Vapor pressure given boiling point and heat of vaporization?

AI Thread Summary
To determine the vapor pressure of benzene at 50.0°C, the Clausius–Clapeyron equation is applicable, which relates vapor pressure to temperature and heat of vaporization. The boiling point of benzene is 80.1°C, and its heat of vaporization is 31 kJ/mol. The equation ln P = -(Hvap)/RT + C can be used, where P is the vapor pressure, R is the gas constant, and T is the temperature in Kelvin. The constant C can be determined using known values at a specific temperature. Understanding the definition of boiling point is also crucial, as it indicates the temperature at which the vapor pressure equals atmospheric pressure.
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Homework Statement



What is the vapor pressure of benzene at 50.0 C? Benzene's boiling point is 80.1 C and its heat of vaporization is 31 kJ/mol

Homework Equations





The Attempt at a Solution



I don't want to be that guy who just says "I don't know" but.. All I know is that:

ln P = -(Hvap)/RT + C (and I don't know what C is besides "constant.")

and

ln(p1/p2) = Hvap/R (T1-T2/T1T2) but this relates the vapor pressure at different temperatures..

I am really not good at these applied things... any help is appreciated.
 
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You are right about trying to use Clausius–Clapeyron relation. What is the definition of a boiling point?
 
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