and other systems that exchange heat with the system inside a large insulating box so that whatever is happening inside the box can be regarded as an adiabatic process. 11.2 The second law and irreversible processes

In the following four sections, we will apply the modern statement of the second law to various irreversible processes and show that they are indeed irreversible according to the second law. Let us then present the modern statement of the second law as precisely as possible.

Suppose we pick an arbitrary pair of equilibrium states 1 and 2 of a macroscopic system. Now the question is whether we can have an adiabatic process going from state 1 to state 2. If there is an adiabatic process connecting these states, then is this adiabatic process reversible (quasi-static) or irreversible? The modern statement provides answers to these questions by using the entropy difference between these states defined by

!S " S2 # S1, where S1 is the entropy of the system in state 1 while S2 is the entropy of the system in state 2. The second law then relates the entropy difference with three possibilities as follows.

An adiabatic process from state 1 to state 2 is:

(i) reversible if and only if (i.e., ! ) !S = S2 " S1 = 0 or S2 = S1 so that the entropy remains

constant; (ii) irreversible if and only if (i.e., ! ) !S = S2 " S1 > 0 or S2 > S1 so that the entropy increases; (iii) impossible if and only if (i.e., ! ) !S = S2 " S1 < 0 or S2 < S1 so that the entropy decreases.

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Quasi-static adiabatic processes are reversible By definition, quasi-static processes are reversible, but using the second law, we can also

show that quasi-static adiabatic processes are reversible. For an infinitesimal quasi-static process in a system, the change dS of its entropy is given by

dS = !Qqs

T,

where !Qqs is an infinitesimal amount of quasi-static heat flowing into the system during the process and T is the temperature of the system. If the process is adiabatic, then !Qqs = 0 so that dS = 0 , which means that the entropy stays constant during the infinitesimal process.

For a quasi-static adiabatic process of finite “length,” the entropy therefore stays constant because

!S = Sf " Si = dS =quasi"static: i# f$%Qqs

T= 0quasi"static: i# f$ .

The modern statement of the second law on the TV diagram

For a given initial state 1, the three cases for the entropy difference !S considered in the modern statement of the second law correspond to three separate regions on the TV diagram as shown below.

Those states that we can reach from state 1 by following a quasi-static adiabatic process form a curve or an “adiabat” through state 1, which corresponds to the solution for the following differential equation,

1

TdT = ! "

CV# T

dV ,

which follows from the condition dS = 0 or S = const so that according to the first dS equation, we find

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dS = CV

TdT + !

" T

dV = 0 .

In the figure below, we have assumed ! > 0 and ! CV" T( ) > 0 so that when the volume

expands or dV > 0 , dT < 0 or the temperature decreases. The region above the adiabat corresponds to the case where the entropy increases, because

the first dS equation with ! > 0 indicates that the entropy S increases if either the temperature or the volume increases. Finally, the region below the adiabat corresponds to the case where the entropy decreases, because the first dS equation with ! > 0 implies that the entropy S decreases if either the temperature or the volume decreases.

“Surrogate” quasi-static process to calculate an entropy difference

To apply the second law to any process, we need to find the entropy difference !S between the initial and final states of the process. Our strategy is to measure or calculate !S using any of the quasi-static processes connecting these two states, even when we are interested in an irreversible process between these two states. We can use any quasi-static process because the entropy difference depends only on the initial and final states and is therefore independent of the choice of a process connecting them. We call such a quasi-static process the “surrogate” quasi-static process. Keep in mind that the surrogate quasi-static process cannot be an adiabatic process if the process in question is irreversible or impossible. If the surrogate quasi-static

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process were adiabatic, then we would find !S = 0 automatically, but we were supposed to show either !S > 0 or !S < 0 . 11.3 Adiabatic free expansion of a low-density gas is irreversible

Consider an insulated box which is originally partitioned into two chambers, one of which is filled with a low-density gas while the other chamber is a vacuum. The wall between the chambers has a small hole with a hatch which is originally closed. We then open the hatch so that the gas can start to escape through the hole. Eventually, the gas will fill up the both chambers and reach an equilibrium state where the temperature and the pressure are uniform throughout the both chambers.

