材料科学基础(武汉理工大学,张联盟版)课后习题及答案 第八章
《材料科学基础》第二版 (张联盟 著)课后习题答案 武汉理工大学出版社
2-13 根据半径比关系，说明下列离子与O2—配位时的配位数各是多少？已 知rO2－＝0.132nm，rSi4＋＝0.039nm，rK＋＝0.131nm，rAl3＋＝0.057nm，rMg2＋＝0.078nm。
2-14 为什么石英不同系列变体之间的转化温度比同系列变体之间的转化温度高得多？
2-15 有效离子半径可通过晶体结构测定算出。在下面NaCl型结构晶体中，测得MgS和MnS的晶胞参数均 为a＝0.520nm（在这两种结构中，阴离子是相互接触的）。
为NaCl型结构时，体积变化的百分数是多少？已
知CN＝6时，rMn
2＋＝0.08nm，rS2
－＝0.184nm；CN
＝4时，rMn
2＋＝0.073nm，r
2 S
－＝0.167nm。
2-25 钛酸钡是一种重要的铁电陶瓷，其晶型是钙钛矿结构，试问：（1）属于什么点阵？（2）这个结构 中离子的配位数为若干？（3）这个结构遵守鲍林规则吗？请作充分讨论。
2-5 依据结合力的本质不同，晶体中的键合作用分为哪几类？其特点是什么？
2-6 等径球最紧密堆积的空隙有哪两种？一个球的周围有多少个四面体空隙、多少个八面体空 隙？
2-7 n个等径球作最紧密堆积时可形成多少个四面体空隙、多少个八面体空隙？不等径球是如何进行堆积 的？
2-8 写出面心立方格子的单位平行六面体上所有结点的坐标。
2-14 为什么石英不同系列变体之间的转化温度比同系列变体之间的转化温度高得多？
答： 石英同一系列之间的转变是位移性转变，不涉及晶体结构中键的破裂和重建，仅是键长、键角的调 整、需要能量较低，且转变迅速可逆；而不同系列之间的转变属于重建性转变，都涉及到旧键的破裂和新键 的重建，因而需要较的能量，且转变速度缓慢；所以石英不同系列之间的转化温度比同系列变体之间转化的 温度要高的多。
材料科学基础习题与参考答案（doc14页）（优质版）
材料科学基础习题与参考答案（doc14页）（优质版）第一章材料的结构一、解释以下基本概念空间点阵、晶格、晶胞、配位数、致密度、共价键、离子键、金属键、组元、合金、相、固溶体、中间相、间隙固溶体、置换固溶体、固溶强化、第二相强化。
二、填空题1、材料的键合方式有四类，分别是（），（），（），（）。
2、金属原子的特点是最外层电子数（），且与原子核引力（），因此这些电子极容易脱离原子核的束缚而变成（）。
3、我们把原子在物质内部呈（）排列的固体物质称为晶体，晶体物质具有以下三个特点，分别是（），（），（）。
4、三种常见的金属晶格分别为（），（）和（）。
5、体心立方晶格中，晶胞原子数为（），原子半径与晶格常数的关系为（），配位数是（），致密度是（），密排晶向为（），密排晶面为（），晶胞中八面体间隙个数为（），四面体间隙个数为（），具有体心立方晶格的常见金属有（）。
6、面心立方晶格中，晶胞原子数为（），原子半径与晶格常数的关系为（），配位数是（），致密度是（），密排晶向为（），密排晶面为（），晶胞中八面体间隙个数为（），四面体间隙个数为（），具有面心立方晶格的常见金属有（）。
7、密排六方晶格中，晶胞原子数为（），原子半径与晶格常数的关系为（），配位数是（），致密度是（），密排晶向为（），密排晶面为（），具有密排六方晶格的常见金属有（）。
8、合金的相结构分为两大类，分别是（）和（）。
9、固溶体按照溶质原子在晶格中所占的位置分为（）和（），按照固溶度分为（）和（），按照溶质原子与溶剂原子相对分布分为（）和（）。
10、影响固溶体结构形式和溶解度的因素主要有（）、（）、（）、（）。
11、金属化合物（中间相）分为以下四类，分别是（），（），（），（）。
12、金属化合物（中间相）的性能特点是：熔点（）、硬度（）、脆性（），因此在合金中不作为（）相，而是少量存在起到第二相（）作用。
13、CuZn、Cu5Zn8、Cu3Sn的电子浓度分别为（），（），（）。
《材料科学基础》课后答案章
第一章8．计算下列晶体的离于键与共价键的相对比例 (1)NaF (2)CaO (3)ZnS解：1、查表得：X Na =0.93,X F =3.98根据鲍林公式可得NaF 中离子键比例为：21(0.93 3.98)4[1]100%90.2%e ---⨯=共价键比例为：1-90.2%=9.8% 2、同理，CaO 中离子键比例为：21(1.00 3.44)4[1]100%77.4%e---⨯=共价键比例为：1-77.4%=22.6%3、ZnS 中离子键比例为：21/4(2.581.65)[1]100%19.44%ZnS e --=-⨯=中离子键含量共价键比例为：1-19.44%=80.56%10说明结构转变的热力学条件与动力学条件的意义．说明稳态结构与亚稳态结构之间的关系。
答：结构转变的热力学条件决定转变是否可行，是结构转变的推动力，是转变的必要条件；动力学条件决定转变速度的大小，反映转变过程中阻力的大小。
稳态结构与亚稳态结构之间的关系：两种状态都是物质存在的状态，材料得到的结构是稳态或亚稳态，取决于转交过程的推动力和阻力(即热力学条件和动力学条件)，阻力小时得到稳态结构，阻力很大时则得到亚稳态结构。
稳态结构能量最低，热力学上最稳定，亚稳态结构能量高，热力学上不稳定，但向稳定结构转变速度慢，能保持相对稳定甚至长期存在。
但在一定条件下，亚稳态结构向稳态结构转变。
第二章1．回答下列问题：(1)在立方晶系的晶胞内画出具有下列密勒指数的晶面和晶向：(001）与[210]，(111）与[112]，(110)与[111],(132)与[123],(322)与[236］(2)在立方晶系的一个晶胞中画出（111)和（112)晶面，并写出两晶面交线的晶向指数。
(3)在立方晶系的一个晶胞中画出同时位于（101).(011)和（112)晶面上的[111]晶向。
解：1、2．有一正交点阵的a=b,c=a/2。
某晶面在三个晶轴上的截距分别为6个、2个和4个原子间距，求该晶面的密勒指数。
材料科学基础课后习题答案
《材料科学基础》课后习题答案第一章材料结构的基本知识4. 简述一次键和二次键区别答：根据结合力的强弱可把结合键分成一次键和二次键两大类。
其中一次键的结合力较强，包括离子键、共价键和金属键。
一次键的三种结合方式都是依靠外壳层电子转移或共享以形成稳定的电子壳层，从而使原子间相互结合起来。
二次键的结合力较弱，包括范德瓦耳斯键和氢键。
二次键是一种在原子和分子之间，由诱导或永久电偶相互作用而产生的一种副键。
6. 为什么金属键结合的固体材料的密度比离子键或共价键固体为高？答：材料的密度与结合键类型有关。
一般金属键结合的固体材料的高密度有两个原因：（1）金属元素有较高的相对原子质量；（2）金属键的结合方式没有方向性，因此金属原子总是趋于密集排列。
相反，对于离子键或共价键结合的材料，原子排列不可能很致密。
共价键结合时，相邻原子的个数要受到共价键数目的限制；离子键结合时，则要满足正、负离子间电荷平衡的要求，它们的相邻原子数都不如金属多，因此离子键或共价键结合的材料密度较低。
9. 什么是单相组织？什么是两相组织？以它们为例说明显微组织的含义以及显微组织对性能的影响。
答：单相组织，顾名思义是具有单一相的组织。
即所有晶粒的化学组成相同，晶体结构也相同。
两相组织是指具有两相的组织。
单相组织特征的主要有晶粒尺寸及形状。
晶粒尺寸对材料性能有重要的影响，细化晶粒可以明显地提高材料的强度，改善材料的塑性和韧性。
单相组织中，根据各方向生长条件的不同，会生成等轴晶和柱状晶。
等轴晶的材料各方向上性能接近，而柱状晶则在各个方向上表现出性能的差异。
对于两相组织，如果两个相的晶粒尺度相当，两者均匀地交替分布，此时合金的力学性能取决于两个相或者两种相或两种组织组成物的相对量及各自的性能。
如果两个相的晶粒尺度相差甚远，其中尺寸较细的相以球状、点状、片状或针状等形态弥散地分布于另一相晶粒的基体内。
如果弥散相的硬度明显高于基体相，则将显著提高材料的强度，同时降低材料的塑韧性。
材科基
NJ University of Technology
NJ University of Technology
NJ University of Technology
NJ University of Technology
NJ University of Technology
第六章、扩散与固相反应
பைடு நூலகம்
NJ University of Technology
第三章、非晶体固体
1、熔体的概念：不同聚合程度的各种聚合物的混合物 硅酸盐熔体的粘度与组成的关系 2、非晶态物质的特点 3、玻璃的通性（4个） 4、 Tg 、Tf ,相对应的粘度和特点 5、网络形成体、网络变化体、网络中间体
NJ University of Technology
第一章、晶体结构基础
1、晶体的基本概念 晶体的本质：质点在三维空间成周期性重复排列 晶体的基本性质：结晶均一性、各向异性、自限性、对称性、 最小内能性 2、对称的概念 物体中的相同部分作有规律的重复 对称要素：对称面、对称中心、对称轴（对称轴的类型和特点） 4次倒转轴不能被其他的对称要素及其组合取代 （L1、L2、L3、L4、L6、C、P、 L4i ） 对称操作:借助对称要素，使晶体的相同部分完全重复的操作 对称要素的组合必须满足晶体的整体对称要求，不是无限的
材料科学基础第8章习题答案
第8章 习题答案8.2 什么叫相变？按照相变机理来划分，可分为哪些相变？相变：随自由能变化而发生的相结构的变化。
按相变机理可分为：成核-生长机理、连续型相变、马氏体相变、有序-无序转变。
8.10为什么在成核一生成机理相变中，要有一点过冷或过热才能发生相变? 什么情况下需过冷，什么情况下需过热，试证明之。
解：由热力学可知，在等温、等压下有：G H T S ∆=∆-∆在平衡条件下，0G ∆=，则有00H T S ∆-∆=，0/S H T ∆=∆式中： 0T 是相变的平衡温度；H ∆为相变热。
若在任意温度T 的不平衡条件下，则有0G H T S ∆=∆-∆≠若H ∆与S ∆不随温度而变化，将上式代入上式得：0000/T T T G H T H T H H T T -∆∆=∆-∆=∆=∆ 可见，相变过程要自发进行，必须有0G ∆<，则0/0H T T ∆∆<。
若相变过程放热（如凝聚、结晶等）0H ∆<。
要使0G ∆<，必须有0T ∆>，00T T T ∆=->，即0T T >，这表明系统必须“过冷”。
若相变过程吸热（如蒸发、熔融等）0H ∆>，要满足0G ∆<这一条件则必须0T ∆<，即0T T <，这表明系统要自发相变则必须“过热”。
8.13 铁的原子量为55.84 ，密度为 7.32 克 /cm 3，熔点为 1593 ℃，熔化热为 11495 J/mol ，固液界面能为 2.04×10-5 J/cm 2，试求在过冷度为 10 ℃、 100 ℃时的临界晶核大小并估计这些晶核分别由多少个晶胞所组成（已知铁为体心立方晶格，晶格常数a =3.05 nm)解: 当过冷度为10℃时337.311495108.0555.841866m V m H Tg cm J mol G J cm MT g mol ρ∆∆-⨯∆==⨯=-℃℃ 52*6322 2.0410 5.110518.05LS V r J cm r cm nm G J cm--⨯⨯=-=-=⨯=∆- 晶核体积：()3*143V r π= 晶胞体积：32V a = 因此，晶胞个数：()()3*373344 3.145133 1.95100.305r n a π⨯⨯===⨯个 当过冷度为100℃时 337.31149510080.555.841866m V m H Tg cm J mol G J cm MT g mol ρ∆∆-⨯∆==⨯=-℃℃ 52*7322 2.0410 5.110 5.180.5LS V r J cm r cm nm G J cm--⨯⨯=-=-=⨯=∆- 因此，晶胞个数：()()()33*43344 3.14 5.133 1.95100.305r n a π⨯⨯===⨯个 8.14 熔体析晶过程在 1000 ℃时，单位体积自由焓变化418 J/cm 3 ；在 900 ℃时是 2090 J/cm 3 。
材料科学基础(武汉理工大学,张联盟版)课后习题及答案 - 副本
第二章答案2-2（1）一晶面在x、y、z轴上的截距分别为2a、3b、6c，求该晶面的晶面指数；（2）一晶面在x、y、z轴上的截距分别为a/3、b/2、c，求出该晶面的晶面指数。
