Ch6Sol - 3“ Chapter 6 - Problem Solutions FIND: Compare...

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Unformatted text preview: 3“ Chapter 6 - Problem Solutions FIND: Compare the structure of a glass to that of the liquid. GIVEN: Both are noncrystalline. SOLUTION: The structure of a glass is essentially that of a frozen liquid. There is SRO but no LRO. Since the glass is at a lower temperature than the liquid of the same composition and most materials contrast as they are cooled, the density of the glass is usually less than that of the liquid. The density of the glass is usually considerably greater than that of the crystal, however. Density is mass per unit volume, and the units we see most often are g/cm3. The most common units of reciprocal density are cm3/g. This is volume per unit mass, or specific volume. It is frequently used by chemical physicists, who study noncrystalline materials. FIND: How does CI, of an amorphous material change as the temperature is increased through the glass transition temperature? SOLUTION: Heat capacity is an intensive property, one that does depend on the bulk properties of the material. Most thermodynamic intensive properties behave exactly the same as (molar) volume in the vicinity of the glass transition temperature. Hence, heat capacity changes slope through the TS. FIND: Estimate the volume thermal expansion coefficient of a glass. GIVEN: Its linear thermal expansion coefficient in the melt is 10 x 10" °C". ASSUMPTIONS: The material behaves typically, so that the thermal expansion coefficient of the glass is about 1/3 of that of the melt. SOLUTION: The linear thermal expansion coefficient is an. and the volumetric thermal expansion coefficient is 0a,. onfiglass) z etu.(melt)/3 and (1.,(glass) 2: 3cm,(glass) Hence, 0L.,(glass) a 3a...(me1t) = 10 x 10*"’(:*1 FIND: Derive the relationship between cm. and av: eta/ca, = 1/3 SOLUTION: consider a cube of materials, length 1 on a side. With heat, the materials arn=t 2 id?" .. l :1 a. 191K idfi)’ 1312 3 VdT 13 dT l: expands isotropically to length 1+5]. We do the problem first using difi‘erentials. Now with deltas: V = (l + 803 = l3 + 3361 + “.2; AV :2 31281. Therefore, 105 10. ll. 12. am: “AT: 1 =1 JAV 1 3 av __'____'___ _ 2 . VAT .3( 0’ FIND: Is Ts a temperature or range of temperatures? SOLUTION: Although we ofien cite a glass transition temperature, the glass transition occurs over a range of temperatures. T; is rate sensitive and structure sensitive; it 7 depends on the rate of heating or cooling and on the local structure which is statistically variable in a glass or melt. FIND: The temperature range for transitions that involve units smaller than a mer. SKETCH: is“? SOLUTION: In a polymer, the repeat unit is a mer. If the repeat unit gains mobility, then the entire molecule and all of its parts are mobile. It requires less thermal input, kT, to excite smaller units, such as rotation of the ring side group in the polystyrene shown in the figure. Hence, such transition are sub—TE. COMMENTS: Transitions in the crystalline regions can occur above T, and below T.... FIND: Design a rubber gasket for use in outer space. GIVEN: Outer space can fluctuate between cold and hot, and the atmosphere depends on the location in space. Solar radiation is very strong in space. ASSUMPTIONS: We'll design for space being a vacuum. SOLUTION: Fortunately the seal will probably never see solar radiation, so this is not a design problem. That is fortunate, since flexible materials (polymers) do not stand up to radiation. (Look at what happens to your skin when you expose it to bright sunshine.) Temperature can also be a problem. High temperatures can soften, melt or degrade polymers. Low temperature can change a rubber to a glass, making a material that is flexible at room temperature to brittle at high temperature. Rubber seals that have become glassy may break or leak. Metal seals will decrease in volume as the temperature is decreased. If your choice is a rubber gasket, then it needs to remain flexible at use temperature. Liquid oxygen is very cold and embn'ttles many materials. FIND: Provide examples of materials that behave like silly putty. GIVEN: Silly putty is used at a temperature that it sometimes behaves in a fluid-like manner and sometimes in a solid-like manner. SOLUTION: There are many such examples, but they are sometimes hard to identify. Consider these materials: 7 l. Plumbers putty 106 l3. 14. 15. 