复旦大学物理化学AII 14-1 Minimum Gibbs free energy principle 2015
4
Chemical equilibrium
Minimum Gibbs free energy principle
Chemical reaction system: A closed single phase system,
no non-expansion work, a reaction occurs:
when a reverse rate which we assume is considerably
smaller than the forward rates, that the reaction can roughly go almost to completion”.
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Physical Chemistry II
Criteria 1： S calculation
aA bB cC dD
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Chem. potential: B
2
Physical Chemistry II
Chemical equilibrium
Minimum Gibbs free energy principle
Chapter 12(14) Thermodynamics of chemical equilibria
§1 Minimum Gibbs free energy principle §2 Chem reaction isotherm and equilibrium constant §3 Calculating the reaction equilibrium constant
§4 Equili. constant of ideal gas from statistical thermodynamics
dD eE fF gG
The changes in the amounts of all substances must fulfill:
0 B B
BAcΒιβλιοθήκη ording to the definition of extent of reaction, one can derive:
Eq. (a) signifies that small changes occur in a finite system; Eq. (b) signifies the change in the extent of reaction by 1 mol occurred in a large system. Concentration of each substance remains basically unchanged, also the corresponding chemical potential
（ r Gm） T , p B B
B
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(b)
6
Physical Chemistry II
Chemical equilibrium
Minimum Gibbs free energy principle
Conditions of applicability of the two equations:
§5 Dependence of equilibrium constants on T and P §6 Condition analysis of gas-phase reactions §7 Condition analysis of liquid-phase and complex reactions
at constant temp. and pressure,
B B
B
（dG） T , p B dnB B B d
(dnB B d )
G ( )T , p B B B
(a)
Just when 1 mol , it follows that:
G ( )T , p , B B 或 ( r Gm )T , p B
(r Gm )T , p 0
(r Gm )T , p 0
(r Gm )T , p 0
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proceeds spontaneously to the right
poceeds spontaneously to the left, not spontaneous in the opposite direction
Chemical equilibrium
Minimum Gibbs free energy principle
Physical Chemistry
Minimum Gibbs free energy principle
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Physical Chemistry II
1
Chemical equilibrium
2015/6/1
Physical Chemistry II
7
Chemical equilibrium
Minimum Gibbs free energy principle
Extent and direction of a chemical reaction
Can be judged from the same criteria
instructiveness and utility?
2015/6/1
Physical Chemistry II
10
Chemical equilibrium
Minimum Gibbs free energy principle
Why do the reactions not always go to completion? Strictly speaking, by allowing all of the reactants and products to be well mixed in the same system, the reaction is essentially reversible and incomplete The Gibbs free energy of mixing is therefore the main factor that determines the particular situation that “only
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Physical Chemistry II
3
Chemical equilibrium
Minimum Gibbs free energy principle
§1 Minimum Gibbs free energy principle
Chemical reaction system
Fundamental equation of thermodynamics
Direction and extent of a chemical reaction
Chem. reactions do not always go to completion?
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Physical Chemistry II
2. It also doesn’t mean reactions stop when equilibrium is reached, but that the concentrations of both reactants and products remain unchanged as a result of the fact that the forward and reverse reaction rates are equal; 3. If r G m< 0 , only considered as the possibility, not the ready occurrence of the reaction at a given situation
11
Chemical equilibrium
Minimum Gibbs free energy principle
Why do the reactions not always go to completion?
The minimum free energy Principle:
Any process, at constant T and P, no non-expansion work, the chemical potential would decrease and approach a minimum value at equilibrium
(1) A chemical reaction with no non-expansion work at constant temp. and pressure:
(2) Chemical potential B for all substances remains unchanged during the reaction
d
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dnB
B
dnB Bd
Physical Chemistry II
5
Chemical equilibrium
Minimum Gibbs free energy principle
Fundamental equation of thermodynamics
dG SdT Vdp B dnB
Minimum Gibbs free energy principle
A Brief Retrospect
First Law
Third Law
Laws of thermodynamics
Criteria 2： G、 A calculation Second Law
Extent and direction of the changes
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Physical Chemistry Peter Atkins(Sixth edition)BilingualProgramPart1:Equilibrium 4.The Second Law:the conceptsBilingual ProgramThis chapter will explain the origin of the spontaneity of physical and chemical change. Two simple processes are examined.It shows that a property,the entropy can be defined,measured,and used to discuss spontaneous changes quantitatively.This chapter also introduces a major subsidiary thermodynamic property,the Gibbs energy.The direction of spontaneous change4.1The dispersal of energy4.2Entropy4.3Entropy changes accompanying specific processes4.4The Third Law of thermodynamics4.5Reaching very low temperatures Concentrating on the system4.6The Helmholtz and Gibbs energies4.7Standard molar Gibbs energiesSome things happen naturally.A gas expands to fill the available volume,a hot body cools to the temperature of its surroundings,and a chemical reaction runs in one direction rather than another. But some things don't.We can confine a gas to a smaller volume,we can cool an object with a refrigerator,and we can force some reactions to go in reverse.However,none of these processes happens spontaneously.The characteristic of these two processes, spontaneous and non-spontaneous is summarized by the Second Law of thermodynamics.4.1The dispersal of energySketch for the distribution of energy4.1The dispersal of energyThe spontaneous changesare always accompanied bya dispersal of energy into amore disordered form.Sketch for the distribution of energy(a)A ball resting on a warm surface;the atoms are under-going thermal motion;(b)For the ball to fly upwards,some of the random vibra-tional motion would have to change into coordinated,directed motion.4.1The dispersal of energySketch for the distribution of energy (a)(b)The direction of spontaneous changeThe spontaneous changes are always accompanied by a dispersal of energy into a more disordered form.The entropy of an isolated system increases in the course of a spontaneous change:In the term of entropytot S ∆>0where S t ot is the total entropy of the system and its surroundings.Thermodynamically irreversible processes are spontaneous processes,and must be accompanied by an increase in entropy.4.1The dispersal of energyThe characteristic of these two process, spontaneous and non-spontaneous is summarized by the Second Law of Thermodynamics.The Second Law may be expressed in terms of the entropy,a state function that lets us assess whether one state is accessible from another by a spontaneous change.the Second Law uses the entropy to identify the spontaneous changes among those permissible changes.The First Law led to the intro-duction of the internal energy, a state function that