热力学第一定律-TTUPhysics课件

Chapter 2The first law of thermodynamicsthermodynamicsdealswith(a)energyconversionand(b)directionofchange.Twolaws,thefirstlawandthesecondlawofthermodynamics,arefoundationofthermodynamics.Theyareanexperiencesummaryofhumanbeingsinlonghistory,thereforecannotbeprovedmathematically,whiletheircorrectnessisindubitable.2.1 Concept and nomenclature in thermodynamics2.1.1 System and surroundingsThreekindsofsystemsOpen systems:exchange of both matter and energyClosed systems:no exchange of matter but some exchange of energy.Isolated systems:neither exchange of matter nor exchange of energy.2.1.2 Extensive property and intensive propertyExtensive propertyits value depends on the extent or size of the system.Theoverallvalueisthesumofvariouspartsofthesystem.Forexample,m,V,U,etc.Intensive propertyItsvalueisindependentoftheextentorsizeofthesystem.Forexample,T,c,etc.anyextensivevariabledividedbythemolesormassbecomesanintensivevariable.2.1.3 State and state functionState状态是指系统的各种内在及外在性质在一定条件下的宏观表现。(化工2004一同学）Statefunctionapropertyofasystemthatisnotdependentonthewayinwhichthesystemgetstothestateinwhichitexhibitsthatproperty.Twopropertiesofstatefunctions(1)theinfinitesimalchangesofastatefunctioncanbeexpressedintotaldifferential.e.g.z=f(x,y)(2)Anychangesbetweentheinitialandfinalstatesdependonlyonthestateofthesystemnotonthepathsthroughwhichthechangetakesplace.T=T2T1，U=U2 U1Y=Y2Y1，X=X2X1 ABXY2.1.4Equilibriumstatethreeconditionshavetobenecessarytoanequilibriumstate:(A)Thermal equilibrium(B)Mechanicalequilibrium(C)Chemical reaction and phase transition equilibrium2.1.5 Steady stateSteadystateisasituationinwhichallstatevariablesareconstantinspiteofongoingprocessesthatstrivetochangethem.Itisdifferentfromtheequilibriumstate.2.1.6 Process and pathProcessisachangeofasystemfrominitialstatetofinialstate.Path istheintermediatestepsbetweentheinitialstateandthefinalstateinachangeofstate.Isobaric:processdoneatconstantpressure,p1=p2=psur.Isochoric:processdoneatconstantvolume,V1=V2.Isothermal:processdoneatconstanttemperature,T1=T2=Tsur.Adiabatic:processwhereQ=0,thatis,noheatexchangesCyclic:processwhereinitialstate=finalstate.Spontaneousandnon-spontaneousAspontaneous processisonethatwillnaturallyoccurintheabsenceofexternaldrivingforces.Forexample,aballrollsoffatableandfallstothefloor.A non-spontaneous processisthereverseofaspontaneousprocess.Thisdoesnotmeanthatnon-spontaneousprocessesdonothappen.Theysimplydonothappenbythemselves.2.1.7 Heat and workHeat(Q)istheexchangeofthermalenergyfromahotbodytoacoldbody.Itisakindofenergytransferredinadrivingforceoftemperaturedifference.the zeroth law of thermodynamicsIftwobodiesareinthermalequilibriumwithathirdbody,theyarealsointhermalequilibriumwitheachother.Sign Convention for heatQ Positiveheatin,negativeheatout.EndothermicQ0,exothermicQ0.Example HoldapieceoficeinyourhanduntilitmeltsSolution ASystem：YouSurroundings：Ice+therestoftheuniverseQ0：Heatflowsintothesystem(ice)fromyou.ExampleH2burnsinaheatinsulated(adiabatic)containerfilledwithO2.whatistheheatsignofthisprocess?Positive,negative,orzero?WorkWork(W):Allthetransferringformofenergyexceptheat.Thereareseveralkindsofwork.Pressure-volume(pV)work,electricalwork,surfacework,andmechanicalwork,etc.non-volume work(W):Exceptpressure-volumework,alltheotherworks.WorksignW0WW绝热pVA(p1,V1)B(p2,V2)C(p3,V2)V1V2 等温线等温线绝热线绝热线 ComparisionofisothermstoadiatatsExample1mol双原子理想气体从25，100kPa突然绝热恒外压减压至50kPa，求终态温度T2及W、U、H。