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Chapter 20 -- Thermodynamics

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ChapterChapter 2020 -- ThermodynamicsThermodynamics AA PowerPointPowerPoint PresentationPresentation byby PaulPaul E.E. TippensTippens,, ProfessorProfessor ofof PhysicsPhysics SouthernSouthern PolytechnicPolytechnic StateState UniversityUniversity © 2007 THERMODYNAMICSTHERMODYNAMICS ThermodynamicsThermodynamics isis thethe studystudy ofof energyenergy relationshipsrelationships thatthat involveinvolve heat,heat, mechanicalmechanical work,work, andand otherother aspectsaspects ofof energyenergy andand heatheat transfer.transfer. Central Heating Objectives:Objectives: AfterAfter finishingfinishing thisthis unit,unit, youyou shouldshould bebe ableable to:to: •• StateState andand applyapply thethe first andand second laws ofof thermodynamics. •• DemonstrateDemonstrate youryour understandingunderstanding ofof adiabatic, isochoric, isothermal, and isobaric processes.processes. •• WriteWrite andand applyapply aa relationshiprelationship forfor determiningdetermining thethe ideal efficiency ofof aa heatheat engine.engine. •• WriteWrite andand applyapply aa relationshiprelationship forfor determiningdetermining coefficient of performance forfor aa refrigeratior.refrigeratior. AA THERMODYNAMICTHERMODYNAMIC SYSTEMSYSTEM •• AA systemsystem isis aa closedclosed environmentenvironment inin whichwhich heatheat transfertransfer cancan taketake place.place. (For(For example,example, thethe gas,gas, walls,walls, andand cylindercylinder ofof anan automobileautomobile engine.)engine.) WorkWork donedone onon gasgas oror workwork donedone byby gasgas INTERNALINTERNAL ENERGYENERGY OFOF SYSTEMSYSTEM •• TheThe internalinternal energyenergy UU ofof aa systemsystem isis thethe totaltotal ofof allall kindskinds ofof energyenergy possessedpossessed byby thethe particlesparticles thatthat makemake upup thethe system.system. Usually the internal energy consists of the sum of the potential and kinetic energies of the working gas molecules. TWOTWO WAYSWAYS TOTO INCREASEINCREASE THETHE INTERNALINTERNAL ENERGY,ENERGY, U.U. ++UU WORKWORK DONEDONE HEATHEAT PUTPUT INTOINTO ONON AA GASGAS AA SYSTEMSYSTEM (Positive)(Positive) (Positive)(Positive) TWOTWO WAYSWAYS TOTO DECREASEDECREASE THETHE INTERNALINTERNAL ENERGY,ENERGY, U.U. WW outout QQout --UU DecreaseDecrease hot hot WORKWORK DONEDONE BYBY HEATHEAT LEAVESLEAVES AA EXPANDINGEXPANDING GAS:GAS: SYSTEMSYSTEM WW isis positivepositive QQ isis negativenegative THERMODYNAMICTHERMODYNAMIC STATESTATE TheThe STATESTATE ofof aa thermodynamicthermodynamic systemsystem isis determineddetermined byby fourfour factors:factors: •• AbsoluteAbsolute PressurePressure PP inin PascalsPascals •• TemperatureTemperature TT inin KelvinsKelvins •• VolumeVolume VV inin cubiccubic metersmeters •• NumberNumber ofof moles,moles, nn,, ofof workingworking gasgas THERMODYNAMICTHERMODYNAMIC PROCESSPROCESS Increase in Internal Energy, U. WWout QQin Initial State: HeatHeat inputinput Final State: P V T n P V T n 1 1 1 1 WorkWork byby gasgas 2 2 2 2 TheThe ReverseReverse ProcessProcess Decrease in Internal Energy, U. WWin QQout WorkWork onon gasgas Initial State: Final State: P1 V1 T1 n1 LossLoss ofof heatheat P2 V2 T2 n2 THETHE FIRSTFIRST LAWLAW OFOF THERMODYAMICS:THERMODYAMICS: •• TheThe netnet heatheat putput intointo aa systemsystem isis equalequal toto thethe changechange inin internalinternal energyenergy ofof thethe systemsystem plusplus thethe workwork donedone BYBY thethe system.system. Q = U + W final - initial) •• Conversely,Conversely, thethe workwork donedone ONON aa systemsystem isis equalequal toto thethe changechange inin internalinternal energyenergy plusplus thethe heatheat lostlost inin thethe process.process. SIGNSIGN