At the normal boiling temperature of iron, TB = 3330 K, the rate of change of the vapor pressure of liquid iron with temperature
2 answers:
The molar latent enthalpy of boiling of iron at 3330 K is ΔH = 342 10^3 J.
<u>Explanation:</u>
Molar enthalpy of fusion is the amount of energy needed to change one mole of a substance from the solid phase to the liquid phase at constant temperature and pressure.
d ln p = (ΔH / RT^2) dt
(1/p) dp = (ΔH / RT^2) dt
dp / dt = p (ΔH / RT^2) = 3.72 10^-3
(p) (ΔH) / (8.31) (3330)^2 = 3.72 10^-3
ΔH = 342 10^3 J.
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