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Physics150 Thermodynamics Chapter15 TheFirstLawofThermodynamics • Let’sconsideranidealgasconfinedinachamberwithamoveablepiston • Ifwepressthepistonèthegasinthechambercompressesèpistonis doingworkèinternalenergyofthegaschanges ΔU = Q + W gas-filled chamber Physics140,Prof.M.Nikolic moveable pistonof massm 2 Thefirstlawofthermodynamics ΔU = Q + W Thechangeininternalenergyofasystemisequaltotheheatflowintothesystem plustheworkdoneonthesystem(conserva]onofenergy). Qistheheat • IfHEATISADDEDtothesystem(gas)èQ>0 • IfHEATISRELEASEDbythesystem(gas)èQ<0 Wisthework • IfWORKISDONEONthesystem(gas)ègascompressesèW>0 • IfWORKISDONEBYthesystem(gas)ègasexpandsèW<0 Physics140,Prof.M.Nikolic 3 Conceptualques]on–Thefirstlawof thermodynamics Q1 Anamountofheatequalto2500Jisaddedtoasystem,and1800J ofworkisdoneonthesystem.Whatisthechangeininternal energyofthesystem? A. B. C. D. 2500J 700J 1800J 4300J ΔU = Q + W IfheatisaddedèQ>0èQ=2500J IftheworkisdoneonèW>0èW=1800J ΔU = 2500J +1800J = 4300J 30 Physics140,Prof.M.Nikolic 4 Thermodynamicprocesses Gasisexpandingètheworkisdoneby thegasètheworkonthegasisnega]ve W = −Fd W = −PAd W = −PΔV Ifworkistobenon-zero,thegasmustexpandorcontract. • WhenVf<VièthegashascontractedandW>0 • WhenVf>VièthegashasexpandedandW<0. Physics140,Prof.M.Nikolic 5 PVDiagrams Pressure Pressure Pi Pi Pf Vi Vf Volume • Thepressureisstayingconstant. • Themagnitudeoftheworkdone istheareaunderneaththe “curve.” Physics140,Prof.M.Nikolic Vi Vf Volume • Boththepressureandthevolume change • Themagnitudeofthetotalworkis s]lltheareaunderthecurve. 6 PVDiagrams isobaricprocess Pressure • Constantpressure(isobaricprocess) isochoricprocess Pi isothermalprocess W = −PΔV = −P(V f −Vi ) • Constantvolume(isochoricprocess) W = −PΔV = 0 Pf • Constanttemperature(isothermal process) Vi V f adiaba]cprocess (steepercurvethanfor isothermalprocess) Physics140,Prof.M.Nikolic Volume • Noheattransfer(adiaba]c process) Q=0 7 Thermodynamicprocessesforanidealgas PV = nR T • Constantvolume(isochoricprocess) W = −PΔV = 0 Q = nCV ΔT ΔU = Q + W = nCV ΔT • Constantpressure(isobaricprocess) W = −PΔV = −nRΔT ΔU = Q + W = nCP ΔT − nRΔT Q = nCP ΔT CV = CP − R • Constanttemperature (isothermalprocess) Physics140,Prof.M.Nikolic ΔU = 0 " Vf % −Q = W = −nRT ln $ ' # Vi & 8 Exercise:Thermodynamicprocesses Howmuchheatisrequiredinanisochoricprocesstoraisethetemperatureof 5molesofanideal,monatomicgasfrom-50oCto50oC? "3 % Q = nCV ΔT = n $ R ' (Tf − Ti ) #2 & 3 Q = 5 8.315(50 − (−50)) = 6236J 2 Howmuchheatisneededtoisothermallychangethevolumeof20molesofanideal gasfrom1.0m3to0.4m3ifthetemperatureremainsat-23.0oC? ! Vf $ Q = nRT ln # & " Vi % First,convertthetemperaturetoKelvins:273+(–23)=250K. " 0.4 % Q = 20 ⋅ 8.315⋅ 250 ⋅ ln $ ' = −38095J # 1 & Physics140,Prof.M.Nikolic 9 Reversibilityvs.Irreversibility Areversibleprocessisaprocessthatcanhappenequallywellfrom stateAtostateBorstateBtostateA. • Perfectlyelas]ccollisionisareversibleprocess AnIrreversibleProcess Physics140,Prof.M.Nikolic 10 TheSecondLawofThermodynamics • Anyprocessthatinvolvesdissipa]onofenergyisnotreversible. • Anyprocessthatinvolvesheattransferfromaholerobjectto acolderobjectisnotreversible. Thesecondlawofthermodynamics(ClausiusStatement): Heatneverflowsspontaneouslyfromacolderbodytoaholerbody. Physics140,Prof.M.Nikolic 11 Heatengines Aheatengineisadevicedesignedtotakeheatandconvertitto mechanicalwork. • takeacertainamountofenergyQHfroma “hotreservoir” • convertsomeofthatenergytoworkW • expeltheremainingenergyQCintoa“cold reservoir.” C QH = W C QH = W + QC Physics140,Prof.M.Nikolic Nolosses! 