1、Chapter 2The first law of thermodynamics thermodynamics deals with(a)energy conversion and(b)direction of change.Two laws,the first law and the second law of thermodynamics,are foundation of thermodynamics.They are an experience summary of human beings in long history,therefore can not be proved mat
2、hematically,while their correctness is indubitable.2.1 Concept and nomenclature in thermodynamics 2.1.1 System and surroundingsThree kinds of systems Open systems:exchange of both matter and energy Closed systems:no exchange of matter but some exchange of energy.Isolated systems:neither exchange of
3、matter nor exchange of energy.2.1.2 Extensive property and intensive property Extensive property its value depends on the extent or size of the system.The overall value is the sum of various parts of the system.For example,m,V,U,etc.Intensive property Its value is independent of the extent or size o
4、f the system.For example,T,c,etc.any extensive variable divided by the moles or mass becomes an intensive variable.2.1.3 State and state function State 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）State function a property of a system that is not dependent on the way in which the system gets to the state
5、in which it exhibits that property.Two properties of state functions(1)the infinitesimal changes of a state function can be expressed in total differential.e.g.z=f(x,y)(2)Any changes between the initial and final states depend only on the state of the system not on the paths through which the change
6、 takes place.yxzzdzdxdyxyT=T2T1，U=U2 U1Y=Y2Y1，X=X2X1 ABXY2.1.4 Equilibrium state three conditions have to be necessary to an equilibrium state:(A)Thermal equilibrium(B)Mechanical equilibrium(C)Chemical reaction and phase transition equilibrium2.1.5 Steady state Steady state is a situation in which a
7、ll state variables are constant in spite of ongoing processes that strive to change them.It is different from the equilibrium state.2.1.6 Process and path Process is a change of a system from initial state to finial state.Path is the intermediate steps between the initial state and the final state i
8、n a change of state.Isobaric:process done at constant pressure,p1=p2=psur.Isochoric:process done at constant volume,V1=V2.Isothermal:process done at constant temperature,T1=T2=Tsur.Adiabatic:process where Q=0,that is,no heat exchanges Cyclic:process where initial state=final state.Spontaneous and no
9、n-spontaneous A spontaneous process is one that will naturally occur in the absence of external driving forces.For example,a ball rolls off a table and falls to the floor.A non-spontaneous process is the reverse of a spontaneous process.This does not mean that non-spontaneous processes do not happen
10、.They simply do not happen by themselves.2.1.7 Heat and work Heat(Q)is the exchange of thermal energy from a hot body to a cold body.It is a kind of energy transferred in a driving force of temperature difference.the zeroth law of thermodynamics If two bodies are in thermal equilibrium with a third
11、body,they are also in thermal equilibrium with each other.Sign Convention for heatQ Positive heat in,negative heat out.Endothermic Q0,exothermic Q0.Example Hold a piece of ice in your hand until it melts Solution A System：You Surroundings：Ice+the rest of the universe Q 0：Heat flows into the system(i
12、ce)from you.Example H2 burns in a heat insulated(adiabatic)container filled with O2.what is the heat sign of this process?Positive,negative,or zero?Work Work(W):All the transferring form of energy except heat.There are several kinds of work.Pressure-volume(pV)work,electrical work,surface work,and me
13、chanical work,etc.non-volume work(W):Except pressure-volume work,all the other works.Work signW0W W绝热pVA(p1,V1)B(p2,V2)C(p3,V2)V1V2 等温线等温线绝热线绝热线 Comparision of isotherms to adiatatsExample1mol 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=n
