工程材料科学与设计(原书第2版)课后习题答案(4—8)

Solutions to Chapter 41. FIND: What material has a property that is hugely affected by a small impurity level?SOLUTION: Electrical conductivity spans a wide range. Incorporation of a few parts per million impurities can change electrical conductivity orders of magnitude. Small cracks in brittle materials decrease their tensile strength by orders of magnitude. Small additions of impurity can change the color of gems.COMMENTS: These are but a few examples.2. COMPUTE: The temperature at which the vacancy concentration is onehalf that of 25o C.GIVEN: C 2 = C C 25v C 35vo oEQUATION:⎪⎪⎭⎫⎝⎛RT Q - = C fv v expwhere C v = vacancy concentrationQ fv = activation energy for vacancy information R = gas constant 8.314 J/mole-KT = absolute temperatureIn the present problem C)25(C = C C);35(C = C o v 2v o v 1vand T1 = 35 + 273 = 308K T2 = 25 + 273 = 298Kalso C v(35o C) = 2C v(25o C)Thus, Solving for Q fv we get Q fv = 52893.5 J/mole.Using this value of Q fv , the C v (25o C) can be calculatedThe problem requires us to calculate the temperature at which thevacancy concentration is ½ C v (25o C).½ C v (25o C) = 2.675 x 10-10Thusfor solving T, we get: T = 288.63K or 15.63o C. 3.COMPUTE:C)80( C 3 = (T) C ov vGIVEN: C) 80( C 41 = C) 25( C o v ovEQUATION:⎪⎭⎫⎝⎛298.R Q - C) 25( C Sv o v expDividing (1) by (2) we get:Solving for Q, we get: Q = 22033.56 J/mole= exp(-7.511)= 5.46 x 10-4The problem requires computing a temperature at which C v = 3C v (80o C).3C v (80o C) = 3 x 5.46 x 10-4= 1.63 x 10-3⎪⎭⎫⎝⎛T x 8.3122033.56- = 10 x 1.633-ex psolving for T, we get: T = 413.05K or 140.05o C 4. 5.FIND: Are Al and Zn completely soluble in solid solution?If Al-Zn system obeys all the Hume-Rothery rules. Then it is expected to show complete solubility.(i)The atomic radii of Al and Zn are 0.143nm and 0.133 nmrespectively. The difference in their radii is 7.5% which is less than 15%.(ii) The electronegativities of Al and Zn are 1.61 an 1.65 respectivelywhich are also very similar.(iii) The most common valence of Al is +3 and +2 for Zn.(iv) Al has an FCC structure where Zn has a HCP structure.It appears that Al-Zn system obeys 3 out of 4 Hume-Rothery rules. In this case they are not expected to be completely soluble.6. SHOW: The extent of solid solution formation in the following systemsusing Hume-Rothery Rules.(a) Al in NiSize: r(Ni) = 0.125nm; r(Al) = 0.143nm difference = 14.4%Electronegativity: Al = 1.61; Ni = 1.91Most Common Valence: Al3+; Ni2+Crystal Structure: Al: FCC; Ni:FCCThe crystal structure of Al and Ni are the same and the most commonvalencies are also comparable. However, the size difference is close to 15% and the difference is electronegativities is rather significant.Based on this, it appears that Ni and Al would not form a solid solution over the entire compositional range.(b) Ti in NiSize: r(Ti) = 0.147 nm, r(Ni) = 0.125nm difference = 17.6%Electronegativity: Ti: 1.54; Ni: 1.91Valence: Ti4+; Ni2+Crystal Structure: Ti:HCP; Ni FCCTi in Ni would not exhibit extensive solid solubility(c) Zn in FeSize r(Zn) = 0.133nm; r(Fe) - 0.124nm difference = 7.25%Electronegativity: Zn = 1.65; Fe = 1.83Most Common Valence: Zn2+; Fe2+Crystal Structure: An: HCP; Fe: BCCSince electronegativities and crystal structures are very different, Zn - Fe will not exhibit extensive solid solubility.(d) Si in AlSize r(Si) = 0.117 nm; r(Al) = 0.143nm; difference = 22.2%Electronegativity: (Si) = 1.90; Al = 1.61Valence: Si4+; A;3+Crystal Structure: Si: Diamond Cubic; Al: FCCSince the size difference is greater than 15%, and the crystalstructures are different, Si-Al would not exhibit extensive solid solubility.