Enthelpy of vaporization of hexane

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SUMMARY

The discussion focuses on estimating the enthalpy of vaporization (ΔHvap) of hexane using the Clausius-Clapeyron equation. The normal boiling point of hexane is 69.0°C, and the user initially calculated ΔHvap as 34 kJ/mol, while the expected value is 29.1 kJ/mol. The discrepancy arises from the assumption that ΔHvap is temperature-independent and the ideal gas behavior of the vapor phase. To improve accuracy, it is suggested to use a reference point closer to the boiling point of hexane.

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  • Understanding of the Clausius-Clapeyron equation
  • Knowledge of thermodynamic concepts such as enthalpy and boiling points
  • Familiarity with the properties of hexane
  • Basic principles of ideal gas behavior
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Homework Statement


The normal boiling point of hexane is 69.0°C. Estimate (a) its enthalpy
of vaporization

The Attempt at a Solution


In order to solve for ΔHvap in the Clausius-Clapeyron equation I need a reference T and p point. I extracted it from the triple point (T=178K and p=1,23Pa).
I solve and I get 34kJ/mol, while the answer is 29,1kJ/mol. What I've done wrong?

Thanks!
 
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The Clausius-Clapeyron equation is just an approximation, because it is assumed in deriving it that the enthalpy of vaporization is temperature independent and that the gas phase behaves as an ideal gas. Like it's said in the question, you are only expected to give an estimate, anyway.
 
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So how would you do to get a more accurate result? One option is to get a (T,p) point closer to the desired 69ºC and 1bar, but if this is not available...?
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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