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Chapter 4

Entropy and the second law of

4.1 engines

In a cyclic transformation thefinal state of a is by definition identical to the initial state. The overall change of the internal energyU hence vanishes,

ΔU=0,ΔW= ΔQ. − A cycle transformation is by definition reversible and the done by the system during a cycle is equal to the heat absorbed. Work. The (negative) of the work

ΔW= P dV = area enclosed. − � corresponds for a reversible cyclic process to the area enclosed by the loop in theV P state diagram. − . Work is converted by a cyclic process into heat, and vice versa. A cyclic process can hence be regarded as an heat engine.

Consider a heat engine operating betweenT 1 > T2. Part of the heat that is transferred to the system from a heat bath with temperatureT 1, Q1, is converted into work,W , and the rest,Q 2, is delivered to a second bath withT 2 < T1 (con- denser). Following thefirst law of thermody- namics, Q Q = W . | 1|−| 2| | | 33 34 CHAPTER 4. AND THE SECOND LAW OF THERMODYNAMICS

4.1.1 The Carnot process is a reversible cycle process bounded by two isotherms and two adiabatic lines.

One Carnot cycle consists of four consecutive thermodynamic processes, which can be realized with an arbitrary working substance. We shall consider here however the case of an ideal .

(1) A B isothermal expansion T=T V V Q absorbed → 1 A → B 1 (2) B C adiabatic expansion T T V V ΔQ=0 → 1 → 2 B → C (3) C D isothermal compression T=T V V Q released → 2 C → d 2 (4) D A adiabatic compression T T V V ΔQ=0 → 2 → 1 C → a Work. We note thatQ 1 > 0 (absorbed from hot bath) andQ 2 < 0 (released to cold bath). Total energy conservation, viz thefirst law of thermodynamics, dictates that

0 = dU= (δQ+δW)=Q+W=Q 1 +Q 2 +W, � � where W is the work performed by the system, equal to the area enclosed in the loop. − Efficiency. Theefficiency of the Carnot engine is defined as

performed work W Q +Q Q Q η = − = 1 2 = 1 −| 2| . ≡ absorbed heat Q1 Q1 Q1

η is 100% if there is no (Q2 = 0). However, we will see that this is impossible due to the second law of thermodynamics.

4.2 Second law of thermodynamics

Definition by Clausius:

“There is no thermodynamic transformation whose soleeffect is to deliver heat from a reservoir of lower to a reservoir of higher temperature.” Summary: heat does notflow upwards. 4.2. SECOND LAW OF THERMODYNAMICS 35

Definition by :

“There is no thermodynamic transformation whose soleeffect is to extract heat from a reservoir and convert it entirely to work.” Summary: a perpetuum mobile of second type does not exist.

Equivalence. In order to prove that both definition are equivalent, we will show that the falsehood of one implies the falsehood of the other. For that purpose, we consider two heat reservoirs with temperaturesT 1 andT 2 withT 1 > T2. If Kelvin’s statement were false, we could extract heat fromT 2 and con- hot reservoir T 1 vert it entirely to work. We could then convert the work back to heat Q 1 Q1 entirely and deliver it toT 1 (there is no law against this). Thus, Clausius’ WW statement would be negated. If Clausius’ statement were false, we could let an amount of heatQ 1 flow Q 2 Q2 fromT 2 toT 1 (T2 < T1). Then, we could connect a Carnot engine be- tweenT 1 andT 2 such as to extract cold reservoir T2 Q1 fromT 1 and return an amount Q2 0, which would mean that an amount of heat Q 1 Q 2 is converted into work,| |−| without| any other effect. This would contradict Kelvin’s| statement.|−| | Order vs. chaos. From the microscopic point of view

- is an exchange of due to the random motion of atoms;

- work’s performance requires an organized action of atoms.

In these terms, heat being converted entirely into work means chaos changing sponta- neously to order, which is a very improbable process.

4.2.1 Universality of the Carnot cycle The second law of thermodynamics has several consequences regarding the Carnot cycle.

– A 100% efficient Carnot engine would convert all heat absorbed from a warm reser- voir into work, in direct contraction to the second law. We hence conclude that η< 1.

– All reversible heat engines operating between heat bath with temperaturesT 1 and T2 have the same efficiency. 36 CHAPTER 4. ENTROPY AND THE SECOND LAW OF THERMODYNAMICS

– No irreversible engine working between two given can be more efficient than a reversible .

