3.00 moles of La and 7.00 moles of Cu are mixed at 1 bar and 1500 °C. The follow

Question

3.00 moles of La and 7.00 moles of Cu are mixed at 1 bar and 1500 °C. The following phase diagram applies. The mixture is cooled at a constant rate of heat loss. At what temperature will a second phase appear? At what temperature will the temperature drop be temporarily halted? At the very start of this halt, what are the compositions and amounts of the two phases present? 4 Cu-La 1500 PressurePa 1.01325E5 1400 500.05 1092.78 1146.62 1357.77 134.84 1193.71 500.05 107248 1014.78 1011.23 1101.61 798.58 550.00 747.98 1134.00 1193.00 LIQUID 1200 1000 FCC A1 FCC A1 800 FCC A1 CULA DHCP FCC A1 CULA 400 300 0.0 0.2 0.4 0.6 0.8 1.0 x La La

Explanation / Answer

Calculate mole fractions of each element

xLa= 3/(3+7)= 0.3

xCu= 7/(3+7)= 0.7

Inital condition

xLa=0.3

T=1500 C

Under constant heat loss we come down from 1500 to lower temp keeping mole fractions constant.

From diagam it can be observed that around 1100 C second phasewill start appearing as our point moves inside the dome on graph

if we keeop collling this material we will come down and encounter a horizontal line,

this is the point where temperatire drop will be halted.

At this point mole fractiomn values of 2 extremes are 0.33 and 0.25 and current mole fraction = 0.3

Using key rule,

xCu2La= (0.33-0.3)/(.33-0.25)

= 0.375

xCu4La= (0.3-.25)/(.33-0.25)

= 0.625

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