Mixing of liquids -regular solution theory

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Thermodynamics of mixing

Consider the mixing of two pure liquids(at constant volume):
Jones Fig 3-1.png
The free energy of mixing is: <math>F_{mix}=U_{mix}-TS_{mix}</math>
or <math>F_{mix}=F_{A+B}-\left( F_{A}+F_{B} \right)</math>
The entropy of mixing is: <math>S=-k\sum\limits_{i}{p_{i}\ln p_{i}}</math>
or <math>S_{mix}=-k\left( \varphi _{A}\ln \varphi _{A}+\varphi _{B}\ln \varphi _{B} \right)</math>
The enthalpy of mixing is: <math>U_{AB}=\frac{z}{2}\left( \varphi _{A}^{2}\varepsilon _{AA}+2\varphi _{A}\varphi _{B}\varepsilon _{AB}+\varphi _{B}^{2}\varepsilon _{BB} \right)</math>
where <math>\begin{align}
 & z\text{ is the coordination number of each site} \\ 
& \varphi _{A}\text{ and }\varphi _{B}\text{ are the 2 volume fractions} \\ 
& \varepsilon _{AA},\varepsilon _{AB},\varepsilon _{BB}\text{ are the enthalpies of intactions} \\ 
& \text{The factor of 2 eliminates double counting}\text{.} \\ 

\end{align}</math>

The enthalpies of interactions in the neat phases are: <math>U_{A}=\frac{z}{2}\left( \varphi _{A}\varepsilon _{AA} \right)\text{and }U_{B}=\frac{z}{2}\left( \varphi _{B}\varepsilon _{BB} \right)</math>
After a little algebra: <math>U_{Mix}=\frac{z}{2}\varphi _{A}\varphi _{B}\left[ 2\varepsilon _{AB}-\varepsilon _{AA}-\varepsilon _{BB} \right]=kT\chi \varphi _{A}\varphi _{B}</math>
where: <math>\chi =\frac{z}{2kT}\left( 2\varepsilon _{AB}-\varepsilon _{AA}-\varepsilon _{BB} \right)</math>
Arriving at the free energy of mixing for the regular solution model: <math>\frac{F_{mix}}{kT}=\varphi _{A}\ln \varphi _{A}+\varphi _{B}\ln \varphi _{B}+\chi \varphi _{A}\varphi _{B}</math>





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Phase separation

The energy of mixing is a function of the <math>\chi </math> parameter and the volume fractions: <math>\frac{F_{mix}}{kT}=\varphi _{A}\ln \varphi _{A}+\varphi _{B}\ln \varphi _{B}+\chi \varphi _{A}\varphi _{B}</math>
<math>\frac{F_{mix}}{kT}</math> for various <math>\chi </math> values can be plotted as a function of concentration. <math>\text{For }\chi <2\text{ the curve has a single minimum}</math>. Consider graph (a). This energy surface shows that for any concentration, <math>\varphi _{0}</math>, the system is at lower free energy fully mixed than separated into two different concentrations.
Jones, Fig. 3.2
However, for <math>\text{ }\chi \ge 2\text{ }</math> the curve has two minima. Consider graph (b), It shows that combinations of the two low energy concentrations,

<math>\varphi _{1}\text{ and }\varphi _{2}</math> will have a lower free energy than the homogeneous mixture, <math>\varphi _{0}</math>. Any composition in this region will spontaneously separate into two phases, a phase separation.

Jones, Fig. 3.3


In the 2-phase region (the concentrations "inside" the phase boundary) the proportions of both phases are calculated by the "Lever Law": <math>\begin{align}
 & \varphi _{0}=\alpha \varphi _{1}+\left( 1-\alpha  \right)\varphi _{2} \\ 
& \text{or }\alpha =\frac{\varphi _{0}-\varphi _{2}}{\varphi _{1}-\varphi _{2}} \\ 

\end{align}</math>

where: <math>\begin{align}
 & \text{ }\varphi _{1}\text{ is the volume fraction of liquid 1 at the left phase boundary,} \\ 
& \text{ }\varphi _{2}\text{ is the volume fraction of liquid 1 at the right phase boundary,} \\ 
& \text{ and }\varphi _{0}\text{ is the given concentration of 1 in the system}\text{.} \\ 

\end{align}</math>

When two phases are formed, the free energy of mixing is: <math>\begin{align}
 & F_{2-phases}=\alpha F_{Mix}\left( \varphi _{1} \right)+\left( 1-\alpha  \right)F_{Mix}\left( \varphi _{2} \right) \\ 
& F_{2-phases}=\frac{\varphi _{0}-\varphi _{2}}{\varphi _{1}-\varphi _{2}}F_{Mix}\left( \varphi _{1} \right)+\frac{\varphi _{1}-\varphi _{0}}{\varphi _{1}-\varphi _{2}}F_{Mix}\left( \varphi _{2} \right) \\ 

\end{align}</math>




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"Stable" and "Unstable" states

