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材料科学基础习题第四章答案与翻译

材料科学基础习题第四章答案与翻译
材料科学基础习题第四章答案与翻译

4.2 根据本章给出的结构,画出下列链节结构:

(1)聚氟乙烯:—CH2—CHF—;

(2)聚三氟氯乙烯:—CF2—CFCl—

(3)聚乙烯醇:—CH2—CHOH—

4.3 计算下列聚合物的链节分子量

(1)聚氯乙烯:

—CH2—CHCl— : m = 2?12.011+2?1.008+35.453=61.491g/mol (2)聚对苯二甲酸乙二醇酯:—OCH2-CH2OCOC6H4CO—

m = 10?12.011+8?1.008+4?15.999=192.17g/mol

(3)聚碳酸酯:

m = 16?12.011+14?1.008+3?15.999=254.285g/mol

(4)聚二甲硅氧烷:C2H6OSi

m = 28.086+2?12.011+6?1.008+3?15.999 = 106.153g/mol 4.4 聚丙烯的数均分子量为1,000,000 g/mol,计算其数均聚合度。答:链节为—CH3CH—CH2—,

其分子量:m = 3?12.011+6?1.008=42.081 g/mol

4.5 (a) 计算聚苯乙烯链节的分子量

答:链节为-CHC6H5-CH2-,

分子量:m = 8?12.011+8?1.008=104.152

(b) 计算重均聚合度为25000的聚苯乙烯的重均分子量

答:= 25000?104.152 g/mol = 2603800 g/mol

4.6 下表列出了聚丙烯的分子量,计算(a) 数均分子量

(b) 重均分子量(c) 数均聚合度(d) 重均聚合度

答:(a)= 0.05?12000+0.16?20000+0.24?28000

+0.28?36000+0.20?44000+0.07?52000 = 600+3200

+6720+10080+8800+3640 = 33040 (g/mol)

(b)= 0.02?12000+0.1?20000+0.20?28000

+0.30?36000+0.27?44000+0.21?52000 = 240+2000

+5600+10800+11880+10920 = 41440 (g/mol)

(c)聚丙烯链节的分子量:m = 42.081 g/mol

(d)

4.7 下表列出了某聚合物的分子量分布。计算(a) 数均分子量

(b) 重均分子量(c) 如果已知这一聚合物的重均聚合度为780,指出此聚合物为表4.3所列聚合物中的哪一个?为什么?(d) 这一材料的数均聚合度为多少?

答:(a)= 0.04?22500+0.07?37500+0.16?52500

+0.26?67500+0.24?82500+0.12?97500+0.08?112500

+0.03?127500 = 900+2625+8400+17550+19800+11700

+9000+3825 = 73800 (g/mol)

(b)= 0.01?22500+0.04?37500+0.11?52500

+0.24?67500+0.27?82500+0.16?97500+0.12?112500

+0.05?127500 = 225+1500+5775+16200+22275

+15600+13500+ 6375 = 81450 (g/mol)

(c)

此聚合物为聚苯乙烯

4.8 根据下面的分子量分布和重均聚合度为585的条件,判断是否为聚甲基丙烯酸甲酯均聚物

答:聚甲基丙烯酸甲酯链节分子式为:C5H8O2

(—CH2CH3COOCH3C—);

其分子量m = 5?12.011+8?1.008+2?15.999=100.117g/mol

重均分子量为:

=0.01?14000+0.05?26000+0.12?38000+0.25?50000

+0.27?62000+0.21?74000+0.09?86000

=140+1300+4560+12500+16740+15540+7740=58520

与条件相符,能形成均聚物

4.9 高密度聚乙烯通过诱导氯原子随机取代氢而被氯化。(a) 确定取代5%的原始氢原子所需Cl (wt%) 的浓度(b)如何区别于传统的聚氯乙烯的制备过程而得到氯化的聚乙烯?