CONVENTIONSCONVENTIONS +W FORFOR FIRSTFIRST LAWLAW +Wout ++QQ in •• HeatHeat QQ inputinput isis positivepositive U • Work BY a gas is positive --WWin • Work ON a gas is negative U •• HeatHeat OUTOUT isis negativenegative --QQout Q = U + W final - initial) APPLICATIONAPPLICATION OFOF FIRSTFIRST LAWLAW OFOF THERMODYNAMICSTHERMODYNAMICS ExampleExample 1:1: InIn thethe figure,figure, thethe gasgas absorbsabsorbs 400400 JJ ofof heatheat andand WW out =120=120 JJ atat thethe samesame timetime doesdoes 120120 JJ ofof workwork onon thethe piston.piston. WhatWhat isis thethe changechange inin internalinternal energyenergy ofof thethe system?system? QQin 400400 JJ Apply First Law: Q = U + W ExampleExample 11 (Cont.):(Cont.): ApplyApply FirstFirst LawLaw Q is positive: +400 J (Heat IN) WWout =120=120 JJ W is positive: +120 J (Work OUT) QQ QQ == UU ++ WW in 400400 JJ UU == QQ -- WW UU == QQ -- WW == (+400(+400 J)J) -- (+120(+120 J)J) U = +280 J == +280+280 JJ ExampleExample 11 (Cont.):(Cont.): ApplyApply FirstFirst LawLaw Energy is conserved: WWout =120=120 JJ TheThe 400400 JJ ofof inputinput thermalthermal energyenergy isis usedused toto performperform QQin 120 J of external work, in 120 J of external work, 400 J increasingincreasing thethe internalinternal 400 J energyenergy ofof thethe systemsystem byby 280280 JJ The increase in internal energy is: U = +280 J FOURFOUR THERMODYNAMICTHERMODYNAMIC PROCESSES:PROCESSES: ••• IsochoricIsochoricIsochoric Process:Process:Process: VVV === 0,0,0, WWW === 000 ••• IsobaricIsobaricIsobaric Process:Process:Process: PPP === 000 ••• IsothermalIsothermalIsothermal Process:Process:Process: TTT === 0,0,0, UUU === 000 ••• AdiabaticAdiabaticAdiabatic Process:Process:Process: QQQ === 000 Q = U + W ISOCHORICISOCHORIC PROCESS:PROCESS: CONSTANTCONSTANT VOLUME,VOLUME, VV == 0,0, WW == 00 0 QQ == UU ++ WW soso thatthat QQ == UU Q QIN QQOUT NoNo WorkWork -U +U DoneDone HEAT IN = INCREASE IN INTERNAL ENERGY HEAT OUT = DECREASE IN INTERNAL ENERGY ISOCHORICISOCHORIC EXAMPLE:EXAMPLE: NoNo ChangeChange inin volume:volume: PP2 BB PP A PP B = TTA TT B PP1 AA VV1 == VV2 400400 JJ HeatHeat inputinput 400400 JJ heatheat inputinput increasesincreases increasesincreases PP internalinternal energyenergy byby 400400 JJ withwith const.const. VV andand zerozero workwork isis done.done. ISOBARICISOBARIC PROCESS:PROCESS: CONSTANTCONSTANT PRESSURE,PRESSURE, PP == 00 QQ == UU ++ WW ButBut WW == PP VV Q QIN QQOUT WorkWork OutOut WorkWork -U +U InIn HEAT IN = Wout + INCREASE IN INTERNAL ENERGY HEAT OUT = Wout + DECREASE IN INTERNAL ENERGY ISOBARICISOBARIC EXAMPLEEXAMPLE ((Constant Pressure): A B P VA VB = TA T B 400 J V1 V2 Heat input 400400 JJ heatheat doesdoes 120120 JJ ofof increases V work,work, increasingincreasing thethe with const. P internalinternal energyenergy byby 280280 JJ. ISOBARICISOBARIC WORKWORK A B P VA VB = TA T B PPA == PPB 400 J V1 V2 Work = Area under PV curve Work P V ISOTHERMALISOTHERMAL PROCESS:PROCESS: CONST.CONST. TEMPERATURE,TEMPERATURE, TT == 0,0, UU == 00 QQ == UU ++ WW ANDAND QQ == WW QQIN QQOUT WorkWork OutOut WorkWork U = 0 U = 0 InIn NETNET HEATHEAT INPUTINPUT == WORKWORK OUTPUTOUTPUT WORKWORK INPUTINPUT == NETNET HEATHEAT OUTOUT ISOTHERMALISOTHERMAL EXAMPLEEXAMPLE (Constant(Constant T):T): A PA B PB UU == TT == 00 V2 V1 Slow compression at constant temperature: P V =PV A A B B ----- NoNo changechange inin UU. ISOTHERMALISOTHERMAL EXPANSIONEXPANSION ((ConstantConstant T)T):: AA PA BB PA VA = PBVB PB T = T V V A B U = T = 0 A B 400 J of energy is absorbed Isothermal Work by gas as 400 J of work is done on gas. V WnRT ln B T = U = 0 VA ADIABATICADIABATIC PROCESS:PROCESS: NONO HEATHEAT EXCHANGE,EXCHANGE, QQ == 00 QQ == UU ++ WW ;; WW == --UU oror UU == --WW W = -U U = -W Work Out Work U +U In Q = 0 Work done at