# Clausius-Clapeyron equation

Tags: clausiusclapeyron, equation
 P: 1,236 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
 P: 1,236 anybody have any ideas? thanks
 P: 599 I'm not sure what your question really is but if you plot lnP vs 1000/T the slope will be $$-\frac{\Delta H_{vap}}{1000 R}$$
 Quote by plugpoint 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 $$\frac{1000}{T}$$.
 P: 1,236 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