How to Calculate Vapor Pressure of Isooctane at 25°C?

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SUMMARY

The vapor pressure of isooctane at 25°C can be calculated using the Clausius-Clapeyron equation, which is ln(P2/P1) = (Hvap)/R * (1/T1 - 1/T2). Given that the normal boiling point of isooctane is 99.2°C and its enthalpy of vaporization is 35.76 kJ/mol, the vapor pressure at 25°C can be determined by isolating P2/P1. At the boiling point, the vapor pressure equals 1 atm, indicating that the vapor pressure at 25°C will be significantly lower than this value.

PREREQUISITES
  • Understanding of the Clausius-Clapeyron equation
  • Knowledge of thermodynamic concepts such as enthalpy of vaporization
  • Familiarity with temperature conversions between Celsius and Kelvin
  • Basic algebra skills for isolating variables in equations
NEXT STEPS
  • Calculate vapor pressure using the Clausius-Clapeyron equation with provided values
  • Research the significance of boiling points in relation to vapor pressure
  • Explore the properties of isooctane and its applications in gasoline
  • Learn about other methods for calculating vapor pressure for different substances
USEFUL FOR

Chemistry students, chemical engineers, and professionals involved in thermodynamics or petroleum engineering will benefit from this discussion on calculating vapor pressure.

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Homework Statement


A handbook lists the normal boiling point of isooctane, a gasoline component as 99.2°C and its enthalpy of vaporization as 35.76 kJ/mol. Calculate the vapor pressure of isooctane at 25°C.


Homework Equations


The only equation that makes sense to use for me is the Clausius-Clapeyron equation:
ln(P2/P1) = (Hvap)/R * (1/T1 - 1/T2)


The Attempt at a Solution


Since I don't have either P2 or P1, how can I solve this problem?
Should I just isolate (P2/P1), and solve?
 
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What does "boiling point" mean in terms of vapor pressure?
 
I think I got it. The vapor pressure should equal to 1 atm since boiling point is the temperature where the vapor pressure equals the surrounding's pressure or the atmospheric pressure. Thanks a lot!
 

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