Example 2.14: Benzene has a vapor pressure of 0.086 bar
and 0.9162 bar at 290 °K and 350 °K,
respectively. Find the vapor pressure
of benzene at 320 °K.
Solution
$
P1
= 0.086 bar = 0.086/1.01325 = 0.08488 atm
$
P2
= 0.9162 bar = 0.9162/1.01325 = 0.9042 atm
$
Constant B (Equation 2.45):
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$
Constant A (Equation 2.45):
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