How Does the Clausius-Clapeyron Equation Derive the Slope -ΔHvap/RT?

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Discussion Overview

The discussion centers on the Clausius-Clapeyron equation and its implications for the relationship between vapor pressure and temperature, specifically how the slope of the plot of \(\ln P\) versus \(\frac{1000}{T}\) is derived. The scope includes mathematical reasoning and conceptual clarification regarding the equation's parameters.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant states that the relationship between vapor pressure and temperature is exponential, leading to the equation \(\ln P = -\frac{\Delta H_{vap}}{RT} + b\), questioning how \(-\frac{\Delta H_{vap}}{RT}\) becomes the slope when plotting \(\ln P\) against \(\frac{1000}{T}\).
  • Another participant suggests that the slope of the plot \(\ln P\) versus \(\frac{1000}{T}\) is actually \(-\frac{\Delta H_{vap}}{1000 R}\).
  • A different participant argues that the slope cannot be \(-\frac{\Delta H_{vap}}{RT}\) unless plotted against \(1\), indicating confusion about the variables involved in the slope calculation.
  • One participant proposes that the correct slope should be \(\frac{\Delta H_{vap}}{R}\) and suggests a reformulation of the equation to \(\ln P = -\frac{\Delta H_{vap}}{R}\frac{1}{T} + b\).
  • A later reply confirms the previous assertion about the slope, relating it to the vapor pressure at 1 atm.

Areas of Agreement / Disagreement

Participants express differing views on the correct interpretation of the slope in the context of the Clausius-Clapeyron equation, with no consensus reached on the correct formulation or the implications of the slope.

Contextual Notes

There are unresolved questions regarding the definitions of variables and the conditions under which the slope is derived, particularly concerning the relationship between the axes in the plot.

courtrigrad
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I know that the relationship between pressure and temperature for an ideal gas is linear. The relationship between vapor pressure and temperature for a liquid, however, is exponential. To make it linear we take the natural log and end up with: \ln P = -\frac{\Delta H_{vap}}{RT} + b. How did we get -\frac{\Delta H_{vap}}{RT} to be the slope? The y-axis is \ln P and the x-axis is \frac{1000}{T}.

Thanks
 
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anybody have any ideas?

thanks
 
I'm not sure what your question really is but if you plot lnP vs 1000/T the slope will be

-\frac{\Delta H_{vap}}{1000 R}
 
plugpoint said:
To make it linear we take the natural log and end up with: \ln P = -\frac{\Delta H_{vap}}{RT} + b. How did we get -\frac{\Delta H_{vap}}{RT} to be the slope? The y-axis is \ln P and the x-axis is \frac{1000}{T}.
The slope will be whatever -Hvap/RT is divided by to get the x axis. If I have y=ab and I plot y vs b, the slope is a. If I plot y vs a, the slope is b. The only (theoretical) way you will get -Hvap/RT as the slope is if you plotted lnP vs 1, but this doesn't make any sense. So in conclusion, you will never get -Hvap/RT as your slope :smile:

Which variable are you trying to solve or prove something for?
 
The slope was actually \frac{\Delta H_{vap}}{R}. I think it was meant to be written as: \ln P = -\frac{\Delta H_{vap}}{R}\frac{1}{T} + b

Is this correct?

Thanks
 
yes, and assuming that we're referring to vapor pressure 1 as 1atm, b is

(DHvap/R)1/T(1atm)
 
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