e dQ 0 The 64257rst law of thermodynamics then reduces to dQ dW dU dU dW 1 where dW is the amount of work done by the gas and dU is the change in internal energy Since the gas is expanding the work being done by the gas is positive Eqn 1 then implie ID: 54051
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PartBAsthegasexpands,theinternalenergyofthesystembeginstodecrease.ItwasfoundinpartA,however,thattheinternalenergyofthesystemmustremainconstantthroughanisothermalprocess.Thiscanonlyoccuriftheamountofenergylostwhilethegasexpandsisrestoredtothesystemintheformofheat.ItisthenthecasethatWprovidesenergyoutput,whileQprovidesenergyinput.Theyareequalinmagni-tude.PartCTheworkdonebyanidealgasisdenedtobeW=ZpdV:(4)Sincetheprocessisisochoric,dV=0,whichimpliesthattheworkdonebythegasiszero.TherstlawofthermodynamicsthenreducestodU=dQ:(5)BecausethetemperatureofthesystemdecreasesfromTA=4cKtoTB=cK,theinternalenergyofthesystemmustdecrease.Thisimpliesthatthesystemislosingenergyintheformofheat.ItisthenthecasethatQprovidesenergyoutput,whileWequalstozero.PartDItwasfoundinPartC,thatingoingfromstateAtostateDtheinternalenergyofthesystemdecreases.However,ingoingfromstateAtostateB,theinternalenergyofthesystemmustremainconstant.Itmustbethecasethatduringtheisobaricportionoftheprocess,thesystemmustgaininternalenergy.Sincethesystemisexpandingduringtheisobaricprocess,theinternalenergyofthesystemwillbefurtherreducedduetotheworkdonebythegas.Thesystemmustthenabsorbenoughenergyintheformofheatthatwillcompensatefortheenergylostduetotheexpansionofthegasandforthatlostduringtheisochoricportionoftheprocess.QandWarethenrelatedasWprovidesenergyoutput,whileQprovidesenergyinput;Qislarger.PartEAswasdiscussedabove,noworkisdonebythegasduringanisochoricprocess.SincethetemperatureincreasesingoingfromstateEtoB,theinternalenergymustalsoincrease.Eqn.(5)thenimpliesthatQmustprovideanenergyinput.2 ThetemperatureofthesysteminstateBisthengreaterthanthetemperatureofsysteminstateO.PartFItisknownthattheinternalenergyofasystemisproportionaltothetemperatureofthesystem.ItfollowsthattheinternalenergyofthesysteminstateBisgreaterthantheinternalenergyofthesysteminstateO.PistoninWaterBathConceptualWork-EnergyProblemPartABecausethepistonisheldinplacethroughouttheprocess,thevolumeofthesystemremainsunchanged.Itfollowsfromeqn.(4)thattheworkdonebythegasisequaltozerosincedV=0.Furthermore,becausethepistonisheldabovea ame,energyisbeingtransferredintothesystemimplyingthattheheattransferredispositive.Finally,therstlawofthermodynamicswithdW=0yieldsdU=dQ:(6)Thechangeininternalenergyisthenpositivesincetheheattransferredispositive.Thismakessensebecauseasyouheatthesystemthetemperatureofthesystemwillriseinturnraisingtheinternalenergy(U/T).(0,+,+).PartBIfthecompressionofthepistonoccursveryslowly,thesystemwillcontinuetoremaininthermalequilibriumwiththewaterbaththroughouttheprocess.Sincethetemperatureofthesystemdoesnotchange,theinternalenergyofthesystemwillalsoremainedunchanged.Furthermore,sincethegasisbeingcompressed,i.e.workisbeingdoneonthegas,theworkisnegative.Itfollowsfromtherstlawthattheamountofheattransferredtothesystemmustbenegative.(-,-,0)PartCOnceagain,becausethepistonisheldinplace,theworkdonebythegasiszero.Furthermore,sincethepistonisplungedintocoldwater,energywillleavethesystemintheformofheatinaccordancewiththezerothlawofthermodynamics.Itfollowsfromeqn.(6)thatthechangeintheinternalenergyofthesystemmustbenegative.(0,-,-)PartDBecausethepistoniswrappedininsulation,noheatcanenterorexitthesystem,i.e.dQ=0.Furthermore,asthepistonispulledup,thegaswillexpandsothattheworkdonebythegasispositive.Itthenfollowsfromtherstlawofthermodynamicsintheformofeqn.(1)thatthechangeininternalenergymustbenegative.