Vapor pressure given boiling point and heat of vaporization?

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SUMMARY

The vapor pressure of benzene at 50.0°C can be calculated using the Clausius–Clapeyron equation. Given that benzene's boiling point is 80.1°C and its heat of vaporization is 31 kJ/mol, the equation ln P = -(Hvap)/(RT) + C is applicable. The constant C can be determined based on known vapor pressures at specific temperatures. This approach allows for the determination of vapor pressure at temperatures below the boiling point.

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  • Understanding of the Clausius–Clapeyron equation
  • Knowledge of vapor pressure concepts
  • Familiarity with thermodynamic principles
  • Basic algebra for solving equations
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  • Research the relationship between boiling point and vapor pressure
  • Explore the concept of heat of vaporization in detail
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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.
 
Physics news on Phys.org
You are right about trying to use Clausius–Clapeyron relation. What is the definition of a boiling point?
 

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