(e) Li in AlSize r(li): 0.152, r(Al): 0.143; difference - 6.29%Electronegativity: Li: 0.98; Al: 1.61Most Common Valence: Li1+; Al3+Crystal Structure: Li:BCC; Al: FCCSince electronegativity and crystal structures are very different, Li-Al will not exhibit extensive solid solubility.(f) Cu in AuSize r(Cu) = 0.125nm; r(au) = 0.144nm; difference = 12.5% Electronegativity: Cu = 1.90; Au = 1.93Most Common Valence: Cu+; Au+Crystal Structure: Cu:FCC; Au:FCCCu-Au will exhibit extensive solid solubility.(g) Mn in FeSize r(Mn) = 0.112, r(Fe) = 0.124 difference = 10.71%Electronegativity: Mn 1.55; Fe 1.83Most Common Valence: Mn2+; Fe2+Crystal Structure: Mn:BCC; Fe BCCThe difference in electronegativity is high but Mn-Fe does obey the other 3 Hume-Rothery rules. Therefore, it will form solid solutions but not over the entire compositional range.(h) Cr in FeSize r(Cr) = 0.125nm, Fe = 0.144nm difference = 12.5%Electronegativity: Cr = 1.66; Fe = 1.83Most Common Valence: Cr3+; Fe2+Crystal Structure: Cr:BCC; Fe:BCCCr in Fe will exhibit extensive solid solubility but not over the entire compositional range since it obeys only 3 of 4 Hume-Rothery rules.(i) Ni in FeSize r(Ni) = 0.125nm, r(Fe) = 0.124nm difference = 0.8%Electronegativity: Ni: 1.91; Fe 1.83Most Common Valence: Ni3+; Fe3+Crystal Structure: Ni:FCC; Fe: BCCNi and Fe obeys 3 of the 4 Hume-Rothery rules therefore, extensive solid solution will be exhibited but not over the entire compositional range.7. (a) When one attempts to add a small amount of Ni to Cu, Ni is thesolute and Cu is the solvent.(b) Based on the relative sizes of Ni and Cu, radius of Ni = 0.128nm,radius of Cu = 0.125nm, these two are expected to form substitutionalsolid solutions.(c) Ni and Cu will be completely soluble in each other because theyobey all four Hume-Rothery rules.8. FIND: Predict how Cu dissolves in Al.DATA: Cu Alatomic radius (A) 1.28 1.43electronegativity 1.90 1.61valence 1+,2+ 3+crystal structure FCC FCCSOLUTION: All of Hume-Rothery's rules must be followed for asubstitutional solution. In this case, the valences do not match. Cuwill not go into substitutional positions in Al to a large extent.COMMENTS: This principle is often used to precipitation harden Al using Cu.9. What type of solid solution is expected to form when C is added to Fe?The radius of carbon atom is 0.077nm and that of an Fe atom is 0.124nm.The size difference between these two is ~61% which is much grater than ~15%. Thus, these two are not expected to form substitutional solid solution.If we compare the size ratio of C to Fe atoms with the size oftetrahedral and octahedral interstitial sites in BCC iron, we find thatC does not easily fit into either type of interstitial position. C,however, forms an interstitial solid solution with Fe but the solubility is limited.10. FIND: Calculate the activation for vacancy formation in Fe.GIVEN: The vacancy concentration at 727 C = 1000K is 0.00022.SOLUTION: We use equation 4.2-2 to solve this problem:C v = exp (-Q fv/RT)Solving for Q fv:Q fv = -RT ln C v = -(8.31 J/mole-K)(1000K) ln 0.00022 = 7.0 x 104 J/mole11. SHOW: A Schottky and Frenkel defect in MgF2 structuresA 2-D representation of the MgF2 structure containing a Schottky defectand a Frenkel defect is shown below.12. Explain why the following statement is incorrect: In ionic solids thenumber of cation vacancies is equal to the number of anion vacancies.In ionic crystals, even in the presence of vacancies, the chargeneutrality must be maintained. Therefore, single vacancies do not occur in ionic crystals since removal of a single ion would lead to chargeimbalance. Instead the vacancies occur in a manner such that the anion: cation vacancy ratio render the solid electrically neutral. This,however, does not mean that the anion vacancies are equal to cationvacancies. For example, a Schottky defect in MgCl2 or MgF2 involves two Cl - or