This is an adiabatic process simply because the entire box is insulated so that there is no heat exchange between the gas inside and the outside. We know, from experience, that this adiabatic “free” expansion is an irreversible process. Let’s see if the second law agrees with our experience.

Before we calculate the entropy difference between the initial and the final equilibrium state, let’s learn a few things about this adiabatic free expansion. First, through this expansion, the volume of the gas increases from its initial value Vi to its final value V f .

As this expansion is adiabatic, there is no heat exchange between the gas and the outside so that

Q = 0 . Does the gas do some work? The answer is “no” because during the expansion, the gas is

pushing nothing: the gas molecules are moving into a vacuum so that they are not exerting forces on anything:

W = 0 .

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If no work is done by the gas and no heat is flowing into or out of the system, then the first law of thermodynamics tells us that the internal energy of the gas does not change by the adiabatic free expansion:

!U = Q +W = 0 .

As we have found in Sec.7.6, the internal energy of a low-density gas depends only on its temperature or

U = nu T( )

so that no change in the internal energy means no change in the temperature or T = const .

A quasi-static isothermal expansion at T can serve as a surrogate process

We are ready to come up with a surrogate quasi-static process for the adiabatic free expansion. Since the temperature of the gas is the same for the initial and the final equilibrium state, the surrogate process must start with state T ,Vi, n( ) and end with state T ,V f ,n( ) , where n

is the mole number of the gas. We can then use a quasi-static isothermal expansion of the gas from state T ,Vi, n( ) to state T ,V f ,n( ) as the surrogate quasi-static process.

Applying the first dS equation to this quasi-static isothermal expansion, we find

dST =!" T

dV =nRVdV . (HW#11.3.1.: show this)

As the volume of the gas increases or dV > 0 , the entropy of the gas increases because dST =

nRVdV > 0 .

This implies that the entropy change of the gas during the surrogate quasi-static isothermal expansion is positive because

!S = Sf " Si = dSTVi

V f

# > 0 .

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More specifically, the entropy change of the gas during the surrogate quasi-static isothermal expansion is given by

!S = dSTVi

Vf

" =nRVVi

V f

" dV = nRlnV f

Vi

#

$ % %

&

' ( ( > 0 , (HW#11.3.2: show this)

where V f > Vi .

The entropy difference between the initial and the final equilibrium state for the original adiabatic free expansion is equal to this entropy change and is therefore also positive. The second law then concludes that the adiabatic free expansion of the low-density gas is an irreversible process.

Answers for the homework questions in Sec.11.3

HW#11.3.1 dST =

CVTdT + !

" T

dV =!"T

dV =PTdV =

nRVdV ,

where dT = 0 , ! =1 T , ! T = 1 P , and P T = nR V . HW#11.3.2

!S = dSTVi

Vf

" =nRVVi

V f

" dV = nR lnV[ ]Vi

Vf = nR lnV f # lnVi( ) = nRlnV f

Vi

$

% & &

'

( ) ) > 0 ,

where V f > Vi and ln x > 0 for x > 1 .

11.4 Friction and Joule’s experiment: adiabatic processes with non-volumic work are

irreversible When a piece of solid is rubbed against another in a vacuum inside an insulated box, friction

heats up or increases the temperatures at the surfaces of the both pieces because some work is done on the surfaces. After a while, each system reaches its own equilibrium state and its temperature gets settled to a value higher than its original temperature. We are implicitly assuming here that these solid pieces hardly change their volumes. This process is known to be irreversible.

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When a paddle immersed in water held in an insulated box is turned and stirs up the water, the temperature of the water increases. After the paddle is stopped, the temperature of the water and that of the paddle reach a final common equilibrium value that is higher than their original temperatures. This is the classic experiment by Joule, where mechanical work is done on the water irreversibly.

What is common in both of these processes? In each of these processes, some mechanical work W is done on a system adiabatically without changing its volume. We can also show that the temperature of the system increases after the process. According to the first law of thermodynamics, the internal energy of the system increases because

!U = W > 0 . Since the internal energy is an increasing function of the temperature if the volume is kept constant,

!U!T

"

# $

%

& ' V ,n

> 0 .

