材料热力学与动力学复习资料+课后习题
材料热力学与动力学复习资料+课后习题
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。
材料热力学与动力学：热力学定律习题
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第一章 热力学定律 习题课
13. Gibbs-Helmholtz方程
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第一章 热力学定律 习题课
1．判断下列说法是否正确： ① 状态给定后，状态函数就有一定的值，反之亦然。
② 状态函数改变后，状态一定改变。
③ 状态改变后，状态函数一定都改变。
解释：若外界条件不变，即状态给定后，所有的状态函数都有定 值。当某一个或某几个状态函数发生变化时，状态一定改变；反 之，当状态发生改变时，状态函数中，一定有某一个或几个发生 变化，而不一定全部的状态函数都发生变化。 如：气体节流膨胀过程，焓不变。
第一章 热力学定律 习题课
10.热力学基本关系式
dU TdS pdV dH TdS Vdp
dA SdT pdV dG SdT Vdp
11.克拉佩龙方程
dT dp
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第一章 热力学定律 习题课
12.克劳修斯—克拉佩龙方程
d ln p dT
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（微分式）
ln
第一章 热力学定律 习题课
⑤ 恒温过程一定是可逆过程。
解释：恒温过程是指系统与环境的温度始终保持相 等且恒定，是一个自始至终保持热平衡的过程，即 为可逆过程。数学表达式dT = 0； ✓若对于一个ΔT = 0的过程，则说明该过程始终态温 度相等，并不一定是恒温过程。 ✓等温可逆过程一定是恒温过程。
可以这么看，一个思想封闭的人遇到烦恼时就 会焦躁不按或者抑郁忧愁，也就是内心混乱度增 加。但是当这个人向他人敞开心扉获得安慰之后 ，他就会趋于平静。所以，同学们都要学会向他 人倾诉，同他人交流，不要把自己封闭在一个狭 小的世界里。
“材料热力学”补充习题参考答案
“材料热力学”补充习题参考答案教材各章习题参考答案 (魏)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 ) 249.46K4.3 1202K4.4 P=5.73?10-6 atm 4.5 0.16P 4.708.10430685ln +-=TP4.8 ( a ) 1180K; ( b ) 695.3K; ( c ) 114.4kJ/mol; ( d ) 7123 J/mol; ( e )4.2J/mol4.9 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 ”5.1atmp H 0005.0=5.2、atmp 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 = 420K 5.4 (a)atmP O 2621014.1-?=, (b) P O2 =2.28?10-10 atm., (c) The equilibriumoxygen pressure remains 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 be bubbled into the melt.5.6（a ）0.880.900.920.940.960.981.00 1.02 1.04 1.06 1.087.27.47.67.88.08.28.48.6l n K a1/T , 10-31/K=-=?ooG kJ H1000;50- 66.6kJ(b) Ja = 3 < Ka, the reaction will proceed from left to right, and theatmosphere will not 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 atm 5.9 5.10．JH o72250=?，the reaction is an endothermic one.5.11. (a),166528J Ho=?the reaction is an endothermic one.;(b) At 1168K, the equilibrium pressure of CO2 equals one atmosphere. 5.12 (a) 略， (b) MgCO P P =; (c) T = 2037 K5.13 (a) 略； (b) 13109.2?=K ; (c) ppm 186.05.14 (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.)(106.08)(atm Pg u -?=6.5 (a) -6252J/mol; (b) 370.0)(=II Cda ; (c) )(42.3mmHg P Cd =;6.67.87?10-4 V 6.7 (a))(22g Cl Mg MgCl+=(b)Pa P Cl 21'1086.82-?=;(c) 2.485V6.8 (a) PaP O 11'2105.5-?=;(b) Anode: e Ni Ni 2+→Cathode: -→+2222/1Oe O ;(c) 0.757V; (d) 0.261V6.10 (a) )(509.3V E o=;(b) 0.074kJ;(c) 4.1?106J;(d) Y es. In this case, the open circuit voltage is 3.648V;(e) In this case, to keep the temperature constant, 3.92?106J heatshould be removed from the battery per hour. 6.11(a)TGCOAl C O Al o26.3211008.12/322/36232-?=+=+Δ(b) The minimum voltage at which the electrolysis may be carriedout at 1250K is 1.172V .7.1 0.117 atm 7.5 ( a ) ,82.52.52.5BA BA BB 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 Mx x V5.2=7.72)1(736.0ln Sn Snx --=γ7.8 The maximum solubility of MgF2 in liquid MgCl at 900?Cis 19mol% .7.9 ( a ) 1121K; ( b ) 1. 8 cal/K 9.69.8 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(compositionis 24% )+91.8%β(85%)补充习题参考答案（魏）1.ΔS m =19.1J/mol.K, ΔG m = -5740 J/mol, ΔF m = -5740 J/molIsothermally expan ds 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、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、JGJ GJ G a a exmix idmix mix B A R B R A 5302)5(;6912)4(;1610)3(;788.1;62.1)2(;894.0;81.0)1()()(=?-=?-=?====γγ6. a endothermic one; b. Y es; c JJ EMn EFe 704;1584==μμd;/9363mol J G m mix -=?ePap Pa p Fe Mn 4;1198==S n P b4578-=ω;418.0=Sn a8. a) Mg boils and which makes oS ?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 chosenThe 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) Pa2110－;g)721063.0/?