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1、一、填充题1 结晶生高聚物熔融后迅速冷却，透明性 ，加工时加成核剂其透明性 。（增加，增加）2 聚四氟乙烯的极性比聚三氟乙烯的 ，熔点前者比后者的 。（低，高）3 大多数高聚物熔体的表现粘度比其真实粘度 。（小）4 聚丙烯的玻璃化温度比聚苯乙烯的 ，流动温度前者比后者的 。（低，低）5 是解决橡胶蠕变的关键措施，克服压力松弛的重要措施是 。（硫化交联，交联）二、是非题（对的在括号内打“”，错的打“”）1 测定高聚的时降温速度越快，测得的值越高；而升温速度越快，测得的值越低。 （ ）2色散力存在于一切高聚物中。（）3 结晶度越高晶粒越大。（）4 用于吸声隔音的高分子材料，希望其内耗大，而用于轮胎

2、的则希望其内耗小。（）5 、都是相变温度。（）6聚甲醛不能作橡胶是因为主链上有氧原子存在易生成氢键的缘故。（）7柔性链高分子宜采用提高温度的办法来改善加工性能，而刚性链宜采用提高剪切速率的办法来改善加工性能。（）三、试分析下列各题中的图或说明是否有错，简单说明原因。1蠕变与温度和外力关系示意图答：图示正确。蠕变与温度高低和外力大小有关，温度过低，外力太小，蠕变很小且很慢，温度过高，外力过大，形变发展过快，也感觉不出蠕变。在适当外力，且在以上不远，则随外力增加，温度增加，形变量增大。2 下图为某结晶性高聚物熔体在快速冷却及缓慢冷却下所得晶体的DSC图，为缓冷下所得晶体的熔融峰，为急冷下所得晶体的

3、熔融峰答：正确。缓冷时结晶高聚物的结晶度高故熔融峰较大。3 下图为时温等效作图法示意图。正确。时温等效原理中升高温度与延长观察时内对分子运动是等效的。而降低频率与延长观察时间也是等效的。故升高温度与降低频率是等效的，可借助移动因子将的两条曲线叠合。四、问答题1 从结构出发排出高聚物顺序（从高低排），并简述理由。聚二甲基硅氧烷：聚氯乙烯，聚1，4-顺丁二烯，聚苯乙烯，聚乙烯。1答：高低顺序：聚苯乙烯聚氯乙烯聚丙烯聚1，4一顺丁二烯聚二甲苯硅氧烷。聚苯乙烯侧醛位阻大，故最高聚氯乙烯极较大，故次之但比聚乙烯高。聚1，4-顺丁二烯由于存在双键时链柔性较单键大，故比聚乙烯低。聚二甲醛硅氧烷由于结构对称，

4、存在醚键，故柔性最，最小。2 如何用物理改性的方法制备下列高分子材料 抗蠕变好的高分子材料 抗冲击聚苯乙烯（即增加PS的韧性） 软聚氯乙烯 强度好的纤维 高聚物单晶2 答：通过加入刚性填料填充或与刚性链高聚物共混等方法可提高材料抗蠕变性，制得抗蠕变性好的材料。可采用橡胶与聚苯乙烯共混的办法制得抗冲聚苯乙烯。在聚氯乙烯中共混入大量增塑剂，可得软氯乙烯。纺丝制得纤维后进行轴向拉伸，提高取向度可制得强度好的纤维。将高聚物配成极稀溶液并在较高温度下结晶可得单晶。3 简述两种实验方法来判断试样是AB无规共聚物，还是A、B均聚物的共混物。3 答：（1）测试样的DSC曲线，如果有两个玻璃化温度则为A、B均聚

5、物的共混物，如果只有一个则为A、B型无规共聚物。（2）测试样的动态粘弹谱。同（1）理。4 列举高聚物粘弹性表现（至少三例，不述理由）4 答：高聚物的粘弹性表现：蠕变，应力松弛滞后和力学损耗等。5 三角胶带用于传动时，为何使用时间越长，打滑现象越烈？5 答：三角胶带用于传动时，会发生蠕变现象，随使用时间的增长，胶带变得越来越长，故打滑现象越突出。一、 Describe and calculation1、Many rubber materials exhibit a time dependence in their mechanical properties (see the following

6、figure). Make a list of possible reasons.Figure 3.21 Results from measurements of continuous stress relaxation of nitrile rubber (low network density). Data from Bjrk (1988).Solution：Physical relaxation caused by disentanglement of chains. The entanglements can be considered as temporary crosslinks.

