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

AI Thread Summary
To calculate the vapor pressure of isooctane at 25°C, the Clausius-Clapeyron equation is applicable, utilizing the normal boiling point of 99.2°C and the enthalpy of vaporization of 35.76 kJ/mol. The boiling point indicates that at this temperature, the vapor pressure equals 1 atm. To solve for the vapor pressure at 25°C, one must isolate the ratio P2/P1 in the equation. The discussion highlights the importance of understanding the relationship between boiling point and vapor pressure. Ultimately, applying the Clausius-Clapeyron equation will yield the desired vapor pressure at the specified temperature.
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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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