=pco p COPure Substance as Standard Statepq（b ）I 、II 、IIIIII:AA x RT T p RT T T ln )(ln)()(**+=+=μμμk A9、① 650oC ，1220 oC and 1520 oC ; ② 1480 oC ;③ When the temperature is equal to or higher that 710 oC ;④ 2/100molO kJ G o-=?⑤ 900 oC; ⑥,102/112,1010'25'2=?=-=?=--G Pa PmolO kJ G Pa P O O ,Pap e O 10')(210-=⑦ 510-=K;⑧ 1220oC10、a) -489120+197.06TlnT J/mol; b) 2.89×10-54 ; c)JG 749429=?; Ni is stable under thiscondition, and NiO is not stable; d)p e o 58')(21046.3?= e) fromthe calculation, we found that at 1000oC,Pap e o 58')(21046.3?=. Soat 1000oC, when the oxygen 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)molkJ G o/2.23=?; b)43.42=eCO CO p p . This is theminimum CO/CO2 ratio required to reduce pure FeO to Fe at 1600oC. c)2.142=eCO CO p p . This is the minimum CO/CO2ratio required to reduce FeO in a slag( 炉渣) to Fe in a metallic iron melt under the given conditions at 1600oC.12.%10?-Ar .3 15 % 100 ) 10。
材料热力学与动力学_2
C 2 C1
Fick’s first law in Cylinder-System
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dm dt 2 Lr1 D C 2 C1 r2 r1
Chapter 2. Diffusion
Prof. Dr. X.B. Zhao
Department of Materials Science and Engineering Zhejiang University
Diffusion Process
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Chapter 2 : Diffusion
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2.2 Steady-State Diffusion
2.2.1 Diffusion through the cylinder wall
For steady-state diffusion the flux through a cylinder wall with radius of r (r1 < r < r2) is a constant.
This means y 2 ( t ) will increase d 2 if the atoms jump one time. If the jump frequency of the atoms is n, an atom will jump nt times from t = 0 to t = t, and then we have: y 2 ( t ) d 2 n t Einstein has demonstrated that d 2n equals to 2D for one dimensional and to 4D and 6D for two- and three-dimensional case respectively.
材料热力学习题集
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例题 6-5 在中碳钢中由于锰的枝晶偏析，定向轧制后，易引起带状组织。为了消除带状组
织，应进行扩散退火，如果锰的浓度按正弦规律分布如下：
C(x,t) C
1 2
C0
sin
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x
exp
例 4-5 碳在 1 个大气压、25℃时，以石墨为稳定相，试求在 25℃由石墨转变为金刚石所需 要的压强。
例题 5-1 实验测得 Cd-Mg 的摩尔体积如下表所示。 Cd-Mg 合金摩尔体积的实验值
例题 5-2 已知某三元溶液的摩尔体积为 Vm=7X1+10X2+12X3-2X1X2+3X1X2X3 (cm3/mol)
个大气压下）。
例 1-4 已知液体铅在 1 个大气压下的比热为： Cp(l)=32.43-3.10×10-3T J/(mol·k) Cp(s)=23.56+9.75×10-3T J/(mol·k)
液体铅在熔点(600K)凝固为固体时放热 4811.6 J/mol，求液体铅过冷至 590K 凝固时熵值的变 化（在一个大气压下）。
6
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六大 板块
第一章 第二章 第三章 第四章 第五章 第六章
物理化学复习纲要
热力学定律（热力学第一定律、热力学第二定律） 自由能（ΔF、ΔG） 热力学状态函数、关系式及应用 相变热力学 溶液 扩散
.
计算锡在绝热器内过冷到 495 K 时能自动凝固的分数。
材料热力学习题解答
《材料热力学》复习思考题解答3. 在1560℃时，C 在液态铁中的活度系数和偏摩尔超额焓由下列式表示： 2l n 0.37711.7c C C X X γ=-++25.415.017.25E C C C H X X =++(K Cal) 其标准态为纯石墨，计算1560℃时液相与石墨平衡的相线的斜率。
解：以石墨为标准态时，C 在液态铁中的化学位为：l n (1)LC CC R T a μμ=+ 石墨 当液相与石墨平衡时，L C Cμμ=石墨。
即ln 0C α=。
又ln ln ln C C C X αγ=+ln ln 0(2)C C X γ∴+=由(2)式得：平衡时0.2067C X =两边取微分得：(ln )(ln )1[](1/)[]0(1/)C C C X T C C C C d T dX dX T X X γγ∂∂++=∂∂ (ln )[](1/)ln ln 1(1/)[()]1()CC X EC C C C C T C TC C CdX H X T d T R X X X X γγγ∂-∂∴==⋅∂∂-++∂∂2(5.415.017.25) 4.1810000.20678.311(723.4)278.6C C CC X X X X ++⨯⨯=-⋅++=- 2C dX T dT=-CdX 又d(1/T)5221278.68.310(1560273)C dX dT T -∴=-==⨯+C dX d(1/T) 1()K - 4. 在1000K 时，A-B 二元溶液中，当0.01B X =时，0.1B a =。
在盛有大量A 的量热计中加入少量的B 组元时，测得吸热7000Cal/mol ，假定2ln ln B A B X γγ=。
求1500K 时，当0.02B X =时，B 组元的活度。
解：在1000K 时，当0.01B X =时，0.1B a =0.1100.01B γ∴== 又022ln ln10ln 2.3490.99B B A X γγ=== 又ln [](1/)ii P H R T γ∂∆=∂15001500010001000l n (1/)BBH d d T Rγ∆∴=⎰⎰1500100011[ln ][ln ]()15001000B B B H R γγ∆∴=+-7000 4.18112.349()8.31150010001.175⨯=+-= 202l n (l n )0.981.175B A B X γγ∴==⨯ 1.128= l n 3.09B γ∴= 3.090.020.0B B B a X γ==⨯=7. 若A-B 二元合金系在液、固态两组元均能无限互溶，且均为理想溶液。
材料热力学与动力学复习资料 课后习题.
