Question 13.6

On the basis of the SiO₂-Al₂O₃ phase diagram (Figure 12.27 in the sixth edition), for each pair of the following list of compositions, which would you judge to be the more desirable refractory? Justify your choices.

1. 99.8 wt% SiO₂-0.2 wt%Al₂O₃ and 99.0 wt% SiO₂-1.0 wt%Al₂O₃
2. 30 wt% SiO₂-70 wt%Al₂O₃ and 26 wt% SiO₂-74 wt%Al₂O₃
3. 10 wt% SiO₂-90 wt%Al₂O₃ and 5 wt% SiO₂-95 wt%Al₂O₃

Answer 13.6

The discussion on p. 429 [sixth edition] is helpful; in particular, it mentions that refractories can go on working even if some small fraction of the brick is liquid. So, although the solidus line is in the same place for each member of the pairs of alloys in (i) and (iii), we should choose whichever member has the higher value of the liquidus line. And by inspection of Figure 12.27, that's the 99.8% SiO₂ and the 95% Al₂O₃ in (i) and (iii) respectively.

In part (ii), going from 70% Al₂O₃ to 74% moves us across the solvus line, from a mixture of mullite and liquid to a mixture of mullite and alumina. So the latter alloy is the better refractory.

Question 13.11

Soda and lime are added to a glass batch in the form of soda ash (Na₂CO₃) and limestone (CaCO₃). During heating, these two ingredients decompose to give off carbon dioxide, the resulting products being soda and lime. Compute the weight of soda ash and limestone that must be added to 100 kg of quartz (SiO₂) to yield a glass of composition 75%wt SiO₂, 15%wt Na₂O and 10% wt CaO.

Answer 13.11

There are two parts to this question: firstly, work out what weights of Na₂O and CaO must be combined with the quartz; secondly, work out what weights of Na₂CO₃ and CaCO₃ are needed to yield these weights of Na₂O and CaO.

For the first part, you do not need to know the molecular weight of quartz. Nor do you need to work out the total mass of glass. You can work out the weight of Na₂O from the fact that you have 15 parts of Na₂O to 75 parts of quartz, that is, they are in the ratio 1:5. So if there are 100 kg of quartz, there will be 20 kg of Na₂O. Similarly, there will be 13.33 kg of CaO.

Next, you calculate that the mass of Na₂CO₃ will be in the ratio MW(Na₂CO₃):MW(Na₂O) to the mass of Na₂O; so you need 34.2 kg of Na₂O. Similarly, you need 23.8 kg of CaCO₃.

Question 13.17

For many viscous materials, the viscosity may be defined in terms of the expression

viscosity = sigma / (d(epsilon)/dt)

where sigma and d(epsilon)/dt are, respectively, the tensile strength and the strain rate. A cylindrical specimen of soda-lime glass of diameter 5 mm and length 100 mm is subjected to a tensile force of 1N along its axis. If its deformation is to be less than 1 mm over a week's time, using Figure 13.6, determine the maximum temperature to which the specimen may be heated.

Answer 13.17

This is really just a question of putting numbers into the formula. If we remember that the area of a circular cross-section is pi r², not pi d², and that sigma is the applied stress, not the applied force, then we get that the viscosity must be less than 3 * 10¹²Pa.s. From Figure 13.6, this is about the strain point for soda-lime glass, and occurs at a temperature of somewhere between 450 and 520 C.

Question 14.3

Compute mer molecular weights for the following:

1. polytetrafluoroethylene
2. polymethyl methacrylate
3. nylon 6,6
4. polyethylene terephthalate

Answer 14.3

The structures for each of these mers are given in Appendix D of Callister. Then it's just a matter of adding up atomic weights. I find the mer molecular weights to be:

1. 100.022
2. 100.119
3. 226.12
4. 192.174

Question 14.8

Is it possible to have a polyvinyl chloride homopolymer with the following molecular weight data and a number-average degree of polymerization of 1120? Why or why not?

Mol. Wt. range (g/mol)	wi	xi
8,000-20,000	0.02	0.05
20,000-32,000	0.08	0.15
32,000-44,000	0.17	0.21
44,000-56,000	0.29	0.28
56,000-68,000	0.23	0.18
68,000-80,000	0.16	0.10
80,000-92,000	0.05	0.03

Answer 14.8

This is exceedingly straightforward: you find the number-average degree of polymerisation of the mixture -- it's 764 -- and note that 764 is not the same as 1120.

Question 14.12

Using the definitions for total chain molecule length L (Equation 14.11, sixth edition) and average chain end-to end distance r, (Equation 14.12), for a linear polytetrafluoroethylene determine:

1. the number-average molecular weight for L = 2000 nm
2. the number-average molecular weight for r = 15 nm

Answer 14.12

The important thing to remember here is that N in Equations 14.11 and 14.12 is the number of backbone bonds, not the total number of mers. So after using the equations to find N, you should multiply by the MW of one-half of a mer, since the mers each have two backbone bonds. For the two cases,

1. N=15,952, MW = 797,600
2. N=9,487, MW = 474,350

Question 14.20

Calculate the number-average molecular weight of a random nitrile rubber [poly (acrylonitrile butadiene) co-polymer] in which the fraction of butadiene mers is 0.3; assume that this concentration corresponds to a number-average degree of polymerization of 2000.

Answer 14.20

The best approach is probably to find the weighted-average molecular weight of butadiene and acrylonitrile:

MWA=0.7 * 53.064 + 0.3 * 54.092 = 53.372.

Multiplying this by the number-average degree of polymerisation gives a MW of 106,745 amu.

Question 14.29

The density of totally crystalline nylon 6,6 at room temperature is 1.213 g/cm³. Also at room temperature the unit cell of this material is tri-clinic with lattice parameters:

a = 0.497 nm, alpha = 48.4 degrees

b = 0.547 nm, beta = 76.6 degrees

c = 1.729 nm, gamma = 62.5 degrees

If the volume of a triclinic unit cell is given by the following expression:

abc(1 - cos²(alpha) - cos²(beta) - cos²(gamma) + 2 cos(alpha)cos(beta)cos(gamma))^0.5

determine the number of mer units per unit cell.

Answer 14.29

Plugging in numbers, we find that the volume of a unit cell is 0.31 * 10^-27m³. The density is 1,213 kg/m³, so the mass of a unit cell is 1213 * 0.3 * 10^-27 kg.

1 amu is 1.67 * 10^-27 kg, so the mass of a unit cell is about 226 amu. By putting atomic weights into the formula for a mer, we find that the mass of 1 mer is also about 226 amu, so we conclude that there is 1 mer per unit cell.