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Equivalent second law of thermodynamics statements - 1Kinetic Theory ofGasesThermodynamicsState VariablesZeroth law ofthermodynamicsReversible andIrreversibleprocessesFirst law ofThermodynamicsSecond law ofThermodynamicsEntropyThermodynamicPotentialThe Clausius statement:It is impossible to construct a device that, operating in a cycle,produces no e ect other than the transfer of heat from a colder to ahotter body.The Kelvin-Planck statement:It is impossible to construct a device that, operating in a cycle,produces no e ect other than the extraction of heat from a singlebody at a uniform temperature and the performance of an equivalentamount of work.Third Law ofThermodynamicsPhase diagram80/120

Equivalent second law of thermodynamics statementsKinetic Theory ofGasesThermodynamicsState VariablesZeroth law ofthermodynamicsReversible andIrreversibleprocessesFirst law ofThermodynamicsSecond law ofThermodynamicsEntropyThermodynamicPotentialThird Law ofThermodynamicsPhase diagramThe R on the diagram of the forbidden device, denotes refrigerator.The E on the diagram of the forbidden device, denotes engine.Note the use of body rather than heat reservoir, meaning thatengines can be considered to operate between two bodies (one asource and the other a sink of heat) of which the hotter one coolsand the colder one heats up whilst the engine is running.81/120

Clausius, Kelvin and PlanckKinetic Theory ofGasesThermodynamicsState VariablesZeroth law ofthermodynamicsReversible andIrreversibleprocessesFirst law ofThermodynamicsSecond law ofThermodynamicsEntropyThermodynamicPotentialThird Law ofThermodynamicsPhase diagram82/120

Assignment/Tutorial - discussionsKinetic Theory ofGasesThermodynamicsState VariablesZeroth law ofthermodynamicsReversible andIrreversibleprocessesFirst law ofThermodynamicsSecond law ofThermodynamicsEntropyThe equivalence of the Clausius and the Kelvin-PlanckstatementsNo engine operating between two reservoirs can be more efficientthan a Carnot engine operating between the same two reservoirs.All Carnot engines operating between the same two reservoirshave the same efficiency (INDEPENDENT of the workingsubstance).ThermodynamicPotentialThird Law ofThermodynamicsPhase diagram83/120

Equivalent second law of thermodynamicsKinetic Theory ofGasesThermodynamicsState VariablesZeroth law ofthermodynamicsReversible andIrreversibleprocessesFirst law ofThermodynamicsSecond law ofThermodynamicsEntropyThermodynamicPotentialThird Law ofThermodynamicsW Q 1 for any heat engine. Heat cannot completely be1converted to mechanical work.Second Law can be formulated as: (A) Heat flows by its ownonly from the hot to cold regions, never into the oppositedirection. (B) There is no periodically acting machine that canconvert heat completely into mechanical work without additionalsupply of energy. OR: The realisation of a perpetuum mobileof second kind is impossible.Example of perpetuum mobile of 2nd kind: A ship with enginesthat receive their energy solely from the heat of the sea. Such aship could move without additional energy and would not needOil or Coal.Phase diagram84/120

Entropy: Basic Definition - 1Kinetic Theory ofGasesThermodynamicsState VariablesZeroth law ofthermodynamicsReversible andIrreversibleprocessesFirst law ofThermodynamicsSecond law ofThermodynamicsNew property energy arose via consideration of the first law ofthermodynamicsEntropy: a thermodynamic property that provides a quantitativemeasure of the disorder of a given thermodynamic state, fromconsideration of the second law of thermodynamicsThe word itself was coined by Clausius who based it on thecombination of - (en-) “to put into,” and o (trope),“turn” or “conversion.”thermal reservoir at Tres , delivers heat Q 0 to engine, the enginedelivers W 0 work and rejects Q heat.EntropyThermodynamicPotentialThird Law ofThermodynamicsPhase diagram85/120

Basic Definition - 2Kinetic Theory ofGaseswe know from Carnot Cycle discussionsThermodynamicsState VariablesZeroth law ofthermodynamicsReversible andIrreversibleprocessesFirst law ofThermodynamicsSecond law ofThermodynamicsEntropyThermodynamicPotentialThird Law ofThermodynamicsPhase diagramThis meansQ0Q TresTorQ0Tres0 TLTH QLQH .QT0From first law we know W Q - dU, which means W 0 Tres TQ - dUHHHfor a thermodynamic cycleW 0 Tres TQ - dUwe cannot convert all the heat to work, Hbut we can convert allthe work to heat: W 0 0. This means Tres TQ 0.HTres 0; TQ 0.If all processesare reversible, we lose the inequality, and getHsimply TQ 0.86/120