10. [ 形核驱动力 ] 碳钢淬火马氏体在进行低温回火时，并不析出该温度下的稳定碳化物Fe3C(θ, 而是首先析出一种碳含量更高的亚稳碳化物相 Fe2.3-2.5C(ε，试分析原因 - 从淬火马氏体低温回火析出和形核驱动力图示可以看出，成分为的过饱和固溶体(淬火马氏体中析出碳化物的相变驱动力实际上比相的相变驱动力要小些。
但是决定那种碳化物优先析出的并不是相变驱动力，而是形核驱动力。
11. [析出相的表面应力效应]在固态相变初期析出的第二相α一般与基体相β 呈共格状态，从表面张力导致的附加压力的角度分析其原因。
假定共格态的表面能(张力为 0.01 J/m2, 半共格态为 0.1 J/m2, 完全非共格态的表面张力为 1 J/m2, 试计算从α 基体相中析出直径为 10 nm 的β 相(球形在上述三种状态下的附加压应力。
P 2 / r - 在压力作用下，其自由能提高为 Gibbs-Thomson 公式： Gm Gm (0 P Vm d 2Gm dxB 2 dX B P Vm XB XB d B A X B X B dx B 2 Vm (1 X B XB r XB XB RT 2 I AB (1 X B X B 12. 假定 H 由纯铁素体和奥氏体组成的双相合金中进行扩散，相区宽度如右图所示。
a1 和 a2 分别表示 H 在单相区边界的活度，ai 表示 H 在/相界的活度；H 在两相中的活度系数为 H 和 H，H 在两相中的体积浓度为 CH 和CH，已知梯度的驱动下发生扩散。
H H 。
H 在活度 ai xi / i JH JH a1 a2 l H l H D D - 对于双相合金的扩散，哪一相对 H 的扩散阻力大，那么该相将对 H 的扩散起控制作用。
已经知道，在相同温度下，H 在相中的扩散系数远大于在相中的扩散系数 DH (100 倍以上，因此扩散作用主要取决于相。
DH。
2012硕士《材料热力学与动力学》复习练习题
Question 16
1） 指出各水平线的三相平衡反应 2） w(SiO2)=0.40 的系统(图中 R 点)从 1700C 冷却到 1000C 时的冷却曲线示意图。 注明每一阶段系统有哪些相？发生哪些 变化？指出各阶段的自由度数？ 3） w(SiO2)=0.10 的系统 12 kg，冷却到 1400C 时，液相中含 MnO 多少 kg？ 4） w(SiO2)=0.60 的系统 1500C 以哪些相存在？计算其相对 量。
4
2012 研究生《材料热力学与动力学》复习练习题（10 月 8 日交，手写完成）
Question 1 进行下述过程时，系统的ΔU、ΔH、ΔS和ΔG何者为零？ 1.1 非理想气体的卡诺循环； 1.2 隔离系统中的任意过程； 1.3 在100C，1大气压下1mol水蒸发成水蒸汽； 1.4 绝热可逆过程。 Question 2 1mol 理想气体等容升温到状态 3，求 Q,W,ΔU,ΔH。 若将理想气体先等压膨胀到状态 2，然后再等温（可 逆）压缩到状态 3，求 Q,W,ΔU,ΔH，并与直接从 1 到 3 的途径相比较。
Question 11 导出液相中 Bi 的活度系数的估算公式。
H m T a Bi exp ( 1) RT Tm
其中，熔化热为 H m 纯 Bi 的熔点为 Tm，R 为气体常数。
Question 12 对下列二元相图，指出其中的错误 (用相律说明原因)
2
Question 13
Trouton's定律为表示为:
1 (V1,T1) 2 (V2,T2) V
H vap 90Tb
单位J/mol, 其中Tb为沸点(K), 汞的沸点为630 K. 计算在
298K液态汞的分压. 用Troutons定律估算汞的汽化热.
