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OpenStudy (anonymous):
Propane has a normal boiling point of -42.0 ∘C and a heat of vaporization (ΔHvap) of 19.04 kJ/mol. What is the vapor pressure of propane at 10.0∘C?
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OpenStudy (aaronq):
Use the Clasius-Clayperion equation: \(\sf ln(\dfrac{P_2}{P_1})=\dfrac{-\Delta H_{vap}}{R}(\dfrac{1}{T_2}-\dfrac{1}{T_1})\)
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