University:
Massachusetts Institute of Technology
Course:
3.091SC | Introduction to Solid State Chemistry
Academic year:

2021
Views:

149

Pages:

3
Author:

Esperanza Lyons

Session #35: Homework Solutions Problem #1 (a) For each of the following Ag-Cu alloys state all phases present at the specified compositions and temperatures. Phase diagram given below. (i) (ii) (iii) (iv) (v) c c c c c = = = = = 20 atomic per cent Cu, T = 900°C 20 atomic per cent Cu, T = 800°C 20 atomic per cent Cu, T = 700°C 5 atomic per cent Cu, T = 700°C 80 atomic per cent Cu, T = 800°C (b) For the Ag-Cu alloy, c = 70 atomic per cent copper, calculate the relative amounts of all phases present at T = 600°C. © source unknown. All rights reserved. Solution (a) (i) all liquid (ii) liquid + Ag-rich solution of Ag-Cu (denoted α Ag) (iii)Ag-rich solid solution of Ag-Cu (denoted α Ag) + Cu-rich solution of Ag-Cu (denoted α Cu) (iv) Ag-rich solid solution of Ag-Cu (denoted α Ag) (v) liquid + Cu-rich solution of Ag-Cu (denoted α Cu) (b) % (Cu-rich solution of Ag-Cu) = % (Ag-rich solution of Ag-Cu) 70 - 5 x 100% = 70% 98 - 5 = 98 - 70 x 100% = 30% 98 - 5 Problem #2 (a) For each of the following Pb-Sn alloys state all phases present at the specified compositions and temperatures. Phase diagram given on the following page. (i) c = 10 atomic per cent Pb, T = 300°C (ii) c = 10 atomic per cent Pb, T = 200°C (iii) c = 10 atomic per cent Pb, T = 100°C (iv) c = 90 atomic per cent Pb, T = 200°C (v) c = 60 atomic per cent Pb, T = 200°C (b) For the Pb-Sn alloy, c = 60 atomic per cent lead, calculate the relative amounts of all phases present at T = 200°C. © source unknown. All rights reserved. Solution (a) (i) all liquid (ii) liquid + Sn-rich solution of Sn-Pb (iii) (Sn-rich solid solution of Sn-Pb) +(Pb-rich solid solution of Sn-Pb) (iv) Pb-rich solid solution of Sn-Pb (v)liquid + (Pb-rich solid solution of Sn-B) (b) % liquid = 73 - 60 x 100% = 31% 73 - 31 % (Pb-rich solid solution of Sn-Pb) = 73 - 60 x 100% = 31% 73 - 31