F- cation vacancies for every Mg2+ anionvacancy to maintain electrical neutrality.The number of cation vacancies equals the number of anion vacancies only for the limiting case where the chemical formula of the compound is MX.13. Calculate the number of defects created when 2 moles of NiO are added to98 moles of SiO2. Also, determine the type of defect created.GIVEN: Neglect interstial vacanciesWe have 2 moles of NiO and 98 moles of SiO2. Since NiO is a 1:1compound there are 2 moles of Ni2+ ions and 2 moles of O2- ions present.SiO2 on the other hand is a 1:2 compound; therefore, there are 98 moles of Si4+ and 196 moles of O2-. The total number of each type of ion is N Ni = 2 molesN Si = 98 molesN O2 = 196 molesThe total number of moles of ions in the system isN T = N Ni + N Si + N O = 2 + 98 + 196 = 196 molesEach substitution of an Ni2+ for Si4+ results in a loss of 2 positivecharges. If no interstitials are created, this loss of positive charge is balanced by the creation of anion vacancies. Charge neutralityrequires one oxygen vacancy created for every Ni2+ ion. Therefore, thenumber of oxygen vacancies isN Ov = N Ni = 2 molesThere are 2 moles of oxygen ion vacancies created with the addition of 2 moles of NiO to 98 moles of SiO2.14. Calculate the number of defects created when 1 mole of MgO is added to99 moles of Al2O3.MgO is a 1:1 compound, therefore there is 1 mole of Mg2+ ions and 1 mole of O2- ions in the system.From Al2O3, there are 198 moles of Al3+ ions and 297 moles of O2- ions inthe system.Each substitution of an Mg2+ ion for Al3+ ion results in a loss of onepositive charge. This loss of positive charge is balanced by oxygenvacancy. Charge neutrality requires one oxygen vacancy to be createdfor every two Mg2+ ion3. Therefore the number of oxygen ion vacanciescreated is0.5 moles of oxygen ion vacancies are created by the addition of 1 moleof MgO to 99 moles of Al2O3.15. COMPUTE: Relative concentration of cation vacancies, anion vacanciesand cation interstitials.GIVEN:Q Cv = 20kJ/moleQ Av = 40kJ/moleQ CI = 30kJ/moleASSUMPTION: assume room temperatureT = 298KConcentration of cation vacancies, C Cv is given bySimilarly for anion vacanciesand for cation interstitials16. (a) Describe a Schottky defect in U2(b) Would you expect to find more cation or anion Frenkel defects inthis compound? Why?UO2 has a fluorite structure with U4+ ions occupying FCC lattice sitesand O2- occupying tetrahedral interstitial sites.(a) A Schottky defect in UO2 will involve one U4+ cation vacancy and 2O2- anion vacancies.(b) In general cation Frenkel defects are more common than anionFrenkel defects because cations are usually smaller. In this case, the radii of U4+ is 0.106nm and that of O2- is 0.132nm. The U4+ cation issmaller than the O2- anion. However, the size difference is not veryhigh. Still, cation Frenkel defects are expected to be more.17. Ionic compound Li2O(a) Describe a Schottky defect(b) Describe a Frenkel defectLi2O has an antifluorite structure. O2- ions occupy FCC lattice sitesand Li+ occupies tetrahedral interstitial sites.(a) A Schottky defect in Li2O involves 2 Li2+ cation vacancies and oneO2- anion vacancy(b) The ionic radii of Li+ and O2- are 0.078nm and 0.132nm respectively.This material is most likely to exhibit cation Frenkel defect since the size of the cation is much smaller than the anion.18. DETERMINE:(a) Interstitial Na+ ions(b) Interstitial O2- ions(c) Vacant Na+ sites(d) Vacant O2- sites in Na2OGIVEN: r(Na+) = 0.098nmr(O2-) = 0.132nmNa2O structure is similar to antifluorite structure. Na+ ions occupytetrahedral interstitial sites and O2- ions occupy FCC lattice sites.Since the ratio of Na:0 is 2:1 for this materials, a Schottky defectresults in 2 cation vacancies for every one anion vacancy.no. of vacant Na+ sites = 2 x no. of vacant O2- sitesA cation Frenkel defect is more likely to occur in this material(a) Interstitial Na+ ions = 1(b) Interstitial O2- ions = 0(c) Vacant Na+ sites = 2(d) Vacant O2- sites = 119. SOLVENT: AuSOLUTE: N, Ag or CsDETERMINE: (a) which element is most likely to form an interstitialsolid solution.