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Clausius-Clapeyron relation

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Let's work out this problem from first principles. Basically, the underlying problem here is about what phase some stuff will be in thermal equilibrium at some temperature and pressure. Suppose we have some stuff that we keep at some constant temperature T and pressure P. Suppose that this stuff can undergo changes, like changes in phase, chemical reactions or whatever. We assume that the thermodynamical variables of the stuff before and after the change are well defined. But during the change the stuff does not have to be in thermal equilibrium.

In case of a change in phase this may sound strange because at a certain temperature and pressure you would expect the stuff to be in some unique phase. However, if you take e.g. pure water and cool it rapidly below 0 °C, then it will remain a fluid. You'll have supercooled water. Supercooled water can undergo a spontaneous change and become ice. But ice cannot become supercooled water via a spontaneous change. Supercooled water does have well defined thermodynamical variables, despite it not being in absolute thermal equilibrium. It is, however, stable against small perturbations. Freezing rain is an example of supercooled water. When rain passes through very cold air it becomes supercooled. When it collides with the ground it freezes instantly, so you'll get a layer of ice on the ground.

In case of chemical reactions this is a more familiar thing. Suppose that the stuff consists of iron and oxygen. The thermodynamical for iron and oxygen are well defined at room temperature and pressure. However iron will slowly react with the oxygen ...

Solution Summary

The Clausius-Clapeyron relation is derived from first principles and a detailed solution to the problem is given.

See Also This Related BrainMass Solution

Differential Equation with the Clausius-Clapeyron Relation

dP/dT = L/T deltaV,

is a differential equation that can, in principle, be solved to find the shape of the entire phase-boundary curve. To solve it, however, you have to know how both L and deltaV depend on temperature and pressure. Often, over a reasonably small section of the curve, you can take L to be constant. Moreover, if one of the phases is a gas, you can usually neglect the volume of the condensed phase and just take deltaV to be the volume of the gas, expressed in terms of temperature and pressure using the ideal gas law. Making all these assumptions, solve the differential equation explicitly to obtain the following formula for the phase boundary curve:

P = (constant) x e-L/RT (the vapor pressure equation).

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