材料热力学 习题答案
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 onemicrometer (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 dropletsare rest (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 an insulated 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),what will 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)()(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 gasfollowing combustion)?(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 surroundingsaverage 400000 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 dT T 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 Hn 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 pf i rf 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 temperaturethe flame 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)/(--∆m o lkJ 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?DATA: 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 HL S SL L P S P LS =⨯-⨯-⨯+⨯+==+++-⎰⎰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) DATA: 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.DATA : 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? DATA; 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. DATA; 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℃ is 1.0021 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:leirreversib g x x x )(138),1000(8018.4)8018.48022261(=-⨯⨯=⨯-⨯+The problems of the second law1 The solar energy flux is about 4J cm 2/min. in no focusing collector the surface temperature can reach a value of about 900℃. If we operate a heat engine using the collector as the heat source and a low temperature reservoir at 25℃, calculate the area of collector needed if the heat engine is to produce 1 horse power. Assume the engine operates at maximum efficiency. (2.1)Solution )(664.0)(74660104273900)25900(24m S W tWP StQ T T T W H H L H ===⨯⨯+-=-=2 A refrigerator is operated by 0.25 hp motor. If the interior of the box is to be maintained at -20℃ ganister a maximum exterior temperature of 35℃, what the maximum heat leak (in watts) into the box that can be tolerated if the motor runs continuously? Assume the coefficient of performance is 75% of the value for a reversible engine. (2.2)Solution:)(114474625.02035202733475.0%75W P P T T T P Q T T T W L LLLH HHLH =⨯⨯+-⨯=-=-=3 suppose an electrical motor supplies the work to operate a Carnot refrigerator. The interior of the refrigerator is at 0℃. Liquid water is taken in at 0℃ and converted to ice at 0℃. To convert 1 g of ice to 1 g liquid. △H=334J/g is required. If the temperature outside the box is 20℃, what mass of ice can be produced in one minute by a 0.25 hp motor runningcontinuously? Assume that the refrigerator is perfectly insulated and that the efficiencies involved have their largest possible value. (2.3)Solution: )(4576033474625.020273g m M m P P T T T P L LLLH ===⨯⨯=-=4 under 1 atm pressure, helium boils at 4.126K. The heat of vaporization is 84 J/mol what size motor (in hp) is needed to run a refrigerator that must condense 2 mol of gaseous helium at 4.126k to liquid at the same temperature in one minute? Assume that the ambient temperature is 300K and that the coefficient of performance of the refrigerator is 50% of the maximum possible. (2.4)Solution: )(52.0)(393'60284216.4216.4300'5.0%50hp W P P T T T P P Q T T T W L L L H LLLH ==⨯⨯-=-==-= 5 if a fossil fuel power plant operating between 540 and 50℃ provides the electrical powerto run a heat pump that works between 25 and 5℃, what is the amount of heat pumped into the house per unit amount of heat extracted from the power plant boiler. (a) assume that the efficiencies are equal to the theoretical maximum values(b) assume the power plant efficiency is 70% of maximum and that coefficient ofperformance of the heat pump is 10% of maximum(c) if a furnace can use 80% of the energy in fossil foe to heat the house would it be moreeconomical in terms of overall fissile fuel consumption to use a heat pump or a furnace ? do the calculations for cases a and b (2.5)solution:1,2,2,1,212,2,2,2,21,1,1,1,198.82527352527354050540)(H H H H H H L H H H L H P P P P P P P T T T P P T T T P a =+-=+-=-=-=.,)(6286.0)(1,2,not is b ok is a c P P b H H =6 calculate △U and △S when 0.5 mole of liquid water at 273 K is mixed with 0.5 mol of liquid water at 373 K and the system is allowed to reach equilibrium in an adiabaticenclosure. Assume that C p is 77J /(mol K) from 273K to 373K (2.6) Solution:)/(933.0)273323ln(5.0)373323ln(5.0)ln()ln()(02211K J C C T T C n T T C n S J U P P E P E P =+=+=∆=∆ 7 A modern coal burning power plant operates with a steam out let from the boiler at 540℃and a condensate temperature of 30℃.