lets us assess whether a change is permissible;only those change may occur for which the U of an isolated system remains constant.First Law uses the internal energy to identify permissible changes;No process is possible in which the sole result is the absorption of heat from a reservoir and its complete conversion into work.The entropy of an isolated system increases in the course of a spontaneous change:tot S ∆>0The spontaneous changes are always accompanied by a dispersal of energy into a more disordered form.4.The Second Law:the concepts The direction of spontaneous change4.1The dispersal of energy4.2Entropy4.3Entropy changes accompanying specificprocesses4.4The Third Law of thermodynamics4.5Reaching very low temperatures Concentrating on the system4.6The Helmholtz and Gibbs energies4.7Standard molar Gibbs energies1).The thermodynamic definition of entropya).Heat stimulates disorderly motion in thesurroundings;work stimulates uniform motion of atoms in the surroundings,does not change the degree of disorder,and so does not change the entropy.b).A change in the extent to which energy isdispersed in a disorderly manner depends on the quantity of energy transferred as heat.1).The thermodynamic definition of entropy Tq S rev d d The thermodynamic definition of entropy is based onThe units of entropy:J K -1.The units of Molar entropy:J K -1mol –1(the same as that of R )1).The thermodynamic definition of entropy For a measurable change between two states i and f this expression integrates to:⎰=fi rev d Tq S ∆To calculate the difference in entropy between any two states of a system,integrate the heat supplied at each stage of the path divided by the temperature at which the heat is supplied.ExampleCalculate the change in entropy when 50kJ of energy is transferred reversibly and isothermally as heat to a large block of copper at(0)℃,and(b)70℃.Example 4.1-Calculating the entropy change for the isothermalexpansion of a perfect gasAnswer:⎰==f i rev rev d Tq q T S 1∆Method:123rev JK 101.8273KJ 1050Δ -⨯=⨯==T q S (a)123rev JK 101.5273)K(70J 1050Δ -⨯=+⨯==T q S (b)1).The thermodynamic definition of entropysurrev sur,sur d d T q S =The change in entropy of the surroundingssursur sur sur T qT q S -==∆Because the temperature and pressure of the surroundings is commonly constant,whatever the change takes place,reversibly or irreversibly,we havesursursur d d T q S =S sur =∆For any adiabatic change,q sur =0,soThe change of entropy of the surroundings can be calculated by dividing the heat transferred by T at which the transfer takes place(regardless of how the change is brought about in the system).IllustrationTo calculate the entropy change in the surroundings when 1.00mol H 2O(l)is formed from its elements under standard conditions at 298K ,we use =-286kJ from a table.The heat released is supplied to the surroundings,now regarded as being at constant pressure,so q sur =+286kJ .Therefore,H ∆1surJKKJ -+=⨯=∆9592981086.25S2).The entropy as a state functionThe entropy is one of the function of states,which implies that:=⎰S dA thermodynamic cycle;the overallchange in a state function is zero.Hot source,T h low source,T c4.2EntropyThe Carnot cycleCarnot cycle–four reversible stages:The structure of a Carnot cycle.1).Reversible isothermal expansion at T h,∆S1=q h/T h,q h>0.4.2Entropy-The entropy as a state function2).Reversible adiabatic expansion∆S2=0.3).Reversible isothermal compressionat T c.∆S3=q c/T c,q c<0.4).Reversible adiabatic compression∆S4=0.2).The entropy as a state functionThe total change in entropy around the closed cycle is:cc h hcc h h0d T q T q T q T q S +=+++=⎰since ch c h T T q q -=then0d =⎰S The structure of a Carnot cycle.h h T q c c T q 00⎪⎪⎭⎫⎝⎛=ABhh VVnRTq ln⎪⎪⎭⎫⎝⎛=BAcc VVnRTq lnchchTTqq-=cchhd=+=⎰TqTqSIn the limit of infinitesimal cycles,the non-canceling edges of the Carnot cycles match the overall cycle exactly;the sum becomes an integral.This result implies that d S is an exact differential and S is state function.∑∑==perimeterrevall rev 0T q T q 0d =⎰S 2).The entropy as a state function4.2Entropy4.The Second Law:the concepts The direction of spontaneous change4.1The dispersal of energy4.2Entropy4.3Entropy changes accompanying specificprocesses4.4The Third Law of thermodynamics4.5Reaching very low temperatures Concentrating on the system4.6The Helmholtz and Gibbs energies4.7Standard molar Gibbs energiesThe Second Law may be expressed in terms of the entropy,a state function that lets us assess whether one state is accessible from another by a spontaneous change.the Second Law uses the entropy to identify the spontaneous changes among those permissible changes.The First Law led to the introduction of the internal energy,a state function that lets us assess whether a change is permissible;only those change may occur for which the U of an isolated system remains constant. First Law uses the internal energy to identify permi-ssible changes;Statements on the Second Law of ThermodynamicThe thermodynamic definition of entropy Tq S rev d d =sursur revsur,sur d d d T q T q S -==For the system surtot S S S ∆+∆=∆For the surroundings For the totalThe efficiency of a heat engine,ε:The efficiency is the workdone divided by the heatsupplied from the hot source.hadsorbed heat performed work q w ε==The structure of a heat enginehc h c h q q q q q ε+=+=1Note that q c <0.The structure of a heat engineThe work performed by the engineis the difference between the heatsupplied by hot source and thatreturned to the cold sink,then:The efficiency of a heat engine,ε:For a Carnot cycle:hc rev T T ε-=1The structure of a heat engine The efficiency of a heat engine,ε:The second law of thermodynamics implies that all reversible engines have the same efficiency regardless of their construction.Suppose an engine that is working reversibly between a hot source at a temperature T h and a cold sink at a temperature T ,then3).The thermodynamic temperaturehT εT )1(-=hh c T T T T ε-=-=113).The thermodynamic temperaturehT εT )1(-=1).The zero of the thermodynamic temperature scale occurs for a Carnot efficiency of 1;2).On the Kelvin scale it is defined by setting the tempera-ture of the triple point of water as 273.16K exactly.3).It is possible to measure temperature on a purely mechanical basis.The expression of the thermodynamic temperature scale defined by Kelvin.4).The Clausius inequalitysur sur d d or 0d d S S S S -≥≥+Consider a system in thermal and mechanical contact with its surroundings at the same temperature,T .Any change of state is accompanied by a change in entropy of the system,d S ,and of the surroundings,d S sur .ThenSince d d sur qq -=d d sur T q S -=It follows that d d Tq S ≥4).The Clausius inequalityd dT qS≥For any change,the Clausius inequality is expressed as:For the isolated system:d≥SIn an isolated system the entropy of the system alone cannot decrease when a spontaneous change takes place.4).The Clausius inequality For the isolated system:d ≥S >irreversible<impossible=reversible0≥∆S4).The Clausius inequality0sur ≥∆+∆S S >spontaneous<impossible=equilibrorus4.The Second Law:the concepts The direction of spontaneous change4.1The dispersal