The temperature of the system then increases after the work W is done on the system because

!T = dTU"U+!U V= const

# =$T$U%

& '

(

) * V ,ndU

U

U+!U

# =1

$U$T

%

& '

(

) * V ,n

dUU

U+!U

# > 0 .

A quasi-static isochoric process can serve as a surrogate quasi-static process:

We know, from experience, both of the processes mentioned above are irreversible. We can even show that any adiabatic process where some mechanical work W is done on a system without changing its volume is irreversible. To show this, we need to come up with a surrogate quasi-static process from an initial state Ti,V, n( ) to a final state Tf ,V,n( ) . If we transfer heat

quasi-statically into the system without changing its volume, we can change the temperature from Ti to Tf . We therefore use this quasi-static isochoric or isovolumic process as a surrogate

process. How much heat should be transferred into the system altogether? To increase the internal energy by !U = W while we keep the volume constant, we must transfer the heat QV ,n

qs that is equal to !U or W:

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QV ,nqs = !QV ,n

qs

Ti"T f

# =W ,

where !QV ,n

qs is an infinitesimal quasi-heat along the surrogate quasi-static isochoric process and is given by !QV ,n

qs = CVdTV ,n so that

dSV ,n =!QV ,n

qs

T=CV

TdTV ,n > 0

for dTV , n > 0 . This shows that the entropy change during the surrogate quasi-static isochoric process is positive because

!S = Sf " Si = dSV , nTi

T f

# > 0 .

The entropy difference between the initial and the final state for the original process is equal to this entropy change and is therefore also positive. The second law then concludes that any adiabatic process where work is done on a system with no volume change is an irreversible process. 11.5 Extending the second law for composite systems consisting of subsystems

To apply the second law to a composite system consisting of subsystems, we must know how to calculate the entropy of the composite system. We can extend the notion of entropy for a single system to the entropy of a composite system by defining the entropy of the composite system in its equilibrium state to be the sum of the entropies of the subsystems in their equilibrium states:

S = Sii!

Note that the entropy of a macroscopic system is a state variable and is therefore well-defined only for its equilibrium states. This definition for the entropy for a composite system is consistent with the fact that entropy is an extensive quantity since if we divide a single system into parts or subsystems, the entropy of the system must be the sum of the entropies of these parts.

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During an infinitesimal quasi-static process, an infinitesimal change in the entropy of the composite system is then the sum of infinitesimal entropy changes of the subsystems:

dS = dSi

i! ,

where dSi is an infinitesimal change in the entropy of the i-th subsystem and is given by

dSi =!Qiqs

Ti,

where !Qi

qs is the infinitesimal quasi-static heat flowing into the i-th subsystem and Ti is the temperature of the subsystem during the infinitesimal process. 11.6 Heat flow from a hot to a cold system is irreversible

Imagine that we place two macroscopic systems in an insulated box. One of the systems is hotter than the other as shown in the figure below on the left, where Th > Tc . After these systems are brought into contact with each other, some heat Q will flow from the hot to the cold system until the temperatures of the both systems reach a common value Teq . This process is

adiabatic because the two systems cannot exchange heat with the outside. To simplify our discussion, we will also assume that the volumes of the two systems are constrained to remain constant throughout the process:

Vh = const and Vc = const .

The final equilibrium temperature

How do we derive the final equilibrium temperature? The final temperature Teq can be

found by solving the following equation:

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! CVh T,Vh,nh( )

Th

Teq

" dT =Q = CVc T,Vc,nc( )

Tc

Teq

" dT ,

where CV

h is the heat capacity at constant volume of the originally hotter system and CVc is the

heat capacity at constant volume of the originally colder system. The left-hand side of this equation is the heat flowing out of the hotter system while the right-hand side is the heat flowing into the colder system.

If both CVh and CV

c are roughly constant in the temperature range of Tc,Th[ ] , then the above

equation becomes !CV

h Teq ! Th( ) = CVc Teq ! Tc( ) so that

Teq =CVhTh + CV

cTcCVh + CV

c .

