The heat of evaporation of copper at its normal boiling point is 74.5 kcal/mole (311.71 kJ/mol). The vapour pressure of copper at 1500°C (1773 K) is 0.256 mm Hg (34 N/m ) . Calculate the boiling point of copper.
Question
The heat of evaporation of copper at its normal boiling point is 74.5 kcal/mole (311.71 kJ/mol). The vapour pressure of copper at 1500°C (1773 K) is 0.256 mm Hg (34 N/m ) . Calculate the boiling point of copper.
Solution
To calculate the boiling point of copper, we can use the Clausius-Clapeyron equation, which is a logarithmic form of the ideal gas law. The equation is as follows:
ln(P2/P1) = -ΔHvap/R * (1/T2 - 1/T1)
Where: P2 and P1 are the vapor pressures at temperature T2 and T1 respectively, ΔHvap is the heat of vaporization, R is the ideal gas constant, and T2 and T1 are the absolute temperatures.
Given in the problem: P1 = 0.256 mm Hg, T1 = 1773 K, ΔHvap = 311.71 kJ/mol = 311710 J/mol (since 1 kJ = 1000 J), R = 8.314 J/(mol*K) (universal gas constant).
We want to find T2 (the boiling point of copper) when P2 is 1 atm = 760 mm Hg.
Substituting the given values into the Clausius-Clapeyron equation:
ln(760/0.256) = -311710 J/mol / 8.314 J/(mol*K) * (1/T2 - 1/1773)
Solving for T2, we get:
T2 = 1 / [1/1773 + ln(760/0.256) * 8.314 J/(mol*K) / 311710 J/mol]
After calculating the above expression, we get the boiling point of copper, T2.
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