of energy4.2Entropy4.3Entropy changes accompanying specificprocesses4.4The Third Law of thermodynamics4.5Reaching very low temperatures Concentrating on the system4.6The Helmholtz and Gibbs energies4.7Standard molar Gibbs energies1).The entropy of phase transitionsBecause a change in the degree of molecular order occurs when a substance freezes or boils,we should expect the transition to be accompanied by a change in entropy.In a reversible processIn an irreversible processƌThe entropy of phase transition at T trs ,Reversible At the normal transition temperature,T trs ,any transfer of heat between the system and its surroundings is reversible because the two phases in the system are in equilibrium.H q trs ∆=Since at constant p ,the change in molar entropy of the system is trstrs trs T H S ∆=∆H 2O(s)1mol 273K 1atmH 2O(l)1mol273K1atm reversible ΔS trs trs trs T H S ∆=∆At T trs =273KƌThe entropy of transition at T ,IrreversibleIf a process changes at which the temperature deviates the normal transition temperature,T trs ,the process is not a reversible and the above equation can not be directly used to calculate the change in entropy.In this caseAt T =263K (T ≠T trs )H 2O(l)1mol263K1atm irreversible H 2O(l)1mol 273K 1atm H 2O(s)1mol 273K 1atmreversible ΔSΔS 1ΔS 3ΔS 2H 2O(s)1mol 263K 1atm ΔS=ΔS 1+ΔS 2+ΔS 3At T =263K (T ≠T trs )H 2O(l)1mol263K1atm irreversible H 2O(l)1mol 273K 1atm H 2O(s)1mol 273K 1atmreversible ΔSΔS 1ΔS 3ΔS 2H 2O(s)1mol 263K 1atm H 2O(s)1mol273K1atm H 2O(l)1mol 273K1atm reversible ΔS trstrs trs T H S ∆=∆ΔS=ΔS 1+ΔS 2+ΔS 3At T trs =273K1).The entropy of phase transition at the T trs If the phase transition is exothermic (Δtrs H <0),the entropy change is negative,Δtrs S <0,which is consistent with the system becoming more ordered.If the transition is endothermic (Δtrs H >0),the entropy change is positive,Δtrs S >0,which is consistent with the system becoming more disordered.2).The isothermal expansion of a perfect gasThe change in entropy of a perfect gas that expands isothermally from V i to V f is⎰==f i rev rev d Tq q T S 1∆The heat absorbed during a reversible isothermal expansion of a perfect gas can be calculated from:△U =q +w and △U =0.For a reversible change q rev =-w revFor a perfect gas in the isothermal processi f rev f i d V V nRT V V nRT w V V T ln ,-=-=⎰It follows that if rev rev V V nRT w q ln =-=if V V nR S ln =∆The change in entropy of a perfect gas that expands isothermally from V i to V f isif V V nR S ln =∆Because S is a state function,this expression applies whether the change of state occurs reversibly or irreversibly .If the change is reversible,the entropy change in the surroundings must be such as to give ΔS tot =0.If the expansion occurs freely (w =0)and irreversibly,and if T remains constant,then q =0.Consequently,ΔS sur =0,and the total entropychange is given by this equation.And in this case:ΔS ≥0,spontaneous process3).The variation of entropy with temperature The entropy of a system at a temperature T f from a knowle-dge of its entropy at a temperature T i and the heat supplied to change its temperature from one value to the other:)()(i f T S T S S -=∆⎰+=f i rev i f d ))Tq T S T S ((From the definition of constant-pressure heat capacity,so long as the system is doing no non-expansion work:。
英语材基试卷
英文原版教材班“材料科学基础”考试试题试卷一Examination problems of the course of “fundament of materials science”姓名：班级：记分：1. Glossary (2 points for each)1) crystal structure:2) basis (or motif):3) packing fractor:4) slip system:5) critical size:6) homogeneous nucleation:7) coherent precipitate:8) precipitation hardening:9) diffusion coefficient:10) uphill diffusion:2. Determine the indices for the planes in the cubic unit cell shown in Figure 1. (5 points)Fig. 13. Determine the crystal structure for the following: (a) a metal with a0 =4.9489 Å, r = 1.75 Å and one atom per lattice point; (b) a metal with a0 = 0.42906 nm, r = 0.1858 nm and one atom per lattice point. (10 points)4-1. What is the characteristic of brinell hardness test, rockwell hardness test and Vickers hardness test? What are the effects of strain rate and temperature on the mechanical properties of metallic materials? (15 points)4-2. What are the effects of cold-work on metallic materials? How to eliminate those effects? And what is micro-mechanism for the eliminating cold-work effects? (15 points)5-1. Based on the Pb-Sn-Bi ternary diagram as shown in Fig. 2, try to(1)Show the vertical section of 40wt.%Sn; (4 points)(2) Describe the solidification process of the alloy 2# with very low cooling speed (includingphase and microstructure changes); (4 points)(3)Plot the isothermal section at 150o C. (7 points)Fig. 25-2. A 1mm sheet of FCC iron is used to contain N2in a heated exchanger at 1200o C. The concentration of N at one surface is 0.04 atomic percent and the concentration at the second surface is 0.005 atomic percent. At 1000 o C, if same N concentration is demanded at the second surface and the flux of N becomes to half of that at 1200o C, then what is the thickness of sheet?(15 points)6-1. Supposed that a certain liquid metal is undercooled until homogeneous nucleation occurs. (15 points)(1)How to calculate the critical radius of the nucleus required? Please give the deductionprocess.(2)For the Metal Ni, the Freezing Temperature is 1453︒C, the Latent Heat of Fusion is 2756J/cm3, and the Solid-liquid Interfacial Energy is 255⨯10-7 J/cm2. Please calculate the critical radius at 1353︒C. (Assume that the liquid Ni is not solidified.)6-2. Fig.3 is a portion of the Mg-Al phase diagram. (15 points)(1)If the solidification is too rapid, please describe the solidification process of Mg-10wt%Alalloy.(2)Please describe the equilibrium solidification process of Mg-20wt%Al alloy, and calculate theamount of each phase at 300︒C.Fig. 37-1. Figure 4 shows us the Al-Cu binary diagram and some microstructures found in a cooling process for an Al-4%Cu alloy. Please answer following questions according to this figure. (20 points)Fig. 4(1)What are precipitate, matrix and microconstituent? Please point them out in the in the figure and explain.(2)Why is need-like precipitate not good for dispersion strengthening? The typical microstructure shown in the figure is good or not? why?(3)Please tell us how to obtain the ideal microstructure shown in this figure.(4)Can dispersion strengthened materials be used at high temperature? Please give the reasons (comparing with cold working strengthening)7-2. Please answer following questions according to the time-temperature-transformation (TTT) diagram as shown in Fig. 5. (20 points)(1)What steel is this TTT diagram for? And what means P, B, and M in the figure? (2)Why dose the TTT diagram exhibi ts a ‘C’ shape?(3)Point out what microconstituent will be obtained after austenite is cooled according to the curves I, II, III and IV .