解：因为绝热，Q=0，U=W=-pambVnCV,m(T2-T1)=-p2(V2-V1)V2=nRT2/p2；V1=nRT1/p1代入上式，解出T2=255.56KU=nCV,mT=(5/2)R(255.56-298.15)=-885.3 JH=nCp,mT=(7/2)R(255.56-298.15)=-1239 JH=U+(pV)=U+RT =-885.3+8.3145(255.56-298.15)=-1239 JExample:4mol双原子理想气体从p1=50kPa,V1=160dm3绝热可逆压缩至p2=200kPa。求末态温度T2及W，U，H。解：先求T1=p1V1/nR=240.53KT2=(p2/p1)R/Cp,mT1=357.43KU=nCv,m(T2-T1)=9.720kJH=nCp,m(T2-T1)W=U2.3 Phase transformationphaseisaportionofasystemthathasuniformpropertiesandcomposition.PhasechangePhasechangeincludesfromaliquidtoagas(vaporization)fromasolidtoaliquid(fusion)fromasolidtoagas(sublimation)crystalformtransitionPhasechangeatconstantpressureThemolarchangeofenthalpyFormeltingandcrystaltransitionprocessatconstantpressureandconstanttemperatureForvaporizationandsublimationprocessesexample100，50dm3真空容器内有一小瓶，瓶内有50g水。将小瓶打破，蒸发到平衡，求Q，W，U，H。已知水的vapHm=40.668kJmol-1。解：水只能部分蒸发。设为nmol。n=pV/RT=1.633mol,即29.42g。H=1.63340.668=66.41kJW=0Q=U=H-(pV)=H-pV(g)=H-nRT=61.34kJTemperaturedependenceofenthalpyofphasechange2.4 Standard molar enthalpy of reaction2.4.1 Stoichiometric coefficients aA+bB=yY+zZ0=aAbByY+zZThenumbers,a,b,y,andz,showingtherelativenumbersofmoleculesreacting,arecalledthestoichiometric coefficients.2.4.2 Extent of reactiond=dnB/Bforasamereaction,iftheequationofchemicalreactioniswrittenindifferentform,Bwillalsobedifferent,andthenextentofreactionwillbedifferenttoo.For example:N N2 2(g)+3H(g)+3H2 2(g)=2NH(g)=2NH3 3(g)(g)N N2 2(g)+3/2 H(g)+3/2 H2 2(g)=NH(g)=NH3 3(g)(g)2.4.3 Molar enthalpy of reactionMolar enthalpy of reaction is an enthalpy change of a reaction.For example:rHm ismolarenthalpyofreaction;*standsforapuresubstance.2.4.4.Standard molar enthalpy of reactionStandard molar enthalpy of reaction:theenthalpychangepermoleforconversionofreactantsintheirstandardstatesintoproductsintheirstandardstates,ataspecifiedtemperature.2.5 Calculation of standard enthalpy of reactions2.5.1 Standard molar enthalpy of formationStandard molar enthalpy of formation Theenthalpychangewhenonemoleofthecompoundisformedat100kPapressureandgiventemperaturefromtheelementsintheirstablestatesatthatpressureandtemperature.thestableformsoftheelementshaveForexample:NotethatAnyformofelementsotherthanthemoststablewillnotbezero;C(diamond),C(g),H(g),andS(monoclinic)areexamples.CalculationofstandardenthalpyofreactionsExample Example Calculatethestandardenthalpyoffollowingreactionat25byusingstandardmolarenthalpyofformationSolutionC2H5OH(g)C4H6(g)H2O(g)H2(g)-235.10110.16-241.810note2.5.2 Standard molar enthalpy of combustionDefinition:Theenthalpychangewhenamoleofsubstanceiscompletelyburntinoxygenatagiventemperatureandstandardpressure.Thegeneralconventionfortheproductsofcombustionisasfollows.CarboninorganiccompoundbecomesCO2(g);HbecomesH2O(l);NbecomesN2(g);SbecomesSO2(g);ClbecomesHCl(aq)andsoon.Allthesecompleteproductshaveanenthalpyofcombustionofzero.For example,under 298.15K and standard pressure:反应物和产物均为相同的氧化产物反应物和产物均为相同的氧化产物 用图解的方法表示燃烧热与反应热的关系用图解的方法表示燃烧热与反应热的关系H1H2H1=Hm+H2即Hm=-(H2-H1)ForexampleABCDDeterminationofheatofcombustion2.5.3 Dependence of standard molar enthalpy of reaction on temperatureaA+bBaA+bByY+zZyY+zZsincethenItiscalledKirchhoff equation2.5.4 The relationship between heat of chemical reaction at constant pressure