A surrogate quasi-static process consists of “short” processes

To show that the above heat-transfer process is irreversible, we must come up with a surrogate quasi-static process. Our surrogate process must quasi-statically transfer the heat Q out of the hotter system and quasi-statically deposit the same amount of heat into the colder system. The surrogate process then takes the whole system from its initial equilibrium state to its final equilibrium state as follows:

Initial : Th ,Vh;Tc ,Vc( ) ! Final: Teq ,Vh;Teq ,Vc( ) .

An infinitesimal portion of this surrogate process can proceed as follows. First, to the hotter

system, we attach another system (for example, a low-density gas) whose temperature is slightly lower than that of the hotter system so that a small amount of heat !Qqs can flow from the hotter system to this extra system. We then detach this extra system off the hotter system and let it expand adiabatically so that its internal energy and therefore its temperature decrease until the temperature becomes slightly higher than the that of the colder system. We now attach this extra system to the colder system so that we can have the heat !Qqs flow into the colder system.

In this infinitesimal quasi-static surrogate process, the heat !Qqs flows out of the hotter system and then flows into the colder system. By repeating this process over and over again, we will eventually transfer the heat Q from the originally hotter system to the originally colder system.

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During each infinitesimal surrogate process, the temperatures of the both systems are roughly constant. If during a particular infinitesimal surrogate process, the temperature of the hotter system is T < Th( ) while the temperature of the colder system is ! T > Tc( ) , then the entropy

change dS in this process is then given by

dS =!"Qqs

T+"Qqs

# T = "Qqs !1

T+1# T

$ % &

' ( ) = "QqsT ! # T

T # T > 0 ,

where !Qqs is assumed to be positive and !"Qqs is the heat flowing out of the hotter system. As the entropy increases in each infinitesimal surrogate process, the entropy increases by the entire surrogate process

!S = Sf " Si = dSi# f$ = %Qqs T " & T

T & T i# f$ > 0 .

The entropy difference between the initial and the final equilibrium state for the original heat transfer process is equal to the entropy change by the surrogate process and is therefore positive. By applying the second law, we now know that the heat transfer process is irreversible.

11.7 The second law and heat engines

Applying the second law to a heat engine, we can arrive at a very important result that the efficiency of a heat engine becomes maximum when all the processes in the heat engine become quasi-static. In other words, for a particular design for the heat engine based on a particular cyclic process, the corresponding quasi-static engine, where all the processes proceed quasi-statically, sets the upper limit for the efficiency of the engine.

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Heat engine with two heat reservoirs As an example, we will look at the heat engine that operates between two heat reservoirs at

two different temperatures, Th and Tc , for which we assume Th > Tc . This heat engine goes through repetitive cycles in each of which it absorbs input heat Qin from the hotter heat reservoir at temperature Th , does work !W on the outside, and discards exhaust heat !Qex to the colder heat reservoir at temperature Tc .

After each cycle, the internal energy of the engine must return to its initial value so that according to the first law of thermodynamics,

!U = Qin +Qex +W = 0

so that !W = Qin +Qex .

By placing the heat engine and the heat reservoirs inside an insulated box, we can regard the cyclic process of the engine and the heat exchange processes between the engine and the reservoirs as an adiabatic process.

Heat reservoirs

A heat reservoir is an idealization of a large system that is so large that its heat capacity CV is almost infinite and therefore it hardly changes its temperature when it exchanges heat with a smaller regular system:

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dT = 1CV

!Q " 0 ,

where dT is the temperature change in the reservoir due to the heat !Q flowing into or out of the reservoir. Therefore, we can regard any heat transfer into or out of the reservoir to be practically quasi-static so that when some heat Q flows into the heat reservoir at temperature T, then its entropy changes by !S = Q

T.

