工程材料科学与设计原书第2版课后习题答案48章复习过程Word下载.docx

1、Thus,Solving for Qfv we get Qfv = 52893.5 J/mole.Using this value of Qfv, the Cv（25oC） can be calculatedThe problem requires us to calculate the temperature at which the vacancy concentration is Cv（25oC）. Cv（25oC） = 2.675 x 10-10Thusfor solving T, we get: T = 288.63K or 15.63oC.3. COMPUTE: Dividing

2、（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 Cv = 3Cv（80oC）. 3Cv（80oC） = 3 x 5.46 x 10-4= 1.63 x 10-3solving for T, we get: T = 413.05K or 140.05oC4.5. FIND: Are Al and Zn completely soluble in solid solut

3、ion?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 nm respectively. 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 respe

4、ctively which 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 s

5、olid solution formation in the following systems using 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 t

6、he most common valencies 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 Ni r（Ti） = 0.147 nm, r（Ni）

7、 = 0.125nm difference = 17.6% Ti: 1.54; 1.91Valence: Ti4+;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% Zn = 1.65; Fe = 1.83 Zn2+; Fe2+ An: HCP; Fe: BCCSince electronegativities and crystal structures are very dif

8、ferent, Zn - Fe will not exhibit extensive solid solubility.（d） Si in AlSize r（Si） = 0.117 nm; r（Al） = 0.143nm; difference = 22.2% （Si） = 1.90; Al = 1.61 Si4+; A;3+ Si: Diamond Cubic; FCCSince the size difference is greater than 15%, and the crystal structures are different, Si-Al would not exhibit

9、extensive solid solubility.（e） Li in AlSize r（li）: 0.152, r（Al）: 0.143; difference - 6.29% Li: 0.98; 1.61 Li1+; Al3+BCC;Since 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 =

10、 12.5% Cu = 1.90; Au = 1.93 Cu+; Au+ Cu:FCC; Au:Cu-Au will exhibit extensive solid solubility.（g） Mn in FeSize r（Mn） = 0.112, r（Fe） = 0.124 difference = 10.71% Mn 1.55; Fe 1.83 Mn2+; Mn: Fe BCCThe difference in electronegativity is high but Mn-Fe does obey the other 3 Hume-Rothery rules. Therefore,

11、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% Cr = 1.66; Cr3+; Cr: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

12、.（i） Ni in FeSize r（Ni） = 0.125nm, r（Fe） = 0.124nm difference = 0.8% 1.91; Ni3+; Fe3+Ni 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 the s

13、olute 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 substitutional solid solutions.（c） Ni and Cu will be completely soluble in each other because they obey all four Hume-Rothery rules.8. FIND: Predic

14、t how Cu dissolves in Al. DATA: Cu Al atomic radius （A） 1.28 1.43 electronegativity 1.90 1.61 valence 1+,2+ 3+ crystal structure FCC FCC All of Hume-Rotherys rules must be followed for a substitutional solution. In this case, the valences do not match. Cu will not go into substitutional positions in

15、 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 muc

16、h 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 of tetrahedral and octahedral interstitial sites in BCC iron, we find that C does not easily fit into either type of interstitial position. C, however

17、, 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 727C = 1000K is 0.00022. SOLUTION: We use equation 4.2-2 to solve this problem:Cv = exp （-Qfv/RT） Solving for Qfv: Qfv = -R

18、T ln Cv = -（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 defect and a Frenkel defect is shown below.12. Explain why the following statement is incorrect: In ionic solids t

19、he number of cation vacancies is equal to the number of anion vacancies.In ionic crystals, even in the presence of vacancies, the charge neutrality must be maintained. Therefore, single vacancies do not occur in ionic crystals since removal of a single ion would lead to charge imbalance. Instead the

20、 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 cation vacancies. For example, a Schottky defect in MgCl2 or MgF2 involves two Cl- or F- cation vacancies for every Mg2+ an

21、ion vacancy 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 to 98 moles of SiO2. Also, determine

22、 the type of defect created. Neglect interstial vacanciesWe have 2 moles of NiO and 98 moles of SiO2. Since NiO is a 1:1 compound 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

23、total number of each type of ion isNNi = 2 molesNSi = 98 molesNO2 = 196 molesThe total number of moles of ions in the system is NT = NNi + NSi + NO = 2 + 98 + 196 = 196 molesEach substitution of an Ni2+ for Si4+ results in a loss of 2 positive charges. If no interstitials are created, this loss of p

24、ositive charge is balanced by the creation of anion vacancies. Charge neutrality requires one oxygen vacancy created for every Ni2+ ion. Therefore, the number of oxygen vacancies isNOv = NNi = 2 molesThere are 2 moles of oxygen ion vacancies created with the addition of 2 moles of NiO to 98 moles of

25、 SiO2.14. Calculate the number of defects created when 1 mole of MgO is added to 99 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 in the system.Each substitution

26、 of an Mg2+ ion for Al3+ ion results in a loss of one positive charge. This loss of positive charge is balanced by oxygen vacancy. Charge neutrality requires one oxygen vacancy to be created for every two Mg2+ ion3. Therefore the number of oxygen ion vacancies created is0.5 moles of oxygen ion vacan

27、cies are created by the addition of 1 mole of MgO to 99 moles of Al2O3.15. COMPUTE: Relative concentration of cation vacancies, anion vacancies and cation interstitials. QCv = 20kJ/moleQAv = 40kJ/moleQCI = 30kJ/moleASSUMPTION: assume room temperature T = 298KConcentration of cation vacancies, CCv 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 d