2. Rubber mounts for car engines and vibrating machines 3. Rubber bumpers 4. Water. (Consider jumping off a 100 foot clifi‘ into a water-filled quarry) 5. Bullet-proof vest (Textile-like in ordinary use and bullet-proof when necessary) COMMENTS: For a number of applications we need materials with similar but difi'erent properties - a material that flows under low stress but does not flow under its own weight (Bingham plastic). Plumbers putty is an example, but it also shows the characteristics of silly putty. FIND: Is it unique that motor oils do not thin with increasing temperature? SOLUTION: Recall equation 6.3-5b: a = no exp (Q/RT). This equation tells you that as temperature increases, viscosity decreases exponentially If motor oil does not behave this way, then it behaves in an unusual manner. COMMENTS: Motor oil, in fact, does behave in an unusual manner. It's viscosity is essentially constant with temperature! Let me try to explain how this is accomplished. Oil contains polymer molecules in a solvent. The molecules do not like the solvent all that much, so they tend to ball up or coil somewhat tightly on themselves. As the temperature is increased, the interaction between solvent and polymer changes. The polymer begins to like the solvent, so it uncoils. The polymer molecules then become entangled in one another, raising the viscosity, which counteracts the normal decrease with increasing temperature. FIND: Whether it is more difficult to obtain a GIVEN: shear strain rate with a high viscosity fluid or a low viscosity fluid. ASSUMPTIONS: The oils behave in a similar fashion in a stress field. SKETCH: shear I x SOLUTION: Newton’s Law of Viscosity states that shear stress is proportional to velocity gradient. The constant of proportionality is viscosity: 1 = n(dv/dx). Since the velocity gradient is invariant in this problem, the shear stress varies with viscosity. The higher viscosity oil will require a larger stress to maintain the plate velocity. Skotch® tape is a polymer film with a thin coating of an oil-like material. It is simply the thinness of the oil film and its viscosity that prevents slippage between substrate and film. COMMENTS: When a large stress is required to shear a fluid, then much work is lost so that heat is generated in the fluid. FIND: Explain how viscosity changes as a material is crystallized or solidifies as a glass. 107 16. 17. SOLUTION: Let us first consider what occurs when a material like molasses or honey is cooled. As cooling proceeds, the material gets "thicker and thicker". Technically, we mean that the viscosity decreases with temperature. Eventually the material is so hard that we say it is frozen. What we mean really is that we are below the glass transition temperature. Thus, viscosity decreases many orders of magnitude as molasses, or any materials that does not crystallize, is cooled from the fluid-like state to the rock-hard state. Now consider a material that crystallizes, say, water, since we are all familiar with it. As water is cooled, it changes viscosity'very little. At 0°C both water and ice coexist. Below 0°C only ice exists. Thus, the viscosity of H20 changes orders of magnitude at 0°C. Unlike materials that do not crystallize, viscosity changes orders of magnitude over a very narrow temperature range — less than a degree. FIND: Show the atactic and isotactic configurations of PP. Which is more likely to be semicrystalline‘? GIVEN: The structure of PP is shown in Table 6.4-1. The structure of atactic and isotactic polymer is shown in Fig. 6.4-5. SKETCH: i-PP has the methyl groups all on the same side. (The B side groups are not shown, but each carbon is bonded to 4 atoms. You should mentally visualize all the H atoms.) /C\ / x / x (0“ IC “0 c C c C c c “\ 'CH3'CH3'CH3’CH3'CH3 a-PP has the methyl groups appearing randomly on one side or the other. /C\ /C\ \9 , c f 9 CH3 CH3 CH3 CH3 CH3 /C\ /C\ ’0 C \ SOLUTION: Since the methyl groups, which are large and bulky, are all on the same side in i-PP, the molecules can be efficiently packed together (when the molecules are stretched out). Hence, i-PP is semicrystalline. It mechanical properties are generally good up to about the melting temperature, 160°C. a—PP is noncrystalline, since the molecules cannot be packed together into a unit cell. It's properties are limited by its glass transition temperature, which is about 0°C. COMMENTS: A-PP is a useless gummy substance. All the PP in use today is i-PP. It is used in huge quantities. ‘ FIND: Show the stereo isomers of PAN. GIVEN:: PAN is poly(vinyl cyanide). 108 18. 19. 20. 