材料热力学与动力学试题2007-2008-v1

判断题：1.由亚稳相向稳定相转变不需要推动力。

X2.压力可以改变材料的结构，导致材料发生相变。

V3.对于凝聚态材料，随着压力升高，熔点提高。

V4.热力学第三定律指出：在0K时任何纯物质的熵值等于零。

X5.在高温下各种物质显示相同的比热。

V6.溶体的性质主要取决于组元间的相互作用参数。

V7.金属和合金在平衡态下都存在一定数量的空位，因此空位是热力学稳定的缺陷。

V8.固溶体中原子定向迁移的驱动力是浓度梯度。

X9.溶体中析出第二相初期，第二相一般与母相保持非共格以降低应变能。

X10.相变过程中如果稳定相的相变驱动力大于亚稳相，一定优先析出。

X1.根据理查德规则，所有纯固体物质具有大致相同的熔化熵。

2.合金的任何结构转变都可以通过应力驱动来实现。

3.在马氏体相变中，界面能和应变能构成正相变的阻力，但也是逆相变的驱动力。

4.在高温下各种纯单质固体显示相同的等容热容。

5.二元溶体的混合熵只和溶体的成分有关，与组元的种类无关。

6.材料相变形核时，过冷度越大，临界核心尺寸越大。

7.二元合金在扩散时，两组元的扩散系数总是相同。

8.焓具有能量单位，但它不是能量，也不遵守能量守恒定律；但是系统的焓变可由能量表达。

9.对于凝聚态材料，随着压力升高，熔点提高,BCC—FCC转变温度也升高。

10.由于马氏体相变属于无扩散切变过程，因此应力可以促发形核和相变。

简答题：1．一般具有同素异构转变的金属从高温冷却至低温时，其转变具有怎样的体积特征？试根据高温和低温下自由能与温度的关系解释此现象。

有一种具有同素异构转变的常用金属和一般金属所具有的普遍规律不同，请指出是那种金属？简要解释其原因？（8分）答：在一定温度下元素的焓和熵随着体积的增加而增大，因此疏排结构的焓和熵大于密排结构。

