Clausius-Clapeyron Equation. As assigned by Mr. Amendola despite the fact that he is no longer our chemistry teacher. The Men Behind the Equation. Rudolph Clausius German physicist and mathematician One of the foremost contributors to the science of thermodynamics
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As assigned by Mr. Amendola despite the fact that he is no longer our chemistry teacher
We will use this diagram in deriving the Clausius-Clapeyron equation.
This diagram represents a generalized phase diagram. The line acts as a phase line, or a coexistent curve, separating phases α and β.
As we know, this indicates that at all points on the line, phases α and β are in equilibrium.
The curved “d” represents the use of a partial derivative.
We previously wrote ∆G=∆H-T∆S. We can also represent this as ∆G=P∆V-T∆S (see the thermodynamics chapter of your book). Taking the derivative of both sides, we find that d∆G=∆VdP-∆SdT
We have two variables in this differential equation: T and P. To solve this, we treat this in two cases. First, we consider P as a constant. Then, we consider T a constant.
By manipulation, we find:
As ∆S=∆H/T, this can be written as:
We integrate this equation, assuming ∆H and ∆V to be constant, to find:
Can be written as:
which clearly resembles the model y=mx+b, with ln P representing y, C representing b, 1/T acting as x, and -∆Hvap/R serving as m. Therefore, the Clausius-Clapeyron models a linear equation when the natural log of the vapor pressure is plotted against 1/T, where -∆Hvap/R is the slope of the line and C is the y-intercept
We can easily manipulate this equation to arrive at the more familiar form of the equation. We write this equation for two different temperatures:
Subtracting these two equations, we find: