Question 6.4: A liquid is at equilibrium with its vapour. The vapour is as...

A liquid is at equilibrium with its vapour. The vapour is assumed to be an ideal gas. The liquid has a molar latent heat of vaporisation \ell_{\ell g} that depends on temperature, with \ell_{\ell g} = A− BT , where A and B are constants. Apply the Clausius–Clapeyron relation (6.50) and consider that the molar volume of the liquid phase is negligible compared to the vapour phase, i.e. ν_\ell \ll ν_g. Use the ideal gas law (5.47) for the vapour phase. Show that at equilibrium at a temperature T, the vapour pressure p depends on temperature according to Dupré’s law,

\frac{dp}{dT} = \frac{\ell _{s\ell }}{T(\nu _\ell -\nu _s)}    and    \frac{dp}{dT} = \frac{\ell _{\ell g}}{T(\nu _g -\nu _\ell)} .                   (6.50)

p V = N R T                (5.47)

\ln \Bigl(\frac{p}{p_0}\Bigr) = \frac{A}{R} \Bigl(\frac{1}{T_0} - \frac{1}{T}\Bigr) - \frac{B}{R} \ln \Bigl(\frac{T}{T_0}\Bigr) .

where p_0 is the vapour pressure at T_0.

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Neglecting the molar volume of the liquid phase compared to the vapour phase and using the ideal gas law, i.e. ν_g = RT/p, the Clausius–Clapeyron relation (6.50) can be written as,

\frac{dp}{dT} = \frac{\ell _{\ell  g}}{T\nu _g} = \frac{A -B T}{T ν_g} = \frac{p(A -B T)}{RT^2}.

which can be recast as,

\frac{dp}{p} = \frac{A}{R} \frac{dT}{T^2} – \frac{B}{R} \frac{dT}{T} .

The integration of the relation in the initial state (p_0, T_0) to the final state (p, T) yields Dupré’s law,

\ln \Bigl(\frac{p}{p_0}\Bigr) = \frac{A}{R} \Bigl(\frac{1}{T_0} – \frac{1}{T}\Bigr) – \frac{B}{R} \ln \Bigl(\frac{T}{T_0}\Bigr) .

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