G=H-TS,低温下，TS项贡献很小，G主要取决于H。

而疏排结构的H大于密排结构，疏排结构的自由能G也大于密排结构。

所以低温下密排结构是稳定相。

高温下，G主要取决于TS项，而疏排结构的熵大于密排结构，其自由能G则小于密排结构。

The problems of the first law1. a lead bullet is fired at a frigid surface. At what speed must it travel to melt on impact, if its initial temperature is 25℃ and heating of the rigid surface of the rigid surface is neglected? The melting point of lead is 327℃. The molar heat of fusion of the lead is 4.8kJ/mol. The molar heat capacity C P of lead may be taken as 29.3J/(mol K) (1.1)Solution: )/(5.112.20721]108.4)25327(3.29[2121)(2322s m V v n n WQ nMv mv W H T C n Q Q Q absorb melting p melt increase absorb ==⨯+-⨯===∆+∆=+=2. what is the average power production in watts of a person who burns 2500 kcal of food in a day? Estimate the average additional powder production of 75Kg man who is climbing a mountain at eh rate of 20 m/min (1.2)Solution )/(24560208.975)/(12160602410467000//)(104670001868.4102500sin 3S J t h mg P S J t Q t W P J Q gincrea Burning Burning =⨯⨯=∆==⨯⨯====⨯⨯=3 One cubic decimeter (1 dm 3) of water is broken into droplets having a diameter of one micrometer (1 um) at 20℃. (1.3)(a) what is the total area of the droplets?(b) Calculate the minimum work required to produce the droplets. Assume that the droplets arerest (have zero velocity)Water have a surface tension of 72.75 dyn/cm at 20℃ (NOTES: the term surface energy (ene/cm 2) is also used for surface tension dyn/cm)Solution)(25.218)106103(1075.72)(103)101(4)101(34)101(232523263631J S W m nS S Single total =⨯-⨯⨯⨯=∆=⨯=⨯⨯⨯⨯⨯⨯⨯⨯==-+----σππ4.Gaseous helium is to be used to quench a hot piece of metal. The helium is in storage in aninsulated tank with a volume of 50 L and a temperature of 25℃, the pressure is 10 atm. Assume that helium is an ideal gas.(a) when the valve is opened and the gas escapes into the quench chamber (pressure=1 atm), whatwill be the temperature of the first gas to hit the specimen?(b) As the helium flows, the pressure in the tank drops. What will be the temperature of thehelium entering the quench chamber when the pressure in the tank has fallen to 1 atm? (1.4)Solution: )(180118298)(1185.229810101325501010101325)5500(1)()(118)101(298)()(0334.0/00K T T T K RR nC W T b K T P PT T Adiabatic a p C R P=-=∆-==⨯⨯⨯⨯⨯⨯⨯-⨯==∆=⨯==--5 An evacuated (P=0), insulted tank is surrounded by a very large volume (assume infinite volume) of an ideal gas at a temperature T 0. The valve on the tank is opened and the surrounding gas is allowed to flow quickly into the tank until the pressure inside the tank is equals the pressure outside. Assume that no heat flow takes place. What is the final tempeture of the gas in the tank? The heat capacity of the gas, C p and C v each may be assumed to be constant over the temperature rang spanned by the experiment. You answer may be left in terms of C p and C vhint: one way to approach the problem is to define the system as the gas ends up in the tank. (1.5)solution 0/000/00)0()(T P P T T P PT T Adiabatic PPC R C R ≈-==6. Calculate the heat of reaction of methane with oxygen at 298K, assuming that the products of reaction are CO 2 and CH 4 (gas)[This heat of reaction is also called the low calorific power of methane] convert the answer into unites of Btu/1000 SCF of methane. SCF means standard cubic feet, taken at 298 and 1atmNOTE: this value is a good approximation for the low calorific powder of natural gas (1.6)DA TA:)()()(224g O H g CO g CH FOR80.5705.9489.17]/[0298---•∆mol g Kcal Hsolution)1000/(9.2610252103048.01101076.191)/(76.191)89.1780.57205.94()2(22333332982982224422SCF Btu mol g Kcal H H H H H OH CO O CH CH O H CO =⨯⨯⨯⨯⨯=•=∆+⨯---=∆-∆+∆-=∆+=+-7. Methane is delivered at 298 K to a glass factory, which operates a melting furnace at 1600 K. The fuel is mixed with a quantity of air, also at 298 K, which is 10% in excess of the amount theoretically needed for complete combustion (air is approximately 21% O 2 and 79% N 2) (1.7) (a) Assuming complete combustion, what is the composition of the flue gas (the gas followingcombustion)?(b) What is the temperature of the gas, assuming no heat loss?(c) The furnace processes 2000kg of glass hourly, and its heat losses to the surroundings average400000 kJ/h. calculate the fuel consumption at STP (in m 3/h) assuming that for gas H 1600-H 298=1200KJ/KG(d) A heat exchanger is installed to transfer some of the sensible heat of the flue gas to thecombustion air. Calculate the decrease in fuel consumption if the combustion air is heated to 800KDA TA STP means T=298K, P=1atm22224O N O H CO CH for 2.82.89.117.1316)/(C mol cal C P •Solution)(210448.1125.9100076.191298)/(25.9)]87.012.72(2.843.179.1171.87.13[01.0)(%87.0%%12.72%%43.17%2%%71.8)11.1(221791.1231%22)(0,,222222224K T T T C mol cal X C C b O N CO O H CO O H CO O CH a i i p p p =⨯⨯+=∆+=•=+⨯+⨯+⨯=======-⨯+⨯⨯+=+=+∑)/(1644)0224.011868.448.11)8001600(48.1125.9189570(102800000)/(189570)298800)](48.1187.8)48.1125.9[(100076.191)()/(87.848.11/]211002.22.816[)()/(3214)0224.011868.448.11)2981600(48.1125.9100076.191(102800000)/(280000040000020001200)(33min ,,,,298,,33min h m V mol g cal dTn C n C H H C mol cal X C C d h m V h KJ P C gConsu i i r p i i p p i i p r p g Consu =⨯⨯-⨯-⨯=•=-⨯-⨯-⨯=--∆=∆•=⨯⨯+===⨯⨯-⨯-⨯⨯==+⨯=⎰∑∑∑8.In an investigation of the thermodynamic properties of a-manganese, the following heat contents were determined:H 700-H 298=12113 J/(g atom) H 1000-H 298=22803 J/(g atom)Find a suitable equation for H T -H 298 and also for C P as a function of temperature in the form (a+bT) Assume that no structure transformation takes place in the given tempeture rang. (1.8)Solution )298(0055.0)298(62.35011.062.35011.062.3522803)2981000(2)2981000(12113)298700(2)298700(]2[2229822222982---=∆-=-===-+-=-+-+=+==∆⎰⎰T T H TC b a ba ba T baT bTdT a dT C H TP T P9.A fuel gas containing 40% CO, 10% CO 2, and the rest N 2 (by volume) is burnt completely with air in a furnace. The incoming and ongoing temperatures of the gases in the furnace are 773K and 1250K,respectively. Calculate (a) the maximum flame temperature and (b) heat supplied to the furnace per cu. ft of exhaust gas (1.9)molJ Hmol J H CO f CO f /393296/1104580,298,0,298,2-=∆-=∆)/(10184.403.29)/(1067.11010.492.19)/(1037.81020.935.44)/(1042.01097.345.283,253,253,253,222molK J T C molK J T T C molK J T T C molK J T T C N P O P CO P CO P -------⨯+=⨯-⨯+=⨯-⨯+=⨯-⨯+=Solution?0)499.0321.018.1()1067.01019.277.28(28.282831067.01038.477.289.0)1019.01058.528.33(2.0282838)()/(1019.01058.528.33722.0278.0)/(1067.01038.477.281.065.005.02.0)()/(282838110458393296%2.72%8.27%10%65%5%20)4/(1122298127332981523733253253298,,,,298,253,,,,,253,,,,,,,0,298,0,298,298,22222222222222==+--⨯+⨯++⨯=⨯-⨯++⨯⨯-⨯+-⨯=--∆=∆⨯-⨯+=+==⨯-⨯+=+++===-=∆-∆=∆========+-----------⎰⎰⎰∑∑⎰∑∑∑∑T T T T T T T dTT T dTT T dT n C n C n H H molK J T T C C n C C molK J T T C C C C n C C a mol J n H n H H N CO production O N CO CO reation then O N air mole need fuel mole when CO O CO T TT i i r p i i p p i i N P CO P i i p p r p O P N P CO P CO P i i p p r p i p f i r f idTT T Q dT T T Q b T T T T T T T dT T T dTT T dT n C n C n H H T TT i i r p i i p p i i 9.0)1019.01058.528.33(2.02828389.0)1019.01058.528.33(2.0282838)(0)499.0321.018.1()1067.01019.277.28(28.282831067.01038.477.289.0)1019.01058.528.33(2.0282838)(253125029812502982531250298125029829812125029815231250253253298,,,,298,⨯⨯-⨯++⨯-=⨯⨯-⨯++⨯-===+--⨯+⨯++⨯=⨯-⨯++⨯⨯-⨯+-⨯=--∆=∆-----------⎰⎰⎰⎰⎰∑∑⎰10. (a) for the reaction 2221CO O CO →+,what is the enthalpy of reaction (0H ∆) at 298 K ? (b) a fuel gas, with composition 50% CO, 50% N 2 is burned using the stoichiometric amount of air. What is the composition of the flue gas?