Clausius-Clapeyron equation


by courtrigrad
Tags: clausiusclapeyron, equation
courtrigrad
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#1
Nov19-05, 10:49 PM
P: 1,239
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].

Thanks
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courtrigrad
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#2
Nov20-05, 07:01 AM
P: 1,239
anybody have any ideas?

thanks
da_willem
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#3
Nov20-05, 10:02 AM
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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]

ShawnD
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#4
Nov20-05, 01:51 PM
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Clausius-Clapeyron equation


Quote Quote by plugpoint
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].
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

Which variable are you trying to solve or prove something for?
courtrigrad
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#5
Nov20-05, 01:58 PM
P: 1,239
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?

Thanks
GCT
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#6
Nov20-05, 04:26 PM
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yes, and assuming that we're referring to vapor pressure 1 as 1atm, b is

(DHvap/R)1/T(1atm)


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