14、RT1/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:4 mol 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
15、(T2-T1)W=U,12222211111p mV mp mV mV mV mp mp mp mV mCCCCCCRCCCCTppppTpppp2.3 Phase transformationphase is a portion of a system that has uniform properties and composition.Phase changePhase change includesfrom a liquid to a gas(vaporization)from a solid to a liquid(fusion)from a solid to a gas(subli
16、mation)crystal form transition,vapmfusmsubmtrsmHHHHPhase change at constant pressurepQH()()HHH/mHH n The molar change of enthalpyFor melting and crystal transition process at constant pressure and constant temperature0pQHWp VUH For vaporization and sublimation processes()()()pQHWp VpV gnRTUHpVHpV gH
17、nRT example100，50dm3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=40.668kJ mol-1。解：水只能部分蒸发。设为n mol。n=pV/RT=1.633mol,即29.42g。H=1.63340.668=66.41kJW=0Q=U=H-(pV)=H-pV(g)=H-nRT=61.34kJTemperature dependence of enthalpy of phase change122121,()()()()()()()()mmTmp mTTmp mTH THH THHCdTHCdT2121p,m,m2m1,C
18、=()-C()()+()()()+p mp mTmmp mTTp mTCHHCdTHTHTCdT 2.4 Standard molar enthalpy of reaction2.4.1 Stoichiometric coefficients aA+bB=yY+zZ 0=aAbByY+zZBB0The numbers,a,b,y,and z,showing the relative numbers of molecules reacting,are called the stoichiometric coefficients.2.4.2 Extent of reaction d=dnB/B f
19、or a same reaction,if the equation of chemical reaction is written in different form,B will also be different,and then extent of reaction will be different too.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 reactionMola
20、r enthalpy of reaction is an enthalpy change of a reaction.For example:a Ab By Yz Z*()()()()rYmZmAmBmHn HYn HZn HAn HB*/()rmrBmBHHHB rHm is molar enthalpy of reaction;*stands for a pure substance.2.4.4.Standard molar enthalpy of reactionStandard molar enthalpy of reaction:the enthalpy change per mol
21、e for conversion of reactants in their standard states into products in their standard states,at a specified temperature.()rmBmBHHB,purepurepurepurermHABLMT pT pT pT pABLM !2.5 Calculation of standard enthalpy of reactions2.5.1 Standard molar enthalpy of formationStandard molar enthalpy of formation
22、 The enthalpy change when one mole of the compound is formed at 100 kPa pressure and given temperature from the elements in their stable states at that pressure and temperature.the stable forms of the elements have For example:fmH222224222C(graphite)+O(g)=CO(g)H(g)+S(rhombic)=2O(g)+H SO(l)2Hg(l)+Cl(
23、g)=Hg Cl(s)正交0fmH!-12(CO,g)393.509 kJ molfmH!Note that Any form of elements other than the most stable will not be zero;C(diamond),C(g),H(g),and S(monoclinic)are examples.Calculation of standard enthalpy of reactions The same stable elements in the same quantities at standard pressureH1H2rABYZ -A +-
24、B Y +Z pure pure pure puremHpppp!12(A)(B)(Y)(Z)AfmBfmYfmZfmHHHHHH!12rmHHH!21 (Y)(Z)(A)(B)rmYfmZfmAfmAfmHHHHHHH!()rmBfmBHHB!Example Example Calculate the standard enthalpy of following reaction at 25 by using standard molar enthalpy of formation(g)HO(g)H2(g)HCOH(g)HC2226452 Solutionfm1kJ molHC2H5OH(g
25、)C4H6(g)H2O(g)H2(g)-235.10 110.16 -241.81 0rmfm46fm2fm25(C H,g)+2(H O,g)2(C H OH,g)HHHH 11molkJ72.96molkJ)10.2352(818.2412110.16=fm2fm2vapm2(H O,g,)(H O,l,)(H O,)HTHTHT note 2.5.2 Standard molar enthalpy of combustionDefinition:The enthalpy change when a mole of substance is completely burnt in oxyg
26、en at a given temperature and standard pressure.The general convention for the products of combustion is as follows.Carbon in organic compound becomes CO2(g);H becomes H2O(l);N becomes N2(g);S becomes SO2(g);Cl becomes HCl(aq)and so on.All these complete products have an enthalpy of combustion of ze
27、ro.cm(,)HB T$For example,under 298.15K and standard pressure:3222-1rmCH COOH(l)2O(g)2CO(g)2H O(l)870.3 kJ molH$-1cm3(CH COOH,l,298.15 K)-870.3 kJ molH$反应物和产物均为相同的氧化产物反应物和产物均为相同的氧化产物 用图解的方法表示燃烧热与反应热的关系用图解的方法表示燃烧热与反应热的关系H1H2H1=Hm+H2 即 Hm=-(H2-H1)rmBcmB(298.15 K)-(B,298.15 K)HH$rmcmcmcm(A)2(B)-(C)HHHH$