(c) If the fuel gas and the air enter there burner at 298 K, what is the highest temperature theflame may attain (adiabatic flame temperature)? DA TA :standard heats of formation f H ∆ at 298 K (1.10))/(393000)/(1100002mol J CO mol J CO -=-=Heat capacities [J/(mol K)] to be used for this problem N 2=33, O 2=33, CO=34, CO 2=57 Solution)(21100)298)(39889.0(222.02830000)/(3975.03325.057)/(33111.034222.033666.033)(%,75%%,251.111002.22%%1.11%%,6.66%%,2.222.0/25.015.0%)()/(283000393000110000)(,0,,,,,,22220,298,0,298,0K T T dT C n H H K mol J X C C K mol J X C C C N CO product O N CO fuel b mol J n H n H H a P p p i P r i P r i P p i P p i P f i r f ==-⨯-⨯=-∆=∆•=⨯+⨯==•=⨯+⨯+⨯====-====+==+-=∆-∆=∆⎰∑∑∑∑11.a particular blast furnace gas has the following composition by (volume): N 2=60%, H 2=4, CO=12%, CO 2=24%(a) if the gas at 298K is burned with the stochiometric amount of dry air at 298 K, what is the composition of the flue gas? What is the adiabatic flame temperature? (b) repeat the calculation for 30% excess combustion air at 298K(C)what is the adiabatic flame temperature when the blast furnace gas is preheated to 700K (the dry air is at 298K)(d) suppose the combustion air is not dry ( has partial pressure of water 15 mm Hg and a total pressure of 760 mm Hg) how will the flame temperature be affected? DA TA(k J/mol) （1.11）2CO CO FOR513.393523.110)/(--∆mol kJ H f 2222,)(O N g O H CO CO FOR34505733]/[K mol J C P •Solution)(1052)(75438286370])295.03450(241604[026.0])335.03457(110523393513[079.0])([%8.66%%,8.6%%,6.2%%,8.15%%,9.72.0/83.110012%)()(1122)(82538313430])295.03450(241604[029.0])335.03457(110523393513[086.0])([%7.65%%,7.5%%,9.2%%,1.17%%,6.82.0/810012%2121)(,,,,,,,02222,,,,,,,0222222222K T K T T n C T T X C dT n C n C H x H N O H CO CO b K T K T T n C T T X C dT n C n C H x H N O H CO CO OH O H CO O CO a i i r P ii P i i r P i i p P i i i i r P ii P i i r P i i p P i i ===∆=∆-∆-⨯--+∆-⨯---=+--∆=∆=====⨯+====∆=∆-∆-⨯--+∆-⨯---=+--∆=∆=====+=→+→+∑∑∑⎰∑∑∑∑∑⎰∑∑)(1419),(11213842594034286.0)402(2.39714.0])295.03450(241604[029.0])335.03457(110523393513[086.0)3(K T K T T T T T H ===∆=∆⨯--∆⨯-∆-⨯--+∆-⨯---=∆12．A bath of molten copper is super cooled to 5℃ below its true melting point. Nucleation of solid copper then takes place, and the solidification proceeds under adiabatic conditions. What percentage of the bath solidifies?DA TA: Heat of fusion for copper is 3100 cal/mol at 1803℃(the melting point of copper) C P,L =7.5(cal/mol ℃), C P,S =5.41+(1.5*10-3T )(cal/mol ℃) （1.12） Solution)/(310355.75.0)17981803(105.1541.5310002231798,1798,17981803,18031798,1803,mol cal H H dT C dT C H L S SL L P S P L S =⨯-⨯-⨯+⨯+==+++-⎰⎰13．Cuprous oxide (Cu 2O) is being reduced by hydrogen in a furnace at 1000K, (a)write the chemical reaction for the reduced one mole of Cu 2O(b)how much heat is release or absorbed per mole reacted? Given the quantity of heat and state whether heat is evolved (exothermic reaction) or absorbed (endothermic reaction)DA TA: heat of formation of 1000K in cal/mol Cu 2O=-41900 H 2O=-59210 （1.13） solution)/(173104190059210222mol cal H OH Cu H O Cu =-=∆+=+,exothermic reaction14. (a) what is the enthalpy of pure, liquid aluminum at 1000K?(b) an electric resistance furnace is used to melt pure aluminum at the rate of 100kg/h. the furnace is fed with solid aluminum at 298K. The liquid aluminum leaves the furnace at 1000K. what is the minimum electric powder rating (kW) of furnace.DA TA : For aluminum : atomic weight=27g/mol, C p,s =26(J/molK), C p,L =29(J/molK), Melting point=932K, Heat of fusion=10700J/mol (1.14)Solution )(28.0)(7.2793600110002727184)/(2718410700)9321000(29)298932(261000932,932298,1000,kW W P mol J H dT C dT C H SLL P S P l ==⨯⨯==+-⨯+-⨯=++=⎰⎰15 A waste material (dross from the melting of aluminum) is found to contain 1 wt% metallic aluminum. The rest may be assumed to aluminum oxide. The aluminum is finely divided and dispersed in the aluminum oxide; that is the two material are thermally connected.If the waster material is stored at 298K. what is the maximum temperature to which it may rise if all the metallic aluminum is oxidized by air/ the entire mass may be assumed to rise to the same temperature. Data : atomic weight Al=27g/mol, O=16g/mol, C p,s,Al =26(J/molK), C p,s,Al2O3=104J/mol, heat formation of Al 2O 3=-1676000J/mol (1.15)Solution;)(600)(3021041029927275.116122711676000K T K T T ==∆∆⨯⨯++⨯⨯=⨯⨯16 Metals exhibit some interesting properties when they are rapidly solidified from the liquid state. An apparatus for the rapid solidification of copper is cooled by water. In the apparatus, liquid copper at its melting point (1356K) is sprayed on a cooling surface, where it solidified and cools to 400K. The copper is supplied to the apparatus at the rate of one kilogram per minute. Cooling water is available at 20℃, and is not allowed to raise above 80℃. What is the minimum flow rate of water in the apparatus, in cubic meters per minute?DA TA; for water: C p =4.184J/g k, Density=1g/cm 3; for copper: molecular weight=63.54g/mol C p =7cal/mol k, heat of fusion=3120 cal/mol (1.16)Solution:min)/(10573.2)2080(1min /min54.631000)]4001356(73120[min /33m V VQ Q Water Copper -⨯=-=⨯⨯-⨯+=17 water flowing through an insulated pipe at the rate of 5L/min is to be heated from 20℃ to 60℃ b an electrical resistance heater. Calculate the minimum power rating of the resistance heater in watts. Specify the system and basis for you calculation. DA TA; For water C p =4.184J/g k, Density=1g/cm 3 (1.17)Solution: )(139476010005)2060(184.4W W =⨯⨯-⨯=18 The heat of evaporation of water at 100℃ and 1 atm is 2261J/mol (a) what percentage of that energy is used as work done by the vapor?(b)if the density of water vapor at 100℃ and 1 atm is 0.597kg/m 3 what is the internal energy change for the evaporation of water? (1.18)Solution: )/(375971822613101%6.71822613101%)/(31010224.0273373101325mol J Q W U mol J V P =⨯+-=+=∆=⨯==⨯⨯=∆19 water is the minimum amount of steam (at 100℃ and 1 atm pressure) required to melt a kilogram of ice (at 0℃)? Use data for problem 1.20 (1.19) Solution )(125,3341000)10018.42261(g m m =⨯=⨯+20 in certain parts of the world pressurized water from beneath the surface of the earth is available as a source of thermal energy. To make steam, the geothermal water at 180℃is passed through a flash evaporator that operates at 1atm pressure. Two streams come out of the evaporator, liquid water and water vapor. How much water vapor is formed per kilogram of geothermal water? Is the process reversible? Assume that water is incompressible. The vapor pressure of water at 180℃is1.0021 Mpa( about 10 atm) Data: C P,L=4.18J/(g k), C P,v=2.00J/(g k), △H V=2261J/g, △H m=334 J/g (1.20)Solution:leirreversibgxxx)(138),1000(8018.4)8018.48022261(=-⨯⨯=⨯-⨯+。