答：（1）h:k:l==3:2:1,∴该晶面的晶面指数为（321）；（2）h:k:l=3:2:1，∴该晶面的晶面指数为（321）。
2-3在立方晶系晶胞中画出下列晶面指数和晶向指数：（001）与[]，（111）与[]，（）与[111]，（）与[236]，（257）与[]，（123）与[]，（102），（），（），[110]，[]，[]答：2-4定性描述晶体结构的参量有哪些？定量描述晶体结构的参量又有哪些？答：定性：对称轴、对称中心、晶系、点阵。
定量：晶胞参数。
2-5依据结合力的本质不同，晶体中的键合作用分为哪几类？其特点是什么？答：晶体中的键合作用可分为离子键、共价键、金属键、范德华键和氢键。
离子键的特点是没有方向性和饱和性，结合力很大。
共价键的特点是具有方向性和饱和性，结合力也很大。
金属键是没有方向性和饱和性的的共价键，结合力是离子间的静电库仑力。
范德华键是通过分子力而产生的键合，分子力很弱。
氢键是两个电负性较大的原子相结合形成的键，具有饱和性。
2-6等径球最紧密堆积的空隙有哪两种？一个球的周围有多少个四面体空隙、多少个八面体空隙？答：等径球最紧密堆积有六方和面心立方紧密堆积两种，一个球的周围有8个四面体空隙、6个八面体空隙。
2-7n个等径球作最紧密堆积时可形成多少个四面体空隙、多少个八面体空隙？不等径球是如何进行堆积的？答：n个等径球作最紧密堆积时可形成n个八面体空隙、2n个四面体空隙。
不等径球体进行紧密堆积时，可以看成由大球按等径球体紧密堆积后，小球按其大小分别填充到其空隙中，稍大的小球填充八面体空隙，稍小的小球填充四面体空隙，形成不等径球体紧密堆积。
2-8写出面心立方格子的单位平行六面体上所有结点的坐标。
答：面心立方格子的单位平行六面体上所有结点为：（000）、（001）（100）（101）（110）（010）（011）（111）（0）（0）（0）（1）（1）（1）。
张联盟《材料科学基础》(第2版)笔记和课后习题答案复习考研资料
张联盟《材料科学基础》（第2版）笔记和课后习题（含考研真题）详解完整版>精研学习网>免费在线试用20％资料全国547所院校视频及题库资料考研全套>视频资料>课后答案>往年真题>职称考试目录隐藏第1章材料引言1.1复习笔记1.2课后习题详解1.3名校考研真题详解第2章晶体结构2.1复习笔记2.2课后习题详解2.3名校考研真题详解第3章晶体结构缺陷3.1复习笔记3.2课后习题详解3.3名校考研真题详解第4章非晶态结构与性质4.1复习笔记4.2课后习题详解4.3名校考研真题详解第5章表面结构与性质5.1复习笔记5.2课后习题详解5.3名校考研真题详解第6章相平衡和相图6.1复习笔记6.2课后习题详解6.3名校考研真题详解第7章基本动力学过程—扩散7.1复习笔记7.2课后习题详解7.3名校考研真题详解第8章材料中的相变8.1复习笔记8.2课后习题详解8.3名校考研真题详解第9章材料制备中的固态反应9.1复习笔记9.2课后习题详解9.3名校考研真题详解第10章烧结10.1复习笔记10.2课后习题详解10.3名校考研真题详解第11章腐蚀与氧化11.1复习笔记11.2课后习题详解11.3名校考研真题详解第12章疲劳与断裂12.1复习笔记12.2课后习题详解12.3名校考研真题详解内容简介隐藏本书是张联盟主编的《材料科学基础》（第2版）的学习辅导书，主要包括以下内容：（1）梳理知识脉络，浓缩学科精华。
本书每章的复习笔记均对该章的重难点进行了整理，并参考了国内名校名师讲授该教材的课堂笔记。
因此，本书的内容几乎浓缩了该教材的知识精华。
（2）详解课后习题，巩固重点难点。
本书参考大量相关辅导资料，对张联盟主编的《材料科学基础》（第2版）的课后习题进行了详细的分析和解答，并对相关重要知识点进行了延伸和归纳。
（3）精选考研真题，培养解题思路。
本书从历年考研真题中挑选具有代表性的部分，并对之做了详尽的解析。
材料科学基础(武汉理工大学,张联盟版)课后习题及答案之欧阳家百创编
第二章答案欧阳家百（2021.03.07）2-1略。
2-2（1）一晶面在x、y、z轴上的截距分别为2a、3b、6c，求该晶面的晶面指数；（2）一晶面在x、y、z轴上的截距分别为a/3、b/2、c，求出该晶面的晶面指数。
答：（1）h:k:l==3:2:1,∴该晶面的晶面指数为（321）；（2）h:k:l=3:2:1，∴该晶面的晶面指数为（321）。
2-3在立方晶系晶胞中画出下列晶面指数和晶向指数：（001）与[]，（111）与[]，（）与[111]，（）与[236]，（257）与[]，（123）与[]，（102），（），（），[110]，[]，[]答：2-4定性描述晶体结构的参量有哪些？定量描述晶体结构的参量又有哪些？答：定性：对称轴、对称中心、晶系、点阵。
定量：晶胞参数。
2-5依据结合力的本质不同，晶体中的键合作用分为哪几类？其特点是什么？答：晶体中的键合作用可分为离子键、共价键、金属键、范德华键和氢键。
离子键的特点是没有方向性和饱和性，结合力很大。
共价键的特点是具有方向性和饱和性，结合力也很大。
金属键是没有方向性和饱和性的的共价键，结合力是离子间的静电库仑力。
范德华键是通过分子力而产生的键合，分子力很弱。
氢键是两个电负性较大的原子相结合形成的键，具有饱和性。
2-6等径球最紧密堆积的空隙有哪两种？一个球的周围有多少个四面体空隙、多少个八面体空隙？答：等径球最紧密堆积有六方和面心立方紧密堆积两种，一个球的周围有8个四面体空隙、6个八面体空隙。
2-7n个等径球作最紧密堆积时可形成多少个四面体空隙、多少个八面体空隙？不等径球是如何进行堆积的？答：n个等径球作最紧密堆积时可形成n个八面体空隙、2n个四面体空隙。
不等径球体进行紧密堆积时，可以看成由大球按等径球体紧密堆积后，小球按其大小分别填充到其空隙中，稍大的小球填充八面体空隙，稍小的小球填充四面体空隙，形成不等径球体紧密堆积。
2-8写出面心立方格子的单位平行六面体上所有结点的坐标。
武汉理工大学材料科学基础课后习题和答案
第一章绪论1、仔细观察一下白炽灯泡，会发现有多少种不同的材料每种材料需要何种热学、电学性质2、为什么金属具有良好的导电性和导热性3、为什么陶瓷、聚合物通常是绝缘体4、铝原子的质量是多少若铝的密度为2.7g/cm3，计算1mm3中有多少原子5、为了防止碰撞造成纽折，汽车的挡板可有装甲制造，但实际应用中为何不如此设计说出至少三种理由。
6、描述不同材料常用的加工方法。
7、叙述金属材料的类型及其分类依据。
8、试将下列材料按金属、陶瓷、聚合物或复合材料进行分类：黄铜钢筋混凝土橡胶氯化钠铅-锡焊料沥青环氧树脂镁合金碳化硅混凝土石墨玻璃钢9、 Al2O3陶瓷既牢固又坚硬且耐磨，为什么不用Al2O3制造铁锤第二章晶体结构1、解释下列概念晶系、晶胞、晶胞参数、空间点阵、米勒指数（晶面指数）、离子晶体的晶格能、原子半径与离子半径、配位数、离子极化、同质多晶与类质同晶、正尖晶石与反正尖晶石、反萤石结构、铁电效应、压电效应.2、（1）一晶面在x、y、z轴上的截距分别为2a、3b、6c，求出该晶面的米勒指数；（2）一晶面在x、y、z轴上的截距分别为a/3、b/2、c，求出该晶面的米勒指数。
3、在立方晶系的晶胞中画出下列米勒指数的晶面和晶向：（001）与[210]，（111）与[112]，（110）与[111]，（322）与[236]，（257）与[111]，（123）与[121]，（102），（112），（213）， [110]，[111]，[120]，[321]4、写出面心立方格子的单位平行六面体上所有结点的坐标。
5、已知Mg2+半径为，O2-半径为，计算MgO晶体结构的堆积系数与密度。
6、计算体心立方、面心立方、密排六方晶胞中的原子数、配位数、堆积系数。
7、从理论计算公式计算NaC1与MgO的晶格能。
MgO的熔点为2800℃，NaC1为80l℃, 请说明这种差别的原因。
8、根据最密堆积原理，空间利用率越高，结构越稳定，金钢石结构的空间利用率很低(只有％)，为什么它也很稳定9、证明等径圆球面心立方最密堆积的空隙率为25．9％；10、金属镁原子作六方密堆积，测得它的密度为1.74克/厘米3，求它的晶胞体积。
材料科学基础 第二版(张联盟 著)课后习题答案武汉理工大学出版
材料科学基础第二版(张联盟著)课后习题答案武汉理工大学出版材料科学基础第二版(张联盟著)课后习题大家找到答案了吗?下面是为大家推荐一些材料科学基础第二版(张联盟著)课后习题答案和下载地址，希望大家有用哦。
2-1名词解释晶系晶胞晶胞参数空间点阵晶面指数晶格能原子半径与离子半径配位数离子极化同质多晶与类质同晶正尖晶石与反正尖晶石反萤石结构铁电效应压电效应热释电效应电光效应2-2(1)一晶面在x、y、z轴上的截距分别为2a、3b、6c，求该晶面的晶面指数;(2)一晶面在x、y、z轴上的截距分别为a/3、b/2、c，求出该晶面的晶面指数。
2-3在立方晶系晶胞中画出下列晶面指数和晶向指数：(001)与[]，(111)与[]，()与[111]，()与[236]，(257)与[]，(123)与[]，(102)，()，()，[110]，[，[]，[]2-4定性描述晶体结构的参量有哪些?定量描述晶体结构的参量又有哪些?2-5依据结合力的本质不同，晶体中的键合作用分为哪几类?其特点是什么?2-6等径球最紧密堆积的空隙有哪两种?一个球的周围有多少个四面体空隙、多少个八面体空隙?2-7n个等径球作最紧密堆积时可形成多少个四面体空隙、多少个八面体空隙?不等径球是如何进行堆积的?2-8写出面心立方格子的单位平行六面体上所有结点的坐标。
2-9计算面心立方、密排六方晶胞中的原子数、配位数、堆积系数。
2-10根据最密堆积原理，空间利用率越高，结构越稳定，金刚石结构的空间利用率很低(只有34.01%)，为什么它也很稳定?2-11证明等径圆球六方最密堆积的空隙率为25.9%。
2-12金属镁原子作六方密堆积，测得它的密度为1.74g/cm3，求它的晶胞体积。
2-13根据半径比关系，说明下列离子与O2—配位时的配位数各是多少?已知rO2-=0.132nm，rSi4+=0.039nm，rK+=0.131nm，rAl3+=0.057nm，rMg2+=0.078nm。
材料科学基础课后习题答案8
SOLUTIONS FOR CHAPTER 81. FIND: When a phase transformation occurs such as a liquid phase transforming to asolid below its melting temperature, what are the two steps involved in the process?Briefly describe each.SOLUTION: During a phase transformation such as a liquid transforming to solid, there are two steps involved in the process. They are:1. nucleation of the new phase, and2. growth of the phase.Nucleation relates to the formation of the new phase and the development of theinterface seperating the two phases. Nucleation can either occur randomly throughout the structure - termed homogeneous nucleation or at specific sites such as interfaces - termed heterogeneous nucleation.Growth - Once the phases has nucleated, it begins to grow. The growth process iscontrolled by diffusion and undercooling. As in the nucleation step, there may becompeting processes that lead to a maximum growth rate at an intermediate