Phase Behaviour of Colloidal Superballs Mixed with Non-adsorbing Polymers

Abstract

Inspired by experimental work on colloidal cuboid–polymer dispersions [Rossi et al., Soft Matter, 2011, 7, 4139–4142] we have theoretically studied the phase behaviour of such mixtures. To that end, free volume theory was applied to predict the phase behaviour of mixtures of superballs and non-adsorbing polymer chains in a common solvent. Closed expressions for the thermodynamic properties of a suspension of hard colloidal superballs have been derived, accounting for fluid (F), face centred cubic (FCC) and simple cubic (SC) phase states. Even though these expressions are approximate for the solid phases, the hard superballs phase diagram semi-quantitatively matches with more evolved methods. The theory developed for the cuboid–polymer mixture reveals a rich phase behaviour, which includes not only isostructural F$_1$–F$_2$ coexistence, but also SC$_1$–SC$_2$ coexistence, several triple coexistences, and even a quadruple phase coexistence region (F$_1$–F$_2$–SC–FCC). The model proposed offers a tool to assess the stability of cuboid–polymer mixtures in terms of the colloid-to-polymer size ratio and superball shape.

Appendices

5.A Superball Properties: Calculation

Superball properties required for the calculation of the second virial coefficient and for the free energy of the solid phases are detailed in this section. The distance $\mathfrak {r}$ between the centre of a superball and an arbitrary point on the surface of the superball is given by [39]:

$$\begin{aligned} \mathfrak {r}(\theta ,\phi ) = R\left( |\cos {\phi }|^m |\sin {\theta }|^m + |\sin {\phi }|^m |\sin {\theta }|^m + |\cos {\theta }|^m \right) ^{-1/m} \text {,} \end{aligned}$$

(5.21)

where $\theta$ and $\phi$ are the polar angle and the azimuthal angle, respectively. The maximum distance ($\mathfrak {r}_\text {max}$) between the centre and the surface of the superball is the distance from the centre to the corner, as shown in Fig. 5.1 for the 2D superball projection. The angles corresponding to this maximum distance are $\theta = \pi /4$ and $\phi = 0$ for a 2D superball and $\theta = \arccos {\left( \sqrt{2/3}\right) }$ and $\phi = \pi /4$ for a 3D superball, which leads to a maximum distance given by:

$$\begin{aligned} \mathfrak {r}^{\mathrm {2D}}_{\mathrm {max}} = \mathfrak {r}\left( \frac{\pi }{4},0\right) = \sqrt{2}R\left( \frac{1}{2}\right) ^{1/m} \text {,} \end{aligned}$$

(5.22)

$$\begin{aligned} \mathfrak {r}^{\mathrm {3D}}_{\mathrm {max}} = \mathfrak {r}\left( \arcsin {\left( \sqrt{2/3}\right) },\frac{\pi }{4}\right) = \sqrt{3}R\left( \frac{1}{3}\right) ^{1/m}\text {.} \end{aligned}$$

(5.23)

The volume of the superball $v_{\mathrm {c}}$ is obtained by integration of Eq. (5.21) [39]:

$$\begin{aligned} v_\text {c} = \frac{8}{3}\int _0^{\pi /2}\int _0^{\pi /2}\sin {\left( \theta \right) }r(\theta ,\phi )^3\mathrm {d}\theta \mathrm {d}\phi \text {,} \end{aligned}$$

(5.24)

where integration is performed over one octant due to symmetry. Equation (5.24) can be solved analytically, resulting in [5, 11]:

$$\begin{aligned} v_\text {c} = \sigma ^3f(m) \text {,} \end{aligned}$$

(5.25)

with

$$\begin{aligned} f(m) = \frac{\left[ \Gamma (1+1/m)\right] ^3}{\Gamma (1+3/m)} \text {,} \end{aligned}$$

