An Exact Argument for a Hypothetical Solute

In section 5.11 we derived an exact expression for the work required to create a cavity of radius r a/2 in a liquid consisting of hard spheres with diameter a and density p. A cavity of radius ajl can be formed by a hard solute of diameter zero. We shall refer to such a particle as a hard point. The solvation Gibbs energy of such a hard point is thus 

Relation (7.13.33) suggests that we use the hard point particle as our test solute to compare the solvation thermodynamics of this solute in different solvents. We immediately see from (7.13.33) that our test solute will be more soluble in a liquid for which the quantity pFf is smaller. In other words, decreasing either the density or the size of the solvent particles causes a decrease in AG. or an increase in solubility. Here we refer to solubility from an ideal gas phase—see for instance relation (7.10.2). 

The corresponding solvation entropy and enthalpy of the hard point are 

Note that since the hard point has no soft interaction, the average binding energy Bs s is zero (as for any hard particle of any size see section 7.11). Therefore all of AH must be due to structural changes induced in the solvent. This can be interpreted in terms of a relaxation term using any method for classifying the solvent molecules into 

This exact relation has been derived here for a hard point particle. For a real solute, the same conclusion was also inferred from the discussion based on the Kirkwood-Buff theory. Note also that the relaxation term in appears likewise in AS, and in the formation of the combination AH - TAS, it cancels out. 

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