21. SKETCH: (H are not shown) isotactic \c/C\C/c\c/C\c/c\c/C\c ‘ 3N I l I l I CN CN CH CN CN sYHdiotactic \E/c\c/C\E/cxc/c\E/C\¢/ CH CH CN éN CH CN atactic \f/C\c/C\c/c\¢/c\f/cxc/ COMMENTS: Commercially available PAN is atactic. ' FIND: Which polymer is more likely to be semicrystalline, PVdF (which has 2 F's) or PVF (a vinyl polymer, which has 1 F)? SOLUTION: Note that the PVdF [poly(vinylidene fluoride)] is symmetric. Symmetric molecules are easier to pack than nonsymmetric ones. PVdF is semicrystalline. PVF, like PVC, is noncrystalline. FIND: Estimate the glass transition temperature of PET. GIVEN: It's melting temperature is about 255°C = 528K. SOLUTION: The melting temperature of nonsymmetric polymers is about 2/3 Tm, when the melting temperature is in absolute degrees. Hence, Tis a: 2/3 is 528K = 353K = 79°C. COMMENTS: The T8 of PET is up to 40°C higher than 80°C, depending on the specific polymer characteristics. FIND: Calculate the number of mers or degree of polymerization is a sample of PP with a molecular weight of 150,000 g/mole. DATA: PP is a vinyl polymer with a methyl side group. SOLUTION: There are 3 C and 6 H per repeat unit :3 MW = 3 x 12 + 6 x1= 42 g/mer. Hence, number of mers = 150,000 g/mole / 42 g/mole of mers = 3571 mers. COMMENTS: We always ignore chain end groups in these sorts of calculations because the effect is negligible. FIND: Calculate the MW of a cellulose mer. If a molecule of cotton has a MW of 9,000 g/mole, how many mers are joined? . GIVEN: The structure of cellulose is shown in Fig. 6.4-3a. DATA: According to Appendix A, the atomic weight of C is 12.01, H 1.01, and 0 16.00 g/mole. SOLUTION: Count the number of atoms of each type per met: 10 0, 12 C, and 8 H. 109 22. 23. 24. 25. 26. 27. Thus, the MW ofa merile x 16 +12 x 12.01 + 8 x 1.01 = 312.20 g/mole. A molecule of cotton with a MW of 9,000 g/mole has 9,000 g/mole / 312.2 g/mole as 29 mers joined. COMMENTS: This is a typical MW of cotton, 9,000 g/mole. It is formed by joining only about 29 mars. FIND: How will the addition of pentaerythritol affect the crystallinity and glass transition temperature of PET? GIVEN: Pentaerythritol is tetra functional. SOLUTION: Any branching agent will disrupt the ability of the molecules to pack efficiently. Hence, crystallinity or the potential for crystallinity will be reduced. Crosslinking makes molecular motion more difficult, so it raises the glass transition temperature. FIND: How does radiation that cleaves covalent bonds effect crystallinity? SOLUTION: Order must be perfect or near perfect in order to have a crystal. Cleaving bonds in crystals destroys the balance of order. Some bonds are broken, creating a difference in the bond arrangement than exists in the virgin crystal. COMMENTS: Organic molecules form crystals readily under appropriate conditions, but the structure of the crystal and the unit cell parameters do not resemble those of similar polymers. FIND: Does the fact that PP, crystallized under quiescent conditions, is characterized by Maltese cross patterned spherulites that fill the entire sample imply 100% crystallinity? SOLUTION: Spherulites are aggregates of crystalline and noncrystalline material. Thus, a completely spherulitic sample is semlcrystalline. ‘ FIND: Explain why amorphous PET that is hot stretched becomes opaque and slowly cold drawn amorphous PET may remain transparent. DATA: The glass transition temperature of PET is about 100°C. SOLUTION: Stretching a limited extent at room temperature does not induce crystallization in PET, whereas stretching above T, and orienting the more mobile molecules into a position similar to those occupied by the molecules in a crystal likely induces crystallization. COMMENTS: In making a very strong PET fiber it is necessary to orient the molecules at first without inducing crystallization. Subsequent drawing steps are typically carried out at progressively higher temperatures. FIND: How many 0 are in a mer of cellulose? GIVEN: The structure of cellulose is shown in Fig. 6.4-3 a. SOLUTION: Count the 0: There are 2 in the other positions, connecting the rings; there are 3 as OH groups on each of the 2 rings; and there is one as part of each of the 2 rings. Addthemup: 2+2x3 +1x2=100permer. COMMENTS: They are extremely important in providing cellulose its properties! FIND: Why are CaO and NaZO added to Si02 in most applications for silicate glasses? 110 28. 29. 30. 31. 32. 