The vapor pressure of liquid aluminum is 400 mm Hg at 2.59 x 103 K. Assuming that its molar heat of vaporization is constant at 296 kJ/mol, the vapor pressure of liquid Al is __ mm Hg at a temperature of 2.57 x 103 K.
This is solved using Clayperon Clausius equation:
ln (P at T1 / P at T2) = (molar heat of vaporization/ R) . ( 1/T2 – 1/T1) since the molar heat of vaporization, units are in kJ/mol, convert this to J/mol and use R-value 8.314 J/K.mol.The unit of pressure calibration does not matter as they form a ratio. The temperature has to be substituted in Kelvin only.
This will finally give you an equation that will look like
ln (400/P2) = – X
Take antilog on both sides
this will give you (400/P2) = e^-X
and then solve for P2