答:(a) 聚乙烯链节为:—CH2CH2—;一个链节含有4个氢,现氯取代其中5%的氢,则被氯取代5%氢的链节的分子量为:

m = 2?12.011+4?0.95?1.008+4?0.05?35.453

=24.022+3.8304+7.0906 = 34.943g/mol

其中氯占重量百分数为:4?0.05?35.453/34.943 = 20.3 wt%

(b) 聚氯乙烯是聚乙烯的改性产品。PVC树脂经过氯化后,分子链排列的不规则性增加,极性增加,使树脂的溶解性增大,化学稳定性增加,从而提高了材料的耐热性及耐酸、碱、盐、氧化剂等的腐蚀的性能。CPVC是一种应用前景广阔的新型工程塑料。在富含氯的氯仿和四氯化碳为溶剂,溶解PVC后进行氯化反应。…..

4.10 关于聚合物链,configuration(构型)和conformation (构象)间的区别是什么?

Conformation is the spatial arrangement of atoms in a molecules that can come about through free rotation of atoms about a single chemical bond. It can be changed without breaking bonds. Example rotation about single bonds produce the cis-trans, and E-Z conformations, especially of organic molecules.由于分子中的某个原子(或原子基团)绕C-C单键自由旋转而形成的不同的暂时性的易变的空间结构形式,不同的构象之间可以相互转变,在各种构象形式中,势能最低,最稳定的构象是优势构象。

On the contrary, configuration is permanent geometry(各个原子特有的固定的空间排列). It refers to spatial arrangement of bonds and can be

changed only by breaking bonds. Example: L- & D and R- &S-configurations of organic molecules can only be changed by breaking one or more bonds connecting the chiral atom.

4.11 对于一个线型聚合物分子,其总链长L依赖于链原子间的键长

d、分子中的总键数N和骨架链中邻近原子间的键角θ,其关系如下:

而且,一系列聚合物分子的平均末端距r为:,线型聚四氟乙烯的数均分子量为500,000g/mol;计算这一材料的平均L和r值。

答:PTFE的链节为:—CF2CF2—,链节的分子量为:

m = 2?12.011+4?18.998 = 100.014 g/mol

聚合物的数均聚合度为:

每个链节包含有2个C原子,因此有2个C—C单键,聚合物中总键数N = 5000?2 = 10000。C—C单键键长:d = 0.154 nm,键角:θ = 109?,则聚合物链总长及末端距各为:

4.12 利用聚合物链分子总长度L及链平均末端距r的定义,关于线型聚乙烯(a) 确定L=2500 nm时的数均分子量;(b) 确定r = 20 nm 时的数均分子量

答:(a) PE的链节为:—CH2CH2—,d=0.154 nm,键角:θ = 109?由关系式得出:

每个链节包含2个C原子,2个C-C单键,∴聚合物的聚合度为:19940/2= 9970

聚乙烯链节的分子量m = 2?12.011+4?1.008 = 28.054 g/mol:

数均分子量

= 9970?28.054 = 279698 g/mol

(b) 由关系式得出:

对应的聚合度为: 16866/2=8433,数均分子量

= 8433?28.054 = 236579 g/mol

4.13 (a) 根据加热后的力学特性及(b) 可能的分子结构比较热塑性和热固性聚合物

答:热塑性聚合物在一定温度范围内,能反复加热软化和冷却硬化的性能,线型或支链型聚合物具有这种性能,很容易进行挤出,注射或吹塑等成型加工。热固性聚合物加热后产生化学变化,逐渐硬化成型,反应不可逆,再加热也不软化,也不能溶解。热固性聚合物其分子结构为体型,它包括大部分的缩合树脂,其优点是耐热性高,受压不易变形,其缺点是机械性能较差。