(a) what is the maximum electrical work that can be produced by the plant per joule of heatprovided to the boiler?(b) How many metric tons (1000kg) of coal per hour is required if the plant out put is to be500MW (megawatts). Assume the maximum efficiency for the plant. The heat of combustion of coal is 29.0 MJ/k g(c) Electricity is used to heat a home at 25℃ when the out door temperature is 10℃ bypassing a current through resistors. What is the maximum amount of heat that can be added to the home per kilowatt-hour of electrical energy supplied? (2.7)Solution:)(3.69)(6937136005000.29)()(89.013054030540)(ton kg m T T T mb J Q T T T W a LH LH H L H ==⨯=-=+-=-=)(9.191102525273)(J Q Q T T T W c H HHLH =-+=-=8 an electrical resistor is immersed in water at the boiling temperature of water (100℃) the electrical energy input into the resistor is at the rate of one kilowatt(a) calculate the rate of evaporation of the water in grams per second if the water containeris insulated that is no heat is allowed to flow to or from the water except for that provided by the resistor(b) at what rate could water could be evaporated if electrical energy were supplied at therate of 1 kw to a heat pump operating between 25 and 100℃data for water enthalpy of evaporation is 40000 J/mol at 100℃; molecular weight is 18g/mol; density is 1g/cm 3 (2.8)solution:)(23.2,2510027310010004000018)()(45.0,10004000018)(g m m b g m ma =-+===9 some aluminum parts are being quenched (cooled rapidly ) from 480℃ to -20℃ byimmersing them in a brine , which is maintained at -20℃ by a refrigerator. The aluminum is being fed into the brine at a rate of one kilogram per minute. The refrigerator operates in an environment at 30℃; that is the refrigerator may reject heat at 30℃. what is them minuspower rating in kilowatts, of motor required to operate the refrigerator? Data for aluminum heat capacity is 28J/mol K; Molecular weight 27g/mol (2.9)Solution:)(5.102)(102474202732030)20480(28271000kW W P P T T T P P L L L L H W L ==---=-=--⨯=10 an electric power generating plant has a rated output of 100MW. The boiler of the plantoperates at 300℃. The condenser operates at 40℃(a) at what rate (joules per hour) must heat be supplied to the boiler?(b) The condenser is cooled by water, which may under go a temperature rise of no morethan 10℃. What volume of cooling water in cubic meters per hour, is require to operate the plant?(c) The boiler tempeture is to be raised to 540℃,but the condensed temperature and electricoutput will remain the same. Will the cooling water requirement be increased, decreased, or remain the same?Data heat capacity 4.184, density 1g/cm 3 (2.10)Solution: )(109.7)(102.21040300273300)(1188J t P Q W P T T T P a H H L H H H ⨯==⨯=-+=-=)(1003.1184.41010)(103.4)(34611m V Q V J Q b L L ⨯==⨯⨯⨯⨯=noW P T T T P c L H H H )(10626.11040540273540)(88⨯=-+=-=11 (a) Heat engines convert heat that is available at different temperature to work. Theyhave been several proposals to generate electricity y using a heat engine that operate on the temperature differences available at different depths in the oceans. Assume that surface water is at 20℃, that water at a great depth is at 4℃, and that both may be considered to be infinite in extent. How many joules of electrical energy may be generated for each joule of energy absorbed from surface water? (b) the hydroelectric generation of electricity use the drop height of water as the energy source. in a particular region the level of river drops from 100m above sea level to 70m above the sea level . what fraction of the potential energy change between those two levels may be converted into electrical energy? how much electrical energy ,in kilowatt-hours, may be generated per cubic meter of water that undergoes such a drop? (2.11)Solution:)/(1006.136001000)()(055.0127320420)(6h kW hmg P b J Q T T T W a H H L H ⨯=⨯∆==+-=-=12 a sports facility has both an ice rink and a swimming pool. to keep the ice frozen during the summer requires the removal form the rink of 105 KJ of thermal energy per hour. It has been suggested that this task be performed by a thermodynamic machine, which would be use the swimming pool as the high temperature reservoir. The ice in the rink is to be maintain at a temperature of –15℃, and the swimming pool operates at 20℃, (a) what is the theoretical minimum power, in kilowatts, required to run the machine? (b) how much heat , in joule per hour , would be supplied t the pool by this machine? (2.12)Solution:)(1014.1101527320273)()(77.33600/10152731520)(555kJ Q b kW P T T T P a H L L L H ⨯=-+==-+=-=13solution:)/(81.6810ln 