temperature.2. FIND: We presented a derivation in Section 8.2.3 showing that the barrier for nucleation,∆G*, decreases with increasing undercooling following the proportionalityBy starting with an expression for the free energy of a distribution of spherical particles of radius r, derive equation 8.2-9a. Explain each step in the derivation. Explain anyassumptions that are made.SOLUTION: To determine the barrier to the nucleation process, ∆G* we begin bynoting that the free energy as a function of particle size for homogeneous nucleation has two terms, one that increases with r2, the interfacial energy per unit volume term, and one that decreases with r3. A maximum occurs in ∆G(r) at some r*. These graphicalrelationships are sketched below.The dependence of the various free energy terms associated with nucleation as a function of temperature: (a) the relationship between cluster radius and surface energy of agrowing spherical solid phase in liquid, (b) the relationship between the cluster radius and(c) the sum of (a) and (b). Instal l Equa tion E ditor a nd do uble -click h ere to view equat ion.The change in free energy can be written as:In this equation we assume that the nuclei can be considered as a random distribution of spheres. To locate the maximum of a function we equate the first derivative of thefunction with respect to the parameter which is the variable to zero. Here we assume thatr is the only variable. The is SL is independent of size and orentation.Using equation 8.2-4 for Instal l Equa tion E ditor a nd do uble -click h ere to view equat ion.we have:In writingwe have assumed that the heat capacity difference between the liquid and solid phases is zero. (Note: Although this may be a reasonable assumption at small undercoolings, i.e.small ∆T’s, at the large under coolings that are typical for homogeneous nucleation thatapproximation may not be valid and a more complex term is required. But for a firstorder approximation this assumption is reasonable.)In order to determine the value of ∆G(r) at r*, we introduce the expressionintoRearranging,If all the terms in parentheses are constant then,3. FIND: Explain the simultaneous influence that undercooling has on the barrier tonucleation and the atomic rearrangements necessary to initiate the transformation. Show how these competing effects lead to classical C-curve behavior in the nucleation ofdiffusional transformations.SOLUTION: With larger undercoolings, both r* and ∆G* decrease, suggesting thatsimply lowering the temperature of the system allows nucleation to occur ever morereadily. However, there are practical kinetic limits to this effect. For example, withdecreased temperature there is a corresponding reduction in atomic mobility. The random fluctuations in the local arrangements of atoms is the process that provides the clusters.Since the formation of the clusters depends on atomic mobility, a reduction in thetemperature reduces the rate of clustering. Thus, as shown in Figure (a) below, theoverall nucleation rate exhibits a maximum at an intermediate temperature. Themaximum in the nucleation rate leads to a minimum in the time required to nucleate aphase, as shown in Figure b. Because of its shape, this curve is known as a C-curve.(a) The influence of temperature on the mobility term and the nucleation barrier term.The opposing processes result in a maximum in the nucleation rate at anintermediate temperature.(b) Since the time for nucleation is inversely related to the nucleation rate, the timecurve exhibits a minimum at an intermediate temperature. Because of its shape,this curve is often referred to as a C-curve.4. FIND: Explain how the value of interfacial energy between the parent phase and thetransforming phase affects the critical radius and the barrier to nucleation.SOLUTION: Equation 8.25 gives the change in free energy as a function of r when aliquid transforms to a solid, for example. In the development of that equation it wasassumed that the transforming phase was spherical and the interfacial energy, γSL, wasisotropic. That equation consists of two terms on the right hand side, i.e.∆G(r) = (4πr2) γSL + 4/3 πr3 (∆G v)Since γSL > 0 and ∆G V < 0 for ∆T > 0, the first term increases with radius, and the second decreases. Figure 8.2-3 illustrates that there is a maximum that occurs at some r wedesignated as r* and a corresponding ∆G, we designated as ∆G* wherer* = (-2γSL)/∆G vandBoth the critical radius, r*, and the barrier to the nucleation process contain γSL in thenumerator. Thinking in terms of the barrier to nucleation, ∆G*, there is a cubicdependence on interfacial energy. The larger the S-L interfacial energy, the larger thebarrier to nucleation and hence the more difficult. The use of nucleating agents is basedon the principle of introducing particles with lower interfacial energies to stimulatenucleation.5. FIND: Compare homogeneous nucleation and heterogeneous nucleation.SOLUTION: The process of homogeneous nucleation occurs at random locations in the parent phase. The distribution of the transforming phase occurs without regard to specific sites, such as mold walls in solidification. Heterogeneous nucleation occurs at specificsites. In the case of solidification they can be at mold walls, unintentional additions such as