(4)What is microstructural difference between the curve I and the curve II? (5)How to obtain the steel with the structure of(a) P+B(b) P+M+A (residual) (c) P+B+M+A (residual)(d) Full tempered martensiteIf you can, please draw the relative cooling curve or the flow chart of heat treatment.Fig. 5III III IV英文原版教材班“材料科学基础”考试试题答案Solution s of the course of “fundament of materials science”1. Glossary (2 points for each)1) The arrangement of the atoms in a material into a repeatable lattice.2) A group of atoms associated with a lattice.3) The fraction of space in a unit cell occupied by atoms.4) The combination of the slip plane and the slip direction.5) The minimum size that must be formed by atoms clustering together in the liquid before thesolid particle is stable and begins to grow.6) Formation of a critically sized solid from the liquid by the clustering together of a largenumber of atoms at a high undercooling (without an external interface).7) A precipitate whose crystal structure and atomic arrangement have a continuousrelationship with matrix from which precipitate is formed.8) A strengthening mechanism that relies on a sequence of solid state phase transformationsin a dispersion of ultrafine precipitates of a 2nd phase. This is same as age hardening. It is a form of dispersion strengthening.9) A temperature-dependent coefficient related to the rate at which atom, ion, or otherspecies diffusion. The DC depends on temperature, the composition and microstructure of the host material and also concentration of the diffusion species.10) A diffusion process in which species move from regions of lower concentration to that ofhigher concentration.2. Solution: A(-364), B(-340), C(346).3. Solution: (a)fcc; (b) bcc.4-1. What is the characteristic of brinell hardness test, rockwell hardness test and Vickers hardness test? What are the effects of strain rate and temperature on the mechanical properties of metallic materials? (15 points)4-2. What are the effects of cold-work on metallic materials? How to eliminate those effects? And what is micro-mechanism for the eliminating cold-work effects? (15 points)5-1. Based on the Pb-Sn-Bi ternary diagram as shown in Fig. 2, try to(1)Show the vertical section of 40wt.%Sn; (5 points)(2) Describe the solidification process of the alloy 2# with very low cooling speed (includingphase and microstructure changes); (5 points)(3)Plot the isothermal section at 150o C. (5 points)Fig. 25-2. A 1mm sheet of FCC iron is used to contain N2in a heated exchanger at 1200o C. The concentration of N at one surface is 0.04 atomic percent and the concentration at the second surface is 0.005 atomic percent. At 1000 o C, if same N concentration is demanded at the second surface and the flux of N becomes to half of that at 1200o C, then what is the thickness of sheet?(15 points)6-1. Supposed that a certain liquid metal is undercooled until homogeneous nucleation occurs. (15 points)(3)How to calculate the critical radius of the nucleus required? Please give the deductionprocess.(4)For the Metal Ni, the Freezing Temperature is 1453︒C, the Latent Heat of Fusion is 2756J/cm3, and the Solid-liquid Interfacial Energy is 255⨯10-7 J/cm2. Please calculate the critical radius at 1353︒C. (Assume that the liquid Ni is not solidified.)6-2. Fig.3 is a portion of the Mg-Al phase diagram. (15 points)(3)If the solidification is too rapid, please describe the solidification process of Mg-10wt%Alalloy.(4)Please describe the equilibrium solidification process of Mg-20wt%Al alloy, and calculate theamount of each phase at 300︒C.Fig. 37-1. Figure 4 shows us the Al-Cu binary diagram and some microstructures found in a cooling process for an Al-4%Cu alloy. Please answer following questions according to this figure. (20 points)Fig. 4(1)What are precipitate, matrix and microconstituent? Please point them out in the in the figure and explain.(2)Why is need-like precipitate not good for dispersion strengthening? The typical microstructure shown in the figure is good or not? why?(3)Please tell us how to obtain the ideal microstructure shown in this figure.(4)Can dispersion strengthened materials be used at high temperature? Please give the reasons (comparing with cold working strengthening)7-2. Please answer following questions according to the time-temperature-transformation (TTT) diagram as shown in Fig. 5. (20 points)(1)What steel is this TTT diagram for? And what means P, B, and M in the figure? (2)Why dose the TTT diagram exhibits a ‘C’ shape?(3)Point out what microconstituent will be obtained after austenite is cooled according to the curves I, II, III and IV .(4)What is microstructural difference between the curve I and the curve II? (5)How to obtain the steel with the structure of(a) P+B(b) P+M+A (residual) (c) P+B+M+A (residual)(d) Full tempered martensiteIf you can, please draw the relative cooling curve or the flow chart of heat treatment.Fig. 5III III IV英文原版教材班“材料科学基础”考试试题试卷二Examination problems of the course of “fundament of materials science”姓名：班级：记分：1. You would like to be able to physically separate different materials in a scrap recycling plant. Describe some possible methods that might be used to separate materials such as polymers, aluminum alloys, and steels from one another. (5 points)2. Plot the melting temperature of the elements in the 1A column of the periodic table versus atomic number (i.e., plot melting temperatures of Li through Cs). Discuss this relationship, based on atomic bonding and binding energy. (10 points)3.Above 882℃, titanium has a BCC crystal structure, with a = 0.332 nm. Below this temperature, titanium has a HCP structure, with a = 0.2978 nm and c = 0.4735 nm. Determine the percent volume change when BCC titanium transforms to HCP titanium. Is this a contraction or expansion? (10 points)4. The density of BCC iron is 7.882 g/cm3and the lattice parameter is 0.2866 nm whenhydrogen atoms are introduced at interstitial positions. Calculate (a) the atomic fraction of hydrogen atoms and (b) the number of unit cells required on average to contain one hydrogen atom. (15 points)5. A carburizing process is carried out on a 0.10% C steel by introducing 1.0% C at the surface at 980℃, where the iron is FCC. Calculate the carbon content at 0.01 cm, 0.05 cm, and 0.10 cm beneath the surface after 1 h. (15 points)6. The following data were collected from a standard 0.505-in.-diameter test specimen of acopper alloy (initial length (t o) = 2.0 in.):Load Gage Length Stress Strain(lb) (in.) (psi) (in/in.)0 2.00000 0 0.03,000 2.00167 15,000 0.0008356,000 2.00333 30,000 0.0016657,500 2.00417 37,500 0.0020859,000 2.0090 45,000 0.004510,500 2.040 52,500 0.0212,000 2.26 60,000 0.1312,400 2.50 (max load) 62,000 0.2511,400 3.02 (fracture) 57,000 0.51After fracture, the gage length is 3.014 in. and the diameter is 0.374 in. Plot the data and calculate (a) the 0.2% offset yield strength, (b) the tensile strength, (c) the modulus of elasticity, (d) the %Elongation, (e) the %Reduction in area, (f) the engineering stress at fracture, (g) the true stress at fracture, and (h) the modulus of resilience. (15 points)7. A 1.5-em-diameter metal bar with a 3-cm gage length is subjected to a tensile test. Thefollowing measurements are made.Change in Force (N) Gage length (cm) Diameter (cm)16,240 0.6642 1.202819,066 1.4754 1.088419,273 2.4663 0.9848Determine the strain hardening coefficient for the metal. Is the metal most likely to be FCC, BCC, or HCP? Explain.