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

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.

Assume all contributing decay rates satisfy ${\nu }_l>0$ and the summed expression is a nonnegative density. ThenThus ${\sum }_l{A}_l=1$ normalizes the first-passage distribution. A zero decay rate would require separate treatment of a nonabsorbing component; eventual absorption is implicit here. Individual spectral coefficients need not all be positive, but the full density must be.

For an unnormalized representation, divide the first moment by its total mass. Integration by parts givesConsequentlyFor normalized amplitudes the denominator is one. A mode with small weight can still dominate the mean if its lifetime $1/{\nu }_l$ is sufficiently large; this distinction matters when interpreting the later graph.

The observed statistic is the conditional mean among passages completed before $T=t_{\mathrm{obs}}$, not a mean in which unfinished trajectories are assigned duration $T$. Its normalization isFor each contribution,Dividing the weighted sum of these first moments by $Z_T$ givesThe denominator records the fraction of trajectories actually observed to finish.

Set $x={\nu }_{l}T$. Expanding $e^{-x}=1-x+x^2/2-x^3/6+\cdots$ gives $1-e^{-x}=x+O(x^2)$ and $1-e^{-x}(1+x)=x^2/2+O(x^3)$. The numerator contribution is therefore $A_l{\nu }_l{T}^2/2+O(T^3)$ and the denominator contribution is $A_l{\nu }_{l}T+O(T^2)$.

For one mode, cancellation yieldsThe curve rises monotonically from zero, initially with slope $1/2$, and rounds towards a horizontal plateau. On logarithmic axes its short-time slope is one. The crossover occurs at observation times of order $1/\nu$.

A disconnectivity graph joins minima when an energy threshold first permits a path between their basins. Taking thresholds in steps of 2 and admitting transition states with energy at most the threshold gives the following merges:

These eight merges connect all nine minima without duplicating a basin. Minimum endpoints lie at their stated energies; branch heights are the discretized superbasin thresholds, not necessarily the exact saddle energies. If a strictly-less-than threshold convention is used, barriers exactly on a grid line join at the next grid level; the physical connectivity is unchanged.

The initial slope near one is the short-observation regime $\mathcal{T}(T)\sim T/2$. Successively longer observations include trajectories that enter progressively longer-lived traps before reaching the absorbing minimum. The four marked shoulders represent successive relaxation timescales, not four distinct equilibrium phases. A plateau appears when the contributing faster modes have largely completed but the next slow component has not yet substantially changed the conditional mean. The ultimate plateau is the true mean first-passage time, approximately $e^{32.3}$ in the plotted time units.

To estimate a rate, locate the rounding timescale at which that mode stops growing with the observation window, using ${\nu }_{l}T\sim 1$; the final completion of a crossover requires several lifetimes. Graphical transitions lie approximately around $\ln T\sim 8$, 23, 29 and 35, giving only order-of-magnitude estimatesThese are roughly $3\times 10^{-4}$, $10^{-10}$, $3\times 10^{-13}$ and $6\times 10^{-16}$, with factors of several uncertainty from reading broad crossovers. Plateau heights are weighted sums $\sum A_l/{\nu }_l$ divided by completed probability, not individually $1/{\nu }_l$; in particular the final plateau at $\ln \mathcal{T}\simeq 32.3$ does not require the slowest lifetime to be $e^{32.3}$. The table fixes barriers and basin connectivity, but supplies no temperature or attempt-frequency convention that would justify exact rates directly from the energies.