ceramic inclusions from crucibles or it may occur at nucleating agents which areintentionally added to control the solidification microstructure.6. FIND: What is the difference between the following interfaces?a. coherentb. partially coherent, andc. incoherentSOLUTION: A coherent interface is one in which there is a one-to-one correspondence of atomic planes across the interface. This type of interface occurs when the latticeparameters in the two phases are the same or very close. When the lattice parameters are different in the two phases, the increase in strain energy that will occur as the particlegrows necessitates the periodic insertion of dislocations at the interface to accommodatethe misfit. This type of interface is referred to as partially coherent. An incoherentinterface occurs between two phases of different crystal structures and sufficientlydifferent atomic spacings that can not be accommodated by dislocations.7. FIND: How does interfacial energy vary with coherency?SOLUTION: Interfacial energy is sensitive to the nature of the interface separating thetwo phases. The interfacial energy increases going from coherent to partially coherent to incoherent, i.e. γincoh. > γpart.coh. . γcoh..8. FIND: Based upon your answer to question 7, explain how the probability forheterogeneous nucleation changes as the type of interface changes from coherent to partially coherent to incoherent.SOLUTION: Since the barrier to nucleation, ∆G*, is related to γα/β in the following way:∆G* ∝γα/β3increasing γα/β would increase the barrier to homogeneous nucleation. Consequently, the probability for heterogeneous nucleation would increase as interfacial energy increases. 9. FIND: Figure 5.3-5 contains a schematic illustration of a twin boundary in a crystal.From the point of view of coherency, what is the nature of the type of twin boundaryillustrated in the figure? Comment on the relative energy of a twin boundary compared with a random grain boundary in a polycrystalline material.GIVEN:Figure 5.3-5 illustrates a schematic of a twin in a matrix showing the two twin boundaries separating the matrix from the twin and Figure 8.2-10 illustrates an incoherent interface separating the matrix from a precipitate.Schematic of a twinMatrixPrecipitateSchematic of an incoherent boundarySOLUTION: Atoms that are on the twin plane are part of the stacking sequence in the matrix above the twin plane as well as the stacking of atoms below the twin plane. Sincea coherent interface is an interface that occurs when there is a one-to-one correspondenceacross the interface, then the twin illustrated in this figure would be classified as acoherent twin boundary. Array The incoherent boundary illustrated above occurs in asystem when there is not amatch across the boundaryseparating two phases. Since a general grain boundaryrepresents a situation where theorientation of two grains across a boundary are not the same, we would therefore expect that there would not be a match ofatoms across the boundary.Thus, a coherent twin plane would have lower interfacial energy than the interfacialenergy associated with grain boundary separating two randomly oriented grains.10. FIND: In certain nickel-base superalloys, a second phase can precipitate coherently fromthe matrix during aging because the lattice parameters of the two phases are very closeand both phases are cubic. For a coherent precipitate in this system, what is the mostlikely relationship between the crystallographic axes in the matrix phase and that of theprecipitate? Explain, using sketches.GIVEN: The matrix and precipitate are both cubic with similar lattice parameters.SOLUTION: The best match between two similar cubic structure with similar latticeparameters occurs when the two cubes are aligned with the two sets of orthogonal axesparallel to one another. Thus, the x-axis in the matrix, x m, would be parallel to the x-axis in the precipitate, x p, and the y-axis in the matrix, y m, parallel to the y-axis in theprecipitate, y p. This orientation relationship is sometimes referred to as cube//cubeorientation. That is, the two cubes defining the two crystal structures are oriented x-axisto x-axis and y-axis to y-axis.Furthermore, because the lattice parameters of the two phases are very close, the distances between the atoms in the matrix and theprecipitate match very closely.11. FIND: The transition precipitate γ'(Al2Ag) in the aluminum-silver system is a hexagonalclose-packed structure. The a axis of the HCP phase is 0.2858 nm, and the c axis is0.4607 nm. What crystallographic plane and direction in the aluminum matrix definesthe coherent interface between matrix and precipitate?GIVEN: Crystal structure of aluminum is fcc and that for γ' is hcp. Atomic radius foraluminum is 0.143 nm from Appendix C and the a and c lattice parameters for γ' are0.2858 nm and 0.4607, respectively.SOLUTION: Both structures are close-packed structures. The two structures have the same packing factor and coordination number. Both have sets of planes with the highest possible planar density, the close-packed planes. In the fcc structure these are the {111} and in the hcp they are the {0001}. Both structures have directions with the highestpossible linear density. In the fcc the highest linear density is in the 〈110〉 