Practice paper 2: statistical mechanics to glasses (Lectures 12 to 25)

Plain words, (a). The curly brackets list how many oscillators sit in level 0, 1, 2 and 3. Weight means “how many microstates give this pattern”. The microstates page shows that $N! = 3! = 6$ is the number of ways to order the oscillators, and you divide by $n_i!$ for each level so that swapping two oscillators in the same level is not counted twice. Every microstate is equally likely, so the probability of a configuration is its weight over the total of 10.

Plain words, (b). Think of tossing coins. With 10 tosses an exactly even split is fairly common. With a thousand it is much rarer to stray far from the middle. The count of microstates piles up so sharply near 50:50 that nothing else is ever seen.

Plain words, (c). The characteristic temperature $\Theta$ is the temperature scale at which the first excited state starts to fill. The ratio is $e^{-\Theta/T}$ on the partition function page. A degenerate level is several states at one energy, so it holds that many times more molecules.

Model answer.

(a) $W\{1,1,1,0\} = 3!/(1!\,1!\,1!\,0!) = 6$. $W\{2,0,0,1\} = 3!/(2!\,0!\,0!\,1!) = 3$. All microstates are equally probable, so the probability of $\{1,1,1,0\}$ is $6/10$.

(b) All microstates are equally probable, so the most likely configuration is the one with the most microstates, which is the roughly 50:50 split. For a very large number of molecules the distribution is sharply peaked, so only this dominant configuration is seen (the thermodynamic limit). The lecture’s figure is that 99% of the time the number in each half varies by less than 1 part in $10^{16}$.

(c) (i) The ratio is about $e^{-\Theta/T}$. At 300 K it is $e^{-1} = 0.37$. At 3000 K it is $e^{-0.1} = 0.90$. (ii) With a doubly degenerate upper level the ratio is $(g_j/g_i)\,e^{-\Delta E/kT} = 2 \times 0.37 = 0.74$.

Very good answer: at 3000 K the system is “hot” ($T \gg \Theta$), with the excited state and many others significantly occupied. At 300 K it is comparable to $\Theta$.

Where this is taught: weights, equal probability and the thermodynamic limit on microstates and the Boltzmann distribution; the population ratio, degeneracy and hot versus cold on the partition function, degeneracy and characteristic temperature.

Plain words, (a). Take the sign convention the lecture uses: energy in is positive. Heat in is $+500$ J. A gas that pushes the surroundings back does work on them, so the work done on the gas is negative. Use $\Delta U = Q + W$ on the first law page, with $W = -p_{\text{ext}}\,\Delta V$.

Plain words, (b). Differentiate $U$. One term changes the energies of the levels. The other changes how many molecules sit in each level. The page on the first law says which is which.

Plain words, (c). The second law for an isolated pair says total entropy cannot fall. Write each entropy change as $dQ/T$. The pair is isolated, so what $y$ loses, $x$ gains.

Plain words, (d). At 0 K a perfect crystal has one arrangement, so $S = k\ln 1 = 0$. If each molecule can still point two ways, there are many arrangements left and the entropy is not zero.

Model answer.

(a) $W = -p_{\text{ext}}\,\Delta V = -(1.0\times10^5)(2.0\times10^{-3}) = -200$ J. $\Delta U = Q + W = 500 - 200 = +300$ J.

(b) Work is the levels shifting, $\sum_i n_i\,d\varepsilon_i$: the energies of the states change while the populations stay put. Heat is the populations changing, $\sum_i \varepsilon_i\,dn_i$: molecules move between levels that stay where they are.

(c) For the isolated pair $dS_{\text{total}} = dS_x + dS_y \ge 0$, so $dQ_x/T_x + dQ_y/T_y \ge 0$. With $dQ_y = -dQ_x$ this becomes $dQ_x\,(1/T_x - 1/T_y) \ge 0$. For $T_x < T_y$ the bracket is positive, so $dQ_x > 0$: heat flows into the colder system $x$. Heat flows spontaneously from hot to cold.

(d) Each CO molecule can take one of two orientations in the lattice, so there are $2^N$ arrangements and $S_{\text{residual}} = k\ln 2^N = Nk\ln 2$. A perfect crystal has only one microstate at $T = 0$, so $S(0) = k\ln 1 = 0$. CO is not a perfect crystal, so it keeps a residual entropy.

Very good answer: part (a) uses the lecture’s convention. If a textbook writes $\Delta U = Q - W$, then $W$ means work done by the system, $+200$ J, and $\Delta U$ is the same $+300$ J.

Where this is taught: $\Delta U = Q + W$, $dW = -p_{\text{ext}}\,dV$, the alternative sign convention, and work as shifting levels and heat as changing populations on the first law; $S = k\ln W$ and the second law on the second law and entropy; heat flowing hot to cold, the third law and CO residual entropy on heat flow, the third law and residual entropy.