The free energy of mixing as a function of composition for one portion of the two-phase region is plotted. Two concentrations, <math>\varphi _{a}\text{ and }\varphi _{b}</math>, are identified. The first (a) is a concentration that is metastable. That is, small fluctuations in concentration, indicated by the line just above it, create a higher energy state. Correspondingly, concentrations like (b) are unstable. That is, small fluctuations in concentration, indicated by the line just below it, create a lower energy state.
Jones, Fig. 3.4
The description just given can be repeated for <math>\chi </math> values greater than two. For each <math>\chi </math> value, we find four characterisitic points: the two concentrations with minumum energy (the outer concentrations) and two concentrations in the 2-phase region corresponding to the transition from metastable concentrations to unstable concentrations, that is, the inflection points on the energy versus concentration curve.
RegularSolutionPhaseDiagram.png
The curve on the right above is magnified and replotted here. This is the phase diagram of a two-component liquid system whose free energy of mixing is described by the regular solution model. When the <math>\chi </math> parameter is greater than two, the system phase separates. The outer curve is the phase boundary. The inner curve is the spinodal - the energy vs concentration relation where the homogeneous mixture passes from a unstable to metastable state.
Jones, Fig. 3.5


The curve just described is a plot of the the regular solution model in terms of the <math>\chi </math> parameter. It is instructive to consider that higher <math>\chi </math> values correspond (roughly) to lower temperatures. We can therefore plot the "isothermal" phase diagram as the inverse of the one in terms of interaction parameter. This diagram is now the familiar one: it shows a critical temperature above which the system is one phase and below which the system phase separates.
RegularSolutionIsotherms.png




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Spinodal instabilities

The existence of spinodals in phase behavior is of both theoretical and practical importance. An effect often found is that the preferred fluctuation near the spinodal has a characteristic length scale (dependent on the system). The "reason" is that large fluctuations which would seem to be the most unstable, are somewhat less likely because they include a substantial change in local concentration. Small fluctuations are less likely to produce large changes. Hence it is likely that the system has a characteristic length scale for phase separation. If relaxations in the system are slow (say, by having a high molecular weight) the characteristic length scale is "frozen".
Jones, Fig. 3.7
An example is the frozen pattern formed by spinodal decomposition of a mixture of two polymers in a common solvent. Confocal micrograph of a mixture of polystyrene, polybutyldiene, and toluene after quenching for 82 seconds.
Jones, Fig. 3.9
Light-scattering curves from a polymer mixture quenched into the unstable region of the phase diagram, showing the maximum in intensity at qmax characteristic of spinodal decomposition.
Jones, Fig. 3.10





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Nucleation in the metastable regime - two components

In the metastable regime, small fluctuations are quenched. Therefore some large fluctuation must occur to generate a region of lower free energy; that is, nucleation must occur. A reasonable question is: how big a fluctuation is needed?

The energy of the nucleus is a volumetric energy and a surface energy. <math>\Delta F\left( r \right)=\frac{4}{3}\pi r^{3}\Delta F_{v}+4\pi r^{2}\sigma </math>
The radius for minimum energy is: <math>r*=\frac{-2\sigma }{\Delta F_{v}}</math>
The nucleation energy is: <math>\Delta F\left( r* \right)=\frac{16\pi }{3}\frac{\sigma ^{3}}{\Delta F_{v}^{2}}</math>
The probability is: <math>\sim \exp \left( -\frac{\Delta F\left( r* \right)}{kT} \right)</math>

The volumetric energy is often taken to be that of the bulk material. The barrier to nucleation is greatly reduced by extraneous nuclei such as solid impurities.








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Nucleation in the metastable regime - solid/liquid

When a liquid freezes, the free energy remains the same (the two phases are in equilibrium at the melting temperature): <math>\begin{align}
 & \Delta G_{m}=\Delta H_{m}-T_{m}\Delta S_{m}=0 \\ 
& \Delta S_{m}=\frac{\Delta H_{m}}{T_{m}} \\ 

\end{align}</math>

Consider a small drop in temperature, the enthalpy would not change much, hence: <math>\Delta G_{b}\simeq -\frac{\Delta H_{m}}{T_{m}}\Delta T</math>
Estimate the free energy to form a nucleus: <math>\Delta G\left( r \right)=-\frac{4}{3}\pi r^{3}\frac{\Delta H_{m}}{T_{m}}\Delta T+4\pi r^{2}\sigma _{sl}</math>
Which has a minimum at a radius: <math>r*=\frac{2\sigma _{sl}T_{m}}{\Delta H_{m}\Delta T}</math>
And needs a free energy: <math>\Delta G\left( r* \right)=\frac{16\pi }{3}\sigma _{sl}^{3}\left( \frac{T_{m}}{\Delta H_{m}} \right)^{2}\left( \frac{1}{\Delta T} \right)^{2}</math>

A strong function of temperature; not usually seen, hence heterogeneous nucleation must predominate.


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