Clapeyron's equation

The Clapeyron equation, which Émile Clapeyron developed in 1834 , provides the slope of all phase boundary lines in the pT diagram of a pure substance , i.e. e.g. also between two solid phases . It is:

${\displaystyle {\frac {\mathrm {d} p}{\mathrm {d} T}}={\frac {\Delta _{\mathrm {trs} }S}{\Delta _{\mathrm {trs} }V}}}$

With

• ${\displaystyle p}$- pressure
• ${\displaystyle T}$- temperature in K
• ${\displaystyle \Delta _{\mathrm {trs} }S}$- change in molar entropy , d. H. the entropy per amount of substance , at the phase transition
• ${\displaystyle \Delta _{\mathrm {trs} }V}$- Change in molar volume

Specification for individual phase transitions

The Clapeyron equation can be specified for different phase boundaries; in particular the following transitions are determined by it:

• solid / liquid, see melting point

• liquid / gaseous ( Clausius-Clapeyron equation , temperature dependence of the saturation vapor pressure ):

${\displaystyle {\frac {\mathrm {d} p}{\mathrm {d} T}}\approx {\frac {\Delta _{\mathrm {vap} }H\cdot p}{R\cdot T^{2}}}}$

with - molar enthalpy of vaporization ${\displaystyle \Delta _{\mathrm {vap} }H}$

and - universal gas constant ${\displaystyle R}$

• solid / gaseous (temperature dependence of the sublimation vapor pressure ):

${\displaystyle {\frac {\mathrm {d} \ln p}{\mathrm {d} T}}\approx {\frac {\Delta _{\mathrm {sub} }H}{R\cdot T^{2}}}}$

with - molar enthalpy of sublimation ${\displaystyle \Delta _{\mathrm {sub} }H}$

Derivation

The slope of the phase boundary lines in the p - T diagram is described by the still unknown function . ${\displaystyle \mathrm {d} p/\mathrm {d} T}$

At a phase boundary line, i.e. H. with the value pair of pressure  p and temperature  T , in which two phases  α and β coexist in thermodynamic equilibrium , these two phases have the same chemical potentials  μ :

${\displaystyle \mu _{\alpha }\left(p,T\right)=\mu _{\beta }\left(p,T\right)}$		(1)

Since equation 1 also applies to infinitesimal changes in  p or  T on the entire phase boundary line , the change in potentials must always remain the same:

${\displaystyle \mathrm {d} \mu _{\alpha }=\mathrm {d} \mu _{\beta }}$		(2)

It is known from the Gibbs-Duhem equation that

${\displaystyle \mathrm {d} \mu =-S_{\mathrm {m} }\cdot \mathrm {d} T+V_{\mathrm {m} }\cdot \mathrm {d} p}$		(3)

Substituting in equation 2 yields

${\displaystyle \Rightarrow -S_{\alpha ,\mathrm {m} }\mathrm {d} T+V_{\alpha ,\mathrm {m} }\mathrm {d} p=-S_{\beta ,\mathrm {m} }\mathrm {d} T+V_{\beta ,\mathrm {m} }\mathrm {d} p}$.		(4)

Factoring out d p and d T as well as subsequent transformation provides the Clapeyron equation:

${\displaystyle \Leftrightarrow {\frac {\mathrm {d} p}{\mathrm {d} T}}={\frac {\Delta _{\mathrm {trs} }S}{\Delta _{\mathrm {trs} }V}}}$		(5)

with resp. ${\displaystyle \Delta _{\mathrm {trs} }S=S_{\beta ,\mathrm {m} }-S_{\alpha ,\mathrm {m} }}$
${\displaystyle \Delta _{\mathrm {trs} }V=V_{\beta ,\mathrm {m} }-V_{\alpha ,\mathrm {m} }}$

For reversible processes , the entropy of conversion can be calculated from the amount of heat  Q _{rev converted} , which in isobaric processes is equal to the change in the molar enthalpy  H _m :

${\displaystyle \Delta S_{\mathrm {m} }={\frac {Q_{\mathrm {rev} }}{T}}={\frac {\Delta H_{\mathrm {m} }}{T}}}$		(6)

This gives the Clausius-Clapeyron equation.