一、单选题1、下列过程中，系统内能变化不为零的是（）A.两种理想气体的等温混合过程B.可逆循环过程C. 纯液体的真空蒸发过程D.不可逆循环过程正确答案：C2、在实际气体的节流膨胀过程中，哪一组描述是正确的（）A.Q ＝0, DH ＝0, Dp <0B.Q <0, DH ＝0, Dp <0C.Q >0, DH ＝0, Dp < 0D.Q ＝0, DH <0, Dp >0正确答案：A3、关于热平衡，下列说法中正确的是（）A.并不是所有热力学平衡系统都必须满足热平衡的条件B.系统处于热平衡时，系统的温度一定等于环境的温度C.在等温过程中系统始终处于热平衡D. 若系统A与B成热平衡，B与C成热平衡，则A与C直接接触时也一定成热平衡正确答案：D4、将1 mol 373 K，标准压力下的水，分别经历：(1) 等温等压可逆蒸发，(2) 真空蒸发，变成373 K，标准压力下的水气。

这两种过程的功和热的关系为：A. W1>W2 Q1<Q2B. W1=W2 Q1=Q2C.W1<W2 Q1>Q2D.W1<W2 Q1<Q2正确答案：C5、对有分子间相互作用的实际气体绝热自由膨胀过程，描述错误的是（）A.一定是热力学能不变的过程B.不一定是温度降低的过程C.一定是温度降低的过程D.一定是体积增大的过程正确答案：B6、下面的说法符合热力学第一定律的是（）A.气体在绝热膨胀或绝热压缩过程中, 其内能的变化值与过程完成的方式无关B.在无功过程中, 内能变化等于过程热, 这表明内能增量不一定与热力学过程无关C.封闭系统在指定的两个平衡态之间经历绝热变化时, 系统所做的功与途径无关D.在一完全绝热且边界为刚性的密闭容器中发生化学反应时,其内能一定变化正确答案：C7、下列过程中, 系统内能变化不为零的是（）A.两种理想气体的混合过程B.纯液体的真空蒸发过程C.可逆循环过程D.不可逆循环过程正确答案：B8、关于焓的性质, 下列说法中正确的是（）A. 焓是能量, 它遵守热力学第一定律B.焓是系统内含的热能, 所以常称它为热焓C. 系统的焓值等于内能加体积功D.焓的增量只与系统的始末态有关正确答案：D9、下列哪个封闭体系的内能和焓仅是温度的函数？（）A.理想气体B.理想溶液C.所有气体D.稀溶液正确答案：A10、关于节流膨胀, 下列说法正确的是（）A.节流膨胀中系统的焓值改变B.节流过程中多孔塞两边的压力不断变化C.节流膨胀中系统的内能变化D.节流膨胀是绝热可逆过程正确答案：C11、关于热力学可逆过程,下面的说法中不正确的是（）A.在等温可逆过程中,系统做功时,系统损失的能量最小B.可逆过程中的任何一个中间态都可从正逆两个方向到达C.在等温可逆过程中,环境做功时,系统得到的功最小D.可逆过程不一定是循环过程正确答案：A12、一定量的理想气体，从同一初态分别经历等温可逆膨胀、绝热可逆膨胀到具有相同压力的终态，终态体积分别为V1、V2。

2007年高考物理试题分类汇编-热学全国卷Ⅰ如图所示，质量为m的活塞将一定质量的气体封闭在气缸内，活塞与气缸之间无摩擦。

a态是气缸放在冰水混合物中气体达到的平衡状态，b态是气缸从容器中移出后，在室温（270C）中达到的平衡状态。

气体从a态变化到b态的过程中大气压强保持不变。

若忽略气体分子之间的势能，下列说法正确的是（）A、与b态相比，a态的气体分子在单位时间内撞击活塞的个数较多B、与a态相比，b态的气体分子在单位时间内对活塞的冲量较大C、在相同时间内，a、b两态的气体分子对活塞的冲量相等D、从a态到b态，气体的内能增加，外界对气体做功，气体对外界释放了热量全国卷Ⅱ对一定量的气体，下列说法正确的是A、在体积缓慢地不断增大的过程中，气体一定对外界做功B、在压强不断增大的过程中，外界对气体一定做功C、在体积不断被压缩的过程中，内能一定增加D、在与外界没有发生热量交换的过程中，内能一定不变北京卷为研究影响家用保温瓶保温效果的因素，某同学在保温瓶中灌入热水，现测量初始水温，经过一段时间后再测量末态水温。

改变实验条件，先后共做了6次实验，实验数据记录如下表：A、若研究瓶内水量与保温效果的关系，可用第1、3、5次实验数据B、若研究瓶内水量与保温效果的关系，可用第2、4、6次实验数据C、若研究初始水温与保温效果的关系，可用第1、2、3次实验数据D、若研究保温时间与保温效果的关系，可用第4、5、6次实验数据四川卷如图所示，厚壁容器的一端通过胶塞插进一只灵敏温度计和一根气针，另一端有个用卡子卡住的可移动胶塞。

用打气筒慢慢向筒内打气，使容器内的压强增加到一定程度，这时读出温度计示数。

打开卡子，胶塞冲出容器后A．温度计示数变大，实验表明气体对外界做功，内能减少B ．温度计示数变大，实验表明外界对气体做功，内能增加C ．温度计示数变小，实验表明气体对外界做功，内能减少D ．温度计示数变小，实验表明外界对气体做功，内能增加上海卷如图所示，一定质量的空气被水银封闭在静置于竖直平面的U 型玻璃管内，右管上端开口且足够长，右管内水银面比左管内水银面高h ，能使h 变大的原因是（）（A ）环境温度升高。

年 秋 季学期研究生课程考试试题考 试 科 目：材料热力学与动力学 学生所在院（系）：材料学院、航天学院 学生所在学科：材料学、材料加工工程 (* 题签与答题纸一起上交)一、仔细阅读下列论述，判断正误，如果错误，请说明该论述违反了哪些热力学原理，并给出正确的论述。

（18分）1．材料（封闭系统）在T=T 0温度发生二级相变，（1）在相变温度T 0，高温相的体积总是比低温相大（2）在相变温度T 0，高温相的熵比低温相大（3）在相变温度T 0，高温相的热容与低温相相同（4）在相变温度T 0，高温相的Gibbs 自由能比低温相小2. 合金中每一组元的化学位相等。

3．封闭体系中出现耗散结构。

二、（1）已知某一相的Gibbs 自由能表达式为： ，请导出该相的焓(H)、熵（S ）和定压热容（Cp ）的表达式。

（6分）（2）请画出以G 为纵轴、T 为横轴的固态纯组元的G-T 曲线的示意图。

(4分)三、 在相同温度和压力下，与金刚石（diamond ）相比，碳的另一种同素异构体石墨（graphite ）的密度低、熵值（S ）高。

（1）请在P-T 相图上，示意画出石墨和金刚石的相界，并说明理由。

（6分）。

（2）并请解释为什么高压下石墨有可能转变为金刚石。

（4分）。

四、简答题：(1) 请说明晶界偏析的平行线法则。

（5分）(2) 简述Calphad 的三要素及其主要功能。

（5分）（3）请解释Onsager 倒易关系、最小熵产生原理。

（5分）五、A-B 二元相图如下图所示，(1) 判断A-B 固溶体α的性质、溶体组元间的相互作用能。

（6分）(2) 假设A-B 两组元形成正规溶体，请推导出溶体中A 组元的活度与成分的关系。

（6分）学院学号 姓名 ln n n G a bT cT T d T =+++∑T六、若A-B 二元系中存在正规固溶体相α，还存在化合物中间相θ，其化学式为AmBn ，其平衡相图如下图所示，请证明在温度T 下，当固溶体α为稀溶体时，θ相在α固溶体中的溶解度 为：其中， 为化合物的形成自由能。