314.877.45277.6282.4)/(152940)()/(67.4977.45277.6282.4)()/(152940)(22)(2molK cal S mol cal H d molK cal S c mol cal H b AlNN Al a -=+-⨯-⨯=∆=∆-=-⨯-⨯=∆=∆=+14solution:)/(2257412000)27340273ln 184.4273336263273ln1.2()(40,010,K J dT T C T H dT T C m S WATER P m mICE P =+++=+∆+=∆⎰⎰- 15)(70428)(2896100077773002J W J Q T T T W L L L H ==-=-=16)(4.3719))2.4300(314.85.13.83(3002.4300)(7.58663.832.42.4300J Q T T T W J Q T T T W H H L H L L L H =-⨯+-=-==-=-=17yesd Q c K J PPnR S b J pdV n W Q OU T a )(0)()/(1.1910ln 314.81ln )()(570410ln 298314.810)(0==⨯⨯==∆=⨯⨯=-=-==∆=∆⎰18)(122233527302033560500g m m m T T T L L H =-=-=⨯教材各章习题参考答案 (魏)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/mol4.2 ( 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/mol4.9 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、atmp 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 equilibriumoxygen pressure remains 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 isneeded to be bubbled into the melt.5.6（a ）l n K a1/T, 10-31/K=∆-=∆o o G kJ H 1000;50- 66.6kJ(b) Ja = 3 < Ka, the reaction will proceed from left to right, and theatmosphere will not 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 atm 5.9 5.10．J H o72250=∆，the reaction is an endothermic one.5.11. (a),166528J H o =∆ the reaction is an endothermic one.; (b) At 1168K, the equilibrium pressure of CO2 equals one atmosphere.)(106.08)(atm Pg u -⨯=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 is696.56kJ.6.5 (a) -6252J/mol; (b) 370.0)(=II Cd a ; (c) )(42.3mmHg P Cd =; 6.67.87⨯10-4 V 6.7 (a))(22g Cl Mg MgCl +=(b) Pa P Cl 21'1086.82-⨯=;(c) 2.485V6.8 (a) Pa P O 11'2105.5-⨯=;(b) Anode: e Ni Ni 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 heatshould be removed 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 carriedout at 1250K is 1.172V .7.1 0.117 atm 7.5 ( a ) ,82.5 2.5 2.5B A BA BB 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。
材料热力学习题解答第1一2章
（3）ΔU=ΔH=0
V W = -Q = -nRT ln 2 = - ×8.314×273×ln2 = -5619(J) V1
（4）Q=0; W=ΔU,即-pe(V2-V1)=nCV,m(T2-T1), -0.5p2V2= nCV,m(T2-T1) ; -0.5nRT2= nCV,m(T2-T1)-0.5×8.314×T2=2.5×8.314×(T2-273) ; T2=227.5K
W =Байду номын сангаасΔU = nCV, m (T2 - T1 ) = ΔH = nCp, m (T2 - T1 ) =
100 ×2.5×8.314×(227.5- 273)= -3378(J) 28
100 ×3.5×8.314×(227.5- 273)= -4729(J) 28
13、在244K温度下，1.00mol单原子气体（1）从1.01MPa、244K等温可逆膨胀到 505kPa,(2) 从1.01MPa、244K绝热可逆膨胀到505kPa，求两过程中的Q、W、ΔU 及ΔH，并作p-V图表示上述气体所进行的两个过程。 解：（1）ΔU=ΔH=0
5、2.00mol的水蒸气在100℃、101325Pa下变为水，求Q、W、ΔU及ΔH。已知水 的气化热为2258J/g。
解：Q Q p H -n vapH m -2 22581810-3 -81.29(kJ) W -pe V p e Vg nRT -2 8.314 37310-3 6.20(kJ) U Q W -81.29 6.20 -75.09(kJ)
4、在101kPa下，气体由10.0dm3膨胀到16.0dm3，吸收了1255J的热，求ΔU、 ΔH、W。
2H Q p 1255J 3U Q W 1255- 606 649J 1W -pe V -101103 (16- 10)10 - 3 -606(J) 解：
材料热力学与动力学复习题答案word版本
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。
材料热力学与动力学005相变热力学与动力学2
4 4 V r 3 ( t ) 3 3 3
所以新相总体积
4 dV 3t 3 I vVa dt 3
相变开始阶段ห้องสมุดไป่ตู้
V V
4 dV 3t 3 I vV dt 3
所以
将V作为１，经过t时间产生新相的 体积分数为x, dx=dVß：
4 t x I v 3t 3dt 3 0
对于曲面晶界，R可由下式求得
P
8R 4R 2
2 R
1 1 1 1 ( ) R 2 R1 R2
R1,R2为曲面晶界的最大及最小半径
上式表明，由界面能提供的作用于单位面积晶界的驱动 力与界面能成正比，与界面曲率半径成反比，力的方向 指向曲率中心。对于平直界面， R 驱动力为零。 在三个不同的晶粒交点处（如A点)，为保持界 面张力平衡，即保持三个交角均为1200，晶界必将 凸向大晶粒一方，出现曲面晶界，在驱动力作用下， 小晶粒中原子越过界面向大晶粒迁移。
2 3 DV C a ( ) r 3 r03 t kt 2 RT
r 0 为粗化开始时颗粒的平均半径；r 为经过t时间粗化后颗粒的平均半径
3
3
考虑颗粒尺寸分布，Wagner公式： 2 9 DV C a ( ) 3 3
r r0 8 RT
t
五、晶粒长大 1.晶粒正常长大：在界面曲率驱动力下，晶粒发生均匀 长大的过程。 母相全部转变成新相后，还将通过晶界的迁移发 生晶粒的粗化。推动晶界迁移的驱动力来自界面能的 降低。50nm晶粒系统晶界能为104J/m3。
根据Gibbs-Thomson方程
Ca (r ) 2VB ln Ca () RTr
材料科学基础复习题
有关动力学与热力学填空：描述反应动力学的阿累尼乌斯方程（P21）表明：反应速率对与温度和（）的变化是极为敏感的。
判断：含有少量位错的晶体的滑移开动容易，体现出实际强度低于理想晶体（P120），因此在热力学上是稳定的。
有关晶体结构：名称解释：原子堆垛因子（P65），配位数（P65），面密度（P63），线密度（P61）填空：布拉菲点阵共有种，归纳为个晶系。
面心立方结构单个晶胞中的原子数为，密排六方结构单个晶胞中的原子数为。
选择：NaCl和金刚石（P71～72）的晶体结构相差很大，但它们都属于（）点阵。
（A）简单立方（B）体心立方（C）面心立方选择：体心立方结构最密排的晶向族（P64）为（）。
（A）<110> （B）<100> （C）<111>填空：体心立方BCC材料中沿[110]方向的线密度是（）。
名称解释：空间点阵（P48）晶胞（P48）点阵常数（P48）滑移系（P129）密勒指数（P56）问答：在面心立方晶胞中画出(101)和[110]，并分析它们能否构成滑移系？填空：面心立方（fcc）晶体的滑移面是，滑移方向是，共有个滑移系。
填空：每个体心立方晶胞中的原子数为，配位数（P33）为；每个面心立方晶胞中的原子数为，配位数为。
填空：离子晶体中的配位数主要受（）决定的（P33～34），而在共价建结合的材料中，最近邻的数目是有每个原子的（）决定的。
简答：体心立方、面心立方、密排六方晶胞中的原子数、配位数、致密度分别是多少？选择：BCC的角上的原子彼此（）。
（A）接触（B）不接触（C）无法判断判断：面心立方、体心立方和密排六方（P54）是金属的三种常见晶体结构，它们都属于空间点阵。
问答：分别计算fcc晶体中[100]、[110]和[111]晶向上的线密度（用点阵常数a 表示），并说明哪个晶向是密排方向。
1)、(421)、[111]。
问答：画出立方晶系的晶面和晶向：(11填空：氯化钠（NaCl）的晶体结构属于空间点阵。
材料中的热力学与动力学12-1
{k is in [moles/(liter sec)]}
0
������ = −������������ + ������
II) First Order Reactions A → products
������ ������ − = k ������ ������������ ������ = ������ 0 ������ −������������
II) Analyzing the data A) Reactions with one reactant: A → products
a) Plot or analyze
[A] vs. t ln[A] vs. t 1/[A] vs. t … and find which gives a straight line.
