Past exam of the chemistry course of the year 4 of the natural sciences course of the University of Cambridge / 2022 / Paper 2 / Section E / 26

For a hard disc configuration, the potential energy is zero if no particles overlap and infinite otherwise. All allowed configurations therefore have the same configurational Boltzmann factor. At fixed $N,A,T$, their relative free energies differ through the volume of accessible configuration space:The kinetic contribution is common to the phases at the same temperature. “Entropy-driven crystallization” does not mean that particles develop an energetic attraction; excluded volume changes the number of possible many-particle arrangements.

At high density, disorder can frustrate efficient local packing and leave particles with very little room to move. An ordered arrangement can give every particle a larger cage of allowed displacements. The extensive gain in vibrational/configurational freedom within these cages can outweigh the loss of distinct spatial arrangements of cages. Visual order is therefore not a reliable measure of entropy. At low density the fluid's freedom to rearrange dominates, whereas sufficiently dense packing can favour a crystal through accessible free volume.

Freeze neighbouring particles and approximate the mobile particle's accessible area by $a_c$. Its one-particle translational contribution to the canonical partition function is $a_c/{\Lambda }^2$ in two dimensions. Neglecting correlations between different cages,There is no additional uncancelled $N!$: assigning particles to cages removes permutations, while restoring all $N!$ assignments cancels the indistinguishability factor. The result is a free-volume approximation; collective cage motion is excluded by construction.

The particle centre can move approximately $\ell -\sigma$ from the cell centre towards each of its four nearest fixed neighbours. Approximating the actual curved boundary by a square givesThis is the area accessible to the centre, not the geometrical area enclosed by the fixed particle centres or an area computed using $\sigma /2$. Two equal particles must keep their centres at least $\sigma$ apart.

For a square lattice, $A=N{\ell }^2$. Substituting $\ell =\sqrt{A/N}$ into the preceding result givesThusAlthough the requested parameterization emphasizes particle and thermal parameters, $B$ necessarily includes $N$ because it is a total close-packed area. At $A=B_{\square }$, neighbours touch, the cell free area vanishes and this classical cell model gives $F\to +\infty$. The two logarithms in the printed form are dimensionful separately but their combined argument is dimensionless. For $A<B_{\square }$ the presumed square lattice is geometrically impossible.

A triangular lattice has area per particle $\sqrt{3}{\ell }_{\mathrm{hex}}^2/2$, so ${\ell }_{\mathrm{hex}}=\sqrt{2A/(\sqrt{3}N)}$. Inserting this into the supplied accessible area yieldsThereforeThe prefactor $C$ remains $2/\Lambda$. The lattice is triangular in terms of particle centres; “hexagonal” here describes its sixfold coordination and cell symmetry.

Since $B_{\mathrm{hex}}<B_{\square }$, the hexagonal lattice has larger accessible cell area at any common allowed total area $A>B_{\square }$. HenceThe kinetic entropies cancel at the same temperature. The triangular lattice is favoured and has higher entropy within this cell model, even though both structures look ordered. More efficient packing leaves more local displacement freedom at a fixed number density. This comparison does not by itself establish whether either crystal is stable against the fluid.

Allow the total area and particles to divide between phases: $A=A_1+A_2$, $N=N_1+N_2$. Minimize $F_1(N_1,A_1,T)+F_2(N_2,A_2,T)$ subject to these constraints. Transferring area requires equal pressure and transferring particles requires equal chemical potential:Equivalently, take the lower convex envelope of free energy, using a common-tangent construction for phase coexistence. The lever rule fixes phase fractions between coexistence densities. Since the hexagonal branch lies below the square branch throughout their common domain here, this particular comparison supplies no stable square–hexagonal coexistence. A separate fluid free energy is needed to predict fluid–ordered coexistence and its endpoints.

For the actual two-dimensional hard-disc system, a hexatic phase can intervene between isotropic fluid and crystal; the one-particle cell model cannot describe that orientational-order physics. It is therefore safe to predict competition between fluid and ordered phases, but not a precise two-phase interval or a direct fluid–crystal transition from these two crystal branches alone.