33. SOLUTION: According to Table 6.5-1, neither CaO nor NaZO are glass forming systems. Rather, they are network modifiers, as stated in Table 6.5-2. They loosen the silicate network, lowering the glass transition temperature significantly. Thus, they are added to silica to reduce the cost of raw materials and, more importantly, the cost of processing. COMMENTS: The Tg of silica in on the order of 1000°C and that of soda-lime -silicate can be on the order of 500°C. FIND: Is lead oxide a good glass fonner? SOLUTION: Zachariasen’s rules state that the metal should have a coordination number of 3 or 4. The valence of lead is such the PhD is the oxide predicted. Hence, PhD is not a good glass former. It is an intermediate. FIND: Are epoxies and thermoset polyesters semicrystalline or noncrystalline? GIVEN: Both are highly crosslinked, transparent, hard and brittle. SOLUTION: Highly crosslinked polymers are always noncrystalline and below T3 at room temperature. They are glasses. The crosslinks prevent crystallization. COMMENTS: 0n of the problems with some cemposite matrices is that they are brittle and not tough. Thermoplastic matrices are being developed for various demanding applications. High molecular weight thermoplastics, which are sendcrystalline polymers, have high viscosity. It is difficult to get them to flow into the spaces between fibers. FIND: How can you detect when a glassy metal crystallizes? SOLUTION: Heat is released when crystallization occurs. Ifthe metal were like a coin in your pocket, it might burn you. Use a calorimeter to quantify the efi'ect. X-ray difii'action will also show when crystallization occurs. The density or specific volume of the material changes with crystallization. COMMENTS: Many other techniques can also be used to detect crystallization. FIND: Will mixtures, actually solutions, of PbO and SlOz be good glass formers? GIVEN: SiOz is a good glass former; PhD is not. SOLUTION: Since they form a solution, we expect the solution, which is even more complex than either component, to be even slower to crystallize. Thus, PbO-SlOz mixtures should be excellent glass formers so long as the PM) content is not high. FIND: Explain the role of B and Si in Metglas®. GIVEN: Metglas® is an iron based amorphous metal alloy. SOLUTION: The additives hinder crystallization, by increasing the viscosity of the melt, thereby reducing the diffiision coefficient, and by increasing the size of the unit cell, thereby making it necessary for atoms to move farther to their crystallographic positions. 111 34. 35. 36. 37. 38. FIND: How does modulus change as molecules are aligned along a fiber's axis? GIVEN: A single molecule is bonded by covalent forces. A collection of molecules is boned by both primary and weaker secondary bonds. In a fiber with all molecules aligned along the fiber axis, forces are transmitted along covalent bonds only. SKETCH: Modulus Molecular Mlsallgnment along Fiber Axle SOLUTION: The figure can best be read by beginning a large values on the abscissa. As the alignment increases, the modulus increases. With near zero misalignment, the molecules . are packed together in a parallel fashion and stresses are carried by covalent bonds. The fiber requires a large stress to achieve significant deformation. COMMENTS: Pound for pound, organic fibers with this morphology have better mechanical properties than do metals. __ ' FIND: What polymer would you select for use as a flexible gasket on a liquid nitrogen tank? GIVEN: Liquid nitrogen boils at a very low temperature. SOLUTION: You need a material that remains flexible even down to liquid nitrogen temperatures and one that does not react with nitrogen. Some silicone rubbers (thermoset elastomers) are currently used for this application. FIND: Why is the modulus of Spectra® PE roughly 30 times greater than that of sandwich bag PE? GIVEN: Both are PE. SOLUTION: The difference is modulus is chiefly a result of the very high molecular orientation in Spectra® PE fiber and virtually no molecular'orientation in sandwich bag PE. FIND: Predict interesting properties of poly(dirnethyl siloxane). SKETCH: CH 3 tifi n 112 39. 40. SOLUTION: The polymer contain no carbon in the backbone. It is inorganic. The methyl side groups preclude crystallization, so the material is amorphous. The molecule is symmetric and its T, is well below room temperature. Hence, it is a rubber. ' Lightly crosslinked, it has excellent elastomeric properties. Because of its chemistry, it is chemically inert in most environments, as well as stable to moderate temperatures. COMMENTS: A low grade of the materials is sold as RTV silicone rubber. FIND: Calculate the end-to-end separation of PS molecules. GIVEN: The DP is 5000 DATA: 1 = 1.54A SKETCH: is. “i n l. SOLUTION: We use equation 6.6-2: L = [m12(1 + cosB’)]V' to approximate the end-to- end separation. There are 2 bonds, each 1 = 1.54A per mer. The backbone is all carbon, so the factor (1 +cosB')/(l—cosB’) is 2. Substituting gives: 1. = [2 x 5000 x1542 x 21'” = 218A COMMENTS: This is a lower bound, largely because of the terms neglected in equation 6.6-2. A more sophisticated calculation shows the value is about 300A. FIND: Calculate the end-to—end separation of 150,000 g/mole a-PS. GIVEN: a-PS does not crystallize. PS is a vinyl polymer with a backbone of all C and a side group, as shown in Fig. 6.4-1. ASSUMPTIONS: The chains have no net molecular orientation. The chain end separation is governed by random flight statistics. DATA: The molecular weight ofthe mer is 8 x C + 5 xH = 8 x 12.01 g/mole + 5 x 1.01 g/mole = 101.13 ymole. 