100%efficiency! 12 Efficiencyofanengine Theefficiencyofaheatengineisequaltohowmuchenergyit convertstoworkfromtheenergyistakesinfromthehotreservoir. net work output Wnet e= = heat input QH QH − QC QC e= = 1− QH QH Efficiency(e)isaunitlessquan]tythatgoesfrom0(W=0)to1(QC=0). Physics140,Prof.M.Nikolic 13 Refrigeratorsandheatpumps HeatpumpsandrefrigeratorsbothoperateundertheprincipleofremovingheatQC fromoneenvironmentandexpellingheatQHintoanotherenvironmentby performingworkW. QC + W + QH = 0 ⇒ W + QC = −QH • QC>0becauseisbeingaddedtothesystem • W>0becauseisbeingaddedtothesystem • QH<0becauseisbeingremovedfromthesystem PerformanceofheatpumpsismeasuredwiththecoefficientofperformanceK: QH heat delivered • Heatpumps K P = net work input = W net • Refrigerators K R = Physics140,Prof.M.Nikolic Q heat removed = C net work input Wnet 14 Exercise:Heatengine Wnet e= QH a) Howmuchheatdoesanenginewithefficiencyof33.3%absorb inordertodeliver1.00kJofwork? Wnet e= QH Wnet QH = e 1000J QH = = 3030.3J 0.33 b)Howmuchheatisexhaustedbytheengine? QH = W + QC QC = QH − W QC = 3030.3J −1000J = 2030.3J Physics140,Prof.M.Nikolic 15 ReversibleEnginesandHeatPumps Let’simagineacombina]onofaheatengineandaheatpump.Theworkgenerated bytheheatengine,WH.E.,canbeusedastheworkweputintotheheatpump, WH.P..Theyoperatebetweenthesametemperatureextremes. Noenginecanhaveanefficiencygreaterthanthatofanidealreversibleengine thatusesthesametworeservoirs.Theefficiencyofthisidealreversibleengineis TC er = 1 − . TH HotReservoir,TH QH,HE QH,HP WHE WHP QC,HE Physics140,Prof.M.Nikolic QC,HP ColdReservoir,TC 16 Carnotcycle Theidealreversibleenginewiththehighestefficiency. Aheatengineiscyclic–itreturnstothesamepressure,volume andtemperaturethatithadaperthecyclehasbeencompleted 1. Theenginesucksinheatfromthehotreservoirbut,byexpanding, keepsthetemperatureconstant 2. Thegascon]nuesexpandingadiaba]cally,loweringthetemperature 3. Thegasexhaustsheatandcompresses,keepingthetemperature constant 4. Thegascon]nuescompressingadiaba]callyandreturnstoitsoriginal configura]on Physics140,Prof.M.Nikolic 17 TheCarnotcycle illustrated Physics140,Prof.M.Nikolic 18 Exercise:Engineefficiency er = 1− TC TH Anengineoperatesbetweentemperatures650Kand350Kat65.0%ofits maximumefficiency. a)Whatistheefficiencyofthisengine? Themaximumpossibleefficiencyis er = 1− TC TH 350 K er = 1− = 0.462 650K Theengineoperatesate=0.65er=0.30or30%efficiency. Physics140,Prof.M.Nikolic 19 Exercise:Engineefficiency e= QH − QC Q = 1− C QH QH Anengineoperatesbetweentemperatures650Kand350Kat65.0%ofits maximumefficiency. b)If6.3×103Jisexhaustedtothelowtemperaturereservoir,howmuchworkdoes theenginedo? QH − QC QC e= = 1− QH QH QC (1− e) = QH QC QH = 1− e QH = Wnet = QH − QC Physics140,Prof.M.Nikolic 6300J = 9000J 1− 0.3 Wnet = 9000 − 6300 = 2700J 20 Entropy Heatflowsfromobjectsofhightemperaturetoobjectsatlowtemperaturebecause thisprocessincreasesthedisorderofthesystem. Entropyisameasureofasystem’sdisorder. IfanamountofheatQflowsintoasystematconstanttemperature,thenthe changeinentropyis Q ΔS = T Everyirreversibleprocessincreasesthetotalentropyoftheuniverse. Reversibleprocessesdonotincreasethetotalentropyoftheuniverse. Physics140,Prof.M.Nikolic 21 Entropyandthesecondlawof thermodynamics Athermallyisolatedsystemwillneverseeitsentropydecrease,i.e., ΔS≥0foranyprocessinathermallyisolatedsystem. Physics140,Prof.M.Nikolic 22 TheThirdLawofThermodynamics Itisimpossibletocoolasystemtoabsolutezero. • Allgasesliquefyatlowtemperaturessonoexperimentaldataavailable forlowtemperatures • Extrapolate! Physics140,Prof.M.Nikolic 23