and in the hcp structure 〈1000〉 . The best potential match between the two structures is across the two highest packing planes. Thus the {0001} of the precipitate and the {111} of thealuminum matrix would form a boundary.Recall that the packing sequence for an fcc structure is ......ABCABCABC....., and thatfor the hcp structure is .....ABABABABAB...... . A schematic of the two structures with a common plane would be:To complete the analysis, the atomic spacings in a 〈111〉 direction of the fcc aluminummatrix must be calculated to determine if a match can be made with these planes andthose of the dense packed planes in the 〈0001〉 directions in the hcp structure of γ'.From Appendix C the atomic radius for aluminum is 0.143 nm, since the atoms touchalong 〈110〉 directions, see accompanying figure, in a fcc structure the lattice parameter, a o , for aluminum is:a o = 0.4045 nm.Schematic showing the dense packed arrangement of atoms in an fcc material like aluminum.For fcc stacking ( .....ABCABCABC.....) the distance between equivalent positions A positions can be determined using the above sketch:= ( 2a o 2 + a o2 ) ½ = a o √3 = 0.7006 nmThus the spacing between the adjacent dense packed layers in the fcc structure is= 0.7006/3 = 0.2335 nm.A schematic showing a comparison between the fcc aluminum and the hcp γ' structures are shown below.Thus the atomic spacing in the dense packed planes are similar, and the spacing between the dense packed planes for the two structures are similar.12. FIND: The expression: X = 1 - exp [ - (kt)n ] , equation 8.2-1, in the text, is a powerfulempirical function that is useful in describing the kinetics of diffusional transformations.In the equation, X is the fraction transformed, k is a rate constant having units ofreciprocal time, t is time and n is a unitless constant. Sketch the behavior of this function over a range of times that demonstrate why this expression is useful for describingmicrostructural changes like recrystallization, or the decomposition of austenite to form pearlite.SOLUTION: At the start of the transformation, t = 0, we expect no transformed phase, hence, X = 0. At very long transformation times, t →∞, we expect thetransformation to be complete, i.e. X= 1. Examining equation 9.2-1 at these twoextremes:At t = 1X = 1 - exp - (k ⋅ 0)n= 1 - 1 = 0, andAt t →∞X = 1 - exp - (k ⋅∞)n= 1 - 0 = 1.Over a range of values the function would behave like the sketch shown below:13.The value of n and k for the decomposition of a particular steel has been investigated by following the fraction transformed versus time. From the experimental data, values for n and k were determined at two temperatures. Plot the fraction transformed as a function ofo o SOLUTION: Using the equation:where X is fraction transformed, and k and n are as given in the table, to calculate the fraction transformed as a function of time.Typical values are summarized in the table below and the behavior of the function is plotted and shown on the next page.-1-114. FIND: Explain how the data from problem 10 could be used to determine the location ofthe start and finish times for the transformation.SOLUTION: Since it is often difficult to decide exactly when a reaction begins and iscompleted, the "start" and "finish" points on an isotherm transformation diagram areusually chosen when, for example, 1% of the parent phase has transformed to indicate thestart of transformation, and 99% has transformed for the finish.The start of the transformation:At 400o C0.01 = 1 - exp [ - (0.028)2 t s2]0.01 - 1 = - exp [ - (0.028)2 t s2]0.99 = exp [ - (0.028)2 t s2]- 0.01 = - (0.028)2 t s20.01 = 0.000784 t s212.75 = t22 ;t = 3.6 sec.At 360o C0.01 = 1 - exp [ - (0.085)2 t s2]0.01 = 0.007 2 t s2t s = 1.4 Sec.For the finish of the transformationAt 400o C0.99 = 1 - exp [ - (0.028)2 t f2]4.605 = 0.0078 t f2t f = 76.6 Sec.At 360o C0.99 = 1 - exp [ - 0.085)2 t f2]4.605 = 0.0072 t f2t f = 25.4 Sec.15. FIND: When fraction transformed as a function of the logarithm of time, plotted at twodifferent temperatures, results in two curves that are parallel (same n but different k), the mechanisms for the transformation are the same. Sometimes we refer to the process asbeing isokinetic. For the kinetic data shown below, plot fraction transformed versus time for the two temperatures. Does the process appear to be isokinetic?Isothermal Transformation Isothermal Transformationo oSOLUTION: The data are plotted in the form of fraction transformed versus time and shown belowSince the curves are parallel the processes appear to be isokinetic.16. FIND: The value of k in equation 8.2-12, for a set of kinetic data, can be determined bynoting:X = 1 - exp - (kt)n, andwhen kt = 1X = 1 - exp (-1) - 1 - 1/e orX = 0.632.What are the values for the rate constants for the data in problem 12 ?SOLUTION: To find the value of k, we need to find the time (t*) at which kt =1. The data is nearly linear at X = .632, therefore we can find this time by@ 415o Ct* = 10.5 sec.kt* = 1 k (10.5) = 1 k = 0.095@ 375o Ct* = 23.9 kt* = 1 k = 0.04217. FIND: When the mechanisms controlling a particular transformation are independent oftemperature, we can define an empirical activation energy, Q, for the process by noting:k = A exp (-Q/RT).Determine the empirical activation energy for the data given in problem 12 .SOLUTION: Since we have two value of k for two different temperatures we can solvetwo equations for Q.from problem 13:k 1 = .042 k 2 = .095 T 1 = 648 K T 2 = 688 KR = 8.314 J/(mole K)Q = 75,600 J/mole.18. FIND: The value of n from a set of data can be found by noting:X = 1 - exp - (kt)n 1-X = exp - (kt)n ln (1 - x) = - (kt)n-ln (1 - x) = ln (1/(1-x)), then ln (1/(1-x)) = (kt)nln ln (1/(1-x)) = n lnt + n lnk.Determine n at the two temperatures from the data in problem 12.SOLUTION: We have data for X vs t and we have already determined k at each temperature (problem 13). Computeand ln(t) for each data point. The slope of a plot of Instal l Equa tion E ditor a nd do uble -click h ere to view equat ion. vs ln(t) is equal ton.