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P1 = 1 atm (pressure of normal boiling point)
T1 = -42.0 °C + 273.15 = 231.15 K
T2 = 30.0 °C + 273.15 = 303.15 K
Propane has a normal boiling point of -42.0oC and a heat of vaporization (Hvap) of 19.04 kJ/mol. What is the vapor pressure of propane at 30.0 oC?
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Based on our data, we think this problem is relevant for Professor Ghosh's class at UA.