Basic Definition - 3consider starting from 1, proceeding on path A to 2, andreturning to 1 via path B.Kinetic Theory ofGasesThermodynamicsState VariablesZeroth law ofthermodynamicsReversible andIrreversibleprocessesFirst law ofThermodynamicsSecond law ofThermodynamicsEntropyThermodynamicPotentialThird Law ofThermodynamicsPhase diagramR2R1 BThis means 1 TQA 2 QT 0.same exercise going from 1 to 2 on path A and returning onR2R1 CR1 Bpath C, 1 TQA 2 Q 0. Subtractung we get 2 QTT R 1 QCT 0.R21 QR1 CR 2 QBR 2 QCB 2 QT , this imples 1 T 1 T2 TR2Because paths B and C are di erent and arbitrary, but 1 TQ isthe same on either path, the integral must be path-independent.Defines a thermodynamic property of the system - entropy, S, anR2extensive quantity S2 - S1 1 TQ .87/120

Basic Definition - 3Kinetic Theory ofGasesThermodynamicsIntroducing this thermodynamic variable, the second law ofthermodynamics can be mathematically formulated.Reduced Heat: dQT - Ration of heat supplied (dQ) to a systemat temperature T .State VariablesZeroth law ofthermodynamicsReversible andIrreversibleprocessesFirst law ofThermodynamicsSecond law ofThermodynamicsEntropyThermodynamicPotentialThird Law ofThermodynamicsPhase diagramA system in Carnot cycle can come to (c) in two ways. (a) )1(b) ) (c) : Heat exchanged during isothermal process QT1(a) ) (d) ) (c) : Heat exchanged during isothermal processQ2T288/120

Entropy DefinitionKinetic Theory ofGasesThermodynamicsState VariablesZeroth law ofthermodynamicsReversible andIrreversibleprocessesFirst law ofThermodynamicsSecond law ofThermodynamicsEntropyThermodynamicPotentialThird Law ofThermodynamicsFor reversible process:Q1T1 Q2T2 .Reduced heat (engines) do not depend on the path but only onthe initial and final state of the system.EntropyIntroduce a thermodynamical variable S with dimension [S] [J]/[K]; dS dQT : Change in entropy is the reduced heat exchangedon an infinitesimal part of a reversible process.As dQT does not depend on the path followed, but solely on theinitial and final state of the system, the entropy S, hence like P,T , V , describes a state of the thermodynamic system.Phase diagram89/120

Isentropic ProcessKinetic Theory ofGasesThermodynamicsState VariablesZeroth law ofthermodynamicsReversible andIrreversibleprocessesFirst law ofThermodynamicsSecond law ofThermodynamicsEntropyThermodynamicPotentialThird Law ofThermodynamicsPhase diagramIn a Carnot Cycle; during isothermal process, TQ R ln VV21 S.For a complete reversible process: TQ1 1 - TQ2 2 ) For acomplete reversible cycle S 0, implies S is a constant. Thisprocess is called isentropic processNow S 0 ) Q 0 and T constant. Hence for anisentropic process T has to be kept constant.Note for an adiabatic process Q 0, but T changes.From First Law: dS dQTrev dU PdV. Now for one mole ofTgas, dU CV dT and PV RT .dVHence, dS CV dTT R VIntegrating and considering isobaric process, P constant:V22Sisobar CV ln TT1 R ln V1 .Similarly considering isochoric process, V constant:P22Sisochor CP ln TT1 R ln P1 .We used: CP CV R, TP11 TP22All these shows S depends on the initial and final state of thesystem.90/120

Clausius inequality - 1Kinetic Theory ofGasesThermodynamicsState VariablesThis considers the heat transfers to a substance as it is takenround an arbitrary cyclic process exchanging heat with anynumber of surrounding bodies. It can be derived by breakingdown the process into equivalent interactions with a largenumber of Carnot engines and refrigerators.Zeroth law ofthermodynamicsReversible andIrreversibleprocessesFirst law ofThermodynamicsSecond law ofThermodynamicsEntropyThermodynamicPotentialThird Law ofThermodynamicsPhase diagram3 di erent bodies: two Carnot refrigerators A and B and anEngine. The refrigerators are adjusted to exactly the sameamount of heat to their hot reservoirs as is taken from thosereservoirs by the Engine: (Q1 Q1A ; Q2 Q2A ).91/120

