# 23.3: The Chemical Potentials of a Pure Substance in Two Phases in Equilibrium

#### The Clapeyron Equation

For an equilibrium between e.g. ice and water at constant $$T$$ and $$P$$ we get

$dG = \cancel{-SdT + VdP} +μ_{ice}dn_{ice}+μ_{water}dn_{water}$

As $$dn_{ice} = -dn_{water}$$, we get

$0 = [μ_{ice}-μ_{water}]dn_{water}$

As $$dn_{water}$$ is not zero, this means that $$Δμ$$ must be zero! This must hold true for any set of points where ice and water are in equilibrium. That is the almost vertical line in the diagram. Its points are not at the same P and T, but we can find out where they should be by considering the thermodynamic potential $$μ$$ as a function of $$T$$ and $$P$$:

$dμ = \left(\dfrac{∂μ}{∂P}\right)_TdT + \left(\dfrac{∂μ}{∂T}\right)_PdP$

Because $$μ= G/n$$, it is not hard to identify the partial derivatives:

$\left(\dfrac{∂μ}{∂P}\right) = \left(\dfrac{∂G/n}{∂P}\right) = \dfrac{V}{n} = V_{molar}$

$\left(\dfrac{∂μ}{∂T}\right) = \left(\dfrac{∂G/n}{∂T}\right) = \dfrac{-S}{n} = -S_{molar}$

This is true for both water and ice. As the $$Δμ=0$$, we can equate the $$dμ$$ expressions for both water and ice:

$\left(\dfrac{∂μ_{ice}}{∂P}\right)_TdT + \left(\dfrac{∂μ_{ice}}{∂T}\right)_P dP=\left(\dfrac{∂μ_{water}}{∂P}\right)_TdT + \left(\dfrac{∂μ_{water}}{∂T}\right)_PdP$

Rearranging and identifying the partials gives:

$V_{molar\,ice}dP -S_{molar\,ice}dT= V_{molar\,water}dP -S_{molar\,water}dT$

Solving for $$dP/dT$$ we get:

$\dfrac{dP}{dT} = \dfrac{ΔS_{molar}}{ΔV_{molar}}$

As $$ΔG= ΔH-TΔS = 0$$, we have:

$ΔS_{molar} = \dfrac{ΔH_{molar}}{T_m}$

So:

$\dfrac{dP}{dT} = \dfrac{ΔH_{molar}}{TΔV_{molar}}$

This expression should be valid for all points on the melt line, in fact it tells use that this line is defined by $$ΔH_{molar}/TΔV_{molar}$$. We immediately see why for water the line runs a little to the left: exceptionally ΔVmolar is negative for ice, because water is actually a little denser that ice.

The above expression(s) are named after Clapeyron. The values of $$ΔH_{molar}$$ and $$ΔV_{molar}$$ do not change much with pressure and can often be considered constants for the melting line. When gases are involved that is not really true.