可编辑修改精选全文完整版一、单选题1.一卡诺热机在两个不同温度的热源之间运转，当工作物质为气体时，热机效率为42%，若改用液体工作物质，则其效率应当（）A.减少B.无法判断C.增加D.不变正确答案：D2.求任一不可逆绝热过程的熵变dS，可通过以下哪个途径求得？（）A.始终态相同的可逆恒温过程。

B.始终态相同的可逆绝热过程。

C.始终态相同的可逆非绝热过程。

D.B和C均可。

正确答案：C3.某非理想气体服从状态方程pV=nRT+bp(b为大于零的常数)，1mol 该气体经历等温过程体积从V1变成V2，则熵变ΔS等于（）A.Rln(V1-b)/(V2-b)B.Rln(V2/V1)C.Rln(V1/V2)D.Rln(V2-b)/(V1-b)正确答案：D4.封闭体系在不可逆循环中，热温商之和Σ(δQ/T)（）A.大于零B.等于零C.小于零D.不可能小于零正确答案：C5.1mol的单原子理想气体被装在带有活塞的气缸中，温度300K，压力为1013250Pa。

压力突然降至202650Pa，并且气体在202650Pa的恒定压下做绝热膨胀，则该过程的ΔS是（）A.ΔS≥0B.ΔS＜0C.ΔS＝0D.ΔS＞0正确答案：D6.一封闭体系进行可逆循环，其热温商之和（）A.总是负值B.总是正值C.是温度的函数D.总是为零正确答案：D7.在隔离体系中发生一个有一定速度的变化，则体系的熵值（）A.总是减少B.可任意变化C.保持不变D.总是增大正确答案：D8.下列过程中系统的熵增加的是（）A.NaCl于水中结晶B.金属工件的渗碳过程C.H2(g)+1/2O2(g)® H2O(g)D.将HCl气体溶于水生成盐正确答案：B9.等温混合过程1molO2(p,V) + 1molN2(p,V) ─→O2+N2(p,2V)的熵变为（）A.不确定B.ΔS＞0C.ΔS＝0D.ΔS＜0正确答案：B10.理想气体经节流膨胀后（）A.ΔS＞0B.ΔS＝0C.ΔS变化不定D.ΔS＜0正确答案：A11.熵是混乱度（热力学状态数或热力学几率）的量度，下列结论不正确的是（）A.同一种物质Sm(g)＞Sm(l)＞Sm(s)B.同种物质温度越高熵值越大C.0K时任何物质的熵值均等于零D.分子内含原子数越多熵值一般越大，分解反应微粒数增多，熵值一般增大正确答案：C12.纯液体在正常凝固点时凝固，下述（）量减少。

材料热力学与动力学复习题答案一、常压时纯Al 的密度为ρ=2.7g/cm 3，熔点T m =660.28℃，熔化时体积增加5%。

用理查得规则和克-克方程估计一下，当压力增加1Gpa 时其熔点大约是多少？ 解：由理查德规则RTm Hm R Tm Hm Sm ≈∆⇒≈∆=∆ …①由克-克方程VT H dT dP ∆∆=…② 温度变化对ΔH m 影响较小，可以忽略，①代入②得 V T H dT dP ∆∆=dT T1V Tm R dp V T Tm R ∆≈⇒∆≈…③ 对③积分 dT T1V T Tm R p d T Tm Tm pp p ⎰⎰∆+∆+∆= 整理 ⎪⎭⎫ ⎝⎛∆+∆=∆Tm T 1ln V Tm R p V T R V Tm R Tm T ∆∆=∆⨯∆≈ Al 的摩尔体积 V m =m/ρ=10cm 3=1×10-5m 3Al 体积增加 ΔV=5%V m =0.05×10-5m 3K 14.60314.810510R V p T 79=⨯⨯=∆∆=∆- Tm’=Tm+T ∆=660.28+273.15+60.14=993.57K二、热力学平衡包含哪些内容，如何判断热力学平衡。

内容：（1）热平衡，体系的各部分温度相等；（2）质平衡：体系与环境所含有的质量不变；（3）力平衡：体系各部分所受的力平衡，即在不考虑重力的前提下，体系内部各处所受的压力相等；（4）化学平衡：体系的组成不随时间而改变。

热力学平衡的判据：（1）熵判据：由熵的定义知dS Q T δ≥不可逆可逆对于孤立体系，有0Q =δ，因此有dS 可逆不可逆0≥，由于可逆过程由无限多个平衡态组成，因此对于孤立体系有dS 可逆不可逆0≥，对于封闭体系，可将体系和环境一并作为整个孤立体系来考虑熵的变化，即平衡自发环境体系总0S S S ≥∆+∆=∆ （2）自由能判据 若当体系不作非体积功时，在等温等容下，有()0d ,≤V T F 平衡状态自发过程上式表明，体系在等温等容不作非体积功时，任其自然，自发变化总是向自由能减小的方向进行，直至自由能减小到最低值，体系达到平衡为止。

热力学第二定律练习（化学2007级）2009-4-9一、选择题1. p $，100℃下，1mol H 2O(l)与 100℃大热源接触，使水在真空容器中汽化为 101.325 kPa 的H 2O(g)，设此过程的功为W ，吸热为Q ，终态压力为p ，体积为V ，用它们分别表示ΔU ，ΔH ，ΔS ，ΔG ，ΔF ，下列答案哪个是正确的？( )2. 1mol 理想气体从p 1,V 1,T 1分别经： (1) 绝热可逆膨胀到p 2,V 2,T 2(2) 绝热恒外压下膨胀到/2p ,/2V ,/2T 若p 2=/2p ,则：( )(A) /2T =T 2, /2V =V 2, /2S =S 2 (B) /2T >T 2, /2V <V 2, /2S <S 2 (C) /2T >T 2, /2V >V 2, /2S >S 2 (D) /2T <T 2, /2V <V 2, /2S <S 23. 在等温等压下进行下列相变：H2O (s,-10℃, p$) = H2O (l,-10℃, p$) 在未指明是可逆还是不可逆的情况下，考虑下列各式哪些是适用的?( )(1) δQ/T= Δfus S(2) Q= Δfus H(3) Δfus H/T= ΔfusS(4) －ΔfusG = 最大净功(A) (1)，(2)(B) (2)，(3)(C) (4)(D) (2)4. 理想气体从状态 I 经自由膨胀到状态 II，可用哪个热力学判据来判断该过程的自发性？( )(A) ΔH (B) ΔG(C) ΔS (D) ΔU5. 单原子分子理想气体的C V, m =(3/2)R,温度由T1变到T2时，等压过程体系的熵变ΔS p与等容过程熵变ΔS V之比是：( )(A) 1 : 1 (B) 2 : 1(C) 3 : 5 (D) 5 : 36. 1 mol理想气体向真空膨胀,若其体积增加到原来的10倍,则体系、环境和孤立体系的熵变应分别为： ( )(A) 19.14 J·K-1, -19.14 J·K-1 , 0(B) -19.14 J·K-1, 19.14 J·K-1 , 0(C) 19.14 J·K-1, 0 , 19.14 J·K-1(D) 0 , 0 , 07. 在101.3 kPa下,110℃的水变为110℃水蒸气,吸热Q p,在该相变过程中下列哪个关系式不成立?()(A) S体> 0(B) S环不确定(C) S体+S环> 0(D) S环< 08. 在一容器(是由体积相等的两部分构成)内,置入 1 mol理想气体,这 1 mol理想气体分子全部处在一方的数学概率等于：( )(A) (1/2)L/[L] (B) 2L/[L](C) (1/2)L/[L] (D) 2L/[L]9. 某气体状态方程为p =f(V)T，f(V) 仅表示体积的函数，恒温下该气体的熵随体积V的增加而：( )(A) 增加 (B) 下降(C) 不变 (D) 难以确定10. 恒温恒压条件下，某化学反应若在电池中可逆进行时吸热，据此可以判断下列热力学量中何者一定大于零？( )(A) ΔU(B) ΔH(C) ΔS (D) ΔG11. 2 mol H2和 2 mol Cl2在绝热钢筒内反应生成 HCl 气体，起始时为常温常压。