Clausius inequality - 2Kinetic Theory ofGasesThermodynamicsState VariablesZeroth law ofthermodynamicsReversible andIrreversibleprocessesFirst law ofThermodynamicsSecond law ofThermodynamicsEntropyThermodynamicPotentialThird Law ofThermodynamicsPhase diagramNo such restriction can be put on the heat transfers to and fromthe cold reservoir and therefore heat owing from it is Q0A Q0B- Q0 .The net work done by the composite system is W - (WA WB ).As no net heat flows into/out of the hot reservoirs. The balanceof heat flow and mechanical work (1st Law) for the compositesystem is Q0A Q0B - Q0 W - (WA WB ).Since the heat flow is from a single reservoir, the Kelvin-Planckstatement of the Second Law would be contradicted if W WA WB . Therefore we must have W WA WB and Q0A Q0B - Q0 0.From the Carnot refrigerator efficiencies: Q0A Q10 Q2 TT2 .T0T1and Q0Bthe heat entering the cold reservoir: Q0 - Q0A - Q0B Q0 - (Q2Q021 Q0 ) QT2 )T0T1 T2 - T0 0.Q1T192/120

Clausius inequality - 3Kinetic Theory ofGasesThermodynamicsState VariablesZeroth law ofthermodynamicsReversible andIrreversibleprocessesFirst law ofThermodynamicsSecond law ofThermodynamicsEntropyThermodynamicPotentialThird Law ofThermodynamicsPhase diagramNow consider heat inputs qi to the working substance of theengine. The engine absorbs heat q1 Q1 from the hotreservoirs, and returns heat to the cold reservoir, so there is achange of sign: q0 - Q0 . Hence: Tq00 Tq11 Tq00 0.If the number of engines and refrigerators is arbitrary, Tqii 0.In the limit where the amounts of heat entering the workingsubstance areH very small at each step, but the number of steps isvery large, Tq 0 - Clausius inequalityfor reversibleH cycles : Tsystem Treservoirs during each heatexchange, Tq0. For both inequalitiesHto be validsimultaneously for reversible cycles only - Tq 0.93/120

Entropy and second law of thermodynamics - 1Kinetic Theory ofGasesThermodynamicsConsider the cycle in the T-S diagram. We start at 1, andproceed to 2 along path I, that represents an irreversible process.We return from 2 to 1 along path R, that represents a reversibleprocess.State VariablesZeroth law ofthermodynamicsReversible andIrreversibleprocessesFirst law ofThermodynamicsSecond law ofThermodynamicsEntropyThermodynamicPotentialThird Law ofThermodynamicsPhase diagramH Q0TThe equality implies all processes are reversible; the inequalityimplies some portion of the process is irreversible. For aR2reversible process we have S2 S1 1 TQ .R1 QOR S1 S2 2 T .R 2 QIR 1 QRR20 2 T 1 TQI S1 S21 T94/120

Entropy and second law of thermodynamics - 1Kinetic Theory ofGasesThermodynamicsState VariablesZeroth law ofthermodynamicsReversible andIrreversibleprocessesFirst law ofThermodynamicsSecond law ofThermodynamicsS2S1R21QITor in more generally S2S1R21QTIf 1 to 2 is reversible, the equality holds; if 1 to 2 is irreversible,the inequality holds. Now, if the system is isolated, there can beno heat transfer interactions and Q 0, hence S2 S1 0.This implies 2 occurs later in time than 1. Thus, for isolatedsystems, the entropy increases as time moves forward.EntropyThermodynamicPotentialThird Law ofThermodynamicsPhase diagram95/120

Principle of Increasing EntropyKinetic Theory ofGasesThermodynamicsState VariablesThe entropy of a thermally isolated system increases in anyirreversible process and is unaltered in a reversible process. Thisis the principle of increasing entropy.Zeroth law ofthermodynamicsReversible andIrreversibleprocessesFirst law ofThermodynamicsSecond law ofThermodynamics2nd Law of d Law ofThermodynamicsPhase diagram96/120

Reversible and Irreversible processes First law of Thermodynamics Second law of Thermodynamics Entropy Thermodynamic Potential Third Law of Thermodynamics Phase diagram 84/120 Equivalent second law of thermodynamics W Q 1 1 for any heat engine. Heat cannot completely be converted to mechanical work. Second Law can be formulated as: (A .

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