@ 415o C n = 2.08@ 375o Cn = 2.0819.FIND: In a diffusion-controlled transformation, the time t at different temperatures to yield the same fraction transformed X is as follows:uble -SOLUTION: By noting that the time for a given fraction transformed is inversely related to the rate of the reaction if the processes have the same activation energy. Then using the Arrhenius:Rate = constant exp (-Q/RT)where Q is the required activation energy, R the gas constant and T the absolute temperature, the activation energy is determined graphically by plotting, ln 1/t versus 1/T.From the graph: slope = - 3629.9Then, Q/R = 3629.9R = 8.314 Joules/mole⋅KQ = 30,180 Joules/mole⋅K20. FIND: Using the data from Problem 19, calculate the time required to yield the samefraction transformed at 125︒C.GIVEN: Data from problem 19.SOLUTION: Least squares fit of the experimental data give an equation of the form y (ln(reciprocal of the time)) = m⋅x (1/T) + bWith the values the expression becomes:y = -3630 ⋅ ( 1/ T ) + 9.8572To calculate the time at 125︒Cy = -3630 ⋅ ( 1/ (125 + 273)) + 9.8572y = 0.7366Taking the inverse ln gives:1/t = 2.089,or, t = .4787 hours21. FIND: The precipitation of carbides in certain steels can increase their strength. Shownbelow are data relating time to reach peak strength and the isothermal hold temperature.From the data, determine the activation energy for the precipitation process. Compareyour results with the activation energy for diffusion of carbon in bcc iron. Explain why the similarities between the two activation energies are not too surprising.T (︒C)Time to Peak Hardness (min.)353 0.9375 3.0400 4.0425 10.0450 20.0Applications to Engineering Materials (8.3)22. FIND: Sketch an isothermal transformation (IT) diagram for a plain carbon eutectoid steel. Label theSOLUTION: An IT diagram for a plain carbon eutectoid steel should contain thefollowing information:1. The temperature of the eutectoid - 727o C.2. The start and finish lines for the diffusional reactions:γ→ pearlite and γ→ bainite.3. The start and finish lines for the athermal martensitetransformation.Shown below is the IT diagram.23. FIND: Sketch a section of the Fe - Fe3C diagram over the composition range 0 to 2 wt%carbon and over the temperature range from 900o C to room temperature. Label the phase fields and compare the information that can be extracted from the phase diagram and the IT diagram.SOLUTION: An IT diagram such as that for a steel provides kinetic information that is, how rapidly austenite decomposes into a particular microstructure. It also provides theinformation regarding the morphology of the phases present. For example, dependingupon the quench temperature either coarse pearlite or fine pearlite may form. These are the same microstructures except for the scale. The lower the transformation temperature, the finer the lamellar spacing. Alternatively, rather than pearlite forming the austenitemight decompose into an alternative two phase structure of finely dispersed carbides in ferrite.Also noted on an IT diagram is the presence of metastable phases such as martensite. On an IT diagram, the temperature range (M s - M f) of the metastable phase is presented.In contrast, the equilibrium diagram provides no information regarding kinetics,morphology or the the presence of metastable phases. Since the diagram is constructed on the requirement of equilibrium, only equilibrate phases will be present.24. FIND: Explain experimentally how you would determine an IT diagram for a particularsteel.SOLUTION: Isothermal transformation (IT) diagrams are constructed by following the isothermal changes in microstructure at different temperatures. Since it is difficult todecide exactly when a reaction starts and is completed the start and finish lines on an IT diagram are arbitrarily defined. The microstructural changes occurring can be followed by monitoring microstructural changes directly using metallographic techniques ortracking a property such as electrical resistivity which is affected by the amount of solute in solution or hardness, which is controlled by the nature of the microstructure. As many microstructure sensitive properties as possible are used to provide sufficient collaborative information for a desired degree of accuracy.25. FIND: Based upon the methods you outlined in question 19, explain the limitations ofusing such a diagram. In particular, explain why the application of an IT diagram isrestricted and may not be applied directly to the production environment.SOLUTION: Since an