热力学第一定律-TTUPhysics课件

1、Chapter 2The first law of thermodynamicsthermodynamics 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 pro

2、ved mathematically, while their correctness is indubitable. 2.1 Concept and nomenclature in thermodynamics2.1.1 System and surroundingsThree kinds of systemsOpen systems: exchange of both matter and energy Closed systems:no exchange of matter but some exchange of energy. Isolated systems:neither exc

3、hange 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. The overall value is the sum of various parts of the system.For example, m, V, U, etc.Intensive propertyIts value is independent of the exte

4、nt or size of 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 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）State functiona property of a system that is not dependent on the way in which the system ge

5、ts to the state 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 thr

6、ough which the change takes place. T=T2T1， U=U2 U1Y=Y2Y1， X=X2X1 ABXY2.1.4 Equilibrium statethree 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 stateSteady state is a situati

7、on in which all 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 pathProcess is a change of a system from initial state to finial state.Path is the intermediate steps between the initial state and the

8、final state in 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 exchangesCyclic: process where initial state =

9、final state. Spontaneous and non-spontaneousA 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-s

10、pontaneous processes do not happen. They simply do not happen by themselves. 2.1.7 Heat and workHeat (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 thermodynamicsIf two bodies ar

11、e in thermal equilibrium with a third 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 meltsSolution ASystem：You Surroundings：Ice + the rest of th

12、e universe Q 0：Heat flows into the system (ice) from you. ExampleH2 burns in a heat insulated (adiabatic) container filled with O2. what is the heat sign of this process? Positive, negative, or zero?WorkWork (W): All the transferring form of energy except heat.There are several kinds of work. Pressu

13、re-volume (pV) work, electrical work, surface work, and mechanical 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

14、、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: 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=p

15、1V1/nR=240.53KT2=(p2/p1)R/Cp,mT1=357.43KU=nCv,m(T2-T1)=9.720kJH=nCp,m(T2-T1)W=U2.3 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

16、a gas (sublimation) crystal form transitionPhase change at constant pressureThe molar change of enthalpyFor melting and crystal transition process at constant pressure and constant temperatureFor vaporization and sublimation processes 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=

17、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 change2.4 Standard molar enthalpy of reaction2.4.1 Stoichiometric coefficients aA + bB = yY + zZ 0=aAbByY+zZThe numbers, a, b, y, and z, sho

18、wing the relative numbers of molecules reacting, are called the stoichiometric coefficients. 2.4.2 Extent of reaction d=dnB/B for 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 examp

19、le: N2(g)+3H2(g)= 2NH3(g) N2(g)+ 3/2 H2(g)= NH3(g)2.4.3 Molar enthalpy of reactionMolar enthalpy of reaction is an enthalpy change of a reaction. For example:rHm is molar enthalpy of reaction; * stands for a pure substance.2.4.4. Standard molar enthalpy of reactionStandard molar enthalpy of reaction

20、:the enthalpy change per mole for conversion of reactants in their standard states into products in their standard states, at a specified temperature.2.5 Calculation of standard enthalpy of reactions2.5.1 Standard molar enthalpy of formationStandard molar enthalpy of formation The enthalpy change wh

21、en 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:Note thatAny form of elements other than the most stable will not be zero; C (diamond), C (g),

22、H (g), and S (monoclinic) are examples. Calculation of standard enthalpy of reactions Example Calculate the standard enthalpy of following reaction at 25 by using standard molar enthalpy of formationSolutionC2H5OH(g) C4H6(g) H2O(g) H2(g) -235.10 110.16 -241.81 0note 2.5.2 Standard molar enthalpy of

23、combustionDefinition: The enthalpy change when a mole of substance is completely burnt in oxygen 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

24、SO2(g); Cl becomes HCl(aq) and so on. All these complete products have an enthalpy of combustion of zero.For example, under 298.15K and standard pressure: 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H1H2H1= Hm+ H2 即 Hm= -(H2 - H1)For exampleA B C DDetermination of heat of combustion2.5.3 Dependence of standar

25、d molar enthalpy of reaction on temperatureaA+bBaA+bByY+zZyY+zZsincethenIt is called Kirchhoff equation2.5.4 The relationship between heat of chemical reaction at constant pressure and volumereactions involving only solids or liquids volume work W0, and (pV)0, then for solids and liquids QHU.reactio

26、ns involving gases the product pV may be replaced by BRT 2.5.5 Hesss Law and reaction enthalpyHesss 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 overall reaction may formally be divided. The enthalpy c

27、hange 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 directlyC(graphite) + 2H2(g) CH4(g) Consider following reactionsThe combination of these three reactions from (a)+2

28、(b)+(c), we get the above studied reaction.Then2.5.6 The maximum temperatures of flames and explosionsThe temperature reached for a combustion reaction at constant pressure and adiabatic system is known as the maximum temperature of flame. Qp=H=0the temperature reached for an explosion in an adiabat

29、ic system and at constant volume is called the 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(

30、g)CO2(g)+2H2O(l)rHmcHm(CH4,g) vapHm(H2O) rHm = cHm(CH4,g)+ 2DvapHm(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) 15.05

31、molrHmH2Qp=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 effectThe experiment by Joule and Thomson showed that H of a real gas is not only the function of T,

32、but also the function of p.1 The experiment by Joule and Thomson p1,V1,T1p2,V2,T2Porous plugAdiabatic wallthermometerthrottle expansion ，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

33、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-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)Throttling：d H=0)Structure of air-conditionerStructure of refrigeratorOperating principle of a refrigeratorAnimation of refrigerationCompression-type refrigerating machine 01Compression-type refriger