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

[courtrigrad]

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

 

[courtrigrad]

anybody have any ideas?

thanks

 

[da_willem]

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

$$-\frac{\Delta H_{vap}}{1000 R}$$

 

[ShawnD]

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

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

 

[courtrigrad]

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

 

[GCT]

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

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