# Problem: 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?

###### FREE Expert Solution

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

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###### Problem Details

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?

What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Clausius-Clapeyron Equation concept. You can view video lessons to learn Clausius-Clapeyron Equation. Or if you need more Clausius-Clapeyron Equation practice, you can also practice Clausius-Clapeyron Equation practice problems.

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