Register to reply 
ClausiusClapeyron equation 
Share this thread: 
#1
Nov1905, 10:49 PM

P: 1,237

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 yaxis is [tex] \ln P [/tex] and the xaxis is [tex] \frac{1000}{T} [/tex].
Thanks 


#2
Nov2005, 07:01 AM

P: 1,237

anybody have any ideas?
thanks 


#3
Nov2005, 10:02 AM

P: 599

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] 


#4
Nov2005, 01:51 PM

Sci Advisor
P: 987

ClausiusClapeyron equation
Which variable are you trying to solve or prove something for? 


#5
Nov2005, 01:58 PM

P: 1,237

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 


Register to reply 
Related Discussions  
Clausius Clapeyron equation  Biology, Chemistry & Other Homework  3  
Clausius Clapeyron equation  Biology, Chemistry & Other Homework  1  
ClausiusClapeyron equation  Classical Physics  4  
ClausiusClapeyron equatioin's constant for methanol  Materials & Chemical Engineering  1  
ClausiusClapeyron Equation, and Cubic Unit Cells  Introductory Physics Homework  4 