IT diagram is constructed from numerous samples transformed under isothermal conditions, they can only be used that way. In general, the data for ITdiagrams are collected using thin specimens in order to assure that the requirement ofinstantaneous quench are realized. When larger pieces are considered, especially when considering a production environment where a variety of cross-sections are cooled,instantaneous temperature changes can not be realized. Depending upon the mass of the part, slow cooling may be achieved at the center, whereas rapid cooling may occur near the surface. If the cooling rate is slow, then some microstructural changes will occur as the sample cools down to the desired temperature. Consequently, the continuous cooling (CC) curves were established. These curves recognize that the effects of time andtemperature must be integrated.26. FIND: In our discussions in Chapter 7, we introduce the concept of a phase, and in thischapter we have been concerned with the microstructure of an alloy and ultimately how a particular microstructure affects properties. Explain the difference between the concept of a phase and the phases that are present, and the microstructure of the material.SOLUTION: A phase is a homogeneous portion of matter. A particular microstructure such as 100% martensite can be a single phase microstructure. The microstructure iscomposed of martensite and only martensite. However, a microstructure can bemultiphase. For example, pearlite is a two-phase microstructure which consists oftwophases that form as alternate lamellae of ferrite and cementite. So the concept of amicrostructure includes information about the phase or phases present and theirdistribution.27. FIND: Using the concepts associated with phases and microstructure, explain what ismeant by the terms:a. austeniteb. ferritec. pearlited. bainitee. martensitef. cementite, andg. spheroiditeSOLUTION:(a) austenite - The face-centered cubic (FCC) form of iron.(b) ferrite - The body-centered cubic (BCC) form of iron.(c) pearlite - A two-phase microstructure of alternate ferrite and cementite lamellaeoccurring in some steels. Pearlite forms by the decomposition of austenite.(d) bainite - A two-phase microstructure of ferrite and cementite. Bainite forms whenan austenitic steel is quenched to a temperature below the pearlite region, butabove the martensite start, M s, temperature.(e) martensite - Metastable body-centered tetragonal (BCT) iron phase that issaturated in carbon. Iron carbon martensite forms when austenite is rapidlyquenched to low temperatures. (Note: A wide variety of martensite-liketransformations occur in other systems. Consequently, the term martensite hastaken on a more general concept.)(f) cementite - An iron carbide phase, Fe3C.(g) spheroidite - A two-phase microstructure consisting of spheroidized carbide inferrite, which is formed by heat-treating pearlite, bainite, or martensite at atemperature below the eutectoid temperature.It should be noted that in this list of terms there is a distinction between single-phasematerials - austenite, ferrite, martensite and cementite and microstructures which can be combinations of phases such as pearlite and spheroidite.Problems 28 - 31 can be solved using Figures 8.3-6 and 8.3-11.28. FIND: Thin specimens of a plain carbon steel having eutectoid composition are held at800o C and have been at that temperature long enough to have achieved a complete andhomogeneous austenitic structure. Describe the phases present and the microstructuresthat would occur using the quench paths given below.a. Instantaneous quench to 650o C, hold at that temperature for 200 seconds, andquench to room temperature.b. Instantaneous quench to 300o C, hold for 1000 seconds, and quench to roomtemperature.c. Instantaneous quench to room temperature.d. Instantaneous quench to 500o C, hold for 3 seconds and quench to roomtemperature.SOLUTION: When a eutectoid steel containing approximately 0.8 wt% carbon is taken to 800o C and held, whatever microstructure was present prior to the 800o C exposure isreplaced by single phase, austenite containing 0.8% C.(a) When austenite containing 0.8 wt% carbon is instantaneously quenched to 650o Cand held for 200 seconds it forms the two-phase pearlite microstructure. Since allthe austenite has transformed to pearlite at this temperature and time, no othermicrostructures can form on the quench to room temperature. Consequently, themicrostructure consists of alternate ferrite and cementite lamellae - termed pearlite.The microstructure contains 100% pearlite.(b) When austenite containing 0.8% C is quenched to 300o C and held for 1000seconds approximately 75% of the austenite has transformed to bainite. Theremaining austenite transforms to martensite when the material is quenched toroom temperature. Consequently, the microstructure contains 75% bainite and25% martensite.(c) When the 0.8% C alloy is quenched from 800o C to room temperatureinstantaneously, all the austenite transforms to martensite. Consequently, themicrostructure is 100% martensitic.(d) When the 0.8% C alloy is instantaneously quenched from 800o C to 500o C andheld for 3 seconds, approximately 50% of the austenite transforms to amicrostructure that consists of fine pearlite. The remaining austenite transformsto martensite when the material is quenched to room temperature. Consequently,the microstructure is a mixture of 50% fine pearlite and 50% martensite.29. FIND: Using the same initial conditions outlined in problem 23, describe the phasespresent and the microstructures that would occur using the quench paths given below.a. Instantaneous quench to 650o C, hold for 15 seconds, and quench to roomtemperature.b. Instantaneous quench to 500o C hold for 60 seconds and quench to roomtemperature.c. Instantaneous quench to 170o C hold for 100 seconds and quench to roomtemperature.d. Instantaneous quench to 170o C and hold at that temperature.SOLUTION:。
