When 1 mole of argon gas is bubbled through a large volume of an Fe–Mn melt of X_Mn = 0.5 at 1863 K, the evaporation of Mn into the Ar causes the mass of the melt to decrease by 1.... When 1 mole of argon gas is bubbled through a large volume of an Fe–Mn melt of X_Mn = 0.5 at 1863 K, the evaporation of Mn into the Ar causes the mass of the melt to decrease by 1.5 g. The gas leaves the melt at a pressure of 1 atm. Calculate the activity coefficient of Mn in the liquid alloy. Read more

Steps to Solve

1. Identify the variables and constants

   • The mole fraction of manganese in the alloy, $X_{Mn} = 0.5$.
   • The mass loss of manganese due to evaporation, $\Delta m = 1.5 , \text{g}$.
   • The molar mass of manganese, $M_{Mn} \approx 54.94 , \text{g/mol}$.
   • The pressure $P = 1 , \text{atm}$.
   • The temperature $T = 1863 , \text{K}$.

2. Calculate the number of moles of manganese lost

   • Use the formula: $$ n_{Mn} = \frac{\Delta m}{M_{Mn}} $$ Substituting the known values: $$ n_{Mn} = \frac{1.5 , \text{g}}{54.94 , \text{g/mol}} \approx 0.0273 , \text{mol} $$

3. Calculate the total moles in the alloy before evaporation

   • Since $X_{Mn} = 0.5$, we assume a total of $n_{total} = 1 , \text{mol}$ (for simplification).
   • Then, the moles of manganese in the initial melt can be expressed as: $$ n_{Mn, initial} = X_{Mn} \cdot n_{total} = 0.5 \cdot 1 , \text{mol} = 0.5 , \text{mol} $$

4. Calculate the new number of moles of manganese after evaporation

   • The moles of manganese remaining after evaporation: $$ n_{Mn, final} = n_{Mn, initial} - n_{Mn} $$ Substituting the known values: $$ n_{Mn, final} = 0.5 , \text{mol} - 0.0273 , \text{mol} \approx 0.4727 , \text{mol} $$

5. Find the new mole fraction of manganese in the alloy

   • The new total moles in the alloy, assuming no significant change in the amount of iron: $$ n_{Fe} = n_{total} - n_{Mn, initial} = 1 - 0.5 = 0.5 , \text{mol} $$
   • The new mole fraction of manganese: $$ X_{Mn, final} = \frac{n_{Mn, final}}{n_{Mn, final} + n_{Fe}} = \frac{0.4727}{0.4727 + 0.5} $$

6. Calculate the activity coefficient using Raoult's law

   • Raoult's law gives us: $$ \gamma_{Mn} = \frac{X_{Mn, initial}}{X_{Mn, final}} $$
   • We’ll compute this value using the mole fractions calculated in the previous step.