熱力學(xué)第一定律-TTUPhysics課件

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狀態(tài)是指系統(tǒng)的各種內(nèi)在及外在性質(zhì)在一定條件下的宏觀表現(xiàn)。(化工2004一同學(xué)）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絕熱pVA (p1,V1)B (p2,V2) C (p3,V2) V1V2 等溫線絕熱線 Comparision of isotherms to adiatatsExample1mol 雙原子理想氣體從25，100kPa 突然絕熱恒外壓減壓至50kPa，求終態(tài)溫度T2及W、U

14、、H。解：因為絕熱，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 雙原子理想氣體從p1=50kPa, V1=160dm3絕熱可逆壓縮至p2=200kPa。求末態(tài)溫度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真空容器內(nèi)有一小瓶，瓶內(nèi)有50g水。將小瓶打破，蒸發(fā)到平衡，求Q，W，U，H。已知水的vapHm=

17、40.668kJ mol-1。解：水只能部分蒸發(fā)。設(shè)為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: 反應(yīng)物和產(chǎn)物均為相同的氧化產(chǎn)物 用圖解的方法表示燃燒熱與反應(yīng)熱的關(guān)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)與理論量二倍的空氣混合，始態(tài)溫度25，在常壓(p100kPa)下燃燒，求燃燒產(chǎn)物所能達(dá)到的最高溫度。設(shè)空氣中氧氣的摩爾分?jǐn)?shù)為0.21，其余為氮?dú)?，所需?shù)據(jù)查附錄。解：甲烷(CH4,g)的燃燒反應(yīng)為 CH4(g)+2O2(g)CO2(g)+2H2O(g)先求反應(yīng)的rHm，可以用各反應(yīng)組分的fHm來計算rHm，我們這里用cHm(CH4,g)來計算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對于含1mol甲烷(CH4,g) 的系統(tǒng)，含氧氣4mol，氮?dú)猓?/0.21）0.79mol=15.05mol，則始態(tài)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恒壓絕熱Qp= H = rHm + H2 =0 將附錄中的CO2(g) , H2O(g) ,O2(g) ,N2 (g)的定壓摩爾熱容Cp,m= a+bT+cT2代入上式 。 再代入方程rHm+ H2= 0 ，解T，得 T =1477K即最高火焰溫度就是恒壓絕熱反應(yīng)所能達(dá)到的最高溫度。而最高爆炸溫度就是恒容絕熱反應(yīng)所能達(dá)到的最高溫度。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點(diǎn)擊圖像可以看動畫 開始，環(huán)境將一定量氣體壓縮時所作功（即以氣體為系統(tǒng)得到的功）為：節(jié)流過程是在絕熱筒中進(jìn)行的，Q=0 ，所以：氣體通過小孔膨脹，對環(huán)境作功為： 在壓縮和膨脹時系統(tǒng)凈功的變化應(yīng)該是兩個功的代數(shù)和。即節(jié)流膨脹過程是個等焓過程。H = 0移項2. 節(jié)流膨脹的熱力學(xué)特征及焦-湯系數(shù) 0 經(jīng)節(jié)流膨脹后，氣體溫

33、度降低。 0 經(jīng)節(jié)流膨脹后，氣體溫度升高。 =0 經(jīng)節(jié)流膨脹后，氣體溫度不變。稱為焦-湯系數(shù)（Joule-Thomson coefficient),它表示經(jīng)節(jié)流過程后，氣體溫度隨壓力的變化率。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