7、2、The rubber in a blown-up balloon is stretched in a biaxial fashion. Derive the force-strain relationship under the assumption that the rubber follows the Gaussian statistical theory of rubber elasticity.Solution:The general expression for the free energy is: Biaxial stretching under constant volum

8、e yields the following deformation matrix: and The force (f) is obtained by taking the derivative of :3、 A miscible blend shows a higher glass transition temperature than the pure polymers. Suggest a possible explanation. Suggest also another experiment which will critically test your hypothesis.(混合

9、物比纯净的聚合物有更高的玻璃化温度，解释这种现象。Solution：There should be strong specific interaction (e.g. hydrogen bonding or charge-transfer complex formation between two groups of widely different polarities) between the two polymers. Infrared spectroscopy may detect a frequency shift of certain absorption bands caused

10、 by the specific interaction.4、 You have a polymer blend which shows one Tg but is opaque. Explain. Another polymer shows two Tgs but this material is transparent. Explain.Solution：A similarity in the two polymers Tg values gives rise to a single glass transition despite the fact that the polymers a

11、re immiscible as judged by their opaqueness. The second case can be given two different explanations. The first is that the polymers are immiscible but the similarity in refractive index of the two polymers results in only negligible light scattering, i.e. a transparent material. The second explanat

12、ion is that the polymers are immiscible but the dispersion is considerably finer than the wavelength of light.5、Present an explanation of the observations relating to the die-swell phenomenon shown in following Fig.Die-swell from three types of extrusion geometry. Drawn after Graessley (1984)Solutio

13、n：The convergent flow causes axial orientation of the chains. The die-swell occurs in response to the axial orientation. During the flow through the narrow pipe section, relaxation of the oriented molecules occurs. The degree of die-swell decreases with increasing length of the narrow pipe section.6

14、、 The entanglement molecular weight for polyethylene is about 1900g/mol and for polystyrene it is about 10X higher. Both polymers are linear flexible chainsuggest two factors to account for the rather larger difference in the molecular weight between entanglements in these two polymer.Solution：Polye

15、thylene: 2 Carbons, 4 Hydrogens, MW 28,Polystyrene: 8 Carbons, 8 Hydrogens, MW 104.Polystyrene has higher entanglement molecular weight than polyethylene. It has bulkier monomers than polyethylene with monomer weight being 104g/mol as opposed to28g/mol in polyethylene. Therefore, even for the same s

16、egment length of polymer between entanglements, polystyrene would have higher weight. Also, because of bulky side groups, polystyrene may take on some preferred conformations, making it more difficult for the molecule to be entangled. Also, because of these bulky side groups, we can look at polystyr

17、ene like a thick rope while polyethylene is a thin piece of string. There would be smaller length and weight between entanglements on thin piece of string than a thick rope.7、 Explain the free volume theory of the glass transition. Describe how molecular weight influences the Tg of a polymer.Solutio

18、n：The free volume is the extra space besides the space occupied by atoms in the materials. And it can be represent by mobile holes jumping around in the structure. The larger the molecular weight, the higher the Tg is. This is because if you have lower molecular weight chains, they are smaller, so y

19、ou have more, and the more you have, the more chain ends you have, which will add to the freedom of motion (chain ends can move easily).8、 What types of molecules are useful plasticizers? Describe the difference between a plasticizer and an antiplasticizer?Solution：Plasticizers and antiplasticizers