(15 points)8. Based on Hume-Rothery’s conditions, which of the following systems would be expected todisplay unlimited solid solubility? Explain. (15 points)(a) Au-Ag (b) Al-Cu (c) Al-Au (d)U-W(e) Mo-Ta (f) Nb-W (g) Mg-Zn (h) Mg-Cd英文原版教材班“材料科学基础”考试试题答案Solutions of the course of “fundament of materials science”1.Steels can be magnetically separated from the other materials; steel (or carbon-containing iron alloys) are ferromagnetic and will be attracted by magnets. Density differences could be used—polymers have a density near that of water; the specific gravity of aluminum alloys is around2.7;that of steels is between 7.5 and 8. Electrical conductivity measurements could be used—polymers are insulators, aluminum has a particularly high electrical conductivity.(5 points)2.T (o C)L i–180.7N a– 97.8K – 63.2R b– 38.9As the atomic number increases, the melting temperature decreases, (10 points)3. We can find the volume of each unit cell. Two atoms are present in both BCC and HCP titanium unit cells, so the volumes of the unit cells can be directly compared.V BCC = (0.332 nm)3 = 0.03659 nm3V HCP= (0.2978 nm)2(0.4735 nm)cos30 = 0.03637 nm3△V=x 100 =×100= -0.6%Therefore titanium contracts 0.6% during cooling. (10 points)4. (a) 7.882 g/cm3 =x = 0.0081 H atoms/cellThe total atoms per cell include 2 Fe atoms and 0.0081 H atoms. Thus:(10 points)(b) Since there is 0.0081 H/cell, then the number of cells containing H atoms is:cells = 1/0.0081 = 123.5 or 1 H in 123.5 cells (5 points)5. D = 0.23 exp[-32,900/(1.987)(1253)] = 42 × 10-8 cm2/sC x= 0.87% CC x= 0.43% CC x= 0.18% C(15 points)6. σ=FI (π/4)(0.505)2 = F/0.2ε = (l-2)/2(a) 0.2% offset yield strength = 45,000 psi(b)tensile strength = 62,000 psi(c) E = (30,000 - 0) / (0.001665 - 0) = 18 x 106 psi(d)%Elongation =(e) %Reduction in area =(f) engineering stress at fracture = 57,000 psi(g)true stress at fracture = 11,400 lb / (TC/4)(0.374)2= 103,770 psi (h) From the graph, yielding begins at about 37,500 psi. Thus:(15 points)7.Force(lb) Gage length(in.) Diameter(in.) True stress(psi) True strain(psi)16,240 3.6642 12.028 143 0.20019,066 4.4754 10.884 205 0.40019,273 5.4663 9.848 249 0.600σt=Kεt2or ln143=ln K + n ln0.2ln 249 = ln K + nln 0.6n=0.51A strain hardening coefficient of 0.51 is typical of FCC metals.(15 points)8.The Au–Ag, Mo–Ta, and Mg–Cd systems have the required radius ratio, the same crystal structures, and the same valences. Each of these might be expected to display complete solid solubility. [The Au –Ag and Mo –T a d o have isomorphous phase diagrams. In addition, the Mg–Cd alloys all solidify like isomorphous alloys; however a number of solid state phase transformations complicate the diagram.] (15 points)英文原版教材班“材料科学基础”考试试题试卷三Examination problems of the course of “fundament of materials science”姓名：班级：记分：1. You would like to be able to identify different materials without resorting to chemical analysis or lengthy testing procedures. Describe some possible testing and sorting techniques you might be able to use based on the physical properties of materials. (5 points)2. Plot the melting temperatures of elements in the 4A to 8-10 columns of the periodic table versus atomic number (i.e., plot melting temperatures of Ti through Ni, Zr through Pd, and Hf through Pt). Discuss these relationships, based on atomic bonding and binding energy, (a) as the atomic number increases in each row of the periodic table and (b) as the atomic number increases in each column of the periodic table. (10 points)3. Beryllium has a hexagonal crystal structure, with a o= 0.22858 nm and c o= 0.35842 nm. The atomic radius is 0.1143 nm, the density is 1.848 g/cm3, and the atomic weight is 9.01 g/mol. Determine (a) the number of atoms in each unit cell and (b) the packing factor in the unit cell.(10 points)4. Suppose we introduce one carbon atom for every 100 iron atoms in an interstitial position in BCC iron, giving a lattice parameter of 0.2867 nm. For the Fe-C alloy, find (a) the density and (b) the packing factor. (15 points)5. Iron containing 0.05% C is heated to 912oC in an atmosphere that produces 1.20% C at the surface and is held for 24 h. Calculate the carbon content at 0.05 cm beneath the surface if (a) the iron is BCC and (b) the iron is FCC. Explain the difference. (15 points)6. The following data were collected from a 0.4-in. diameter test specimen of poly vinyl chloride(l0 = 2.0 in):Load(lb) Gage Length(in.) Stress(psi) Strain(in/in.)0 2.00000 0 0.0300 2.00746 2,387 0.00373600 2.01496 4,773 0.00748900 2.02374 7,160 0.011871200 2.032 9,547 0.0161500 2.046 11,933 0.0231660 2.070 (max load) 13,206 0.0351600 2.094 12,729 0.0471420 2.12 (fracture) 11,297 0.06After fracture, the gage length is 2.09 in. and the diameter is 0.393 in. Plot the data and calculate (a) the 0.2% offset yield strength, (b) the tensile strength, (c) the modulus of elasticity, (d) the %Elongation, (e) the %Reduction in area, (f) the engineering stress at fracture, (g) the true stress at fracture, and (h) the modulus of resilience. (15 points)7. A titanium alloy contains a very fine dispersion of tiny Er203 particles. What will be the effectof these particles on the grain growth temperature and the size of the grains at any particular annealing temperature? Explain. (15 points)8. Suppose 1 at% of the following elements is added to copper (forming a separate alloy witheach element) without exceeding the solubility limit. Which one would be expected to give the higher strength alloy? Is any of the alloying elements expected to have unlimited solid solubility in copper?(a) Au (b) Mn (c) Sr (d) Si (e) Co (15 points)英文原版教材班“材料科学基础”考试试题答案Solutions of the course of “fundament of materials science”1.Steels can be magnetically separated from the other materials; steel (or carbon-containing iron alloys) are ferromagnetic and will be attracted by magnets. Density differences could be used—polymers have a density near that of water; the specific gravity of aluminum alloys is around2.7;that of steels is between 7.5 and 8. Electrical conductivity measurements could be used—polymers are insulators, aluminum has a particularly high electrical conductivity.