28、233 22(COOH)(s)2CH OH(l)(COOCH)(s)2H O(l)For exampleA B C DDetermination of heat of combustion2.5.3 Dependence of standard molar enthalpy of reaction on temperature)(1mrTH1H2HaA+bBaA+bByY+zZyY+zZ3H4H)(2mrTHT,CyHT T p,(Y)d21m3T,CzHT T p,(Z)d21m4112234rmrmHTHHHTHH sinceT,CaHT T p,(A)d21m1T,CbHT T p,(B
29、)d21m2then21 21,()()()TBrmrmp mTHTHTCB dT It is called Kirchhoff equation2.5.4 The relationship between heat of chemical reaction at constant pressure and volume reactions involving only solids or liquids volume work W0,and(pV)0,then for solids and liquids QHU.reactions involving gases the product p
30、V may be replaced by BRT ()rmrmBBHURTg()pVBBQQRTgwhen()0,then,or when()0,then,or BrmrmpVBBrmrmpVBgHUQQgHUQQ 2.5.5 Hesss Law and reaction enthalpy Hesss law states that the enthalpy change of any reaction may be expressed as the sum of the enthalpy changes of a series of reactions into which the over
31、all reaction may formally be divided.The enthalpy change of a reaction at constant pressure or constant volume depends only on the final and initial states,and not on the path connecting them.Example some reactions can not be studied directly C(graphite)+2H2(g)CH4(g)Consider following reactions The
32、combination of these three reactions from(a)+2(b)+(c),we get the above studied reaction.-14222,-11222,222,CH(g)+2O(g)CO(g)+2H O(l)890.35kJ mol (a)H(g)+O(g)H O(l)285.84kJ mol (b)C(graphite)+O(g)CO(g)393.5rm arm brm cHHH -11kJ mol (c)Then,2 rmrm arm brm cHHHH -1(890.35)2(285.84)(393.51)74.84kJ molrmH
33、2.5.6 The maximum temperatures of flames and explosions The temperature reached for a combustion reaction at constant pressure and adiabatic system is known as the maximum temperature of flame.Qp=H=0 the temperature reached for an explosion in an adiabatic system and at constant volume is called the
34、 maximum 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(CH
35、4,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=0TCCCCHTTppppdg,N05.15g,O2gO,H2g,CO02m,2m,2
36、m,2m,2 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 effectThe experiment by Joule and Thomson showed that H of a real gas is not only the function of T,but also the function of p.1 The experiment
37、by Joule and Thomson p1,V1,T1p2,V2,T2Porous plugAdiabatic wallthermometerthrottle expansion，p1p2点击图像可以看动画11WpV 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：21UUUW 气体通过小孔膨胀，对环境作功为：22WpV1 111 (=0)pVVVV2222 (=0)p VV VV 在压缩和膨胀时系统净功的变化应该是两个功的代数和。121 122WWWpVp V即211 122UUpVp V节流膨胀过程是个等焓过程。H=021HH移
38、项22211 1Up VUpV2.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。T-JJ-T()HTpT-J0 经节流膨胀后，气体温度升高。T-J=0 经节流膨胀后，气体温度不变。称为焦-汤系数（Joule-Thomson coefficient),它表示经节流过程后，气体温度随压力的变化率。Show that for ideal gases H=f(T,p)0TJ()()0pTHHd Hd Td pTp()()()()TTJTHpPHHTppHpCT()0THp ideal gases 0TJThrottling：d H=0)()()()0pHTHTHTpp()()0pJ
39、TTHHTpStructure of air-conditionerStructure of refrigeratorOperating principle of a refrigeratorAnimation of refrigeration Compression-type refrigerating machine 01 Compression-type refrigerating machine 02 Linde cycle for liquefaction of gas02压缩式制冷01.swf02蒸汽压缩式制冷02.swf02带预冷的节流循环.swfJames Prescott J
40、oule 18181889 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（Lord Kelvin 18241907）19世纪英国卓越
41、的物理学家。原名W.汤姆孙（William Thomson），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 1890 SummaryU=Q+W21Vex
42、VWp dV 21VVWpdV(reversible process)0 02 21 111211111pVpVpVWVV(adiabatic reversible)pV=const (adiabatic reversible for ideal gas)Heat21,Tpp mTQCdT21T,TVV mQnCdTpVTpUVCCpVT理想气体：pVCCnRHUpV21T,Tpp mHnCdT21T,TVV mUnCdTReaction heat()rmBfmBHHB!()rmBCmBHHB 2121,()()Trmrmrp mTHTHTCdT J THTpThrottling expansion：Home work Page 72 2.1(basic concept)2.2(basic concept)2.3(calculation of work)2.4(calculations of work and heat)2.9(Calculate Q,W,U,and H)2.12(Calculation of Wr,Q,U,H of phase change process)2.20(Calculation of adiabatic process)2.25(Calculation of the enthalpy of formation)2.24(reaction heat)
