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Unformatted text preview: Problem 2.5
J. F. Shackelford, Introduction to Materials Science for Engineers, 7th Edition, Prentice Hall, New Jersey (2009) 2.5 An Al2O3 whisker is a small single crystal used to reinforce metalmatrix composites. Given a cylindrical shape, calculate the number of Al atoms and the number of O atoms in a whisker with a diameter of 1 µm and a length of 30 µm. (The density of Al2O3 is 3.97 Mg/m3.) SOLUTION Begin by calculating the volume of the cylindrical whisker (area of base × length),
V =π 1 µm 2 2 × 30 µm and converting units to those used in the density expression (m),
V = π (0.5 × 10−6 m)2 × (30 × 10−6 m) to obtain
V = 23.6 × 10−18 m3 Next, from the given density (mass/volume), solve for the mass of the whisker,
mAl2 O3 = (3.97 × 106 g/m3 ) × (23.6 × 10−18 m3 ) = 9.35 × 10−11 g Now note that one mole of Al2O3 weighs
(2[26.98] + 3[16.00]) g = 101.96 g and that one mole of Al2O3 contains 2 × (Avogadro's number) of Al atoms and 3 × (Avogadro's number) of O atoms. Problem 2.5 Solution Professor R. Gronsky page 1 of 2 Problem 2.5
J. F. Shackelford, Introduction to Materials Science for Engineers, 7th Edition, Prentice Hall, New Jersey (2009) Consequently, the number of Al atoms is calculated from the mass ratio 9.35 × 10−11 g × 2 (6.023 × 1023 atoms) 101.96 g therefore NAl atoms = 1.11 × 1012 Al atoms and the number of O atoms is just 3/2 times the number of Al atoms, or NO atoms = 1.66 × 1012 O atoms Problem 2.5 Solution Professor R. Gronsky page 2 of 2 Problem 2.14
J. F. Shackelford, Introduction to Materials Science for Engineers, 7th Edition, Prentice Hall, New Jersey (2009) 2.14 Make an accurate plot of Fc versus a for an Na+ – O2– pair. SOLUTION The approach to solving this problem is described in Example 2.5 of the text on page 36, which is to calculate the Coulombic force of attraction acting between a pair of ions, in this case sodium and oxygen. Note that the operating equation is given in the text as Equations 2.1 and 2.2 on page 29, which can be combined to generate the expression for the Coulombic force of attraction.
Fc = − k0 ( Z1 q ) ( Z2 q ) a2 The parameters in this equation are the proportionality constant k0 = 9 × 109 V•m/C (given on page 29 of the text), the valence Z of the ions, the elementary unit of charge q, which is also the charge on a single electron, or 0.1602 10–18 C (inside the front cover of the text), and the separation distance a between the ions. When all variables are expressed in these units, Example 2.5 shows that the resulting Coulombic force Fc bears units of Newtons (N). The charges Z on the ions are given as +1 for Na and –2 for O, so the master equation to be used for generating the requested plot is
Fc = −[9 × 109 V · m/C] ([+1] [0.16 × 10−18 C]) ([−2] [0.16 × 10−18 C]) a2 Generating any style spreadsheet (such as Microsoft's Excel™) with reasonably small increments of a will produce a plot to the desired level of accuracy. In the plot that follows, only six points are used, and the plot shows that a smooth interpolation is possible, rendering it “accurate.” Problem 2.14 Solution Professor R. Gronsky page 1 of 2 Problem 2.14
J. F. Shackelford, Introduction to Materials Science for Engineers, 7th Edition, Prentice Hall, New Jersey (2009) Coulombic force vs Interionic separation distance 15 Coulombic Force x 10^9 (N) 12 9 6 3 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 Interionic separation distance, a (nm) Problem 2.14 Solution Professor R. Gronsky page 2 of 2 Problem 2.38
J. F. Shackelford, Introduction to Materials Science for Engineers, 7th Edition, Prentice Hall, New Jersey (2009) 2.38 The monomer for the common fluoroplastic, polytetrafluoroethylene, is F C F F C F (a) Sketch the polymerization of polytetrafluoroethylene. (b) Calculate the reaction energy (per mole) for this polymerization. (c) Calculate the molecular weight of a molecule with n = 500. SOLUTION The polymerization of the tetrafluoroethylene molecule proceeds by the breaking of the double carbon bond and its replacement by two single carbon bonds as the backbone of the chain is formed, as illustrated in the sample problem shown as Example 2.9 on page 42 of the text. (a) The polymerization reaction can be sketched following Example 2.9. A single “mer” in this chain contains 2 C and 4 F, the same as a single molecule, but the carbons are bonded by a single bond instead of a double bond, and the chain extends indefinitely. The result, polytetrafluoroethylene is also known by the trade name Teflon™ and appears as shown in the following sketch.
