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Clausius-Clapeyron equation

  1. Nov 19, 2005 #1
    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: [tex] \ln P = -\frac{\Delta H_{vap}}{RT} + b [/tex]. How did we get [tex] -\frac{\Delta H_{vap}}{RT}[/tex] to be the slope? The y-axis is [tex] \ln P [/tex] and the x-axis is [tex] \frac{1000}{T} [/tex].

    Last edited: Nov 19, 2005
  2. jcsd
  3. Nov 20, 2005 #2
    anybody have any ideas?

  4. Nov 20, 2005 #3
    I'm not sure what your question really is but if you plot lnP vs 1000/T the slope will be

    [tex] -\frac{\Delta H_{vap}}{1000 R} [/tex]
  5. Nov 20, 2005 #4


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    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?
  6. Nov 20, 2005 #5
    The slope was actually [tex]\frac{\Delta H_{vap}}{R} [/tex]. I think it was meant to be written as: [tex] \ln P = -\frac{\Delta H_{vap}}{R}\frac{1}{T} + b [/tex]

    Is this correct?

  7. Nov 20, 2005 #6


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

    Last edited: Nov 20, 2005
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