and volumereactions involving only solids or liquidsvolumeworkW0,and(pV)0,thenforsolidsandliquidsQHU.reactions involving gasestheproductpVmaybereplacedbyBRT2.5.5 Hesss Law and reaction enthalpyHessslawstatesthattheenthalpychangeofanyreactionmaybeexpressedasthesumoftheenthalpychangesofaseriesofreactionsintowhichtheoverallreactionmayformallybedivided.Theenthalpychangeofareactionatconstantpressureorconstantvolumedependsonlyonthefinalandinitialstates,andnotonthepathconnectingthem.ExamplesomereactionscannotbestudieddirectlyC(graphite)+2H2(g)CH4(g)ConsiderfollowingreactionsThecombinationofthesethreereactionsfrom(a)+2(b)+(c),wegettheabovestudiedreaction.Then2.5.6 The maximum temperatures of flames and explosionsThetemperaturereachedforacombustionreactionatconstantpressureandadiabaticsystemisknownasthemaximum temperature of flame.Qp=H=0thetemperaturereachedforanexplosioninanadiabaticsystemandatconstantvolumeiscalledthemaximum temperature of explosion.QV=U=0例甲烷(CH4,g)与理论量二倍的空气混合，始态温度25，在常压(p100kPa)下燃烧，求燃烧产物所能达到的最高温度。设空气中氧气的摩尔分数为0.21，其余为氮气，所需数据查附录。解：甲烷(CH4,g)的燃烧反应为CH4(g)+2O2(g)CO2(g)+2H2O(g)先求反应的rHm，可以用各反应组分的fHm来计算rHm，我们这里用cHm(CH4,g)来计算rHmCH4(g)+2O2(g)CO2(g)+2H2O(g)CO2(g)+2H2O(l)rHmcHm(CH4,g)vapHm(H2O)rHm=cHm(CH4,g)+2vapHm(H2O)=802.286kJ对于含1mol甲烷(CH4,g)的系统，含氧气4mol，氮气（4/0.21）0.79mol=15.05mol，则始态始态T0=298.15KCH4(g)1mol,O2(g)4molN2(g)15.05mol TCO2(g)1mol,H2O(g)2molO2(g)2mol,N2(g)15.05mol T0=298.15KCO2(g)1mol,H2O(g)2molO2(g)2mol,N2(g)15.05molrHmH2Qp=H=0恒压绝热Qp=H=rHm+H2=0将附录中的CO2(g),H2O(g),O2(g),N2(g)的定压摩尔热容Cp,m=a+bT+cT2代入上式。再代入方程rHm+H2=0，解T，得T=1477K即最高火焰温度就是恒压绝热反应所能达到的最高温度。而最高爆炸温度就是恒容绝热反应所能达到的最高温度。2.6 Joule-Thomson effectTheexperimentbyJouleandThomsonshowedthatHofarealgasisnotonlythefunctionofT,butalsothefunctionofp.1TheexperimentbyJouleandThomsonp1,V1,T1p2,V2,T2Porous plugAdiabaticwallthermometerthrottleexpansion，p1p2点击图像可以看动画开始，环境将一定量气体压缩时所作功（即以气体为系统得到的功）为：节流过程是在绝热筒中进行的，Q=0，所以：气体通过小孔膨胀，对环境作功为：在压缩和膨胀时系统净功的变化应该是两个功的代数和。即节流膨胀过程是个等焓过程。H=0移项2.节流膨胀的热力学特征及焦-汤系数 0 经节流膨胀后，气体温度降低。0 经节流膨胀后，气体温度升高。=0 经节流膨胀后，气体温度不变。称为焦-汤系数（Joule-Thomsoncoefficient),它表示经节流过程后，气体温度随压力的变化率。ShowthatforidealgasesH=f(T,p)Throttling：d H=0)Structureofair-conditionerStructureofrefrigeratorOperatingprincipleofarefrigeratorAnimationofrefrigerationCompression-typerefrigeratingmachine01Compression-typerefrigeratingmachine02LindecycleforliquefactionofgasJamesPrescottJoule18181889焦耳是英国物理学家。1818年12月24日生于索尔福。他父亲是酿酒厂的厂主。焦耳从小体弱不能上学，在家跟父亲学酿酒，并利用空闲时间自学化学、物理。他很喜欢电学和磁学，对实验特别感兴趣。后来成为英国曼彻斯特的一位酿酒师和业余科学家。焦耳可以说是一位靠自学成才的杰出的科学家。焦耳最早的工作是电学和磁学方面的研究，后转向对功热转化的实验研究。1866年由于他在热学、电学和热力学方面的贡献，被授予英国皇家学会柯普莱金质奖章。1872年1887年焦耳任英国科学促进协会主席。1889年10月11日焦耳在塞拉逝世。开尔文（LordKelvin18241907）19世纪英国卓越的物理学家。原名W.汤姆孙（WilliamThomson），1824年6月26日生于爱尔兰的贝尔法斯特，1907年12月17日在苏格兰的内瑟霍尔逝世。由于装设大西洋海底电缆有功，英国政府于1866年封他为爵士，后又于1892年封他为男爵，称为开尔文男爵，以后他就改名为开尔文。1846年开尔文被选为格拉斯哥大学自然哲学教授，自然哲学在当时是物理学的别名。开尔文担任教授53年之久，到1899年才退休。1904年他出任格拉斯哥大学校长，直到逝世。Sir William Thomson working on a problem of science in 1890SummaryU=Q+W(reversibleprocess)(adiabaticreversible)pV=const(adiabaticreversibleforidealgas)Heat理想气体：ReactionheatThrottlingexpansion：HomeworkPage722.1(basicconcept)2.2(basicconcept)2.3(calculationofwork)2.4(calculationsofworkandheat)2.9(CalculateQ,W,U,andH)2.12(CalculationofWr,Q,U,Hofphasechangeprocess)2.20(Calculationofadiabaticprocess)2.25(Calculationoftheenthalpyofformation)2.24(reactionheat)