《材料科学基础》第二版 (张联盟 著)课后习题答案 武汉理工大学出版社
2-26 硅酸盐晶体结构有何特点？怎样表征其化学式？
2-27 硅酸盐晶体的分类依据是什么？可分为那几类，每类的结构特点是什么？ 2
2-28 下列硅酸盐矿物各属何种结构类 型：Mg2[SiO4]，K[AlSi3O8]，CaMg[Si2O6]，Mg3[Si4O10]（OH）2，Ca2Al [AlSiO7]。
2-9 计算面心立方、密排六方晶胞中的原子数、配位数、堆积系数。
2-10 根据最密堆积原理，空间利用率越高，结构越稳定，金刚石结构的空间利用率很低（只 有34.01％），为什么它也很稳定？
2-11 证明等径圆球六方最密堆积的空隙率为25.9％。
2-12 金属镁原子作六方密堆积，测得它的密度为 1.74g/cm3，求它的晶胞体积。
1
若CaS （a＝0.567nm）、CaO（a＝0.480nm）和MgO（a＝0.420nm）为一般阳离子－阴离子接触，试求 这些晶体中各离子的半径。
2-16 氟化锂（LiF）为NaCl型结构，测得其密度为 2.6g/cm3，根据此数据计算晶胞参数，并将此值与你从 离子半径计算得到数值进行比较。
2-17 Li2O的结构是O2－作面心立方堆积，Li＋占据所有四面体空隙位置，氧离子半径为0.132nm。求： （1）计算负离子彼此接触时，四面体空隙所能容纳的最大阳离子半径，并与书末附表Li＋半径比较，说明此 时O2－能否互相接触；（2）根据离子半径数据求晶胞参数；（3）求Li2O的密度。
2-1 名词解释
第二章 晶体结构
晶系 晶胞 晶胞参数 空间点阵 晶面指数 晶格能 原子半径与离子半径 配位数 离子极化 同质多晶与类质同晶 正尖晶石与反正尖晶石 反萤石结构 铁电效应 压电效应 热释电效应 电光效应
2-2 （1）一晶面在x、y、z轴上的截距分别为 2a、3b、 6c，求该晶面的晶面指数；（2）一晶面 在x、y、z轴上的截距分别为a/3、b/2、c，求出该晶面的晶面指数。
材料科学基础(武汉理工大学,张联盟版)课后习题及答案 第八章
材料科学根底〔武汉理工大学，张联盟版〕课后习题及答案第八章第八章答案8-1假设由MgO和Al2O3球形颗粒之间的反响生成MgAl2O4是通过产物层的扩散进行的，〔1〕2画出其反响的几何图形，并推导出反响初期的速度方程。
〔2〕假设1300℃时DAl3＋＞DMg2＋，O－根本不动，那么哪一种离子的扩散控制着MgAl2O4的生成？为什么？解：〔1〕假设:a〕反响物是半径为R0的等径球粒B，x为产物层厚度。
b〕反响物A是扩散相，即A总是包围着B的颗粒，且A，B同产物C是完全接触的，反响自球外表向中心进行。
c〕A在产物层中的浓度梯度是线性的，且扩散截面积一定。
反响的几何图形如图8-1所示：根据转化率G的定义，得将〔1〕式代入抛物线方程中，得反响初期的速度方程为：〔2〕整个反响过程中速度最慢的一步控制产物生成。
D小的控制产物生成，即DMg小，Mg2+2+扩散慢，整个反响由Mg的扩散慢，整个反响由Mg的扩散控制。
8-2镍〔Ni〕在0.1atm的氧气中氧化，测得其质量增量〔μg/cm〕如下表：时间时间温度 1〔h〕 2〔h〕 3〔h〕 4〔h〕 1〔h〕 2〔h〕 3〔h〕 4〔h〕13 23 15 29 20 36 650℃ 29 700℃ 56 41 75 50 88 65 106 22+2+温度 550℃ 9 600℃ 17 〔1〕导出适宜的反响速度方程；〔2〕计算其活化能。
解：〔1〕将重量增量平方对时间t作图，如图8-2所示。
由图可知，重量增量平方与时间呈抛物线关系，即符合抛物线速度方程式。
又由转化率的定义，得将式〔1〕代入抛物线速度方程式中，得反响速度方程为：图8-2重量增量平方与时间关系图〔2〕取各温度下反响1h时进行数据处理拟合，如图8-3所示，1/T G(%) ln[1-(1-G)] -3-1(×10K) 1/3T(℃) 550 600 650 700 图8-3数据处理9 17 29 56 1.22 1.14 1.08 1.03 -3.475 -2.810 -2.227 -1.430 由杨德尔方程可得，对数据作线性回归，得(相关系数为0.98839)由上式得活化能kJ/mol8-3由Al2O3和SiO2粉末反响生成莫来石，过程由扩散控制，如何证明这一点？扩散活化能为209kJ/mol，1400℃下，1h完成10％，求1500℃下，1h和4h 各完成多少？〔应用杨德方程计算〕解：如果用杨德尔方程来描述Al2O3和SiO2粉末反响生成莫来石，经计算得到合理的结果，那么可认为此反响是由扩散控制的反响过程。
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第八章答案
8-1若由MgO和Al2O3球形颗粒之间的反应生成MgAl2O4是通过产物层的扩散进行的，（1）画出其反应的几何图形，并推导出反应初期的速度方程。
（2）若1300℃时D Al3＋＞D Mg2＋，O2－基本不动，那么哪一种离子的扩散控制着MgAl
2O4的生成？为什么？
解：（1）假设:
a）反应物是半径为R0的等径球粒B，x为产物层厚度。
b）反应物A是扩散相，即A总是包围着B的颗粒，且A，B同产物C是完全接触的，反应自球表面向中心进行。
c）A在产物层中的浓度梯度是线性的，且扩散截面积一定。
反应的几何图形如图8-1所示：
根据转化率G的定义，得
将（1）式代入抛物线方程中，得反应初期的速度方程为：
（2
）整个反应过程中速度最慢的一步控制产物生成。
D小的控制产物生成，即D Mg2+
小，
Mg2+
扩散慢，整个反应由Mg2+的扩散慢，整个反应由Mg2+的扩散控制。
8-2镍（Ni）在0.1atm的氧气中氧化，测得其质量增量（μg/cm2）如下表：
温度时间
温度
时间
1（h）2（h）3（h）4（h）1（h）2（h）3（h）4（h）
550℃600℃9
17
13
23
15
29
20
36
650℃
700℃
29
56
41
75
50
88
65
106
（1）导出合适的反应速度方程；（2）计算其活化能。
解：（1）将重量增量平方对时间t作图，如图8-2所示。
由图可知，重量增量平方与时间呈抛物线关系，即符合抛物线速度方程式。
又由转化率的定义，得
将式（1）代入抛物线速度方程式中，得反应速度方程为：
图8-2重量增量平方与时间关系图
（2）取各温度下反应1h时进行数据处理拟合，如图8-3所示，
T(℃)
G(%)
1/T
(×10-3K-1)
ln[1-(1-G)1/3] 550 9
1.22 -3.475
600 17 1.14 -2.810
650 29 1.08 -2.227
700 56 1.03 -1.430
图8-3数据处理
由杨德尔方程可得，
对数据作线性回归，得(相关系数为0.98839)
由上式得活化能kJ/mol
8-3由Al2O3和SiO2粉末反应生成莫来石，过程由扩散控制，如何证明这一点？已知扩散活化能为209kJ/mol，1400℃下，1h完成10％，求1500℃下，1h和4h各完成多少？（应用杨德方程计算）
解：如果用杨德尔方程来描述Al2O3和SiO2粉末反应生成莫来石，经计算得到合理的结果，则可认为此反应是由扩散控制的反应过程。
由杨德尔方程，得
又，故
从而1500℃下，反应1h和4h时，由杨德尔方程，知
所以，在1500℃下反应1h时能完成15.03%，反应4h时能完成28.47%。
8-4比较杨德方程、金斯特林格方程优缺点及适应条件。
解：两个方程都只适用稳定扩散的情况。
杨德尔方程在反应初期具有很好的适应性，但杨氏模型中假设球形颗粒截面始终不变。
因而只适用反应初期转化率较低的情况。
而金斯格林方程考虑了在反应进程中反应截面面积随反应过程变化这一事实，因而金氏方程适用范围更广，可以适合反应初、中期。
8-5粒径为1μm球状Al2O3由过量的MgO微粒包围，观察尖晶石的形成，在恒定温度下，第1h有20％的Al2O3起了反应，计算完全反应的时间。
（1）用杨德方程计算；（2）用金斯特林格方程计算。
解：(1)用杨德尔方程计算：
代入题中反应时间1h和反应进度20%，得
h-1
故完全反应（G=1）所需的时间h
(2)用金斯格林方程计算：
同理，代入题中反应时间1h和反应进度20%，得
h-1
故完全反应（G=1）时，
所以完全反应所需的时间h
8-6当测量氧化铝－水化物的分解速率时，发现在等温反应期间，质量损失随时间线性增加到50％左右，超过50％时质量损失的速率就小于线性规律。
速率随温度指数增加，这是一个由扩散控制的反应，还是由界面一级反应控制的反应？当温度从451℃增至493℃时，速率增大到10倍，计算此过程的活化能。
（利用表8-2及图8-22进行分析）
解：根据表8-2部分重要的固相反应动力学方程及图8-22各种类型反应中G-t/t0.5曲线分析，由题意，知
G<50%，G-t呈线性关系
G>50%，G-t小于线性规律，是由扩散控制的反应，且G2=kt
又因为，代入T1=451℃，T2=493℃，G1=G，G2=10G，得
kJ/mol
8-7由Al2O3和SiO2粉末形成莫来石反应，由扩散控制并符合扬德方程，实验在温度保持不变的条件下，当反应进行1h的时候，测知已有15％的反应物发生了反应。
（1）将在多少时间内全部反应物都生成产物？（2）为了加速莫来石的生成，应采取什么有效措施？
解：（1）由杨德尔方程，得
反应完全(G=1)所需的时间为h
（2）可以采用一切有利扩散的因素来加速莫来石的生成：减小粒度，采用活性反应物(如Al2O3·3H2O)，适当加压等等
8-8试分析影响固相反应的主要因素。
解：（1）反应物化学组成与结构的影响:反应物中质点作用力越大，反应能力越小；同一反应体系中，固相反应速度与各反应物间的比例有关；矿化剂的特殊作用。
（2）颗粒度和分布影响：粒径越小，反应速度越快；同一反应体系中由于物料尺寸不同，反应速度会属于不同动力学范围控制；少量较大尺寸的颗粒存在会显著延缓反应过程的完成。
（3）反应温度的影响：温度越高，质点热运动增强，反应能力和扩散能力增强。
（4）压力、气氛的影响：两固相间的反应，增大压力有助颗粒的接触面积，加速物质传递过程，使反应速度增加；对有液汽相参加的固相反应，提高压力不表现积极作用，甚至适得其反。
（5）矿化剂的影响：晶格能越大，结构越完整和稳定，反应活性越低。
加入矿化剂可以提高固相反应。
8-9如果要合成镁铝尖晶石，可供选择的原料为MgCO3、Mg(OH)2、MgO、Al2O3·3H2O、γ-Al2O3、α-Al2O3。
从提高反应速率的角度出发，选择什么原料较好？请说明原因。
解：应选用MgCO3，Mg(OH)2和Al2O3·3H2O作原料较好。
因为MgCO3，Mg(OH)2在反应中可以发生热分解，Al2O3·3H2O发生脱水反应和晶型转变，将获得具有较大比表面和晶格缺陷的初生态或无定形物质从而提高了反应活性，加剧了固相反应的进行。