20、are both small, low molecular wt molecules which you can add to modify a material glass transition temp. Plasticizers spread chains apart increase free volume lower Tg. Plasticizers can also disrupt dipole attractions b/w polymer chains giving them greater mobility lower Tg. Antiplasticizers are so

21、small that they fit ino the spaces b/w chains lower free volume decrease mobility increase Tg。9、 The glass transition temperature of polyisobutene is Tg = 202K. The characteristic values C1、C2 are taken as C1 = 16.6, C2 = 104K in its WLF equation. If one modulus measured at 80 0C for 1 hour is equal

22、 to the other modulus at certain temperature and measured for 10-6 hour. At which temperature is the modulus measured?Solution：Let T2 be the measuring temperature, and Ts = Tg = 202K be the reference temperature，then C1 = 16.6，C2 = 104K. According to the WLF equation, it is known that:, where , and

23、thus Substituting the known data in the above equation, By solving the equations （1） and（2）, T2 = 239.82 K, i.e., 33.3 0C is obtained. A阐述题1 请谈谈聚合物的力学松弛现象及其有关主要数学描述方法。答：高聚物的力学性质随时间的变化统称为力学松弛，根据高分子材料受到外部作用的情况不同，可以观察到不同类型的力学松弛现象，最基本的现象有蠕变、应力松弛、滞后和力学损耗等。蠕变是指在一定的温度和较小的恒定外力(拉力、压力或扭力等)作用下，材科的形变随时间的增加而逐渐增大

24、的现象。应力松弛是指在恒定温度和形变保持不变的情况下，高聚物内部的应力随时间增加而逐渐衰减的现象。滞后现象是指由于链段在运动时要受到内摩擦力的作用，当外力变化时，链段的运动还跟不上外力的变化，所以形变落后于应力，存在相位差的现象。当应力的变化和形变的变化不一致，形变的变化落后于应力的变化，发生滞后现象，则每一循环变化中就要消耗功，称为力学损耗。（8 分）。主要数学描述方法有Maxwell模型、Kelvin模型和多元件模型。Maxwell模型适合描述应力松弛，Kelvin模型适合描述蠕变，多元件模型适合描述各种力学松弛现象。2. 请说说您对聚合物玻璃化转变现象的理解和玻璃化转变的三种确定方法与原

25、理？答：是链段（对应于50100个主链碳原子）运动开始发生（或冻结）的温度。（2分）。对于塑料来说，是使用的最高温度即耐热性指标；而对于橡胶来说，是使用的最低温度，是其耐寒性指标。可见的重要性。（4分）。除了前述的温度-形变曲线（或模量-温度曲线）外，比容、比热、内耗、折射率、黏度（所有聚合物在时的黏度均为1012Pas，据此测定聚合物的黏度称为“等黏度法”）、膨胀系数、扩散系数和电学性能等在时的突变均可用来测定。（4分）。例如，常用的膨胀计法是测定聚合物比容随温度的变化，以拐点为；而差示扫描量热法（DSC）是在等速升温的条件下，连续测定热流速率与温度的关系，时比热发生突变而在热谱图上表现为基

26、线的突然变动。3. 请说说聚合物在升温过程中的物态转变问题。答：对于非结晶性聚合物，随着温度的升高，聚合物依次经历玻璃态、玻璃化转变区、橡胶态、橡胶态-粘流态转变区和粘流态。如图中实线曲线所示。同时分别对各种物态的分子运动物理机制予以说明。对于结晶性聚合物，当结晶度低于40%时，同非结晶性聚合物情形。当结晶度大于40%时，如果分子量低，低于熔点时为玻璃态，超过熔点后即进入粘流态。如果分子量高，低于熔点时为玻璃态，超过熔点后依次经历橡胶态、橡胶态-粘流态转变区和粘流态，如图中虚线曲线所示。同时适当说明。对于高硫化橡胶，聚合物依次经历玻璃态、玻璃化转变区和橡胶态，如图中点线曲线所示。同时适当说明。