(b) which element is most likely to form a substitutional solid solution.r(Au) = 0.144nmr(N) = 0.071nmr(Ag) = 0.144nmr(Cs) = 0.265nm(a) Based on atomic radii N is most likely to form are interstitialsolid solution with Au as solvent.(b) Ag is most likely to form a substitutional solid solution becausethe size difference between Au & N and Au & Cs is more than 15%.In addition, Au and Ag have similar valence, and crystal structure. The electronegativities are not quite similar, but since Ag-Au system obeys3 out of4 of the Hume-Rothery rules, Ag is the most likely element withwhich Au forms a substitutional solid solution.Section 4.4 Diffusion20. Under what condition c an Fick’s first law be used to solve diffusionproblems.The Fick’s first law can be used to solve diffusion problems provided the concentration gradient does not change with time.21. GIVEN: 1 wt% B is added to Fe.FIND: (a) if B would be present as an interstitial impurity orsubstitutional impurity, (b) fraction of sites occupied by B atoms, (c) if Fe containing B were to be gas carburized, would the process befaster or slower than for Fe which has no B? Explain.r(B) = 0.097nmr(Fe) = 0.124nm(a) Based on the atomic radii B would be present as an interstitial impurity(b) amount of B present = 1 wt%As a basis of calculation assume 100gms of material.Determine the no. of moles of Fe and B present.Total no. of moles of Fe and B = 1.773 + 0.092 = 1.865 moles. Fraction of sites occupied by B atoms = 1.8650.092 = mole fraction of B = 0.049Thus, B roughly occupies 5% of the sites.(c) If Fe containing B were to be gas carburized the process would beslower than for Fe which has no B simply because the presence of B atoms already in interstitial sites leave fewer sites for interstitial C todiffuse through.22. Determine which type of diffusion would be easier(a) C in HCP Ti(b) N in BCC Ti(c) Ti in BCC Tir(C) << r(Ti) so we can predict that diffusion occurs via aninterstitial mechanism r(N)<<r(Ti). In this case the diffusion alsooccurs via interstitial mechanism.Ti in BCC Ti is a case of self-diffusion and self-diffusion occurs via a vacancy mechanism. In general the activation energy for self diffusion is higher than interstitial mechanism because vacancy mechanism involves two steps. One is to create a vacancy and second is to promote avacancy/atom exchange. Thus Ti in BCC Ti will be the slowest.The activation energy for diffusion via interstitial mechanism is justthe energy necessary to move an atom into a neighboring interstitialsite. An open crystal structure, as opposed to a dense structure,should have a lower activation energy. Between BCC Ti and HCP Ti, BCCTi has a more open structure (lower APF) than HCP Ti.Thus, N in BCC Ti diffusion would be the easiest by virtue of its lowest activation energy.23. GIVEN: C1 = 0.19 at % at surfaceC2 = 0.18 at % at 1.2mm below the surfaceD = 4 x 10-14 m2/seca o = 4.049 A oCOMPUTE: Flux of copper atoms from surface to interior.We must first calculate the concentration gradient in terms of [copperatoms/cm3/cm]. It can be calculated as follows:The concentration gradient is then24. FIND: Predict whether diffusion is faster in vitreous or crystalline silica.GIVEN: Diffusion is the movement of atoms through the material one step at a time. The ease of movement is in part determined by theamount of space that surrounds each atom. In more open or less densestructures, atoms have an