For the last two statements we consider two enginesC andX (withX not necessarily reverisble) working between the baths atT 1 (warm) andT 2 (cold). We run the Carnot engineC in reverse, as a refrigerator C¯, and feed the work output ofX to C¯.

Work. The total work output of such a system is

W = Q� Q � Q Q . tot | 1|−| 2| − | 1|−| 2| � � � � – If we adjust the two engines such that Q � = Q , no net heat is extracted from | 1| | 1| the heat bath atT 1.

– In this case, an amount of heat Q 2 Q 2� is extracted from the heat bath atT 2 and converted entirely to work,| with|− no| other| effect. This would violate the second law of thermodynamics, unless

Q Q � . | 2|≤| 2|

Efficiencies. We divide this inequality by Q and, using the fact that Q = Q � , get | 1| | 1| | 1|

Q2 Q2� Q1 Q 2 Q1� Q 2� | | | |, | |−| | | |−| | , ηC η X . Q ≤ Q� Q ≥ Q� ≥ | 1| | 1| | 1| | 1|

The opposite inequalityη C η X is also true if bothX andC are reversible. In that case X could be run as a heat pump≤ andC as a heat engine. Universality. We made here use only of the fact that the Carnot machine is reversible. All reversible engines working between two heat baths have hence the same efficiency, sinceX could be, as a special case, a Carnot engine.

The Carnot engine is universal. ⇒ It depends only on the temperatures involved and not on the working substance. 4.3. ABSOLUTE TEMPERATURE 37

Efficiency of irreversible heat engines. All reversible heat engines have the same efficiencyη C . Is it then possible that a heat engineX exist such thatη X >η C ? In this case we would have forQ 1� =Q 1 thatW � > W , as shown above, and hence a violation of the second law. The efficiency of irreversible heat engines is lower that that of any reversible engine.

Irreversible heat pumps ExchangingX andC we may consider the case of irreversible heat pumps. One is then interested in thefigure of merit Q heat absorbed at low temperature 2 = . W work required

Repeating the arguments forQ � =Q wefind that the second law requiresW � W 2 2 ≤ and henceQ 2� /W � Q 2/W . Thefigure of merit of reversible heat pumps is consequently larger than thefigure≥ of merit of a irreversible heat pump.

4.3 Absolute temperature

The Carnot cycle is universal and may hence be used to define the temperatureθ in an absolute way, i.e. independent of working substances.

θ Q Q Q 2 | 2| ,1 η= | 2| = − 2 , (4.1) θ ≡ Q − Q Q 1 | 1| | 1| 1 whereη is the efficiency of a Carnot engine operating between the two reservoirs.

– The second law of thermodynamics implies that Q 2 is strictly greater than zero, Q > 0. The same holds for Q , which is anyhow| larger| (or equal) than Q . | 2| | 1| | 2| Using (4.1) for defining the temperature via

θ Q , θ >0 , i=1,2, i ∝| i| i leads hence to strictly positive temperatureθ.

– This means that the absolute zeroθ = 0 is a limiting value that can never be reached since this would violate the second law of thermodynamics.

Ideal gas. The proportionalityθ Q is defined such that ∼| | PV T= θ. NkB ≡ when the an is used as the working substance in a Carnot engine. 38 CHAPTER 4. ENTROPY AND THE SECOND LAW OF THERMODYNAMICS

4.3.1 Clausius’s inequality The temperature defined by (4.1) allows to transform the differential of the heat,δQ, which is not reversible into a reversible expression.

The inequality δQ 0 T ≤ �Pc holds for arbitrary cyclic processesP . The equality holds whenP c is reversible.

