Entropyentropy Andand 22 Nd Lawlaw Ofof Thermodynamicsthermodynamics

Entropyentropy Andand 22 Nd Lawlaw Ofof Thermodynamicsthermodynamics

1 LectureLecture 66:: EntropyEntropy andand 22 nd LawLaw ofof ThermodynamicsThermodynamics CanCan wewe makemake heatheat work?work? 22nd LawLaw ofof ThermodynamicsThermodynamics Entropy.Entropy. MaximumMaximum entropyentropy principle.principle. HeatHeat Engines.Engines. CarnotCarnot cycle.cycle. ComputingComputing EntropyEntropy SummarySummary ofof Processes.Processes. Process Type Constant Quantity QQ WW Isobaric p Isothermal T Isochoric V 0 γγγ Adiabatic pV 0 Can we find a better variable? Can we make heat work? StirlingStirling engine.engine. Work in cycle A Efficiency = Absorbed Heat p T2 B D T 1 C (absorbed heat) (released heat) V1 V V2 Given amount of absorbed heat, work depends on “shape” (=efficiency) of cycle Can we do better? Can we do better? 4 Limitations.Limitations. Reversible/IrreversibleReversible/Irreversible Processes.Processes. 1.1. GlassGlass ofof winewine dropsdrops fromfrom thethe table.table. PiecesPieces willwill nevernever assembleassemble backback onon theirtheir own.own. 2.2. MiddleMiddle wallwall removedremoved allowingallowing gasgas toto expandexpand freelfreely.y. MoleculesMolecules willwill ““ nevernever ”” spontaneouslyspontaneously returnreturn backback toto leftleft partpart oo ff container.container. Limitations.Limitations. PlayingPlaying withwith 11 st LawLaw ofof TD.TD. 1.1. DirectionDirection ofof HeatHeat FlowFlow BUT!BUT! 11st LawLaw ofof TDTD doesdoes notnot forbidforbid 22 nd processprocess asas longlong asas energyenergy isis conserved.conserved. 2.2. ConsiderConsider thethe WorldWorld Ocean.Ocean. ItIt isis warmwarm afterafter all!all! TakeTake allall itsits heatheat andand convertconvert intointo work.work. ---- stillstill OKOK withwith 11 st LawLaw ofof TDTD CommonCommon sensesense tellstells usus otherwise.otherwise. Second Law of Thermodynamics Second Law of Thermodynamics 6 SecondSecond LawLaw ofof Thermodynamics.Thermodynamics. ManyMany formulations.formulations. ConsiderConsider thethe followingfollowing two.two. 1.1. SpontaneousSpontaneous (i.e. without external agent) transfertransfer ofof heatheat fromfrom coldcold objectobject toto hothot objectobject isis impossible.impossible. 2.2. NoNo PerpetualPerpetual MotionMotion MachineMachine ofof secondsecond type.type. (device that realizes full conversion of heat into useful work, without any “side effect” , in contrast with real engines.) LetLet usus formulateformulate thisthis mathematically.mathematically. 7 Entropy.Entropy. InIn 11 st LawLaw ofof TDTD (for(for infinitesimalinfinitesimal process)process) QQ == notnot functionfunction ofof statestate ,, dependsdepends onon processprocess ThereThere existexist oneone moremore functionfunction ofof statestate == ENTROPYENTROPY soso thatthat •• IfIf TT=const=const :: ∆∆QQ==TT∆∆SS •• ∆∆SS == 00 forfor adiabaticadiabatic processprocess ∆∆SS ≠≠ 00 forfor nonnon --adiabaticadiabatic Note:Note: nownow havehave 44 variablesvariables (( p,p, V,V, T,T, SS ).). (any)(any) twotwo areare stillstill sufficientsufficient toto describedescribe statestate ofof systemsystem 8 ReversibleReversible andand IrreversibleIrreversible Processes.Processes. SummarySummary REVERSIBLE IRREVERSIBLE • Series of quasi-static • Series of complex, non- (equilibrium) steps equilibrium steps • No dissipation (e.g. no • Dissipation may be friction) present - Examples? Any process that is sudden, • Examples? spontaneous, uncontrolled, or Any process that can be involves dissipation represented as a path on a • Free expansion PV diagram; e.g • Explosion • Isochoric • Heat transfer between a hot • Isobaric and a cold object • Isothermal • Volume changes of gas with a • Adiabatic piston that has friction 9 CalculatingCalculating EntropyEntropy Changes.Changes. ReversibleReversible ProcessesProcesses •• AdiabaticAdiabatic process?process? •• GeneralGeneral process:process: (( TTi,, VVi,, ppi)) ---- >> (( TTf,, VVf,, ppf)) WhatWhat isis entropyentropy changechange for:for: ••IsothermalIsothermal process?process? ••IsochoricIsochoric process?process? PrinciplePrinciple ofof MaximumMaximum Entropy.Entropy. Intuitively:Intuitively: EntropyEntropy == MeasureMeasure ofof DisorderDisorder With time, entropy of isolated system: increases in irreversible process, remains the same in reversible process DisorderDisorder grows!grows! IfIf systemsystem isis NOTNOT atat thermalthermal equilibrium,equilibrium, butbut consisconsiststs ofof equilibriumequilibrium subsystemssubsystems withwith SSi,, introduceintroduce SSΣ== ΣΣiSSi In isolated system (fixed internal energy ), state of thermal equilibrium corresponds to absolute maximum of total entropy, i.e. S= (SΣ)max IfIf youyou reachreach “thermal“thermal equilibrium”equilibrium” withwith untidyuntidy roomroommate,mate, thatthat wouldwould correspondcorrespond toto maximummaximum possiblepossible disorder.disorder. 