The vapour pressure of copper over Cu-Fe alloys at 1550°C (1823 K), as a function of composition, is given below. xCu PCu (mm Hg) ; (pCu(N/m2) : 1.0 0.547 72.93 0.792 0.486 64.79 0.476 0.449 59.86 0.217 0.399 53.20 ) Calculate the activity of iron at x~ = 0.217, assuming that the alpha-func tion istthlesame forboth iron andcopper atthis concentration at1550°C (1823 K).
Question
The vapour pressure of copper over Cu-Fe alloys at 1550°C (1823 K), as a function of composition, is given below. xCu PCu (mm Hg) ; (pCu(N/m2) : 1.0 0.547 72.93 0.792 0.486 64.79 0.476 0.449 59.86 0.217 0.399 53.20 ) Calculate the activity of iron at x~ = 0.217, assuming that the alpha-func tion istthlesame forboth iron andcopper atthis concentration at1550°C (1823 K).
Solution
To calculate the activity of iron (aFe) at xCu = 0.217, we need to use Raoult's Law which states that the partial pressure of a component in a mixture is equal to the vapour pressure of the pure component multiplied by its mole fraction in the mixture.
Given that the vapour pressure of copper (PCu) over the Cu-Fe alloy is 0.399 mm Hg (or 53.20 N/m^2) at xCu = 0.217, we can write Raoult's Law for copper as:
PCu = xCu * P°Cu
where P°Cu is the vapour pressure of pure copper.
Rearranging for P°Cu gives:
P°Cu = PCu / xCu = 0.399 / 0.217 = 1.839 mm Hg (or 245.32 N/m^2)
Now, we can write Raoult's Law for iron (Fe) as:
PFe = xFe * P°Fe
where PFe is the vapour pressure of iron, xFe is the mole fraction of iron, and P°Fe is the vapour pressure of pure iron.
Given that the mole fractions of copper and iron in the alloy add up to 1, we have xFe = 1 - xCu = 1 - 0.217 = 0.783.
Assuming that the alpha-function is the same for both iron and copper, we have P°Fe = P°Cu.
Substituting these values into Raoult's Law for iron gives:
PFe = 0.783 * 1.839 = 1.440 mm Hg (or 192.06 N/m^2)
Finally, the activity of iron is defined as the ratio of the vapour pressure of iron to the vapour pressure of pure iron, so we have:
aFe = PFe / P°Fe = 1.440 / 1.839 = 0.783
So, the activity of iron at xCu = 0.217 is 0.783.
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