Gibbs-Thomson effect:1.The Gibbs–Thomson Effect, in common physics usage, refers to variations in vaporpressure or chemical potential across a curved surface or interface. The existence of a positive interfacial energy will increase the energy required to form small particles with high curvature, and these particles will exhibit an increased vapor pressure.See Ostwald–Freundlich equation.2.More specifically, the Gibbs–Thomson effect refers to the observation that small crystalsare in equilibrium with their liquid melt at a lower temperature than large crystals. In cases of confined geometry, such as liquids contained within porous media, this leads to a depression in the freezing point / melting point that is inversely proportional to the pore size, as given by the Gibbs–Thomson equation.Why at a relatively lower temperature solute transport tends to become more effective via grain boundary than through the lattice or through dislocation? Please have an example material to clear.（为什么在一个相对较低的温度，溶质趋向于通过晶界的运输比晶格和位错运输更有效？请举一种材料作为例子详述）答：在多晶体中的扩散除了再晶粒点阵内部进行之外，还会沿表面、晶界、位错等缺陷部位进行。

1、At 300K, 1 mole ideal gas expands from p =10⨯pΘ to p= pΘ isothermally and reversibly calculate (1) Calculate the q, w, ∆H m, ∆U m, ∆G m, ∆F m and ∆S m; (2) If the gas expands isothermally to a vacuum until the pressure reaches p= pΘ, calculate q, w, ∆H m, ∆U m, ∆G m, ∆F m and ∆S m.2. Calculate the equilibrium vapour pressure (atm) of sodium for an aluminum melt containing 0.005 mol% sodium(Na). The activity coefficient of sodium in aluminum is 320 and the vapor pressure of pure sodium at 750 °C is 0.23 atm.3、At 413.15K，the vapor pressure of pure C6H5Cl and C6H5Br are 125.238kPa and 66.104kPa. Given that the two pure liquids are mixed and form ideal solution. If a solution formed by the two pure liquids boils at 413.15K、101.325kPa, please calculate the composition of the solution and the vapor above it.4、Given that when a specie A in a binary solution, its vapor pressure varies with its concentration in the pattern illustrated below. Make a table to indicate the activity, activity coefficient and chemical potential of A in different concentration sectionsI 、II and III，using its pure substance as standard state.III III5、At 300K, the vapor pressure of liquid A and liquid B are 37.33kPa and 22.66kPa.When 2 moles of A and 2 moles of B are mixed to form a solution, the vapor pressure above the solution is 50.66kPa, and the molar fraction of A in the vapor is 0.60. Given that vapors can be taken as ideal gases. ①Calculate a A( R )and a B( R) in the solution, ②γA and γB , ③∆mix G , ④ If the solution is an ideal solution, what is the value of ∆mix G id ? ⑤ What is the value of ∆mix G ex of this solution?6、The variation, with composition, of G E for Fe-Mn alloys at 1863K is listedbelow:X Mn 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 G m E ,Joules 395703 925 1054 1100 1054925 703 395a 、Is the process to form Fe-Mn alloy at 1863K an exothermic one or an endothermic one ?b. Does the system exhibit regular solution behavior?c. Calculate E Feμ and EMn μ at X Mn = 0.6; d. Calculate m mix G ∆ atX Mn = 0.4; e. Calculate the partial pressures of Mn and Fe exerted by the alloy of X Mn = 0.27、Melts in the system Pb-Sn exhibit regular solution behavior. At 473︒C, a Pb =0.055 in a liquid solution of X Pb = 0.1. Calculate the value of PbSn ωfor the system and calculate the activity of Sn in the liquid solution of X Sn = 0.5 at 500︒C.8、With respect to the Ellingham diagram, answer the following questions:93.23ln 27.145390)(ln *+--=T Tatm pFe68.37ln 02.333440)(ln *+--=T Tatm p Mna) Explain the slope changes for the reaction 2Mg + O 2 = 2MgO; b) You want to heat up and a piece of silicon metal to 1600︒C, decide on a suitable crucible material;c) What is the value of ∆H Θ of formation of TiO 2 ? d) Find ∆G Θfor the reaction Fe + 0.5O 2 =FeO at 1200 ︒C;e) Find ∆G Θ for the reaction 3Mg + AlO 3 = 3MgO + 2A1 at 1500 ︒C; f) What is the equilibrium oxygen pressure when metallic titanium is in equilibrium with TiO 2 at 1000 ︒C?g) If you want to reduce pure TiO 2 to pure metallic titanium at 1000︒C using a CO/CO 2 gas mixture, what is the minimum CO/CO 2 ratio that can achieve such a reduction.9、Answer the following questions according to Ellingham diagram:① At what temperature(s) C can reduce SnO 2(s)、Cr 2O 3(s) and SiO 2(s) ？ ② At what temperature, the decomposition pressure of CuO reaches 1.01325⨯105 Pa ?③ The temperature(s) at which Fe 3O 4 can be reduced to FeO by H 2 ? ④ ∆G Θ when Mg reduces Al 2O 3 at 1000︒C,⑤ At what temperature, for the reaction )(322)(3234S S O Cr O Cr =+,Pa 1019'2－（平）is p O ⑥ Calculate the ∆G when Fe reacts with O 2 at 10－5Pa and 10－10Pa respectivelyat1000︒C, and '(2平）O p as well. ⑦ Calculate the equilibrium constant of reaction 2)()(CO Mn CO MnO s s +=+ at1100︒C (CO CO p p K /2=)⑧ At what temperature, for reaction )(2)(2)(g s s O H Mn H MnO +=+, the（平）)/(22O H H is 104/1 ? 10、The standard Gibbs free energy change for reaction I:Ni （s ） + 1/2 O2 == NiO （s ）is -244560 + 98.53TlnT J/ mol , question: a) How much is the standard Gibbs free energy change for reaction II : 2Ni （s ） + O2 == 2 NiO （s ）b) Calculate the equilibrium constants for reaction I and reaction II respectively at 1000︒ C.c) At 1000︒ C, when oxygen pressure is maintained at 10-4 atm, how much is theGibbs free energy change for reaction I ? Can reaction I proceed forward ? Is Ni stable under this condition ? Is NiO stable under this condition ? d) At 1000︒ C, how much should be the oxygen pressure if we want the Gibbs free energy change for reaction I to be 0, and how much should be the oxygen pressure if we want a Ni-NiO-O 2 system to be at equilibrium ?e) At 1000 C, what is the condition to prevent Ni from being oxidized ? and whatis the condition to reduce NiO ?11、Liquid FeO is reduced to metallic iron at 1600 °C with CO(gas) accordingto the following reaction:FeO(liquid) + CO(g) = Fe(liquid) + CO 2a) Calculate ∆G Θ at 1600 °C for this reactionb) Detennine the minimum CO/C02 ratio required to reduce pure liquid FeO topure metallic iron at 1600 °C.c) Determine the minimum CO/CO2 ratio required to reduce FeO dissolved in a liquid slag to metallic iron at 1600 °C. The metallic iron formed has a purity of 96 mole % iron. The activity of FeO in the liquid