Proof of Clausius’s ’s inequality. For a derivation we divide the cycleP c inton segments so that on each segment its temperatureT i (i=1, . . . , n) is constant. Reference reservoir. We consider now a reservoir at reference temperatureT > T ( i) 0 i ∀ and introduce Carnot engines between the reservoir atT 0 andT i. Energy is conserved both by the Carnot engine and by the cycleP c,

n W =Q Q (0),W= Q , i i − i − i i=1 � where ( W ) is the work performed byP . Note thatW is negative/positive for a heat − engine/pump and thatQ i is the heatflowing intoP c atT i, viz out ofC i. Temperature. The definition Q(0) Q i = i T0 Ti (0) of absolute temperature allows us then to rewrite the total heatQ T absorbed from the reservoir atT 0 as n n Q Q(0) = Q(0), Q(0) =T i . (4.2) T i T 0 T i=1 i=1 i � � 4.3. ABSOLUTE TEMPERATURE 39

Global balance. The workW T performed by the overall system composed of cycleP c and of then Carnot engines is given by the overall energy balance

n n n n W =W+ W = Q Q(0) Q = Q(0) = Q (0) . T i − i − i − i − i − T i=1 i=1 i=1 i=1 � � � � � � (0) IfQ T > 0, the combined machine would convert heat from the reservoir atT 0 completely into mechanical work. – Kelvin’s principle state, that no reversible or can convert heat fully into mechanical work.

– There is no law forbidding to convert work into heat, that isQ (0) 0 is allowed. T ≤ Clausius’s inequality Using (4.2) we obtainfinally with

n Q δQ T i =Q (0) 0, 0 0 T T ≤ T ≤ i=1 i P � � Clausius’s inequality. Reversible processes. We may reverse a process, if it is reversible. This implies that both δQ/T 0 and δQ/T 0 are valid. This implies that Pc ≤ Pc ≥ � � δQ = 0 . (4.3) Pc T � Pc reversible

(0) Both the total workW T and the total heatQ T extracted from the reference reservoirT 0 then vanish.

4.3.2 Entropy The equality (4.3) implies that

B δQ S(B) S(A) (4.4) T ≡ − �A depends only on the end pointsA andB and not on the particular path, as long as it reversible, and that δQ dS= (4.5) T is an exact differential. Entropy. Eq. (4.4) states that there exists a state functionS, defined up to an additive constant, whose differential is dS=δQ/T . It is denoted entropy, 40 CHAPTER 4. ENTROPY AND THE SECOND LAW OF THERMODYNAMICS

Irreversible processes. We consider that the statesA andB occurring in ( 4.4) are connected both by a reversible pathP R and by an irre- versible pathP I ? Clausius’ inequality for the combined cycle yields

B δQ A δQ + 0 T T ≤ �A � �R �B � �I and hence A δQ B δQ =S(A) S(B) . T ≤ − T − �B � �I �A � �R Therefore, in general A δQ S(A) S(B) , (4.6) T ≤ − �B where the equality holds for a reversible process. Thermally isolated . From (4.6) it follows with

ΔS 0 ≥ δQ=0 that the entropy can only increase for thermally isolated systems which does not exchange heat with a reservoir. “The entropy of an can only increase.”

Notes.

– The joint system of a system and its environment is called ””. Defined in this way, the ”universe” is an isolated system and, therefore, its entropy never decreases. However, the entropy of a non-isolated system may decrease at the expense of the system’s environment.

– Since the entropy is a ,S(B) S(A) is independent of the path, re- gardless whether it is reversible or irreversible.− For an irreversible path, the entropy of the environment changes, whereas for a reversible one it does not.

– Remember that the entropy difference is

B δQ S(B) S(A) = − T �A only when the path is reversible; otherwise the difference is larger that the integral. 4.4. ENTROPY AS A THERMODYNAMIC VARIABLE 41 4.4 Entropy as a thermodynamic variable

For a reversible process in a thefirst law of thermodynamics can be written as δU=δQ+δW, dU= T dS P dV ,δQ= T dS . (4.7) − This representation implies that the entropyS and the volumeV are the two state vari- ables for which the differential of the internal energyU=U(S,T ) becomes exact. In this section we transform one basic pair of state variables to another, namely from (S,V ) to (T,V ), when one of the variables, here the entropyS, takes also the role of a thermody- namic potential. Differential of the entropy. Inserting dS and dU/T, ∂S ∂S ∂U ∂U dS= dT+ dV, dU= dT+ dV , ∂T ∂V ∂T ∂V � �V � �T � �V � �T into (4.7) leads to ∂S ∂S 1 ∂U 1 ∂U dT+ dV= dT+ dV+P dV , ∂T ∂V T ∂T T ∂V � �V � �T � �V �� �T � c ≡ V where we have used (3.7), namely that (�δQ/��δT)�V = (∂U/∂T) V =C V . Comparing coeffi- cients we thenfind with ∂S C = V ∂T T � �V (4.8) ∂S 1 ∂U = +P ∂V T ∂V � �T �� �T � the partial derivatives of the entropy, as a state function, with respect, with respect toT andV.