(5 points)2. Ti –1668 Zr – 1852 Hf – 2227V –1900 Nb –2468 Ta – 2996Cr –1875 Mo–2610 W–3410Mn–1244 Tc –2200 Re–3180Fe –1538 Ru –2310 Os–2700Co –1495 Rh –1963 Ir –2447Ni –1453 Pd –1552 Pt –1769For each row, the melting temperature is highest when the outer “d” energy level is partly full. In Cr, there are 5 electrons in the 3d shell; in Mo, there are 5 electrons in the 4d shell; in W there are 4 electrons in the 5d shell. In each column, the melting temperature increases as the atomic number increases—the atom cores contain a larger number of tightly held electrons, making the metals more stable. (10 points)3.V= (0.22858 nm)2(0.35842 nm)cos 30 = 0.01622 nm3 = 16.22 × 10-24 cm3(a)From the density equation:1.848 g/cm3 =x = 2 atoms/cell(b)The packing factor (PF) is:PF == 0.77 (10 points)4. There is one carbon atom per 100 iron atoms, or 1 C/50 unit cells, or 1/50 C per unit cell:(a)(b)(15 points)5. t= (24 h)(3600 s/h) = 86,400 sD BCC = 0.011 exp[-20,900/(1.9871185)] = 1.54 × 10-6 cm2/sD FCC = 0.23 exp[-32,900/(1.987)(1185)] = 1.97×10-7 cm2/sBCC: = erf[0.0685] = 0.077C x= 1.11% CFCC: = erf[0.192] = 0.2139C x = 0.95% CFaster diffusion occurs in the looser packed BCC structure, leading to the higher carbon content at point “x”. (15 points)6. σ=F /(π/4)(0.4)2 = F/0.1257ε = (l-2)/2(a)0.2% offset yield strength = 11,600 psi(b) tensile strength = 12,729 psi(c) E= (7160 - 0) / (0.01187 - 0) = 603,000 psi(d)%Elongation =(e) %Reduction in area =(f) engineering stress at fracture = 11,297 psi(g)true stress at fracture = 1420 lb / (TC/4)(0.393)2= 11,706 psi (h) From the figure, yielding begins near 9550 psi. Thus:(15 points)7. These particles, by helping pin die grain boundaries, will increase the grain growth temperature and decrease the grain size. (15 points)8.The Cu-Sr alloy would be expected to be strongest (largest size difference). The Cu-Au alloy satisfies Hume-Rothery ’s conditions and might be expected to display complete solid solubility—in fact it freezes like an isomorphous series of alloys, but a number of solid state transformations occur at lower temperatures.(15 points)英文原版教材班“材料科学基础”考试试题试卷四Examination problems of the course of “fundament of materials science”姓名：班级：记分：1.Aluminum has a density of2.7 g/cm3. Suppose you would like to produce a compositematerial based on aluminum having a density of 1.5 g/cm3. Design a material that would have this density. Would introducing beads of polyethylene, with a density of 0.95 g/cm3, into the aluminum be a likely possibility? Explain. (5 points)2. (a) Aluminum foil used for storing food weighs about 0.3 g per square inch. How many atomsof aluminum are contained in this sample of foil?(b) Using the densities and atomic weights given in Appendix A, calculate and compare thenumber of atoms per cubic centimeter in (i) lead and (ii) lithium. (10 points)3. The density of potassium, which has the BCC structure and one atom per lattice point, is0.855 g/cm3. The atomic weight of potassium is 39.09 g/mol. Calculate (a) the latticeparameter; and (b) the atomic radius of potassium. (10 points)4. The density of a sample of HCP beryllium is 1.844 g/cm3 and the lattice parameters are a0=0.22858 nm and c0= 0.35842 nm. Calculate (a) the fraction of the lattice points that containvacancies and (b) the total number of vacancies in a cubic centimeter. (15 points)5. A ceramic part made of MgO is sintered successfully at 1700℃in 90 minutes. To minimizethermal stresses during the process, we plan to reduce the temperature to 1500℃. Which will limit the rate at which sintering can be done: diffusion of magnesium ions or diffusion of oxygen ions? What time will be required at the lower temperature? (15 points)6. (a) A thermosetting polymer containing glass beads is required to deflect 0.5 mm when aforce of 500 N is applied. The polymer part is 2 cm wide, 0.5 cm thick, and 10 cm long. If the flexural modulus is 6.9 GPa, determine the minimum distance between the supports. Will the polymer fracture if its flexural strength is 85 MPa? Assume that no plastic deformation occurs.(b) The flexural modulus of alumina is 45 x 106 psi and its flexural strength is 46,000 psi. Abar of alumina 0.3 in. thick, 1.0 in. wide, and 10 in. long is placed on supports 7 in. apart.Determine the amount of deflection at the moment the bar breaks, assuming that no plastic deformation occurs. (15 points)7. Based on the following observations, construct a phase diagram. Element A melts at 850°Cand element B melts at 1200°C. Element B has a maximum solubility of 5% in element A, and element A has a maximum solubility of 15% in element B. The number of degrees of freedom from the phase rule is zero when the temperature is 725°C and there is 35% B present. At room temperature 1% B is soluble in A and 7% A is soluble in B. (15 points)8.Suppose that age hardening is possible in the Al-Mg system (see Figure 10-11). (a)Recommend an artificial age-hardening heat treatment for each of the following alloys, and(b) compare the amount of the precipitate that forms from your treatment of each alloy. (i)Al-4% Mg (ii) Al-6% Mg (iii) Al-12% Mg (c) Testing of the alloys after the heat treatment reveals that little strengthening occurs as a result of the heat treatment. Which of the requirements for age hardening is likely not satisfied? (15 points)英文原版教材班“材料科学基础”考试试题答案Solutions of the course of “fundament of materials science”1. In order to produce an aluminum-matrix composite material with a density of 1.5 g/cm 3, we wouldneed to select a material having a density considerably less than 1.5 g/cm 3. While polyethylene’s density would make it a possibility, the polyethylene has a very low melting point compared to aluminum; this would make it very difficult to introduce the polyethylene into a solid aluminum matrix —processes such as casting or powder metallurgy would destroy the polyethylene .Therefore polyethylene would NOT be a likely possibility.One approach, however, might be to introduce hollow glass beads .Although ceramic glasses have densities comparable to that of aluminum, a hollow bead will have a very low density. The glass also has a high melting temperature and could be introduced into liquid aluminum for processing as a casting. (5 points)2. (a) In a one square inch sample:number ==6.69 × 1021 atoms(b) (i) In lead:= 3.3 × 1022 atoms/cm 3(ii) In lithium:= 4.63 × 1022 atoms/cm 3 (10 points)3. (a) Using Equation 3-5:0.855 g/cm 3 =a o 3 = 1.5189 × 10-22 cm 3 or a o = 5.3355 × 10-8 cm(b) From the relationship between atomic radius and lattice parameter:r == 2.3103 × 10-8cm (10 points)4. V u.c.= (0.22858 nm)2(0.35842 nm)cos30 = 0.01622 nm 3= 1.622 x 10~23 cm 3 (a) From the density equation:x = 1.9984fraction =29984.12 = 0.0008(b) number == 0.986 x 1020 vacancies/cm 3 (15 points)5. Diffusion of oxygen is the slower of the two, due to the larger ionic radius of the oxygen.D 1700= 0.000043 exp[-82,100/(1.987)(1973)] = 3.455 × 10-14 cm 2/sD1500= 0.000043 exp[-82,100/(1.987)(1773)] = 3.255 × 10-15 cm2/st1500 = D1700 t1700/D1500== 955 min = 15.9 h (15 points)6. (a) Solution:The minimum distance L between the supports can be calculated from the flexural modulus.L3 = 4w/z3δ(flexural modulus)/3FL3 = (4)(20 mm)(5 mm)3(0.5 mm)(6.9 GPA)(1000 MPa/GPa) / 500 NL3 = 69,000 mm3 or L = 41 mmThe stress acting on the bar when a deflection of 0.5 mm is obtained isσ= WL/2wh2 = (3)(500 N)(41 mm) / (2)(20 mm)(5 mm)2 = 61.5 MPaThe applied stress is less than the flexural strength of 85 MPa; the polymer is not expected to fracture.(b) Solution:The force required to break the bar isF = 2w/z2(flexural strength)/3LF= (2)(1 in)(0.3 in)2(46,000 psi / (3)(7 in.) = 394 lbThe deflection just prior to fracture is8 = FZ3/4wh3(flexural modulus)8 = (394 lb)(7 in)3/(4)(l in)(0.3 in)3(45 x 106 psi) = 0.0278 in. (15 points)7.