User talk:Paul Wormer/scratchbook: Difference between revisions

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{{Image|Clapeyron.png|right|300px|The red line in the P-T diagram is the coexistence curve of two phases: I and II. For instance, ''II'' is the vapor and ''I''  the liquid phase of the same compound.}}


The '''Clausius–Clapeyron relation''', is an equation for a [[phase transition]] between two phases of a single compound. In a [[pressure]]–[[temperature]] (P–T) diagram, the line separating the two phases is known as the coexistence curve.  The Clausius–Clapeyron relation gives the [[slope]] <math>dP/dT\,</math> of this curve:
:<math>
\frac{\mathrm{d}P}{\mathrm{d}T} = \frac{Q}{T\,(V^{II} - V^{I})} ,
</math>
where  ''Q'' is the molar heat of transition (heat necessary to bring one mole of the compound from phase ''I'' into phase ''II''), ''T'' is the absolute temperature (the abscissa of the point where the slope is computed), ''V''<sup>''I''</sup> is the molar volume of phase ''I'' at the pressure and temperature of the point where the slope is considered and ''V''<sup>''II''</sup> is the same for phase ''II''. The quantity ''P'' is the absolute pressure.
The equation is named after [[Émile Clapeyron]], who derived it around 1834, and [[Rudolf Clausius]].
==Derivation==
The condition of thermodynamical equilibrium at constant pressure ''P'' and constant
temperature ''T'' between two phases ''I'' and ''II'' is the equality of the molar [[Gibbs free energy|Gibbs free energies]] ''G'',
:<math>
G^{I}(P,T) = G^{II}(P,T). \,
</math>
The molar Gibbs free energy of phase &alpha; (''I'' or ''II'') is equal to the [[chemical potential]]
&mu;<sup>&alpha;</sup> of this phase. Hence the equilibrium condition can be
written as,
:<math> \mu^{I}(P,T) = \mu^{II}(P,T), \;
</math>
which holds along the red (coexistence) line in the figure.
If we go reversibly along the lower and upper green line in the figure, the chemical potentials of the phases change by  &Delta;&mu;<sup>''I''</sup> and &Delta;&mu;<sup>''II''</sup>, for
phase ''I'' and ''II'', respectively, while the system stays in equilibrium,
:<math>
\mu^{I}(P,T)+\Delta \mu^{I}(P,T) = \mu^{II}(P,T)+\Delta \mu^{II}(P,T)
\;\Longrightarrow\;
\Delta \mu^{I}(P,T) = \Delta \mu^{II}(P,T)
</math>
From classical thermodynamics it is known that
:<math>
\Delta \mu^{\alpha}(P,T) = \Delta G^{\alpha}(P,T)= -S^{\alpha} \Delta T
+V^{\alpha} \Delta P, \quad\hbox{where}\quad \alpha = I, II.
</math>
Here  ''S''<sup>&alpha;</sup> is the molar entropy ([[entropy]] per [[mole]]) of phase
&alpha; and ''V''<sup>&alpha;</sup> is the molar volume (volume of one mole) of this
phase. It follows that
:<math>
-S^{I} \Delta T +V^{I} \Delta P = -S^{II} \Delta T +V^{II} \Delta P
\;\Longrightarrow\;
\frac{\Delta P}{\Delta T} = \frac{S^{II} - S^{I}}{V^{II} - V^{I}}
</math>
From the second law of thermodynamics it is known that for a reversible phase transition it holds that
:<math>
S^{II} - S^{I} = \frac{Q}{T}
</math>
where ''Q'' is the amount of heat necessary to convert one mole of
compound from phase ''I'' into phase ''II''. Clearly, when phase ''I'' is liquid and phase
''II'' is vapor then ''Q'' &equiv; ''H''<sub>v</sub> is the molar [[heat of vaporization]]. Elimination of the entropy and taking the limit of infinitesimally small changes in ''T'' and ''P'' gives the ''Clausius-Clapeyron
equation'',
:<math>
\frac{dP}{dT} = \frac{Q}{T(V^{II} - V^{I})}
</math>
==Approximate solution==
The Clausius-Clapeyron equation is exact. When we make the following assumptions we may perform the integration:
* The [[molar volume]] of phase ''I'' is negligible compared to the molar volume of phase ''II'', ''V''<sup>''II''</sup> >> ''V''<sup>''I''</sup>
* Phase ''II''  satisfies the  [[ideal gas law]]
::<math>
PV^{II} = R T \,
</math>
* The transition heat ''Q''  is constant over the temperature integration interval. The integration runs from the lower temperature ''T''<sub>1</sub> to the upper temperature ''T''<sub>2</sub> and from ''P''<sub>1</sub> to  ''P''<sub>2</sub>.
Under these condition the Clausius-Clapeyron equation becomes
:<math>
\frac{dP}{dT} = \frac{Q}{\frac{RT^2}{P}}
\;\Longrightarrow\;
\frac{dP}{P} = \frac{Q}{R}\; \frac{dT}{T^2}
</math>
Integration gives
:<math>
\int\limits_{P_1}^{P_2} \frac{dP}{P} = \frac{Q}{R}\; \int\limits_{T_1}^{T_2}
\frac{dT}{T^2}
\;\Longrightarrow\;
\ln\frac{P_2}{P_1} = -\frac{Q}{R}\;\left( \frac{1}{T_2} - \frac{1}{T_1} \right)
</math>
where ln is the natural (base ''e'') [[logarithm]].

Revision as of 07:08, 11 September 2009

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