ENGR 220 Homework Problems: Week 3 - Chapter 4 1. The equilibrium fraction of lattice sites that are vacant in gold at 800°C is 2.5 x 10-5. Calculate the number of vacancies (per meter cubed) at 800°C. Assume a density of 18.45 g/cm3 for Au (at 800?). Nt = (6.022*10^23*10.45*10^6)/197.97 = 5.64 *10^28atoms/m^3 Nv= (2.5*10^-5)(5.64*10^28) = 1.410*10^24atoms/m^3 2. For some hypothetical metal the equilibrium number of vacancies at 750°C is 2.8 x 1024 m-3. If th density and atomic weight of this metal are 5.60 g/cm3 and 65.6 g/mol, respectively, calculate the fracti of vacancies for this metal at 750°C. N= (6.022*10^23(5.60)(10^6))/(66.6) N= 5.063*10^28 atoms/m^3 Nv/N=(2.2*10^24m^3)/(5.063*10^28atm/m^3) = 0.55*10^-4 3. Which of the following systems (i.e., pair of metals) would you expect to exhibit complete solid solubility Explain your answers. (a) Cr-V (b) Mg-Zn (c) Al-Zr Aluminum and zirconium have different atomic sizes and crystal structures. Aluminum is a light metal and zirconium is a transition metal, belonging to different groups in the period table. Their differences in properties make complete solid solubility unlikely. Magnesium and zinc have different atomic sizes and crystal structures. Magnesium is an alkaline earth metal (Group 2), while zinc is a transition metal (Group 12), leading to significant differences in atomic properties. Complete solid solubility between these two metals is less likely due to their distinct characteristics. Chromium and vanadium have relatively similar atomic sizes and crystal structures. Both chromium and vanadium belong to the same group in the periodic table (Group 6), which suggests some similarity in their atomic properties. There's a higher likelihood of complete solid solubility between Cr and V due to their similarities 4. Calculate the number of vacancies per cubic meter at 1000°C for a metal that has an energy for vacan formation of 1.22 eV/atom, a density of 6.25 g/cm3, and an atomic weight of 37.4 g/mol. (A) 1.49 × 1018 m-3 (B) 7.18 × 1022 m-3 (C) 1.49 x 1024 m-3 (D) 2.57 × 1024 m-3 5. Aluminum-lithium alloys have been developed by the aircraft industry to reduce the weight and improve the performance of its aircraft. A commercial aircraft skin material having a density of 2.55 g/cm3 is desired. Compute the concentration of Li (in wt%) that is required. Cli = (100Pli(Pal - Pave))/(Pave(Pal - Pli)) Cli = (100(0.534)(2.71 - 2.55))/(2.55(2.71 - 0.534)) Cli = (52.4*0.16)/(2.55*2.176) Cli = 1.54wt