4.14 聚酯有些是热塑性有些是热固性。指出其中的原因之一。

答:聚酯是由多元醇和多元酸缩聚而得,并在大分子主链的重复单元

中含有结构的高聚物的统称。如果由饱和的直链结构的多元羧酸和多元醇通过缩聚反应可制得线型聚合物,柔韧性好,如果采用苯环的多元酸与多元醇反应,合成得到含有苯环结构的树脂,苯环的刚性特征赋予树脂以硬度,而苯环的稳定的结构特征赋予树脂以耐化学性。如果由不饱和多元羧酸与多元醇缩聚,则得到的聚合物在加热时可发生交联,成为热固性聚合物。

4.15 (a) 能否粉碎和再利用酚醛树脂?说明相应原因。

(b) 能否粉碎再利用聚丙烯,说明相应原因。

答:酚类和醛类缩聚而成的合成树脂的总称。通常指由苯酚或其同系物(如甲酚、二甲酚)和甲醛作用而得的液态或固态产品,根据所用原料的类型、酚与醛的配比、催化剂的类型的不同,可制得热塑性和热固性两类不同的树脂。热塑性酚醛树脂(诺伏腊克树脂)受热时仅熔化而不能变为不溶不熔状态。但在加入固化剂(如六亚甲基四胺)后则能转变为热固性。

4.16 画出(a) 间同立构(b) 无规立构(c) 全同立构的线型聚苯乙烯分子的草图。

H H H

—C—C—C—C—C—C—C—全同异构

H H H H H H

H H H H H H H H

C—C—C—C—C—C—C—C—C—C 间同异构

H H

H H H H H

H H H H

H H

—C —C —C —C —C —C —C —C —C —C 无规异构 H H H H H

H

H H

4.17 画出(a)丁二烯和 (b)氯丁二烯的顺式和反式链节结构图 答:(a) C C 顺式结构 C C 反式结构

(b) C C 顺式结构

C C 反式结构

4.18 画出下列交替共聚物中每个链节结构草图 (a) 聚(丁二烯-氯丁二烯) (b) 聚(苯乙烯-甲基丙烯酸甲酯)和 (c) 聚(丙烯氰-氯乙烯)

4.19 聚(苯乙烯-丁二烯)交替共聚物的数均分子量为CH 2

H CH 2 H H CH 2 CH 2

H Cl CH 2

CH 2 H Cl CH 2

H CH 2

1,350,000g/mol,计算一个分子链中苯乙烯和丁二烯链节的平均数量答:因为是交替共聚物,一个分子链中两种链节的数量相等。因此两种合在一起考虑成一个链节。苯乙烯含有8个C和8个H,丁二烯含有4个C和6个H,则链节的分子量为:

m = 12?12.011+14?1.008 = 158.244 g/mol

∴在一个分子链中平均有8531个两种链节。

4.20 计算丁二烯链节占0.3的无规丁腈橡胶[聚(丙烯氰-丁二烯)共聚物]的数均分子量?假定这一浓度对应的数均聚合度为2000。

答:按题意,丙烯氰链节占0.7

m(丁二烯)= 4?12.011+6?1.008 = 54.092 g/mol

m(丙烯氰) = 3?12.011+3?1.008 +14.007 = 53.064 g/mol

数均分子量= 2000?0.3?54.092+ 2000?0.7?53.064

=32455.2+74289.6 = 106,745 (g/mol)

4.21 某一交替共聚物的数均分子量为250,000 g/mol,数均聚合度为

3420。如果其中的一种链节为苯乙烯,另一个链节为乙烯、丙烯、四氟乙烯和氯乙烯中的哪一个?

答:因为是交替共聚物,因此链节的平均分子量中,两种链节摩尔分数相同。

m(苯乙烯) = 8?12.011+8?1.008 = 104.152 (g/mol)

m(乙烯) = 2?12.011+4?1.008 = 28.054 (g/mol)

m(丙烯) = 3?12.011+6?1.008 = 42.081 (g/mol)

m(四氟乙烯) = 2?12.011+4?18.998 = 100.014 (g/mol)

m(氯乙烯) = 2?12.011+3?1.008+35.453= 62.499 (g/mol)