(+,0,-)4 PartCAsmentionedinPartA,Cycle1isacyclicprocesswhichimpliesthatdU=0.Therstlawofthermodynamicsthensuggestthattheamountofheattransferredtothegasmustequaltheamountofworkdonebythegas.SincetheamountofworkdonebythegaswasfoundtobepositiveinPartB,theamountofheattransferredmustalsobepositive.PartDWhenCycle1istraversedcounterclockwise,thegasdoespositiveworkbyexpandingduringportion4andnegativeworkwhenitiscompressedduringportion2.Onceagain,sincetheamountofworkdoneduringportion2isgreaterthantheamountofworkdoneduringportion4,thetotalworkdonebytraversingCycle1counterclockwisewillbenegative.ItimmediatelyfollowsthatWclockwise�Wcouterclockwise.PartEUsingthesameargumentasinPartC,Qclockwise�Qcounterclockwise.PartFThenetworkdonebyagasduringacycliccycleisequaltotheareaenclosedbythelinesdescribingthecycleonapVdiagram.SincetheareasofCycle1andCycle2arethesame,andthecyclesaretraversedclockwiseinbothcases,theworkdonebythegasinbothcasesisthesame.Wclockwise,Cycle1=Wclockwise,Cycle2PartGFollowingthesameargumentmadeinPartCandPartF,theheattransferredinbothprocessesisequal.Qclockwise,Cycle1=Qclockwise,Cycle2AFlexibleBallonPartAInitiallythegasexpandsisobaricallymeaningthatthepressureofthegasremainsconstantthroughouttheprocess.Inorderforthepressuretoremainconstant,however,heatmust owintothesystem,otherwisetheequationofstatedictatesthatthepressurewoulddecreasewithincreasingvolume.Thegasthenexpandsadiabaticallywhich,bydenition,meansthatnoheatitaddedorremovedfromthesystem.Inordertocalculatetheamountofheatsuppliedtothesystem,itisonlynecessarytoconsidertherstprocess.Underconstantpressure,theheatsuppliedtothesystemisgivenbyQ=nCp(Tf�Ti):Itisthennecessarytoknowthenaltemperatureofthesystemoncetheexpansionhastakenplace.Thiscanbefoundbyusingtheequationofstateandnotingthatp=nRwill6 AnExpandingMonatomicGasAsthegasexpands,theamountofheatthatisabsorbedbythesystemisequaltodQ=1160JwhiletheamountofworkthatisdonebythegasisdW=2160J.TherstlawofthermodynamicscanthenbeusedtorelatethesequantitiestothechangeintheinternalenergyofthesystemasdQ=dU+dW)dU=Uf�Ui=dQ�dW:(7)TheEquipartitionTheoremstatesthatforanidealgaswiththreedegreesoffreedomtheinternalenergyisgivenbyU=3 2nRT(8)wherenisthenumberofmoles,RistheidealgasconstantandTisthetemperatureofthesystem.Initially,thesystemisatatemperatureTi=400:15KsotheinternalenergyisthenUi=3 2nRTi:Similarly,theinternalenergyofthesystemoncethegashasnishedexpandingisUf=3 2nRTf:SubstitutingUiandUfintoeqn.(7),thenaltemperatureofthegasis3 2nRTf�3 2nRTi=dQ�dWTf=Ti+2 3nR(dQ�dW)=384:2637K=111:1137oCACarnotRefrigeratorPartATheCarnotcycleisacyclicprocesswhichmeansthattheinternalenergyofthesystemremainsunchangedafterthesystemhasundergonethefullcycle.Eqn.(3)thenstatesthattheamountofworkdonebythegasaftercompletingthecyclewillbeequaltothenetenergytransferredintothesystemW=jQhj�jQcj:UsingtherelationjQcj jQhj=Tc Th;andtheexpressionfortheworkdonebythegas,thecoecientofperformancecanberewrittenasK=jQcj jQhj�jQcj=1 jQhj jQcj�18 FromHottoCool:AChangeinEntropyAstheicemelts,itstemperatureremainsconstantatTice=273:15K,i.e.theprocessisisothermal.Foranisothermalprocess,thechangeintheentropyofthesystemisfoundasS=Q T:(11)KnowingthattheamountofheatrequiredtomelttheiceisQice=miceLf,thechangeintheentropyisthenSice=miceLf Tice=(:001kg)3:34105J/kg 273:15K=1:2227J/K:Theamountofheatthatwasrequiredtomelttheicewasprovidedbytheicewater,i.e.Qice=�Qwater.Sincethetemperatureoftheicewateralsoremainsconstant,thechangeinentropyiscalculatedtobeSwater=�Qice Twater=�(:001kg)3:34105J/kg 293:15K=�1:1393J/K:ThenetchangeinentropyofthesystemisthenSsys=Sice+Swater=:084J/K:CarnotHeatEnginePressureversusVolumeGraphConceptualQuestionPartALetQhbetheamountofenergyabsorbedbytheCarnotengineinonecycleandQctheenergyexpelled.ThetotalchangeinentropyisthenS=jQhj Th�jQcj Tc:Usingeqn.(10),andsubstitutingforThintheaboveequation,thechangeofentropyisthenS=jQhjjQcj jQhjTc�jQcj Tc=0PartBInaccordancewiththezerothlawofthermodynamics,thetemperatureofthesurroundingsmustbegreaterthanthetemperatureofthegas.10