Chapter 5

# Therefore hmix 0 which implies that for an ideal

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Unformatted text preview: P and PB = yB P (Dalton’s law) Using the expressions given above for the mole fractions in terms of nA and nB and the total number of moles n = nA + nB, we get for the free energy of mixing: Δmix G = nRT y Aln y A +yBln yB ( () ( )) Please, note that the free energy of mixing is negative, since all mole fractions are between 0 and 1. Since mixing took place at constant pressure Marand’s Notes: Chapter 5 - The Properties of Simple Mixtures 165 and temperature, and since the free energy of mixing is negative, mixing is therefore a spontaneous process. Mixing of liquids to form solutions: We will now consider the mixing of two liquids, A and B. We consider that the substance A in the liquid phase is in equilibrium with substance A in the vapor phase, above the liquid. Similarly, we can also consider component B in the liquid phase to be in equilibrium with component B in the vapor phase. Conditions of equilibrium are written mathematically as: µAL (T,P) = µAV (T,P) and µBL (T,P) = µBV (T,P) Before we consider the case of a mixture, let us first review what this tells us about pure components (A or B). A liquid in a closed container is in equilibrium with its vapor, which at temperature, T, has a vapor pressure, PA*. For the liquid- vapor equilibrium, the chemical potential of component A in the liquid phase is equal to that of component A in the vapor phase. We can then write: µA*L(T, PA*) = µA*V(T, PA*) where the * indicates that the component is pure (both in liquid and vapor phases). Mara...
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