73?2 -104 = 42 (g/mol), 另一个链节为丙烯。

4.22 (a) 丁二烯与苯乙烯共聚的共聚物的重均分子量为

350,000g/mol,重均聚合度为4425,确定共聚物中丁二烯

与苯乙烯链节之比

(b) 上述共聚物属于无规,交替,接枝和嵌段共聚物的哪一种

类型

答:

m(苯乙烯) = 104.152 (g/mol);m(丁二烯) = 54.092 g/mol

104.152x+54.092y=79; x+y=1

尝试法得出:x=y=0.5, ∴链节之比为1,属于交替共聚物。4.23 由60 wt%的乙烯和40 wt%的丙烯组成的交联共聚物具有类似于天然橡胶的弹性。根据这一共聚物的组成,确定两种链节的摩尔分数。

答:m(乙烯) = 28.054 (g/mol);m(丙烯) = 42.081 (g/mol)

本题为质量百分数换算为摩尔分数的问题,

各占一半,为交替共聚物。

4.24 无规聚(异丁烯-异戊二烯)具有200,000 g/mol的重均分子量,重均聚合度为3000,计算共聚物中两种链节的分数。

答:链节平均分子量为:

m(异丁烯C4H8) = 4?12.011+8?1.008 = 56.108 (g/mol)

m(异戊丁烯C5H8) = 5?12.011+8?1.008 = 68.119 (g/mol)

各占一半时的平均分子量为:62.1135 g/mol,比计算结果偏小,表明分子量大的异戊二烯的量占的相对多一些,尝试法求得异戊二烯占0.88,异丁烯占0.12

4.25 (a) 比较金属和聚合物中的结晶态。(b) 比较聚合物和陶瓷玻

璃中非晶态的应用。

4.26 简要解释聚合物随着其分子量的增加,结晶趋势减小的原因。答:分子量大,意味着分子链大,分子之间大的作用力使得熔体粘度增大,链段的运动能力降低,限制链段向晶核的扩散和排列,结晶能力降低。

4.27 对下列每对聚合物指出(1) 能否确定一种聚合物比另一种更易结晶(2) 如果可能,指出哪个更容易结晶并说明其原因;(3) 如果不能判断,说明其原因。(a) 线型和间同立构的聚氯乙烯;线型和

全同立构的聚苯乙烯(b) 网状酚醛树脂;线型和重度交联的顺式-异戊二烯(c) 线型聚乙烯;轻度接枝的全同立构聚丙烯(d) 交替苯乙烯-乙烯共聚物;无规氯乙烯-四氟乙烯共聚物

4.28 计算完全结晶聚乙烯的密度。聚乙烯正交晶胞示在图4.10中;晶胞包含了两个等价的乙烯链节单元。

答:根据图4.10给出的晶格常数,晶胞体积为:

V C = 0.741?0.494?0.255 = 0.0933 nm3

乙烯链节含有2个C,4个H,分子量为

m(乙烯) = 28.05 (g/mol);

∴密度为

4.29 室温下,完全结晶聚丙烯密度为0.946 g/cm3。并且室温下其晶胞为单斜晶系,晶格常数如下:a=0.666 nm, b=2.078 nm, c=0.650 nm; α=90?, β=99.62?, γ=90?。如果单斜晶胞的体积为晶格常数的函数,

V mono = abc sinβ,计算每个晶胞中的链节单元数。

答:V mono = abc sinβ=0.666?2.078?0.650?sin99.62=0.887 nm3,根据聚丙烯密度,可以求出晶胞中丙烯链节的质量为:

0.887 ?10-21cm3?0.946g/cm3 = 8.39?10-22 g

m(丙烯) = 42.08 (g/mol),单个丙烯分子的质量为:

42.08/(6.023?1023) = 6.987?10-23 g

∴单个晶胞所含的独立链节数为:8.39?10-22/6.987?10-23=12

4.30 聚四氟乙烯材料,其两种结晶度百分比下的的密度列于如下

表:

ρ ( g/cm3) 结晶度(%)