take [A]0 << [B]0 , [C]0 e.g. flood system with B and C
Then [B] ~ [B]0 and [C] ~ [C]0
So that Where
������ ������ − ������������
≈ ������ ′ ������
������ 0
������0 ′ ������0 ������0 ′ ������0 ������0 ′ ������0 ������0 ′ ������0
=1
⇒ ������ = 1
= 2 ⇒ ������ = 1/2 =2 =4 ⇒ ������ = 1 ⇒ ������ = 2
etc… b) Flooding or Isolation (goal is to try to make problem look like a onereactant system)
材料热力学和动力学答案
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。
材料热力学与动力学动力学部分复习
3、速率方程的积分形式
A：一级反应 反应速率与反应物浓度一次方成正比的反应称一级反应(first order reaction)。如： AB
dc A kcA dt
dcA c A k dt
dc A kcA dt
t dc A k dt C A,0 c 0 A CA
2、化学反应的速率方程 表示反应速率与参加反应的各种物质浓度及影响反应的各种因 素之间的关系，也称为化学反应的动力学方程。 元反应：
aA bB lL mM
V ∝ caAcbB 或 V=KcaAcbB
（质量作用定律） 反应速率
K：反应速率常数(rate constant of reaction)或反应比速。 K 物理意义：各反应物质皆为单位浓度时的反应速率。 影响K值因素：反应类型、温度、溶剂、催化剂等。
（1）计算此反应的活化能？ （2）欲使A在10min内转化率达到90%，反应温 度应控制在多少？
二、相变过程动力学
1、形核率 单位时间单位体积母相中形成的新相晶核数，用 I表示 （1）均匀形核
G * Q I B exp( ) exp( ) kT kT
(2)成核速率随温度变化的关系
2 3 T 64 1 * * m G As 3 (H m ) 2 (T ) 2 3
c
A,0
dy kdt yc B,0 y
或
c A,0
dy 1 d y kdt c B,0 c B,0 y c A,0 y
由t = 0到t = t积分上式，得
c A0
c B,0 c A,0 y 1 ln kt c B,0 c A,0 c B,0 y
活化能Ea： 为了能发生化学反应，普通分子（具有平均能量的分子）必须吸 收足够能量先变成活化分子 ，在此变化过程中所要吸收的最小
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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。
-Debye(晶体振动热容理论)：晶体点阵中原子在相互间力作用下振动的频率是连续变化的4. 两相平衡(Clausius-Clapeyron 方程)- 相平衡→两相的摩尔自由能相等 - Clausius-Clapeyron 方程（可逆相变）：-液-固、固-固平衡→压力和温度呈线性关系→对于凝聚态, 如果压力改变不大（or 低压）, 固-固(液)转变的Vm 变化非常小, 相变Sm 对温度和压力不是很敏感，dP/dT 接近常数。
- 液-气、固-气平衡→压力和温度呈指数关系→对于有气体参加的两相平衡，压力改变时摩尔体积的变化较大 -Fe （压力改变晶体结构转变特性）：随着压力升高, 熔点提高, 高压倾向于使固态稳定；BCC －FCC 转变温度下降(扩大FCC 区域)，高压倾向于FCC(密排)相稳定；在室温下压力高于13GPa 时, 出现BCC －HCP(密排相)转变。
-5. 磁性转变的自由能- 磁有序度：物质从绝对零度到居里温度过程中磁有序的变化过程 - 磁转变热容：磁矩有序排列被破坏所决定的热容223(/)(1)E ET V E T e C R T e θθθ=-334234512)(⎪⎪⎭⎫ ⎝⎛≈⎪⎪⎭⎫ ⎝⎛=∆=D D V TR T R dT U d C θθπ0=∆→βαGβαβαdG dG G G d ==-,0)(）RT H A P m /exp(∆-=αβαβ→→∆∆=mm V T H dTdP- 物质由铁磁态变化到顺磁态的过程伴随着自由能的变化- 当x = 1/2时，理论磁矩有序度为0，最大焓变： - 晶体中出现反平行排列带来的熵变： - -在T C 温度以上, 金属处于磁矩完全无序状态，x = 1/2，二组元材料热力学0. 二元系：纯组元相、溶体相（溶液相和固溶体相）和化合物中间相 1. 