increased chance of being able to squeeze pasta neighbor into a new position.SOLUTION: Diffusion can be thought of as an Arrhenius process. Theactivation energy is that required to move an atom from one position to another, as shown in Fig. 2.3-2. In a crystal the activation energy will be greater than in a glass, since the density is higher and there is less free, or unoccupied, volume. Thus, we expect diffusion to be slower in crystal than in glasses at the same temperature.COMMENTS: When a noncrystalline material is raised to a temperature above the glass transition temperature, diffusion increases enormously. Inmetals this brings about rapid crystallization. In some ceramic and polymer systems, crystallization may be slow or absent.25. FIND: Do textile dyes more readily penetrate crystalline ornoncrystalline regions?GIVEN: Most textile fibers are semicrystalline, containing bothcrystalline and noncrystalline regions. The density of thenoncrystalline regions is less than that of the crystalline regions. Often dyeing is conducted at a temperature at which the noncrystalline regions are above their glass transition temperature.SOLUTION: Dye penetration through the glass will be greater than that through the crystal; however, the rate of dyeing is not sufficiently high to be commercially feasible. The temperature must be raised so that the noncrystalline polymer is in the rubber state. Diffusionbecomes rapid (radially inward) into the small fibers.COMMENTS: One of the key lessons that dye houses learn is that asufficient amount of noncrystalline poorly oriented polymer must bepresent in the fiber. The temperature of the dye bath needs to be above the glass transition temperature. Sometimes water and carriers are used to swell the noncrystalline regions to get yet a greater diffusion rate. The dyes may attach to the polymer using ionic bonds or covalent bonds. Unattached dye may wash out later.26. CALCULATE: The factor by which the diffusion coefficient of Al in Al 2O 3change when temperature is increased from 1800o C to 2000o CGIVEN: T 1 = 1800o C = 2073KT 2 = 2000o C = 2273KEQUATION: ⎪⎭⎫ ⎝⎛RT Q - D = D o ex pdividing (1) by (2), we getfrom table 4.4-1 of the text Q = 477kJ/mole and R = 8.31 J/mole-KThus, the diffusion coefficient of Al in Al2O3 changes by a factor of11.43 when the temperature is increased from 1800o C to 2000o C.27. FIND: Temperature at which a specimen of Fe must be carburized for twohours to achieve the same diffusion result as at 900o C for 15 hrs.GIVEN:T1 = 900o C = 1173K; Q = 84000 J/molet1 = 15 hrs; D o = 2.00 x 10-6 m2/sec.t2 = 2 hrs; R = 8.31 J/mole-KThe value of flux J is in units of cm 2 per sec. Flux per cm 2 J f = Jx time 3600 x 15 x dx dcD - = J 11f(1) We need the same result in 2 hours.,J = J 2f 1f dividing (1) by (2).28. GIVEN: D = 4 x 10-4 m2/s @ 20o CC1 = 2.2 x 10-3 k mol/m3wall thickness = 3mm, diameter = 50cmheight = 10cmCOMPUTE: Initial rate of mass loss through cylinder.Initially the concentration of He outside the cylinder, C2, is zero.First, we need to convert the concentration of He from kmol/m3 into (atoms/cm3)/cm.C1 = 2.2 x 10-3 kmol/m3 = 2.2mol/m3 = 2.2 x 10-6 mol/cm3 = 2.2 x 10-6 mole/cm3In terms of (atoms/cm3)/cmThe concentration gradient isThe flux of atoms per second per cm2is obtained by using Fick’s firstThe rate of mass loss is 1.766 x 1019 atoms/cm 2 sec. The total surface area of the cylinder is 2r(r+h) where r = radius and h = height.Total surface area = 2x 25 (25 + 10) = 5497.79 cm 2The rate of mass loss per secondNote:(i ) The steady state mass loss is calculated because the initial rate of mass loss (i.e., rate of mass loss at time t = 0) is 0. (ii ) It is assumed that the curvature of the cylinder is large enough to calculate J using the