(5.26)

with $\sigma$ the diameter of the superball ($\sigma =2R$) and $\Gamma$ the Euler Gamma function. Exact equations for the surface area $s_{\mathrm {sb}}$ and for the mean curvature $c_{\mathrm {sb}}$ of a superball are not known, but they can be calculated numerically using the surface integral and the integral of mean curvature [39]:

$$\begin{aligned} s_\text {c} = 8\int _0^{\pi /2}\int _0^{\pi /2} \mathrm {d}\theta \mathrm {d}\phi \left|\left|\frac{\partial \vec {x}}{\partial \theta }\times \frac{\partial \vec {x}}{\partial \phi }\right|\right|\text {,} \end{aligned}$$

(5.27)

$$\begin{aligned} \begin{aligned} c_\text {c} =&\frac{8}{4\pi }\int _0^{\pi /2}\int _0^{\pi /2} \mathrm {d}\theta \mathrm {d}\phi \\ {}&\Big \{(\vec {x}_\theta \cdot \vec {x}_\theta )[(\vec {x}_\theta \times \vec {x}_\phi )\cdot \vec {x}_{\phi \phi }]+ (\vec {x}_\phi \cdot \vec {x}_\phi )[(\vec {x}_\theta \times \vec {x}_\phi )\cdot \vec {x}_{\theta \theta }] - \\&2(\vec {x}_\theta \cdot \vec {x}_\phi )[(\vec {x}_\theta \times \vec {x}_\phi )\cdot \vec {x}_{\theta \phi }]\Big \}\times \\ {}&\Big \{2(\vec {x}_\theta \cdot \vec {x}_\theta )(\vec {x}_\phi \cdot \vec {x}_\phi )-2(\vec {x}_\theta \cdot \vec {x}_\phi )^2\Big \}^{-1} \text {,} \end{aligned} \end{aligned}$$

(5.28)

where subscripts denote partial derivatives and $\vec {x}$ represents a vector from the centre of the superball to the surface, given by:

$$\begin{aligned} \vec {x} = \{\mathfrak {r}(\theta ,\phi )\sin {\theta }\cos {\phi },\mathfrak {r}(\theta ,\phi )\sin {\theta }\sin {\phi },\mathfrak {r}(\theta ,\phi )\cos {\theta }\} \text {,} \end{aligned}$$

(5.29)

with $\mathfrak {r}$ the distance from the centre of the superball (Eq. (5.21)). In view of the complicated forms of Eqs. (5.28) and (5.29), it is not surprising that formal solutions for the surface and mean curvature of superballs are not available.

5.B Close Packing and Free Volume of FCC and SC Crystals

In this section, clarification on the close packing fraction of the two solid states considered is provided. The general equation for the close packing fraction of a superball crystal is given by:

$$\begin{aligned} \phi _\text {c}^\text {cp} = \frac{N_\text {c} v_\text {c}}{V_\text {UC}^\text {cp}} \text {,} \end{aligned}$$

(5.30)

with $N_\text {c}$ the number of superballs in the crystal unit cell and $V_\text {UC}^\text {cp}$ the volume of the unit cell at the close packing fraction.

For the FCC crystal, the number of particles inside the unit cell is 4 and the volume of the unit cell at the close packing fraction is given by:

$$\begin{aligned} V_\text {UC}^\text {FCC,cp}&= \left[ 4\mathfrak {r}_\text {max}^\text {2D}\sin \left( \frac{\pi }{4}\right) \right] ^3 = (4R)^32^{-3/m} \text {,} \end{aligned}$$

which, combined with Eq. (5.30), gives the close packing fraction of superballs in the FCC crystal [Eq. (5.12)]. For the SC crystal, there is only a single particle inside the unit cell and the volume of the unit cell at the close packing fraction is simply given by:

$$\begin{aligned} V_\text {UC}^\text {SC,cp}= (2R)^3 \text {,} \end{aligned}$$

(5.31)

which results in the close packing fraction of superballs in a SC crystal given by Eq. (5.16).