slag is 0.3.CO(g) at 1600 °C: ∆GΘ= -274.9 kJ/molCO2(g) at 1600 °C: ∆GΘ = -396.3 kJ/molFeO at 1600 °C: ∆GΘ = -144.6 kJ/molR= 8.314 J/ mol.K= 1.987 ca1/mol.K12、In an experiment, it was found that the Ar was not pure enough. So a setup was devised in an attempt to purify the Ar, as illustrated below. Ar which was at 2 atm was let to flow through a glass tube and the Cu powder pile in it. Given that the temperature in the glass tube is 600︒C and gas pressure is constant at 2 atm.. Calculate the purity of the outgoing Ar in percentage.Ellingham Diagram习题参考答案3.2 ΔG = -108.9 J/mol; ΔS = -21.42 J/(mol.K)3.6 (a ) 22.09/(.)S J mol K ∆=；(b) At 0︒C, ∆G =0; (c) ∆H = 5841.9 J;(d) ∆S =21.39J /(mol.K)，∆G = 109.38 J/mol4.1 (a ) 2898.28J/mol; ( b ) No; ( c ) 345 J/mol; ( d ) 14939 atm; ( e ) 4921 J/mol 4.2 ( a ) 272.8K; ( b ) Pa P 610345⨯≈∆ ; ( c ) 249.46K 4.3 1202K4.4 P=5.73⨯10-6 atm 4.5 0.16P4.7 08.10430685ln +-=TP 4.8 ( a ) 1180K; ( b ) 695.3K; ( c ) 114.4kJ/mol; ( d ) 7123 J/mol; ( e ) 4.2J/mol 4.9 In the initial state: 4.06 mol %; in the final state: 5.3 mol% 4.10 ( a )348 kJ; ( b ) 2.3×10-3Pa ;( c ) “ solution not possible”; (d ) “solution not possible”5.1 atm p H 0005.0=5.2、atm p o 1221007.1-⨯= If the error in enthalpy is 500cal, the uncertainty in the pressure calculated is 28.6%, and if the error in enthalpy is -500cal, the uncertainty is -22.1%5.3、(a) T =462K; (b) T = 420K5.4 (a) atm P O 2621014.1-⨯=, (b) P O2 =2.28⨯10-10 atm., (c) The equilibrium oxygen pressureremains the same when the total pressure increases, which means a higher purity level of N 2 .5.5 (a) 略； (b) Pa atm P H 8.181013056.1800019.0)('2=⨯==; (c) 21.5L Ar is needed to bebubbled into the melt.5.6（a ）l n K a1/T, 10-31/K=∆-=∆ooG kJ H 1000;50- 66.6kJ(b) Ja = 3 < Ka, the reaction will proceed from left to right, and the atmosphere willnot oxidize Ni.5.7 略5.8. (a) P SiO = 8.1⨯10-8 (atm) (b) ∆H o = 639500J; ∆So =334.9J/K (c ) PO2 =10-30 atm5.9 5.10．J H o72250=∆，the reaction is an endothermic one. )(106.08)(atm P g u -⨯=5.11. (a),166528J H o =∆ the reaction is an endothermic one.; (b) At 1168K, the equilibrium pressure of CO2 equals one atmosphere. 5.12 (a) 略 ， (b) Mg CO P P =; (c) T = 2037 K 5.13 (a) 略； (b) 13109.2⨯=K ; (c) ppm 186.0 5.14 (a) 略； (b) kJ H 52.267=∆; (c) K T 1592= 5.15 (a) )(106.13atm -⨯≈; (b) )(1028.210)(2atm P g O H -⨯=5.16 (a) 97.9=K ; (b) atm x 14.4=; (c) if the temperature is increased, the fraction of water reacted will increase since the equilibria constant increases with increasing temperature.6.2 （a ）1.287V;(b) When the water impure, the voltage will go higher; (c) 1.219V 6.4 (a) 145.3kJ;(b) The maximum work that could be derived is 702.36kJ;(c) In this case, the maximum work that could be derived is 696.56kJ. 6.5 (a) -6252J/mol; (b) 370.0)(=II Cd a ; (c) )(42.3mmHg P Cd =; 6.6 7.87⨯10-4 V 6.7 (a))(22g Cl Mg MgCl +=(b) Pa P Cl21'1086.82-⨯=; (c) 2.485V6.8 (a) Pa P O 11'2105.5-⨯=;(b) Anode:e NiNi 2+→Cathode:-→+2222/1O e O ; (c) 0.757V; (d) 0.261V6.10 (a) )(509.3V E o=;(b) 0.074kJ;(c) 4.1⨯106J;(d) Yes. In this case, the open circuit voltage is 3.648V;(e) In this case, to keep the temperature constant, 3.92⨯106J heat should beremoved from the battery per hour.6.11(a) TG CO Al C O Al o 26.3211008.12/322/36232-⨯=+=+Δ(b) The minimum voltage at which the electrolysis may be carried out at1250K is 1.172V .7.1 0.117 atm 7.5 ( a ) ,82.5 2.5 2.5B A B A B B T PV V V x x x x x ⎛⎫∂=+=--⎪∂⎝⎭ ,102.5 2.5 2.5A B A A B A T PV V V x x x x x ⎛⎫∂=+=-- ⎪∂⎝⎭( b) B A M x x V 5.2=7.7 2)1(736.0ln Sn Sn x --=γ7.8 The maximum solubility of MgF2 in liquid MgCl at 900 C is 19 mol% .7.9 ( a ) 1121K; ( b ) 1. 8 cal/K9.6Temperature(ºC ) Phase Composition Fraction1300 Liquid 0.59 0.64 α 0.06 0.36 β ---- 01000+Liquid 0.8 0.43 α 0.1 0.57 β ---- 01000-Liquid ---- 0α 0.1 0.65 β 0.95 0.359.8 Solution:(a) 90 mol%B is the composition of the first solid to form;10 mol % is the composition of the last liquid drop.(b) solid (60 mol%B is the composition) is about 77% ; liquid (15 mol%B is the composition) is 23%9.9 (a) 2900℃, α(12%) (b) 2300℃, liq(95%) (c) 8.2%α(composition is24% )+91.8%β(85%)习题参考答案1.ΔS m =19.1J/mol.K, ΔG m = -5740 J/mol, ΔF m = -5740 J/molIsothermally expands to a vacuum: w = 0, ΔH m =0 , ΔU m = 0,ΔS m =19.1J/mol.K, ΔG m = -5740 J/mol, ΔF m = -5740 J/mol2. 3.68 × 10-3 atm3、Pa x x Br H C Cl H C 406.0;594.05556==Pa p Pa p Br H C Cl H C 26838;744445556==4.5、JG J G J G a a ex mix id mix mix B A R B R A 5302)5(;6912)4(;1610)3(;788.1;62.1)2(;894.0;81.0)1()()(=∆-=∆-=∆====γγ 6. a endothermic one; b. Yes; c J J EMn E Fe 704;1584==μμd ;/9363mol J G m mix -=∆e Pa p Pa p Fe Mn 4;1198==7. J SnPb 4578-=ω; 418.0=Sn aPure Substance as Stand ard Statepq（b ）I 、II 、IIIAAA AAA A AAAA x R T T p x k R T T p p R T T T 0'ln )(ln)(ln )()(γμμμμ+=+=+=*****III:I:II:AA AAAA a RT T p p RT T T ln )(ln )()('+=+=***μμμAA AA AA x RT T p p RT T T ln )(ln)()(*'*+=+=*μμμk A8. a) Mg boils and which makes o S ∆more negative, so the slope changes for larger; b) Firstly, we should avoid using metallic material for this purpose since the melting points of metals are mostly too low. Ceramic materials, usually composed of oxides and having high melting points can be chosen The material should not be reduced by pure silicon at 1600ºC. By examing Ellingham diagram, crucibles (坩埚) made of Al 2O 3 .c ) -890kJ /molO2;d ) -170kJ /molFeO; e) -30kJ; f) Pa 2110－; g)721063.0/⨯=pco p CO 9、⑨ 650ºC ，1220 ºC and 1520 ºC ; ⑩ 1480 ºC ;⑪ When the temperature is equal to or higher that 710 ºC ; ⑫ 2/100molO kJ G o -=∆ ⑬ 900 ºC; ⑭0,102/112,1010'25'2=∆=-=∆=--G Pa P molO kJ G Pa P O O , Pa p e O 10')(210-= ⑮ 510-=K ;⑯ 1220ºC10、a) -489120+197.06TlnT J/mol;b) 2.89×10-54 ; c) J G 749429=∆; Ni is stable under this condition, and NiO is not stable; d) Pa p e o 58')(21046.3⨯= e) from the calculation, we found that at 1000ºC,Pa p e o 58')(21046.3⨯=.So at 1000ºC, when theoxygen pressure is less than 3.46×1058Pa, Ni is stable and can not be oxidized, and NiO will be reduced to Ni under this condition.11. a) mol kJ G o /2.23=∆; b)43.42=⎪⎭⎫ ⎝⎛eCO COp p. This is the minimumCO/CO2 ratio required to reduce pure FeO to Fe at 1600ºC. c)2.142=⎪⎭⎫ ⎝⎛eCO CO p p . This is the minimum CO/CO2 ratio required to reduce FeO in a slag( 炉渣) to Fe in a metallic iron melt under the given conditions at 1600ºC.12.%100)1015.3%10⨯⨯-＝（Ar。