expression for plate geometry.29. Diffusion across a polymer membrane depends not only on size of thediffusing species but also the polarity of the diffusing species. A polar membrane may pass nonpolar species but serve as a barrier to polar species.Saran wrap contains highly polar atoms making it a polar membrane which serves as a barrier to water which is a polar compound. thus, there is no diffusion of water through the package unlike polyethylene, which is a nonpolar membrane and allows diffusion of water molecules which form ice.30. COMPUTE: Temperature required to yield a carbon content of 0.5% at adepth of 0.4mn below the surface of the rod in 48 hours.GIVEN: Carbon concentration the interior = 0.2w/oCarbon concentration in the furnace = 1.0w/o secsec secmoles 0.16 = .atoms 10 x 9.709 =cm 5497.79 x - cm atoms 10 x 1.766 =222219Base material: HCP TiEQUATION: In this problem c(x, t) = 0.5wt%c o = 0.2 wt%c s = 1.0 wt% From figure 4.4-11, when From Metals Handbook, Desk Edition, Pg. 28.66 for C diffusion in Ti, D o = 3.02 x 10-3 cm 2/sec, Q = 20,000 cal/mole = 83682 J/mole.31. The diffusion process through vacancy-interchange mechanism depends oncreation of vacancies and vacancy/atom interchange.At comparable homologous temperatures, for Ge and Cu the diffusioncoefficient for that material which has a higher vacancy concentration would be higher.A covalent bond as opposed to a metallic bond is stronger anddirectional. It is also difficult to create vacancies in a covalently bonded material due to its strong bonding. Therefore, the activation energy for vacancy creation in a covalently bonded material such as Ge is larger than Cu which has a weak metallic bond.The directional nature of a covalent bond places geometricalrestrictions on the vacancy atom interchange which again results in an increase in the activation energy.Therefore, at comparable temperatures the diffusion coefficient for Ge will be larger.32. FIND: Describe the energy and entropy in Fig. 4.4-5a, b, and c.SOLUTION: The order in part a is high. The materials is perfect. There is only one way to arrange the atoms in such a system. The entropy is low. In part b there is less order, more disorder, and the entropy hasincreased. Part c is nearly random. It has low order and high entropy. Energy contains a contribution from entropy: E = H -TS, where E is energy, T is absolute temperature, and S is entropy. Assuming all othercontributions to energy change negligibly (T and H), the energy of part c is the low, part a is high and part b is intermediate.COMMENTS: What is shown in going from a to c is the entropy of mixing.33. GIVEN: After 10 hrs at 550o C an oxide layer of thickness 8 m is formed.COMPUTE: Thickness after 100 hrs.Using the definition of effective penetration distance and equation 4.4-11 of text, with = 2 we have Dt 2 x eff ≈.In this case34. GIVEN: D w = 1.0 x 10-12 m 2/s (water)m25.3 = x 100hr 10hr = x m8t t = t D t D = x x100hr = t 10hr = t? = x m 8 = x 2eff 2eff2122111eff 1eff 212eff 1eff μμμD dc = 1.0 x 10 (dye carrier)D d = 1.0 x 10-14 m 2/s (dye)COMPUTE:(a) Times required for the water, dye and carrier to penetrate to the center of the fiber.(b) Same as (a) but fiber diameter doubles (c) If thermal diffusivity of PET is sec m 10 x 828-how long will it take for the heat to penetrate to the center of a 50m diameter fiber.