11 HowHow cancan wewe extractextract heat?heat? HeatHeat Engines.Engines. WantWant engineengine toto workwork repeatedlyrepeatedly WantWant cycliccyclic process!process! Heat engine takes some substance (e.g. gas) through cyclic process during which: Heat QH absorbed from hot reservoir Work W+ done by engine Heat QC dumped into cold reservoir “Negative” work W- done by engine to bring gas to original state. In a cyclic process, it is impossible to convert all the energy extracted from an equilibrium system into work. Some part of this energy must be transferred to a colder system. InternalInternal combustioncombustion engine.engine. TypicalTypical cyclecycle consistsconsists ofof 44 processesprocesses a) Burning of fuel (isothermal expansion) b) Fast expansion (adiabatic) c) Compression to original pressure (isothermal process) d) Compression to original temperature (adiabatic process) 13 CarnotCarnot HeatHeat EngineEngine Carnot heat engine takes gas through reversible cyclic process consisting of 4 steps: Isothermal expansion at T=TH Adiabatic expansion from T=TH to T=TC Isothermal compression at T=TC Adiabatic compression from T=T to T=T C H T K = 1− C < 1 TH CarnotCarnot heatheat engineengine andand 22 nd LawLaw ofof TD.TD. •• AA corollarycorollary toto thethe secondsecond lawlaw ofof thermodynamics:thermodynamics: •• NONO REALREAL ENGINEENGINE CANCAN EXCEEDEXCEED THETHE EFFICIENCYEFFICIENCY OFOF AA CARNOTCARNOT ENGINEENGINE OPERATINGOPERATING BETWEENBETWEEN THETHE SAMESAME HEATHEAT RESERVOIRSRESERVOIRS •• Q:Q: cancan anan engineengine (real(real oror ideal)ideal) everever havehave anan efficefficiencyiency ofof 11 (I.e.(I.e. 100100 %)?%)? Why/whyWhy/why not?not? CalculatingCalculating EntropyEntropy Changes.Changes. IrreversibleIrreversible ProcessesProcesses • An irreversible process is characterized by a series of non -equilibrium states -- thermodynamic parameters such as T,p are ill -defined • So: how to calculate ∆S? • Entropy is a state function -- so ∆S only depends on the initial and final equilibrium states -- NOT THE PATH! • Strategy: choose a reversible path connecting the initial and fi nal states and determine ∆S. 16 SampleSample problem.problem. FreeFree expansions.expansions. 2 moles of a gas in a thermally insulated box freely expands, tripling its volume in the process. What is the change in entropy? 3V0 dQ 3V0 pdV p ∆S = ∫ = ∫ , , T T p1 V1 T1 V0 V0 nRT 3V0 dV = ∫ T V V p2, V2, T1 0 nRT 3V0 dV V = T ∫ V •Q = 0; W = 0; so, ∆∆∆U = 0 V0 •Final T = initial T = nR ln 3 •Choose a reversible, isothermal path = 2R ln 3 = 18 3. J/K SampleSample problem.problem. HeatHeat transfer.transfer. Two identical 500 g copper blocks A and B of mass are initially at 20 0C and 80 0C. They are placed in thermal contact and allowed to come to thermal equilibrium. What is the change in entropy? (Specific heat capacity of copper = 400 J/kg. K) 323 mcdT What reversible process can we use? ∆S = ∫ A 293 T 323 ∆SB = mc ln 323 353 = mc ln 293 = -17.8 J/K = (0.5kg)(400J/kg.K)(0.098) =19.6 J/K ∆Stotal = (19 .6 −17 .8)J/K = +1.8 J/K WhatWhat wewe learnedlearned • Reversible/Irreversible Processes • Entropy Tf V f ∆S ≥ 0 ∆S = nC v ln + nR ln Ti Vi • 2nd Law • Heat Engines (Carnot) W T K = K = 1− C QH TH “The law that entropy always increases, holds, I think, the supreme position among the Laws of Nature. If someone points out to you that your pet theory of the Universe is in disagreement with Maxwell’s Equations — then so much the worse for Maxwell's equations. If it is found to be contradicted by observation — well, these experimentalists do bungle things sometimes. But if your theory is found to be against the second law of thermodynamics I can give you no hope; there is nothing for it but to collapse in deepest humiliation.” --Sir Arthur Stanley Eddington, The Nature of the Physical World (1927) 1882 -1944.

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