F C F F C F F C F F C F mer F C F F C F F C F F C F (b) The reaction energy associated with this polymerization is the difference between the bond energies of a double carbon bond and two single carbon bonds, and the calculation is repeated from the sample problem shown in Example 2.10. Reaction Energy = (740 – 680) kJ/mol = 60 kJ/mol. Problem 2.38 Solution Professor R. Gronsky page 1 of 2 Problem 2.38
J. F. Shackelford, Introduction to Materials Science for Engineers, 7th Edition, Prentice Hall, New Jersey (2009) (c) Using the data contained in Appendix 1 on the atomic mass of the constituents in polytetrafluoroethylene, with 2 carbon atoms and 4 fluorine atoms per molecule, and n = 500 molecules in the polymeric chain (n is defined in Example 2.11 on page 44), Molecular Weight = 500 (2 [12.01 amu] + 4 [19.00 amu]) = 50,000 amu. Problem 2.38 Solution Professor R. Gronsky page 2 of 2 Problem 2.43
J. F. Shackelford, Introduction to Materials Science for Engineers, 7th Edition, Prentice Hall, New Jersey (2009) 2.43 Superimpose on the plot generated for Problem 2.42 the metaloxide bond lengths for the same elements. 2.42 In order to explore a trend within the periodic table, plot the bond length of the group II A metals (Be to Ba) as a function of atomic number. (Refer to Appendix 2 for the necessary data.) SOLUTION The periodic table of the elements appears on the inside cover of the text and again on page 25, from which the Group II A elements are found. Because this problem asks for a “plot,” it would be frugal to enter the appropriate data directly into a graphing application, such as Microsoft Excel™. And the plot must show the trends of both Problem 2.42 and Problem 2.43. For Problem 2.42, bond lengths of the pure metals are requested, and these are calculated from the atomic radii tabulated in Appendix 2. Assuming the atoms are closelypacked (touching one another with no gaps), the bond length is simply twice the atomic radius. For Problem 2.43, the metaloxide bond lengths are requested. These can also be calculated on the same assumption that the metal ion and the oxygen ion are "touching." The bond length is then the sum of the appropriate ionic radii, namely the divalent +2 metal ion (the only one given in each case) the divalent –2 oxygen ion (r = 0.132 nm in Appendix 2) to which each metal ion is bonded. First generate a table of the appropriate values
Atomic number Atomic radius (r) in nm Ionic Radius (r2+) in nm Metallic bond length (2r) MetalOxide Bond Length (r2+ + 0.132 nm) Be Mg Ca Sr Ba 4 12 20 38 56 0.114 0.160 0.197 0.215 0.217 0.054 0.078 0.106 0.127 0.143 0.228 0.320 0.394 0.430 0.434 0.186 0.210 0.238 0.259 0.275 Problem 2.43 Solution Professor R. Gronsky page 1 of 2 Problem 2.43
J. F. Shackelford, Introduction to Materials Science for Engineers, 7th Edition, Prentice Hall, New Jersey (2009) Using these values, the requested plot can be obtained. Metal Bond Lengths 0.5 MetalOxide Bond Lengths 0.375 Bond Length (nm) 0.25 0.125 0 0 15 30 Atomic Number 45 60 As to the “trend” (see problem statement) exhibited by these metals of increasing atomic number, it is clear that our approximation to a “bond length” shows an early monotonic increase, following by a flattening of the curve. Among higher atomic number metals in this group, the bond lengths do not change significantly. A similar trend is shown by their oxides, but with a much smaller overall increase as atomic number increases. Problem 2.43 Solution Professor R. Gronsky page 2 of 2 Problem 2.49
J. F. Shackelford, Introduction to Materials Science for Engineers, 6th Edition, Prentice Hall, New Jersey (2005) 2.49 The secondary bonding of gas molecules to a solid surface is a common mechanism for measuring the surface area of porous materials. By lowering the temperature of a solid well below room temperature, a measured volume of the gas will condense to form a monolayer coating of molecules on the porous surface. For a 100 g sample of fused copper catalyst, a volume of 9 × 103 mm3 of nitrogen (measured at standard temperature and pressure, 0°C and 1 atm) is required to form a monolayer upon condensation. Calculate the surface area of the catalyst in units of m2/kg. (Take the area covered by a nitrogen molecule as 0.162 nm2 and recall that, for an ideal gas, pV = nRT where n is the number of moles of the gas.) SOLUTION Using the guidance given in the problem statement that the nitrogen gas can be treated as an ideal gas (for which the ideal gas law applies), the number of molecules of nitrogen in the condensate is computed after some needed unit conversion (see Appendix 3, p. 801 of the text for units of pressure).
n= PV (1 atm) (9 × 10−6 m3 ) 1 N/m2 = × × 0.6023 × 1024 molecules/mol RT (8.314 J/K) (273 K) 9.869 × 10−6 atm Solving, n = 2.42 × 1020 molecules N2. The total surface area (S) covered by this number of molecules is next computed using the information on the area covered by one molecule given in the problem statement. S = (0.162 × 1018 m2/molecule) (2.42 × 1020 molecules) = 39.2 m2. This is the surface area of a sample with 100 g mass. The answer is requested in units of m2/kg, requiring more unit conversion,
S= 39.2 m2 1000 g × 100 g 1 kg which yields the final solution S = 392 m2/kg. E 45 Professor R. Gronsky page 1 of 1 ...
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