(a) using equation 4.4-11 of text with = 2.for water,for dye carriersimilarly for dye t = 6.25secs. Dt 2 = x eff minutes2.60 or secs 156 = tt x 10 x 1.0 = 10 x 625.00t x 10 x 1.0 = 2)10 x (25.0m 10 x 25 = x/s m 10 x 1.0 = D 12-12-12-6-26-eff 2-12w ⎪⎪⎭⎫ ⎝⎛ minutes26.04 = secs 15.63 = tm 10 x 25 = x/s m 10 x 1.0 = D 6-eff 2-13dc(b) If the diameter fiber is doubled x eff = 50 x 10-6 mfor water,similarly for dye carrier t = 6250 secsand for dyet = 6.25 x 104 secs(c) Substituting D with D th , we can use the same equation to calculate the time required for heat to penetrate the center of fiber diameter = 50m.Note: The units of thermal diffusivity is m 2/sec and notK - m Watt as printed in text secs.625 = 10 x 1.0)10 x 50 ( = tt x D = 2)10 x (5010 x 50 = x/s m 19 x 1.0 = D 12-26-26-6-eff 2-12w35. FIND: How long will it take to case carburize a steel chain to a depth of 1/16 inch?GIVEN: It requires 4 hours to carburize a plate of similarcomposition to a depth of 1/16 inch.ASSUMPTIONS: All carburization conditions are the same in bothtreatments.SOLUTION: Equation 4.4-11 is used to solve the problem:x eff = γ(Dt)1/2.In this situation we set up a ratio for the plate (1) and the chain (2):x x Dt Dt 121122=γγ, where γ is 1 for the plate and 2 for the chain. Subscripts were omitted for the D's, since diffusion is the same in the two cases. Similarly, x 1 = x 2 = 1/16 inch. Reducing the equation gives:112411122212212=⇒= ☺= ☺=γγγγt t t t hrs hr 36. GIVEN: D m = 1 x 10-11m 2/s for moisture through wool or cottond w = 25md c = 2m secs0.00195 = t10 x 81 x 210 x 25 = _tt x 10 x 8 = 210 x 25t x D 2 = x 8-6-28-6-2th eff ⎪⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛using equation 4.4-11 of textwe need to find t required to reach equilibriumWoolCotton (b) for a tightly packed cubic baleof fiberwith side length of 1 meter.A cube has six sides and diffusion is expected to occur through all six sides. The time required to reach equilibrium Dx = t 2eff Here x eff = 0.5mNote: This solution overestimates time. A more precise solution can be obtained by setting up the problem in 3D and solving the Fick’s secondlaw for a cube geometry.37. EXPLAIN: Why fizz reduction is higher in plastic bottles compared toglass bottles and metal cans.Plastic bottles are soft and can be squeezed. Upon squeezing the airinside the bottle is driven out and to maintain the equilibrium pressure over the soda, CO2 comes out of the liquid.Such is not the case for bottles and metal cans. It is for this reason, the fizz reduction is high in plastic bottles.38. Compare the diffusion coefficient of methane in rubber at 239K to thediffusion of Cu in Ag at the same temperature.The diffusion coefficient of methane in rubber at 293K is 1.515 x 10-6m2/sec whereas that of Cu in Ag is 1.20 x 10-4m2/sec. The difference isexpected because the diffusion process in polymers such as rubberdepends on the size of the diffusing species such as CH4 in this example.Compared to the diffusion process of Cu in Ag, which occurs due tovacancy-interchange, the size of the methane molecule is large, also,the individual mers in the polymer chains are not free to moveindependently.39. FIND: Make a schematic plot of D vs. T in natural rubber.GIVEN: Natural rubber has a T g of about 0 C.SKETCH:TemperatureSOLUTION: The sketch above is diagrammatic. We know that before and after the glass transition temperature the diffusion coefficient behaves as an Arrhenius function:D = D o exp (-Q/RT),so that ln D is linearly proportional to 1/T. Q is large below T g.40. EXPLAIN: Will PTFE work as a membrane to separate water vapor from benzene?PTFE is Teflon.Based on the structure of PTFE (Teflon), it appears to be a non-polar membrane.Non-polar membranes allow passage of polar molecules such as water.Furthermore, the size difference between water vapor molecule andbenzene also plays an important role. A water vapor molecule is muchsmaller than a benzene molecule and can easily pass through a Teflonmembrane whereas benzene being a large molecule will not pass.Thus, PTFE will work as an effective membrane to separate water vapor and benzene.41.The solution to Fick’s second law for a thick plate is。