4.4.1 Entropy of the ideal gas

We recall (3.9) and (3.6), namely that the specific heatC V and the free energyU of the ideal gas are 3 3 ∂U C = nR U= nRT = 0. V 2 2 ∂V � �T Using the partial derivatives (4.8) and the equation of statePV= nRT of the ideal gas then leads with 3 dT dV dS= nR + nR 2 T V to the entropy difference 3nR T V S(T,V) S(T ,V ) = log + nR log . (4.9) − 0 0 2 T V � 0 � � 0 � Note that the number of molesn is here constant. 42 CHAPTER 4. ENTROPY AND THE SECOND LAW OF THERMODYNAMICS

4.4.2 Maxwell equations The commutativity of differentiation operations, Schwarz’s theorem, can be used to derive relations between thermodynamic quantities. For the case of the differential (4.7) this implies that∂ 2U/(∂S∂V)=∂ 2U/(∂V∂S) and hence

∂T ∂P ∂V ∂S dU= T dS P dV, = , = , (4.10) − ∂V − ∂S ∂T − ∂P � �P � �T

where we used an inversion for the last step. The relation (∂V/∂T) P = (∂S/∂P) T is denoted a Maxwell equation. −

4.4.3 Energy equation The entropy is not an experimentally controllable variable, in contrast toT,V andP, which allow to the thermal

P=P(T,V). (4.11)

We however use (4.7) to deduce an relation, the energy equation, which allows to determine the caloric equation of stateU=U(T,V). Energy equation. We use the commutativity of differentiation operations, as in Sect. 4.4.2, but this for the derivatives of the entropy:

∂ 1 ∂U ∂ ∂S = ∂V T ∂T ∂V ∂T � � � � ∂ ∂S ∂ 1 ∂U = = +P ∂T ∂V ∂T T ∂V � � � � �� 1 ∂U 1 ∂2U ∂P = +P + + . − T 2 ∂V T ∂T∂V ∂T � � � � Canceling identical terms we get

∂U ∂P =T P energy equation. (4.12) ∂V ∂T − ⇒ � �T � �V

The derivative of the is written with (4.12) in terms of measurable quan- tities. It is fulfilled for an ideal gas, for whichPV= nRT and (∂U/∂V) T = 0.

4.5 Eulers’s cyclic chain rule

The partial derivative (∂P/∂T) V entering the energy equation (4.12) may be related to thermodynamic coefficients as well. The involved type of variable transformation is can be applied to a large set of thermodynamic quantities. 4.5. EULERS’S CYCLIC CHAIN RULE 43

Implicit variable dependencies. We are dealing in general with a set of variables, e.g. P,V andT , which are related we a equation of state ∂f ∂f ∂f f(P,V,T)=0, dP+ dT+ dV= df(P,V,T)=0. ∂P ∂T ∂V From the relative partial derivatives of the state variables

∂P ∂f/∂T ∂T ∂f/∂V ∂V ∂f/∂P = , = , = ∂T − ∂f/∂P ∂V − ∂f/∂T ∂P − ∂f/∂V � �V � �P � �T it follows that ∂P ∂T ∂V = 1 (4.13) ∂T ∂V ∂P − � �V � �P � �T The importance of (4.13), Euler’s chain rule, lies in the fact that one does not know the equation-of-state functionf(P,V,T ) explicitly, only that it exists. Expansion and compression coefficients. Using (4.13), we get

∂P 1 (∂V/∂T) ∂P α = = P , = , (4.14) ∂T − (∂T/∂V) (∂V/∂P) − (∂V/∂P) ∂T κ � �V P T T � �V T

where have made use of the thermodynamic coefficients

1 ∂V α= (coefficient of ) V ∂T � �P 1 ∂V κ = (isothermal ) T − V ∂P � �T 1 ∂V κ = (adiabatic compressibility) S − V ∂P � �S

Using (4.14) the energy equation (4.12) reads

∂U ∂P α +P=T =T . (4.15) ∂V ∂T κ � �T � �V T

Mayer’s relation betweenC P andC V . The energy equation (4.15) can be used to rewrite Mayer’s relation (3.12), which we derived considering the chain rule for (∂U/∂T) P , as

∂U ∂V C =C + P+ , P V ∂V ∂T � � �T � � �P Tα/κ T Vα � �� � � �� � 44 CHAPTER 4. ENTROPY AND THE SECOND LAW OF THERMODYNAMICS

which leads then to

α2 CP C V = T V >0 . (4.16) − κT

Note that the thermal expansionκ T is normally positive. close to the freezing point has however an anomalousκ T < 0.

