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We know that some solutions show positive deviations and some show negative deviation from Raoult's law.The former is endothermic and the latter is exothermic. And again, the volume change show positive and negative terms. In which case entropy change is more? And why is that?

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The detailed calculation for $\Delta G$ is here, Derive expression for internal energy of mixing and entropy of mixing using statistical thermodynamics which gives $$\Delta G = RT[n_1\ln(x_1)+n_2\ln(x_2)] +(n_1+n_2)x_1x_2w$$

and as $\Delta S= -d\Delta G/dT$ then

$$\Delta S = -R[n_1\ln(x_1)+n_2\ln(x_2)] - (n_1+n_2)x_1x_2\frac{dw}{dT}$$

where $w \approx (2E_{12}- E_{11} - E_{22})$ and $E_{12}$ is the interaction energy between molecules of type $1$ and $2$ when they are adjacent and similarly for $E_{11}$ and $E_{22}$ all of which are negative. From a consideration of the energies involved $w$ is expected to be positive.

The first term is the Raoult's law expression for an ideal solution and the second term the deviation, which is sometimes also called the excess entropy of mixing. For benzene cyclohexane when $n_1=n_2=x_1=x_2=1/2$ the excess entropy $\displaystyle -\frac{1}{4}\frac{dw}{dT}\approx 1.7$ J/mol/K at $298$ K and $1.2$ at $343$ K.

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