Suppose we have a closed container with mobile walls, so it will change its volume in order to equal the external pressure(constant) to the internal pressure. This container has inside it a liquid phase only of component 1 and a vapor phase of 1 and the inert gas(2) that is insoluble in the liquid phase. If the external pressure is constant how will change the vapor pressure of the component 1 changing the temperature? (suppose we know the molar partial enthalpy of the the components in gas phase and of 1 in liquid phase at the required temperature) $$\left( \frac{\partial P_{vap}}{\partial T} \right)$$

I have tried to find this relation by equaling the chemical potentials, and proceeding in the same manner of the demonstration of the Clausius-Clapeyron equation. However I am not sure of the result. Can anybody help me in the search of this partial derivative?


1 Answer 1


If the gas phase can be treated as an ideal gas mixture and the Poynting correction for the liquid can be neglected, then the partial pressure of component 1 in the gas phase will be the equilibrium vapor pressure of component 1 at the temperature of the system. The partial pressure of the non-condensible component 2 will be determined by the ideal gas law, based on the number of moles of this component, the temperature of the system, and the volume of head space.

If the gas phase cannot be treated as an ideal gas mixture but can still be regarded as an ideal solution, then the Lewis Randall law can be used to determine the fugacity of component 1 in the vapor, and this can be matched to that of component 1 for the liquid, with the Poynting correction included.

Does this come anything close to answering what you are asking?


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