4.5.1 Entropy differentials The rewritten energy equation (4.15) can be used to rewrite also the the differential (4.8) of the entropy as

∂U α T dS=C dT+ +P dV=C dT+T dV . (4.17) V ∂V V κ �� �T � T

T andV as independent variables. Expression ( 4.17) for the differential of the entropy implies that the absorbed heatδQ can be expressed likewise in terms of directly measurable coefficients,

α δQ= T dS=C V dT+ T dV , (4.18) κT

whereT andV are here the independent variables.

T andP as independent variables. We use the Maxwell equation ( 4.10), (∂V/∂T) P = (∂S/∂P) , − T ∂S ∂S C ∂V dS= dT+ dP= P dT dP , ∂T ∂P T − ∂T � �P � �T � �P and obtain T dS=C dT αT V dP , (4.19) P − where the independent variables are nowT andP. V andP as independent variables. We note that dT can be rewritten in terms of dV and dP as ∂T ∂T 1 κ dT= dV+ dP= dV+ T dP , ∂V ∂P αV α � �P � �V where we have used (4.14), viz (∂P/∂T) V =α/κ T , in the last step. Inserting dT into (4.19) we obtain C C κ T dS= P dV+ P T αTV dP (4.20) αV α − � � for T dS, where the independent pair of state variables is nowV andP. 4.6. THIRD LAW OF THERMODYNAMICS (NERNST LAW) 45 4.6 Third law of thermodynamics (Nernst law)

Analyzing experimental data, Nernst has concluded that in the limitT 0 the entropy becomes a constant independent of other thermodynamic parameters such→ as and , ∂S ∂S = = 0. (4.21) ∂V T 0 ∂P T 0 � � → � � → The entropy is defined via (4.4) only up to constant, which can be selected hence such that lim S(T)=0 . (4.22) T 0 → This equation is equivalent in , as we will discus in Sect. 8.2, that nearly all states of matter are characterized by a unique . Macroscopically degenerate ground state leading tofiniteT = 0 are observed only for exotic phases of matter. Heat capacities vanish forT 0. The heat capacities disappear atT = 0 as a consequence of (4.22): → ∂S lim CV = lim T = 0, T 0 T 0 ∂T → → � �V ∂S lim CP = lim T = 0. T 0 T 0 ∂T → → � �P

The ideal gas does not fulfill the third law. The heat capacities (3.9) and (3.11) of the ideal gas are constant,

3 5 C = nR, C = nR , V 2 P 2 in contradiction with the third law. This is because the idea gas corresponds the high- temperature limit of the , which undergoes further gas transitions upon cooling. → →

No thermal expansion forT 0. The Maxwell equation ( 4.10), (∂V/∂T) P = (∂S/∂P) , implies → − T 1 ∂V 1 ∂S lim α = lim = lim − = 0, T 0 T 0 V ∂T T 0 V ∂P → → � �P → � �T where the last step follows from the fact that any derivative of a constant vanishes. The absoluteT=0 (zero point) is unattainable. We analyze what happens when we are trying to reach low and lower temperatures by subsequently performing adiabatic and isothermal transformations. 46 CHAPTER 4. ENTROPY AND THE SECOND LAW OF THERMODYNAMICS

Using a gas as a working substance, cooling is achieved by the Linde method through a sequence isothermal and adiabatic transformations.

A B isothermal compression → Work is performed on the gas and an amount of heatQ 1 < 0 is transferred from the substance to be cooled (characterized by a low temperatureT 1), to the reservoir (having a higher temperature) in a reversible process. The entropy of the substance being cooled diminishes consequently by

Q1 ΔS1 = . T1

B C adiabatic expansion → The gas cools by performing work. The entropy remains however withδQ=0 constant.

Note that all entropy curves converge toS(T 0) 0. The process becomes hence progressively ineffective and an infinite number→ of→ Linde iterations would be needed to reach the limitT 0. →