Enthalpy of vaporization from Clausius–Clapeyron

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Homework Equations



d ln(Psat) / d (1/T) = -ΔHvap/R (Clausius-Clapeyron)

The Attempt at a Solution



When solving for the enthalpy of vaporization using the Clausius-Clapeyron equation, is the resulting value an average over the temperature range? I assume this is the case, because enthalpy of vaporization is dependent on temperature.

I found the enthalpy of vaporization for carbon dioxide from 220K to 300K to be 16.59 kJ/mol (using the Clausius-Clapeyron equation) and tabulated saturation pressures.

I'm comparing this to the value I get from Hv-Hl=ΔHvap. When I take an average (arithmetic mean) of these values over the same temperature range, I get a value of 11.4 kJ/mol.

Why is there such a difference between using these two methods?
 

Answers and Replies

  • #2
Borek
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When solving for the enthalpy of vaporization using the Clausius-Clapeyron equation, is the resulting value an average over the temperature range? I assume this is the case, because enthalpy of vaporization is dependent on temperature.

It is not a simple average. On the left you have a derivative, ΔH is a function of the temperature - so you need to integrate, and integration can be thought of as a weighted average.
 
  • #3
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But if the values are over evenly spaced intervals (say 5K)?

I remember my prof talking about how carbon dioxide can behave unusually (I don't recall under what circumstances). Could this unusual behavior void some of the assumptions that are made in deriving the Clausius-Clapeyron equation?

Which according to my course notes are
-liquid volume<<vapor phase
-vapor phase acts as an ideal gas
 

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