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1、2-1.Please explain the following thermodynamic terms:,1).internal energy(p2-2 L1-3,L6-7)Answer:Internal energy is the energy of the molecules making up the substance,including kinetic energy and potential energy.2).closed system(P2-3,para5,L1-3);Answer:If the boundary of a system does not permit the
2、 transfer of matter between the system and surroundings,the system is said to be closed,and its mass is necessarily,the system is said to be closed.,3).state functions(p.2-5,L.13-17);Answer:These properties do not depend on the past history of the substance or on the path it has before or on the pat
3、h it has followed in reaching a given state.They depend only on the present conditions.4).critical point(p.3-1,L.9-11);Answer:The coordinates of this point are called the critical pressure Pc and the critical temperature Tc.These represent the highest temperature and temperature at which a pure mate
4、rial can exist in vapor-liquid equilibrium.Homogeneous fluids are normally divided into two classes,liquids and gases.However,the distinction cannot always be sharply drawn,because the two phases become indistinguishable at what is called the critical point.,5).ideal gas(p.3-8,para2,L6-10);Answer:Th
5、eir actual volume becomes infinitesimal compared with the total volume cccupied by the gas,and the intermolecular forces of attraction approach zero.A gas which meets these conditions is known as an ideal gas.6).compressibility factor(p.3-9,L.11).Answer:The ration PV/RT is called the compressibility
6、 factor,2-2 Give the statement and formulation of the first law of thermodynamics(P2-2,L25-26),Answer:1.The first law of thermodynamics may be stated formally in many ways.one of these is as follows:the total quantity of energy is constant,and when energy disappears in one form it appears simultaneo
7、usly in other forms 2.（Energy of the system）+(energy of surroundings)=0,2-3 What is the difference between the state function and work or heat?(p2-5,2-6),Answer:1.State function are properties of the system,while work and heat are not state propertities,they are depended on the path followed.2.The v
8、alues of state function can always be represented by points on a graph,but work and heat can represented by points on a graph,rather are represented by areas,3.The differential of a state function is spoken of that as an infinitesimal chang in the property,and is not referred to as a quantity.The di
9、fferentials of heat and work are not referred to as changes,but are regarded as infinitesimal quantities of heat and work.4.Another difference between state functions and heat or work follows from the fact that a state function represents an instantaneous(瞬间的)property of a system and always has a va
10、lue(有一确定值).Work and heat appear only when changes are caused in a system by a process,which requires time.,2-4.What are the necessary conditions for the steady state flow process?(p.2-12)Answer:The term steady state implies that conditions at all point in the apparatus are constant with time.It mean
11、s that all rates must be constant,and there must be no accumulation of material of energy within the apparatus over the period of time considered.Moreover,the total mass flow rate must be the same at all point along the path of flow of the fluid.,2-5.Please indicate in which condition that the expre
12、ssion of first law U=W Q or H=W Q can be used respectively?Answer:The expression ofH=W Q is used for a steady-state flow process,and the expression of U=W Q is used for a nonflow process.(p.2-14,upward 1st line;p.2-15,L.1-3),2-6.What is the word equilibrium denoting?Answer:Equilibrium is a word deno
13、ting a static condition,the absence of change.In thermodynamics it is taken to mean not only the absence of change but the absence of any tendency toward change on a macroscopic scale.(p.2-18,upward 7-10th line),2-7.What is the meaning of a phase in thermodynamics?(p.2-20,L.4-5)Answer:A phase is a h
14、omogeneous region of matter.It is not necessary that a phase be continuous.,2-8.What are the characteristics of reversible process?Answer:In summary,a reversible process is frictionless;it is never more than differentially removed from equilibrium;the driving forces are differential in magnitude;and
15、 the process can be reversed,leaving no more than an infinitesimal change in the system or surroundings.(p.2-26,upward 6-9th line),2-1 An insulated and nonconducting container filled with 9.07(kg)of water at 20 is fitted with a stirrer.The stirrer is made to turn by lowering a weight having a mass o
16、f 22.68(kg).The local acceleration of gravity is 9.75(m)/(s2).The weight slowly falls a distance of 9.144(m)while turnning the stirrer.Assumig that all the work done by gravity on the weight is transferred to the water,determine:,2l 一装有20C9.07Kg水的隔离又不导热的容器,配备的搅拌器靠质量为22.6kg的重物下落来带动。当地重力速度为9.75(m)/(s2