(15 points)8. (a) The heat treatments for each alloy might be:Al-4% Mg Al-6% Mg Al-12% MgT Eutectic451°C 451°C 451°CT Solvs210°C 280°C 390°CSolutionTreat at: 210-451°C 280-451°C 390-451°CQuench Quench QuenchAge at: <210°C <280°C <390°C(b) Answers will vary depending on aging temperature selected. If all threeare aged at 200°C, as an example, the tie line goes from about 3.8 to 35% Mg:A1-4% Mg: %β = (4− 3.82)/(35 − 3.82) X 100 = 0.6%Al-6% Mg: %β = (6 − 3.82)/(35 − 3.82) X 100 = 7.1%Al-12% Mg: %β = (12 −3.82)/(35− 3.82) X 100 = 26.8%(c) Most likely, a coherent precipitate is not formed; simple dispersionstrengthening, rather than age hardening, occurs. (15 points)英文原版教材班“材料科学基础”考试试题试卷五Examination problems of the course of “fundament of materials science”姓名：班级：记分：1. You would like to design an aircraft that can be flown by human power nonstop for adistance of 30 km. What types of material properties would you recommend? What materials might be appropriate? (5 points)2. Boron has a much lower coefficient of thermal expansion than aluminum, even though bothare in the 3B column of the periodic table. Explain, based on binding energy, atomic size, and the energy well, why this difference is expected. (10 points)3. Determine the ASTM grain size number if 20 grains/square inch are observed at amagnification of 400. (10 points)4. We currently can successfully perform a carburizing heat treatment at 1200o C in 1 h. In aneffort to reduce the cost of the brick lining in our furnace, we propose to reduce the carburizing temperature to 950℃. What time will be required to give us a similar carburizing treatment? (15 points)5.The data below were obtained from a series of Charpy impact tests performed on foursteels, each having a different manganese content. Plot the data and determine (a) the transition temperature (defined by the mean of the absorbed energies in theductile and brittle regions) and (b) the transition temperature (defined as the temperature that provides 50 J absorbed energy). Plot the transition temperature versus manganese content and discuss the effect of manganese on the toughness of steel. What would be the minimum manganese allowed in the steel if a part is to be used at 0°C?Test temperature°C Impact snergy (J)0.30% Mn 0.39% Mn 1.01% Mn 1.55% Mn-100 2 5 5 15-75 2 5 7 25-50 2 12 20 45-25 10 25 40 700 30 55 75 11025 60 100 110 13550 105 125 130 14075 130 135 135 140100 130 135 135 140(15 points)。
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西安电子科技大学课后答案/bbs/viewthread.php?tid=1083fromuid=1010194《编译原理》课后习题答案/bbs/viewthread.php?tid=175fromuid=1010194《常微分方程》王高雄高等教育出版社课后答案/bbs/viewthread.php?tid=567fromuid=1010194##################【物理/光学/声学/热学/力学类--答案】#################### 理论力学第六版 (哈尔滨工业大学理论力学教研室著) 高等教育出版社课后答案/bbs/viewthread.php?tid=932fromuid=1010194理论力学第六版 (哈尔滨工业大学理论力学教研室编著) 高等教育出版社【khdaw】/bbs/viewthread.php?tid=461fromuid=1010194《热力学统计物理》汪志诚（第三版）高教出版社 (手抄版)习题答案【khdaw】 /bbs/viewthread.php?tid=84fromuid=1010194原子物理学褚圣麟版课后答案【khdaw】/bbs/viewthread.php?tid=368fromuid=1010194《物理学教程》 (马文蔚著) 高等教育出版社【khdaw】/bbs/viewthread.php?tid=2782fromuid=1010194《光学》姚启钧第三版高等教育出版社课后答案【khdaw】/bbs/viewthread.php?tid=178fromuid=1010194大学物理实验报告与部分范例陈金太厦门大学【khdaw】/bbs/viewthread.php?tid=2350fromuid=1010194梁昆淼数学物理方法第三版的课后答案/bbs/viewthread.php?tid=2600fromuid=1010194《理论力学教程》周衍柏高等教育出版社完整版课后答案【khdawlxywyl】/bbs/viewthread.php?tid=676fromuid=1010194固体物理 (黄昆版) 课后习题答案【khdaw】/bbs/viewthread.php?tid=339fromuid=1010194哈工大《理论力学》第6版 (赵诒枢尹长城沈勇著) 华中科技大学出版社课后答案/bbs/viewthread.php?tid=1033fromuid=1010194热力学统计物理汪志诚第三版高等教育出版课后答案【khdaw】 /bbs/viewthread.php?tid=289fromuid=1010194《量子力学教程》周习勋课后习题答案【khdaw】/bbs/viewthread.php?tid=388fromuid=1010194《原子物理学》杨福家版部分答案高等教育出版社【khdaw】/bbs/viewthread.php?tid=1065fromuid=1010194/bbs/viewthread.php?tid=566fromuid=1010194《固体物理教程》王矜奉山东大学出版社课后答案【khdaw】##################【化学/环境/生物/医学/制药类--答案】#################### 物理化学 (董元彦著) 科学出版社课后答案/bbs/viewthread.php?tid=412fromuid=1010194化工原理 (陈敏恒著) 化学工业出版社课后答案【khdaw】/bbs/viewthread.php?tid=704fromuid=1010194生物化学第三版 (王镜岩朱圣庚著) 高等教育出版社课后答案/bbs/viewthread.php?tid=241fromuid=1010194遗传学第三版 (朱军著) 农业大学出版社课后答案/bbs/viewthread.php?tid=39fromuid=1010194有机化学 (汪小兰著) 高等教育出版社课后答案/bbs/viewthread.php?tid=841fromuid=1010194武汉大学版《无机化学》(第三版) 上册【khdaw】/bbs/viewthread.php?tid=196fromuid=1010194有机化学 (徐寿昌著) 高教出版社课后答案/bbs/viewthread.php?tid=1752fromuid=1010194物理化学习题及答案【khdaw】/bbs/viewthread.php?tid=965fromuid=1010194有机化学第二版 (胡宏纹著) 高等教育出版社课后答案/bbs/viewthread.php?tid=41fromuid=1010194分析化学第三版武汉大学课后答案/bbs/viewthread.php?tid=199fromuid=1010194武汉大学版无机化学(第三版) 下册【khdaw】/bbs/viewthread.php?tid=200fromuid=1010194物理化学第四版 (傅献彩著) 高等教育出版社课后答案/bbs/viewthread.php?tid=3611fromuid=1010194##################【土建/机械/车辆/制造/材料类--答案】#################### 西工大机械原理配套作业题答案/bbs/viewthread.php?tid=570fromuid=1010194机械设计基础(第五版) 杨可桢程光蕴李仲生高教版课后答案/bbs/viewthread.php?tid=2316fromuid=1010194材料力学第4版（刘鸿文）答案（有附件）/bbs/viewthread.php?tid=1931fromuid=1010194材料力学课后答案/bbs/viewthread.php?tid=96fromuid=1010194材料力学 (范钦珊主编著) 高等教育出版社课后答案机械设计基础(第五版) 答案7-18章杨可桢程光蕴李仲生/bbs/viewthread.php?tid=2570fromuid=1010194《结构力学习题集》课后答案【khdaw】/bbs/viewthread.php?tid=3016fromuid=1010194电工学第六版秦曾煌高等教育出版社课后答案/bbs/viewthread.php?tid=2986fromuid=1010194机械原理学习指南（第二版） (孙恒著) 课后答案/bbs/viewthread.php?tid=569fromuid=1010194机械原理高等教育出版社课后答案【khdaw_cola】/bbs/viewthread.php?tid=664fromuid=1010194电力电子技术试题习题考题及答案题解【khdaw】/bbs/viewthread.php?tid=1169fromuid=1010194机械原理习题＋答案【khdaw_cola】/bbs/viewthread.php?tid=1210fromuid=1010194材料力学第四版 (刘鸿文著) 高等教育出版社课后答案/bbs/viewthread.php?tid=2461fromuid=1010194机械设计及答案【khdaw_cola】/bbs/viewthread.php?tid=1172fromuid=1010194材料力学(i)第四版（孙训方）高等教育出版社课后答案/bbs/viewthread.php?tid=5342fromuid=1010194##################【经济/金融/营销/管理/电子商务类--答案】####################高鸿业版西方经济学习题答案(微观.宏观)【khdaw】/bbs/viewthread.php?tid=92fromuid=1010194西方经济学(微观部分) (高鸿业著) 中国人民大学出版社课后答案 /bbs/viewthread.php?tid=2817fromuid=1010194袁卫统计学(第二版)习题答案【khdaw】/bbs/viewthread.php?tid=98fromuid=1010194曼昆《经济学原理》题目及课后答案/bbs/viewthread.php?tid=162fromuid=1010194统计学(贾俊平第二版）中国人民大学出版社课后答案/bbs/viewthread.php?tid=42fromuid=1010194运筹学教程第三版 (甘应爱胡运权等著) 清华大学出版社课后答案 /bbs/viewthread.php?tid=7016fromuid=1010194高鸿业版西方经济学习题答案（第三版）/bbs/viewthread.php?tid=1277fromuid=1010194西方经济学（宏观部分）第四版 (高鸿业著) 中国人民大学出版社课后答案 /bbs/viewthread.php?tid=7171fromuid=1010194财务管理学课后答案荆新王化成中国人民大学出版社/bbs/viewthread.php?tid=3433fromuid=1010194西方经济学课后答案 (高鸿业著) 人民大学出版社/bbs/viewthread.php?tid=6189fromuid=1010194克鲁格曼_国际经济学(第六版)的教师手册(含习题答案)【篇二：物理化学(第四版)公式】一．基本概念和基本公式1．q、w不是体系的状态函数 we???pdv ev1v21）恒外压过程 we??pe(v2?v1) 2）定压过程we??p(v2?v1) 3）理想气体定温可逆过程 we?nrtlnv1p?nrtln2 v2p12．u、h、s、f、g都是体系的状态函数，其改变值只与体系的始终态有关，与变化的途径无关。
复旦物理化学典型的复杂反应精品文档
思考
如何求最佳反应温度？
返回
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三．连续反应 (Consecutive reaction )
1–1 级连续反应 t=0 t=t
A k1 a x
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c
k1/k2较大时 cH c
k1/k2较小时 cH
cG cA
cG
cA
t
t
返回
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三．连续反应 (Consecutive reaction )
3．连续反应的温度和时间问题
A k1 E1
G
k2 E2
H
若中间产物为目标产物，尽可能使 k1，k2
温度选择
返回
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三．连续反应 (Consecutive reaction )
1．速率方程
dx dt k1x
dy dt
k1xk2y
dz dt
k2
y
积分式
对A：
ln a x
k1t
xaek1t
对G：
dy dt
k1aek1t
k2y
ddyt k2yk1aek1t
一阶常微分方程 y'+P(x) y = Q(x)
返回
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四．链反应 (Chain reaction )
链传递形式 直链传递 支链传递
复旦大学无机化学IC02
Periodic table is a reflection of the electronic structure of atoms. To understand the properties of an element it forms, we should know how electrons are arranged in atoms.