2.144 51.3

2.215 74.2

(a)计算完全结晶和完全无定形聚四氟乙烯的密度

(b)确定密度为2.26 g/cm3聚合物的结晶度百分比

答:(a) 根据公式:

代入已知表中的数据,得联立方程:

求解得出:ρc(完全结晶) = 2.301g/cm3;ρa(完全非晶)=2.001g/cm3

(b) 代入上述公式中:

4.31 尼龙6,6材料,其两种结晶度百分比下的密度如下列表所示:

ρ ( g/cm3) 结晶度(%)

1.188 67.3

1.152 43.7

(c)计算完全结晶和完全无定形尼龙6,6的密度

(d)确定具有55.4%结晶度的样品密度

答:(a) 根据公式:

代入已知表中的数据,得联立方程:

求解得出:ρc(完全结晶)=1.24g/cm3;ρa(完全非晶)=1.09g/cm3

(b)

求得密度为:ρ =1.168 g/cm3

第五章

合金(Alloy);原子百分数(Atom percent);原子振动(Atomic vibration);玻尔兹曼常数(Boltzmann’s constant);柏格斯矢量(Burgers vector);组成(Composition);缺陷结构(Defect structure);位错线(Dislocation line);刃型位错(Edge dislocation);电中性(Electroneutrality);弗兰克尔缺陷(Frenkel defect);晶粒尺寸(Grain size);不完美性或缺陷(Imperfection);间隙固溶体(Interstitial solid solution);显微镜(Microscopy);微观结构,组织(Microstructure);混合位错(Mixed dislocation);光学显微镜(Photomicrograph);点缺陷(Point defect);扫描电子显微镜(Scanning electron microscope);扫描探针显微镜(Scanning probe microscope);肖脱基缺陷(Schottky defect);螺型位错(Screw dislocation);自间隙(Self-interstitial);固溶体(Solid solution);溶质(Solute);溶剂(Solvent);化学计量比(Stoichiometry);置换固溶体(Substitutional solid solution);透射电子显微镜(Transmission electron microscope, TEM);空位(Vacancy);质量百分数(Weight percent);

5.1 计算在熔点327?C ,Pb 的原子空位百分数。空位形成能为0.55 eV/atom 。

答: = exp[-0.55?1.6?10-19/(1.38?10-23?600)] = exp(-10.628) = 2.42?10-5

5.2 计算每立方米铁在850?C 中的空位数,其空位形成能为 1.08 eV/atom 。Fe 的密度和原子量各为7.65 g/cm 3和55.85 g/mol 。 答:1 m 3铁的质量为:7.65 g/cm 3 ? 106 cm 3 = 7.65? 106 g /m 3

1 m 3铁中的原子数目为:

7.65?106 ? 6.02?1023 /55.85 = 8.25?1028/m 3

= 8.25?1028?exp[-1.08?1.6?10-19/(1.38?10-23?1123)]

= 8.25?1028?exp[-11.15] =1.186?1024 /m 3

5.3 计算银的空位形成能,已知800?C 时的平衡空位数为3.6?1023 m -3,800?C 下的银的原子量和密度分别为107.9 g/mol 和 9.5 g/cm 3。 答:1 m 3银的质量为:9.5 g/cm 3 ? 106 cm 3 = 9.5? 106 g

1 m 3银中的原子数目为:

9.5?106 ? 6.02?1023 /107.9 = 5.3?1028/m 3

= 5.3?1028?exp[-Q V ?1.6?10-19/(1.38?10-23?1073)]

= 5.3?1028?exp[-Q V ?10.81] = 3.6?1023 /m 3

求得:

5.4 计算每立方米铝中的原子数

)exp(kT

Q N N v v -=)exp(kT

Q N N v -=ν)exp(kT

Q N N v -=ν

5.5 离子陶瓷中可否存在相对大浓度的弗兰克尔缺陷,为什么?

5.6 用你的语言简要定义化学计量组成

5.7

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