理想溶体模型：- A 、B 混合后无热效应或体积效应，U AB ＝(U AA ＋U BB )/2 -- -固溶体的总熵值 = 热熵（振动熵）+ 组态熵 -组态熵：2max 41N k H m =∆)1(4max x x H H -∆=∆[])1ln()1(ln x x x x R S --+-=∆[])1ln()1(ln )1(2x x x x RT x x RT G C --++-=∆Pm G G G -=∆[][]2ln )1ln()1(ln 2/1)1(2+--++--=∆x x x x RT x x RT G C m mA AB Bm A A B Bm A A B BV X V X VU X U X U H X H X H =+=+=+0m mixG G G =+∆m A A B B mix S S S X S X S S =+=++∆振组()ln ln mix A A B B S R X X X X ∆=-+max 115.736 mix S J mol K --∆=⋅⋅00(ln ln )m A A B B A A B B G X G X G RT X X X X =+++2. 规则溶体模型： - 规则溶体的非理想性完全由混合热效应引起 - ΔG E : 过剩Gibbs自由能 - ΔG E =ΔH mix =ΔU mix - I AB 在+16.7 kJ/mol 以上, 自由能-成分曲线上出现拐点。
发生同类原子将偏聚的失稳分解(Spinodal 分解)；这种现象也称为溶解度中断或者溶解度间隙。
- 如果I AB <<0, 异类原子则更倾向于聚合在一起(异类原子偏聚),这称为有序化(有序-无序转变)3. 溶体的性质- 与相互作用能I AB 有关：[][]R ID Em m EAB A BG G G G I X X =+∆∆=-如果异类原子间的吸引力很强，则有序相可以扩展到液相为止（Al-Ni ）；在低温下出现原子的有序排列，系统的内能会降低；在高温下由于混合熵项的作用,有序排列（或偏聚作用）消失。
-异类原子间具有更大的结合强度→ 合金更难熔化，在异类原子键分数最大的成分处有熔点的极大值；异类原子间具有较小的结合强度 → 这个系统中所有合金的熔化温度都在两个纯组元熔点以下，且液固相线出现极小值；A 、B 原子混乱、随机分布，不出现任何类型的原子偏聚，形成无序固溶体，液、固相线都没有极值。
4. 亚规则溶体- 为了修正I AB ≠ 0所带来的混合熵偏差，Gaggenheim 提出一种修正方法-计算溶体混合熵与理想溶体混合熵的偏差∆S E ，考虑了温度和I AB 对混合熵的影响。
- 当温度足够高时，这个偏差很小，那么溶体的混合熵和理想溶体的混合熵接近相等：- I AB 与温度有关：处于晶格结点上的原子间结合能是取定于原子间距离的，因此温度变化时，原子间距变化；I AB 与溶体成分有关：溶体成分变化时，每个原子周围的异类原子的数目要发生变化，如果两种原子的尺寸不同，则溶体成分的变化也要影响原子间的距离；u AA 、u AB 和 u BB 都要随之变化。
- 两种原子混合时振动频率将发生变化．因此，混合焓及混合熵中的线性项是不能严格成立的。
- 亚规则溶体模型的思想：保留规则溶体模型原来的形式，即仍保留I AB 这一参数，并对它进行修正，使之成为成分和温度的函数，同样可以达到准确描述实际溶体的摩尔自由能的目的。
- 这种牺牲物理意义而强调描述效果的亚规则溶体模型在实际的相图计算、相变模拟、化学反应模拟等方面发挥了很大的作用，取得了许多非常重要的成果。
5. 化合物相- 固溶体：晶体结构与其某一组元相同的相，含溶剂和溶质。
- 化合物：组成原子有固定比例，其结构与组成组元均不相同的相。
- 化合物相的主要热力学参数：化合物的生成焓，生成自由能- 等温、等压条件下化学反应的热效应(∆H )等于生成物焓的总和与反应物焓的总和之差，规定用标准摩尔生成焓。
- 根据Kirchhoff 定律，并借助于等压热容，可以求得其它温度的生成焓22)(zRT I X X S AB B A E=∆111.72 5.44 FeCr Cr I X KJ mol αα-=-⋅[][][][][][]⋅⋅⋅+-+-+++=2120211011000))(())((B A AB AB B A AB AB AB AB AB X X T I I X X T I I TI I I dTC T H T H T T P ⎰∆=∆-∆21)()(21差反应物与生成物的热容温度的标准焓温度的标准焓-∆-∆-∆PC T T H T T H 2211)()(-Gibbs-Helmholtz 方程获得生成自由能与温度关系：- 许多金属间化合物材料具有: 高熔点、双相、低比重、高比强等特点；- 金属间化合物可以作为基体(或者强化相)，构成航空航天材料的重要组成部分。
如：中高温Ti-Al 系; 高温Ni-Al 系; 超高温Nb-Si 系；- 并不是所有的金属间化合物都具有高脆性，如TiNi 系合金；-高比强金属间化合物在发动机上应用前景：高、低压涡轮（Ni-Al ，Nb-Si ）；高压压气机（Ti-Al ）6. 化学势与活度- 描述二组元及多组元溶体时，考虑某组元浓度变化对自由能的影响引出偏摩尔量的概念；描述摩尔自由能时，经常采用化学势与活度等偏摩尔热力学函数 - 摩尔热力学函数属强度性质- 纯组元组成溶体后, 由于各组元之间的相互作用以及质点间排列的变化, 溶体的热力学函数已不再是各组元热力学函数的简单加和, 各组元在溶体中的热力学性质也不能用他们独立存在时的纯组元热力学函数来描述。
- 为描述各组元在溶体中的性质，需要引入偏摩尔量的概念 - 偏摩尔量物理意义是：在等温、等压、保持i 物质以外所有组分的摩尔数不变，改变 dn i 所引起广度性质M 的变化值，或在等温、等压条件下，在大量的定组成体系中加入1 mol 的i 组元所引起广度性质M 的变化值 - 化学势 －偏摩尔自由能-上坡扩散：固溶体中原子定向迁移的驱动力是化学势梯度，而不是浓度梯度。