1.3.1 Ground-state Electronic Configuration
Electronic configuration of an atom is denoted by the principal quantum number (n), the azimuthal quantum number (l), the magnetic quantum number (m) and the spin quantum number (s).
Zeff = Z –σ= Z*
In order to estimate the extent of shiቤተ መጻሕፍቲ ባይዱlding, a set of
empirical rules has been proposed by Slater.
To calculate the shielding constant for an electron in an np or ns orbital:
Because the average field is taken to be centrosymmetric, the angular components of the wave functions will be the same as those of the hydrogenic orbitals. However, because the effective nuclear charge changes with distance from the nucleus, the radial wavefunctions will be markedly different from those in a hydrogenic atom but the energy levels can be simply designed as:
2015源资培训班-VASP上机练习讲解(MedeA-LAMMPS-GIBBS-MOPAC模块的使用)
GIBBS
MedeA®——蒙特卡洛计算方法
超临界CO2流体中声速和密度计算
创建分子模型 MedeA-GIBBS
通过Monte Carlo计算能够获得的性质
Calculation of first order derivative of the thermodynamic potential (i.e. pressure, molar volume, enthalpy)
•OPLS-AA •PCFF •PCFF+ •COMPASS •CFF91 & CFF93 •CVFF •AUA-4 （仅用于Gibbs）
MedeA® LAMMPS-EAM
嵌入原子法（Embedded Atom Method, EAM）力场模拟是一种 描述金属体系结构、力学性质、 热性质的有效方法。 支持计算多种金属体系的性质： 结构、缺陷的结构和能量、力学 性质及动力学性质（熔点等）。
t-mer
-357.90 1.45 8.29
MedeA®——半经验量化计算方法
UV-vis光谱预测
MedeA®——创建热固性材料模型
关键优势： •完全集成到MedeA® Flowcharts中，并且能够直 接与以下模型结合： 与MedeA® Amorphous Builder结合创建未 经处理的原料结构 与MedeA® LAMMPS结合生成动力学轨迹 与其他性质计算模块结合，使用精确的力 场来预测密度（shrinkage），力学性质和热 导率 •在交联过程中，对材料特性进行严格检验： 网状结构和凝胶点的演变 粘结应力 自动缺陷检测（避免非物质的环连锁） •研究有/没有外加溶剂的情况下，多种树脂和固 化剂的结合
MedeA Molecular Buiห้องสมุดไป่ตู้der MedeA Polymer Builder
现代化学原理复旦大学普通化学共51页文档
谢谢！
51、 天 下 之 事 常成 于困约 ，而败 于奢靡 。——陆 游 52、 生 命 不 等 于是呼 吸，生 命是活 动。——卢 梭
53、 伟 大 的 事 业，需 要决心 ，能力 ，组织 和责任 感。 ——易 卜 生 54、 唯 书 籍 不 朽。——乔 特来
现代化学原理复旦大学普通化学
36、如果我们国家的法律中只有某种 神灵， 而不是 殚精竭 虑将神 灵揉进 宪法， 总体上 来说， 法律就 会更好 。—— 马克·吐 温 37、纲纪废弃之日，便是暴政兴起之 时。— —威·皮 物特
38、若是没有公众舆论的支持，法律 是丝毫 没有力 量的。 ——菲 力普斯 39、一个判例造出另一个判例，它们 迅速累 聚，进 而变成 法律。 ——朱 尼厄斯
复旦大学车静光教授固体物理课件sec14
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* 大多数情况可以略去，晶格振动能级在10-3eV量级
http://10.107.0.68/~jgche/ 单电子近似
6
如何描写电子之间的相互作用？
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http://10.107.0.68/~jgche/ 单电子近似
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• 基本事实：原子核比电子重得多 • 绝热近似：考虑电子运动时可不考虑原子核得 运动。原子核固定在它的瞬间位置
http://10.107.0.68/~jgche/
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2023年物理化学面试常问简答题
1.一隔板将一刚性容器分为左、右两室，左室气体的压力大于右室气体的压力。
现将隔板抽去，左、右室气体的压力达成平衡。
若以所有气体作为系统，则U、Q、W 为正？为负？或为零？答：由于容器时刚性的，在不考虑存在其他功德作用下系统对环境所作的功0 ；容器又是绝热的，系统和环境之间没有能量互换，因此Q=0，根据热力学第一定律U = Q ＋W，系统的热力学能（热力学能）变化U = 0。
3.若系统经下列变化过程，则 Q、W、Q + W 和U 各量是否完全拟定？为什么？（1）使封闭系统由某一始态通过不同途径变到某一终态（2）若在绝热的条件下，使系统从某一始态变化到某一终态【答】（1）对一个物理化学过程的完整描述，涉及过程的始态、终态和过程所经历的具体途径，因此仅仅给定过程的始、终态不能完整地说明该过程。
Q、W 都是途径依赖（path-dependent）量，其数值依赖于过程的始态、终态和具体途径，只由于 Q + W =U，只要过程始、终态拟定，则U 拟定，因此 Q + W 也拟定。
（2）在已经给定始、终态的情况下，又限定过程为绝热过程，Q = 0，Q 拟定；W =U，W和U 也拟定。
4.试根据可逆过程的特性指出下列过程哪些是可逆过程？（1）在室温和大气压力（101.325 kPa）下，水蒸发为同温同压的水蒸气；（2）在 373.15 K 和大气压力（101.325 kPa）下，蒸发为同温同压的水蒸气；（3）摩擦生热；（4）用干电池使灯泡发光；（5）水在冰点时凝结成同温同压的冰；（6）在等温等压下将氮气和氧气混合。
【答】（1）不是可逆过程。
（一级）可逆相变过程应当是在可逆温度和压力下的相变过程，题设条件与之不符，室温下的可逆压力应当小于101.325 kPa，当室温为298 K 时，水的饱和蒸气压为3168 Pa。
（2）也许是可逆过程。
（3）不是可逆过程。
摩擦生热是能量耗散过程。
（4）不是可逆过程。
无穷小的电流不能将灯泡点亮。
2016复旦大学化学系721物理化学真题回忆
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