材料热力学与动力学（复习资料）一、 概念•热力学基本概念和基本定律1. 热0：一切互为热平衡的物体，具有相同的温度。

2. 热1： - 焓：恒压体系→吸收的热量=焓的增加→焓变等于等压热效应 - 变化的可能性→过程的方向；限度→平衡3. 热2：任何不受外界影响体系总是单向地趋向平衡状态→熵+自发过程+可逆过程→隔绝体系的熵值在平衡时为最大→熵增原理（隔离体系）→Gibbs 自由能：dG<0，自发进行（同T ，p ： ）4. 热3：- (H.W.Nernst ，1906)： - (M ．Plank ，1912)：假定在绝对零度时，任何纯物质凝聚态的熵值为零S*(0K)=0 - (Lewis ，Gibson ，1920)：对于过冷溶体或内部运动未达平衡的纯物质，即使在0K 时，其熵值也不等于零，而是存在所谓的“残余熵” - Final ：在OK 时任何纯物质的完美晶体的熵值等于零• 单组元材料热力学1. 纯金属固态相变的体积效应- 除非特殊理由，所有纯金属加热固态相变都是由密排结构（fcc ）向疏排结构（bcc ）的转变→加热过程发生的相变要引起体积的膨胀→BCC 结构相在高温将变得比其他典型金属结构（如FCC 和HCP 结构）更稳定（除了Fe ）- 热力学解释1→G ：温度相同时，疏排结构的熵大于密排结构；疏排结构的焓大于密排结构→低温：H ；高温：TS - 热力学解释2→ Maxwell 方程： - α-Fe →γ-Fe ：磁性转变自由能- Richard 规则：熔化熵-Trouton 规则：蒸发熵 （估算熔沸点）2. 晶体中平衡状态下的热空位- 实际金属晶体中空位随着温度升高浓度增加，大多数常用金属（Cu 、Al 、Pb 、W 、Ag …）在接近熔点时，其空位平衡浓度约为10-4；把高温时金属中存在的平衡空位通过淬火固定下来，形成过饱和空位状态，对金属中的许多物理过程（例如扩散、时效、回复、位错攀移等）产生重要影响3. 晶体的热容- Dulong-Petit ：线性谐振动子+能量均分定律→适应于较高温度及室温附近，低温时与实验不符U Q W∆=-dH PV U d Q =+=)(δRd Q S Tδ=()d dH TdS G H d TS =--=00lim()lim()0p T T T GS T→→∂∆-=∆=∂()()V T T P V V S ∂∂=∂∂//()()()T T T V P V V S T V H ∂∂+∂∂=∂∂///RK mol J T H S mm m ≈⋅≈∆=∆/3.8/K mol J T H S b v v ⋅≈∆=∆/9.87/3V V VQ dU C RdT dT δ⎛⎫⎛⎫=== ⎪ ⎪⎝⎭⎝⎭-Einstein(固体振动热容理论)：晶体总共吸收了n 个声子，被分配到3N 个谐振子中；不适用于极低温度，无法说明在极低温度时定容热容的实验值与绝对温度的3次方成比例。