Question

2. Determine (a) the solubility of tin in solid lead at 100 C, (b) the maximum solubility of lead in solid tin, (c) the amount of B that forms if a Pb-10% Sn alloy is cooled to 0 C, (d) the masses of tin contained in the a and B phases, and (e) mass of lead contained in the a and B phases. Assume that the total mass of the Pb-10% Sn alloy is 100 grams. Weight percent in

Answer 1

© The loốc temperature milerseets the savus curve at 5% sn. - The solubility of tin (sn) in lead (Pb) at looc there fore is 5%. The maximum sdebility of lead CPD in tin (sn), which is found from the tin-rich side of the phase diagram, occurs at the eutectic temperature of 188€ and is 97.5) Sn. o At oc, the 10% sn alloy is in af region of the phase diagram. By drawing a tie line at oc & applying the lever rale 48 = 10-2 x100 = 8:27 100-2 100- 8.27 7.2 = 100% -8% = - For lck = 91-84 100g of 10% Sn, Amount of 8 = 100 x 8.2 = 8129 ( 91.8 Amount of a = 100x 91.8 - a = 91.89 100 100 & Note that loog of the alloy will consist of log of Sn and gog of pb. The pb and so are distributed in two phases Crea and B. The mass of Sn in a phase = 2% Sn x 91.89 of a phase = 0.02 x 91.8 = 1.8369

1 - Since tin (sn) appears in bolts 2 and @ phases, the mass of Sn in the e phase will be, tes test mp = (10-1836)9 = 8.164 g @ The mass of pb in a phase will be equal to the mass of the a phase minus lite mass of Sn in a-phase. mpbind = 91.89- 1-8369 mpb ind = 89.964 g Alternatively, we can also found by mobina = 98%. Snx 91.8 g of d- phase = 0.98 x91-8 9 = 89.964 g Sao mass of the lead (mph) = qog mpb = m ob ind + mpb in e 90 = 89.964+ m pb in e Mass of ples in & phase al mobine = 00036g