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Chapter 4 Entropy and the Second Law of Thermodynamics

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Chapter 4 Entropy and the Second Law of Thermodynamics Chapter 4 Entropy and the second law of thermodynamics 4.1 Heat engines In a cyclic transformation thefinal state of a system 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 work 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. − Heat engine. 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. ENTROPY AND THE SECOND LAW OF THERMODYNAMICS 4.1.1 Carnot cycle 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 gas. (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 waste heat (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 temperature to a reservoir of higher temperature.” Summary: heat does notflow upwards. 4.2. SECOND LAW OF THERMODYNAMICS 35 Definition by Kelvin: “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 <Q 1 back toT 2. The net work heat engine heat pump output| | of such an engine would be Q1 Q 2 > 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 - heat transfer is an exchange of energy 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 temperatures can be more efficient than a reversible thermodynamic process. 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 ideal gas 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.
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