EXPENSE of internal energy INPUT Work INCREASES internal energy ADIABATICADIABATIC EXAMPLE:EXAMPLE: A PP A B PP B VV1 VV2 Insulated ExpandingExpanding gasgas doesdoes Walls: Q = 0 workwork withwith zerozero heatheat loss.loss. WorkWork == --UU ADIABATICADIABATIC EXPANSION:EXPANSION: A PP A B PP AV VA PP BV VB = PP B TT A TT B Q = 0 VVA VVB 400 J of WORK is done, DECREASING the internal energy by 400 J: Net heat PVA ABB PV exchange is ZERO. QQ == 00 MOLARMOLAR HEATHEAT CAPACITYCAPACITY OPTIONAL TREATMENT TheThe molarmolar heatheat capacitycapacity CC isis defineddefined asas thethe heatheat perper unitunit molemole perper CelsiusCelsius degree.degree. CheckCheck withwith youryour instructorinstructor toto seesee ifif thisthis moremore thoroughthorough treatmenttreatment ofof thermodynamicthermodynamic processesprocesses isis required.required. SPECIFICSPECIFIC HEATHEAT CAPACITYCAPACITY RememberRemember thethe definitiondefinition ofof specificspecific heatheat capacitycapacity asas thethe heatheat perper unitunit massmass requiredrequired toto changechange thethe temperature?temperature? Q c mt ForFor example,example, copper:copper: cc == 390390 J/kgJ/kgKK MOLARMOLAR SPECIFICSPECIFIC HEATHEAT CAPACITYCAPACITY TheThe ““molemole”” isis aa betterbetter referencereference forfor gasesgases thanthan isis thethe ““kilogram.kilogram.”” ThusThus thethe molarmolar specificspecific heatheat capacitycapacity isis defineddefined by:by: QQ CC == nn TT ForFor example,example, aa constantconstant volumevolume ofof oxygenoxygen requiresrequires 21.121.1 JJ toto raiseraise thethe temperaturetemperature ofof oneone molemole byby oneone kelvinkelvin degreedegree.. SPECIFICSPECIFIC HEATHEAT CAPACITYCAPACITY CONSTANTCONSTANT VOLUMEVOLUME HowHow muchmuch heatheat isis requiredrequired toto raiseraise thethe temperaturetemperature ofof 22 molesmoles o o ofof OO 2 fromfrom 00 CC toto 100100 C?C? Q = nCv T QQ == (2(2 mol)(21.1mol)(21.1 J/molJ/mol K)(373K)(373 KK -- 273273 K)K) Q = +4220 J SPECIFICSPECIFIC HEATHEAT CAPACITYCAPACITY CONSTANTCONSTANT VOLUMEVOLUME (Cont.)(Cont.) SinceSince thethe volumevolume
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  • The First Law of Thermodynamics Continued Pre-Reading: §19.5 Where We Are

    The First Law of Thermodynamics Continued Pre-Reading: §19.5 Where We Are

    Lecture 7 The first law of thermodynamics continued Pre-reading: §19.5 Where we are The pressure p, volume V, and temperature T are related by an equation of state. For an ideal gas, pV = nRT = NkT For an ideal gas, the temperature T is is a direct measure of the average kinetic energy of its 3 3 molecules: KE = nRT = NkT tr 2 2 2 3kT 3RT and vrms = (v )av = = r m r M p Where we are We define the internal energy of a system: UKEPE=+∑∑ interaction Random chaotic between atoms motion & molecules For an ideal gas, f UNkT= 2 i.e. the internal energy depends only on its temperature Where we are By considering adding heat to a fixed volume of an ideal gas, we showed f f Q = Nk∆T = nR∆T 2 2 and so, from the definition of heat capacity Q = nC∆T f we have that C = R for any ideal gas. V 2 Change in internal energy: ∆U = nCV ∆T Heat capacity of an ideal gas Now consider adding heat to an ideal gas at constant pressure. By definition, Q = nCp∆T and W = p∆V = nR∆T So from ∆U = Q W − we get nCV ∆T = nCp∆T nR∆T − or Cp = CV + R It takes greater heat input to raise the temperature of a gas a given amount at constant pressure than constant volume YF §19.4 Ratio of heat capacities Look at the ratio of these heat capacities: we have f C = R V 2 and f + 2 C = C + R = R p V 2 so C p γ = > 1 CV 3 For a monatomic gas, CV = R 3 5 2 so Cp = R + R = R 2 2 C 5 R 5 and γ = p = 2 = =1.67 C 3 R 3 YF §19.4 V 2 Problem An ideal gas is enclosed in a cylinder which has a movable piston.