6 in equation 6.6-2 is 1095" and l is 1.54 A, as shown in Table A as twice the covalent radius of C. SOLUTION: The number of mers in a 150,000 g/mole samples of PS is: 150,000 g/mole + 101.13 g/mole of mers = 1483 mers. For each met there are 2 C-C bonds, as shown in Fig. 6.4—1. Equation 6.6-2 can be used to estimate the end-to-end separation: (The molecule will look as is depicted in Fig. 6.6- 8b.) end to-end separation = Ni? 1+ “’30 = 1.54AJ2966 W = 168A. l—cosH 1—cos109.5 113 41. 42. 43. FIND: Show the change in modulus with temperature for a semicrystalline polymer. GIVEN: T3 = 0°C and Tlrn a 160°C. Log Modulus l 160 Temperature (C) SOLUTION: The modulus will decrease at T3 and fall to a very low value at Tm. COMMENTS: This is how PP behaves. To lessen the impact of TE, polymer scientists and engineers attempt to increase crystallinity as much as possible. FIND: Show how the molecular weight between crosslinks affects the mechanical properties of an ep0xy. SOLUTION: Increasing the molecular weight between crosslinks is equivalent to decreasing the crosslink density. As the crosslink density decreases, the modulus decreases and the elongation-to—break increases. Strength changes are not straightforward to predict. Usually, strength increases with crosslink density up to a point. COMMENTS: Too high or Too low a crosslink density leads to a mechanically inferior product. FIND: How might you make a fiber fiom crosslinked rubber? from a thermoplastic elastomer? GIVEN: Crosslinked rubber is one molecule. It does not melt and flow. The molecules cannot slide past one another irreversibly. ThennOplastic elastomers have temporary crosslinks. Upon heating, the molecules can slide past one another irreversibly. SOLUTION: It can indeed be difficult to make a fiber from crosslinked rubber. In fact, to make a fiber using latex (natural) rubber, the crosslinking is induced after fiber formation. Rubber bands, which are essentially thick fibers, are generally not round in cross-section. They are slit sheets, so the fibers are square or rectangular in cross-Section. Thermoplastic fiber can be melt- or solution-formed directly. Lycra is a thermoplastic elastomer. FIND: Why do fabrics shrink when washed in hot water? GIVEN: Fiber are composed of aligned polymer molecules. 114 45. 46. SKETCH: SOLUTION: When heat is applied the aligned molecules seek to crystallize or coil on themselves. When they move to coil, the fiber shrinks. _ COMMENTS: Most of the shrinkage occurs during the first heat. Hence, the fiber can be heat set to subsequent shrinkage. FIND: Explain why the modulus of rubber is much lower than that characteristic of ceramic or oxide glass. SOLUTION: Rubbers are polymers that undergo conformational changes or straightening out of the coiled polymer chains in the fluid-like state with stress. Ceramics and oxide glasses respond to stress by attempting to push or pull atoms out of their energy wells. Conformation changes are much easier to achieve. Rubber elasticity is entropy driven. Hookean elasticity is energy driven. FIND: Calculate the end—to-end separation of PP (a) coiled on itselfand (b) completely stretched out. GIVEN: The molecular weight is 150,000 g/mole. DATA: 1 = 1.54A fcnz 1311‘} SKETCH: CH3 n SOLUTION: We need to determine the number of bonds in the molecule for both parts and b. MW....,.«curs/IVIWmgr = number of mers or DP. MW..m = 3 x 12 + 6 x 1 z 42 g/rnole. Therefore, DP «1 150,000/42 = 3571 :> twice that many bonds (a) Using equation 6.6-2, L = [m12(1+cose')/(1-cose')” = [2 x 3571 x 1.542 x 2]“2 = 184A Note that in the hydrocarbon chain the entire cos factor is 2 and that there are 2 bonds per mer in all vinyl polymers. (b) Using equation 6.6-], Ln = mlcos(6/s) m 2 x 357] x 1.54 x cos(lO9.5”/2) = 6348A 115 ...
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This note was uploaded on 09/22/2011 for the course MSE 2001 taught by Professor Tannebaum during the Fall '08 term at Georgia Institute of Technology.

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Ch6Sol - 3“ Chapter 6 - Problem Solutions FIND: Compare...

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