17、)。当搅拌器转动时，重物缓慢下降9.144(m)，假设重力力对重物所作的功都传给了水，试确定，c水的最终温度(C)d使水恢复到原始温度需要移走的热量 e下还各过程的总能量变化：(1)重物下降过程(2)使水冷却到原始温度的过程 3）统一考虑这两个过程,(c)The final temperature of the water in()c.solution:Since P and V are constant,so(PV)=0,H=U=m C pT,Cp=4.18J/(gK)H=9.074.18103(T2-T1)=2022.1,but,T1=20,so,T2=20.053(d)The amoun
18、t of heat which must be removed from the water to return it to its initial temperature.d.solution:Q=(20-20.053)9.071034.18=-2022.1 J,(e)The total energy change of the universe because of(1)the process of lowering the weight,(2)the process of cooling the water back to its initial temperature,and(3)bo
19、th of these processes taken together.e.(1)Because the energy given by stirrer is equal to the energy received by water.The total energy change of the universe is equal to zero.(2)The heat given by water is equal to the heat received by surrounding,the total energy change of the universe is still equ
20、al to zero.(3):(1)+(2)=0+0=0,2-3 It has been suggested that the kitchen in your house could be cooled in the summer by closing it off from the rest of the house and opening the door to the electric refrigerator.Comment on this.State clearly and consicely the basis for your conclusion.,23 建议在夏天将你住宅中的
21、厨房与其他房间隔离开，并打开电气冷冻机的门以冷却之。试评论之。并简要地阐明你的结论。,solution:If the kitchen is closed with the rest of the house,then the refrigerator wouldnt be able to cool the kitchen temperature down.Because the heat generated from refrigerator motor and radiator is more than that the heat absorbed inside the refrigerat
22、or.Finally the temperature of the kitchen must be raised.,2-4.a.Liquid water at 100()and 1(atm)has an internal energy(on an arbitrary basis)of 418.8(kJ)/(kg).What is its enthalpy?The specific volume of liquid water at these conditions is 0.0010438(m3)/(kg).a.100()和1(atm)的液体水的内能为418.8(kJ)/(kg)(根据适宜的基
23、淮)。焓为多少?在此条件下液体水的比容为0.0010438(m3)/(kg)solution:a.We know that:t=100,U=418.8kJ/kg V=0.0010438 m3so,H1=U+PV=418.8103+1013250.0010438=418.9 kJ/kg,b.The water of part a is brought to the vapor state at 204.44()and 6.895(10)5(N),where its specific volume is 0.3083(m)3/(kg)and its enthalpy is 2856.9(kJ)/(
24、kg).Calculate U and H for the process.ba部分的水变为204.44()和6.895(10)5(N)的蒸汽，其比容为0.3083(m)3/(kg)，焓为2856.9(kJ)/(kg)。计算此过程的U和H,b.The enthalpy of steam:H2=2856.9 kJ/kgH=H2-H1=2856.9-418.9=2438 kJ/kg U=H-(PV)P2=6.895105 N/m2 V2=0.3083 m3/kg thereforeU=2438103-(6.8951050.3083-1.013105 0.0010438)=2225.5kJ/kg,2
25、-5.a.How much difference in elevation will result in a change in potential energy equivalent to 2326 J/kg of the 1kg substance considered?a.一物质势能的变化等于2326 J/kg，改变了多少位差？a.solution:Ep=mgh=2326J/kg,because,m=1 kg so,h=2326/9.8=237.3m,b.A fluid has a velocity of 30.48(m/s)when entering a piece of appara
26、tus.With what velocity must the fluid leave the apparatus so that the difference in entering and leaving kinetic energies is equivalent to 2326 J/kg of the fluid?b流体进设备时的速度为30.48(m/s)。当流体进出口动能差相当于2326 J/kg 流体时，离开设备的速度应为多少？b.solution:Ek=(1/2)m(V22-V12)=2326kJ/kgbut,V1=30.48m/s,so V2=74.71m/s,c.What c
27、onclusions are indicated by these examples?Solution:Energy can transfer from one form to another,from above examples we may know that they are not easy to change potential or kinetic energy for increasing 2326J/kg internal energy of fluid.But it will be easy to do with supplying heat or doing work t
28、o a system.,2-6.Calculate the work done by 1(mol)of a gas during a reversible,isothermal expansion from an initial volume V1 to a final volume V2 when the equation of state is P(V b)=RTWhere b is a positive constant.If the gas were ideal,would the same process produce more or less work?,26 试计算1(mol)
29、气体可逆等温由V1膨胀到V2时作的功，其状态方程为 P(V b)=RT式中b是正数。如果气体是理想的，进行同样的过程，作的功是多还是少?,2-6(1)by Eq.1-5,W=PdV(2)Because the process is a reversible and isothermal expansion,and P(V-b)=RT,then P=RT/(V-b),W=RTln(v2-b)/(v1-b)for ideal gas:W=RTln(V2/V1),since V2V2-b,but V1V1-b,so W W,2-7.In a natural gasoline fractionatio
30、n system there are usually six components present in appreciable quantities:methane,ethane,propane,isobutene,n-butane,and n-pentane.If a mixture of these components is placed in a bomb and the temperature and pressure are fixed so that a liquid and a gas phase exist at equilibrium,how many additiona
31、l phase-rule variables must be chosen to fix the composition of both phases?If the temperature and pressure are to remain the same,is there any way that the composition of the total contents of the bomb can be changed(by adding or removing material)without affecting the composition of the gas and li
32、quid phases?,2-7 在天然气液化汽油分馏系统中，通常存在可估量的组分有六种二甲烷、乙烷、丙烷、异丁烷、正丁烷和正戊烷。如果将这些组分的混合物放进一个高压弹中，并固定其温度和压力使之气液两相下平衡。要想确定两相的组成，必须另外选定几个相律变数?如果保持相同的温度和压力，有没有某种方法能改变弹中物料的总组成(加进或移去物料)，而不影响气相和液相的组成?,2-7a.by phase rule,F=2-+N,=2 N=6-2=4 so F=4It is necessary to know 4 variables of gas or liquid components to fix the composition of system.b.When the composition of gas and liquid are fixed,F=0,therefore it is impossible to change the composition of the total contents of the bomb without affecting the composition of the gas and liquid phases.,