Kelvin’s statement: a full conversion of heat into work is impossible

Applying the modern statement of the second law, we can show that Kelvin’s statement of the second law is true or that a full conversion of heat into work is impossible. If a heat engine receives heat Q from a heat reservoir and converts it fully into work in its cycle, then the entropy of the system returns to its initial value so that there is no net change in the engine’s entropy whereas the entropy of the heat reservoir must decrease by !Sreservoir = "

QT

,

where T is the temperature of the reservoir. By placing the heat engine and the heat reservoir inside an insulated box, we can regard the cyclic process of the engine and the heat exchange process between the engine and the reservoir as an adiabatic process. The total entropy, which is the sum of the engine’s entropy and the entropy of the heat reservoir, therefore must decrease. According to the modern statement, this process of full conversion of heat into work is then impossible. Carnot’s inequality for the efficiency of a heat engine:

We define the efficiency of a heat engine by

! "#WQin

=Qin +QexQin

= 1##Qex( )Qin

,

where we have used !W = Qin +Qex . A higher efficiency therefore means that we get more work output from a given amount of input heat. Since the invention of steam engines, we have been trying to develop a heat engine with a higher efficiency. Historically, Carnot was the first to

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discover that the efficiency for a heat engine between tow heat reservoirs has an upper limit that depends only on the temperatures of the heat reservoirs. Applying the second law, we will show the following inequality for the efficiency by Carnot: ! "1 # Tc

Th= !Carnot,

where the equality holds when the engine is reversible (i.e., each cycle of the engine is a reversible or quasi-static Carnot cycle discussed in Sec.10.1.1). Derivation of Carnot’s inequality for the efficiency

After one cycle, the engine returns to its initial equilibrium state so that there is no net change in the entropy of the engine: !Se = 0 . After the cycle, the entropy of the hotter reservoir must decrease by

!Sh = "

QinTh

,

while the entropy of the colder reservoir must increase by !Sc =

"QexTc

.

After the cycle, the total entropy change for the combined system of the engine and the two

heat reservoirs is therefore !Stotal = "

QinTh

+"QexTc

.

We now apply the second law to the combined system to find that the total entropy change must be either zero or positive: !Stotal = "

QinTh

+"QexTc

# 0

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so that !

!QexQin

" !TcTh

,

which leads to

! = 1""Qex( )Qin

#1 "TcTh

= !Carnot,

where the equality holds when there is no change in the total entropy so that the cycle proceeds reversibly or quasi-statically. The maximum work theorem

We can also obtain the above result for the efficiency of a heat engine by applying what is called the “maximum work theorem,” which is a generalization of Carnot’s finding and is discussed in Appendix #6.

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SUMMARY FOR SEC.11.1 THROUGH SEC.11.7

1. All the macroscopic processes are divided into two classes: reversible processes and irreversible processes. A reversible process of a macroscopic system is a process for which we can find some way of restoring the initial states for both the system and its surroundings.

2. Quasi-static processes and purely mechanical processes are reversible processes 3 The “modern” statement of the second law distinguishes between reversible and irreversible

processes using entropy as follows: An adiabatic process from state 1 to state 2 is: (i) reversible if and only if !S = S2 " S1 = 0 or S2 = S1 so that the entropy remains constant; (ii) irreversible if and only if !S = S2 " S1 > 0 or S2 > S1 so that the entropy increases; (iii) impossible if and only if !S = S2 " S1 < 0 or S2 < S1 so that the entropy decreases.

4. A quasi-static adiabatic process in a system is reversible as its entropy remains constant

during the process. 5. An adiabatic free expansion of a low-density gas is irreversible as its entropy increases in the

process. To show the increase in the entropy, we can use a quasi-static isothermal process of the gas as an surrogate process to calculate the entropy difference between the initial and the final states.

6. An adiabatic process where work is done on a system with no volume change is irreversible

as its entropy increases in the process. To show the increase in the entropy, we can use a quasi-static isochoric process of the system as an surrogate process to calculate the entropy difference between the initial and the final states.

7. A heat flow from a hot to a cold systems is irreversible as the total entropy of the two

systems increases in the process. To show the increase in the total entropy, we can use a series of infinitesimal quasi-static heat transfer processes between the systems as an surrogate process to calculate the entropy difference between the initial and the final states.

8. A full conversion of heat from a heat reservoir into work by a cyclic process in a heat engine

is impossible as the total entropy of the engine and the reservoir decreases. 9. By applying the modern statement of the second law to a heat engine that operates between

two heat reservoirs at two different temperatures, Tc , and Th , that satisfy Tc < Th , we can show Carnot’s inequality for the efficiency of the engine:

! "1 # Tc

Th= !Carnot.