Plain words. Each experimental set-up fixes some things and lets others go. Look up the row (volume or pressure fixed) and the column (adiabatic or isothermal) on the free energy page. Surroundings that are huge stay at equilibrium whatever the system does, so their entropy change can be written in terms of the system’s own heat.

Model answer.

(a) (i) Constant $T$ and $V$: the Helmholtz free energy, $dA \le 0$. (ii) Constant $T$ and $p$: the Gibbs free energy, $dG \le 0$.

(b) The surroundings are a large reservoir that is always close to equilibrium, so the entropy change of the universe can be written using only properties of the system. The condition $dS_{\text{universe}} \ge 0$ then becomes $dA \le 0$ at constant $T$ and $V$, or $dG \le 0$ at constant $T$ and $p$. The universe’s entropy is still being maximised, but only the system’s properties are tracked. At constant $T$, $A$ is the maximum work that can be extracted (A is for Arbeit, German for work).

(c) The temperature $T$, the pressure $p$ and the chemical potential $\mu$ of the system equal those of the surroundings.

Where this is taught: reservoirs, equal $T$, $p$ and $\mu$, the Clausius form, $A$ as maximum work, $H$, $G$ and the spontaneity table all on free energy and spontaneity.

Plain words. $\Theta$ is the same idea as in Q1. If $T$ is far above $\Theta$ the oscillator has many levels in play and behaves classically. If $T$ is far below, nearly everything sits in the ground state and the vibration cannot soak up heat. The ideal gas page and the Einstein solid page carry the numbers.

Model answer.

(a) $E = NkT^2\,\dfrac{d\ln q}{dT} = \tfrac{3}{2}NkT = \tfrac{3}{2}nRT$. The internal energy of an ideal gas depends only on its temperature. $C_V = (dU/dT)_V = \tfrac{3}{2}nR$, so $\tfrac{3}{2}R$ per mole.

(b) At 300 K, $T \gg \Theta_{\text{vib}}$ for Ar(s) (40 K), so it is in the high-temperature limit. For the SO₂ stretch (2000 K), $T \ll \Theta_{\text{vib}}$, so it is barely excited. For a diatomic gas, $C_V$ rises from $5R/2$ (translation and rotation) towards $7R/2$ as the vibration adds $R$.

(c) Each atom is an independent 3D quantum harmonic oscillator, and all atoms oscillate with the same frequency. At high $T$ the molar heat capacity approaches $3R = 24.9$ J K⁻¹ mol⁻¹ (the Dulong-Petit law). The model fails at very low temperature, where interactions give collective motion (phonons). The Debye model is more accurate there.

Very good answer: for (c), the factor of 3 arises because each atom vibrates in three directions and each contributes the heat capacity of one oscillator. Differences between materials come from their bond strengths, that is, their vibrational frequencies $\nu$.

Where this is taught: $q_{\text{trans}}$, $E = \tfrac{3}{2}NkT$, $U$ depending only on $T$ and $C_V$ on the ideal gas from statistical mechanics; $\Theta_{\text{vib}}$ values, the diatomic $C_V$, the Einstein assumptions, $3R$ and the low-temperature failure on vibrations and the Einstein solid.

Plain words, (a). Match each defect to the crystal it fits. The defects page gives the rule: a similar ion size and high coordination, or very different ion sizes.

Plain words, (b). There is a trade. Each defect adds energy, but the number of ways to scatter $n_d$ defects over $N$ sites is huge, and that adds entropy. Minimising $G = H - TS$ lands on a small non-zero number of defects.

Plain words, (c). An ion sits in its site and can only move if there is an empty one next to it. Extra vacancies mean more jumps. The ionic conductivity page has the plot and the slopes.

Model answer.

(a) NaCl has Schottky defects (pairs of oppositely charged vacancies), because it is highly coordinated and its ions are similar in size. AgCl has Frenkel defects (a vacancy with an interstitial), because its ions differ in size and have large interstitial holes, so the smaller cation moves into an interstice.

(b) Intrinsic defects form spontaneously because they increase entropy. $\Delta G = \Delta H - T\Delta S$, with $\Delta H$ rising linearly with $n_d$ and $T\Delta S$ curving, so $G$ has a minimum at a non-zero defect concentration, $n_d/N = \exp(-\Delta H_d/kT)$. Number of vacant sites $= 10^{-17} \times 10^{20} = 10^3$, about 1000.

(c) $\mathrm{Na}_{1-2x}\mathrm{Mn}_x\mathrm{V}_x\mathrm{Cl}$: each Mn cation added comes with one cation vacancy. More vacancies mean more sites for Na⁺ to jump into, and each jump has an activation energy $E_a$, so $\sigma = \sigma_0 e^{-E_a/kT}$. On a plot of $\log\sigma$ against $1/T$, at low $T$ the conductivity is set by the impurity concentration (flatter lines, slope set by $E_m$). At high $T$ intrinsic defects dominate (the steepest line, slope $E_m + E_d$).

Very good answer: at room temperature there are very few intrinsic defects, yet they are what make ionic conduction possible.

Where this is taught: Schottky and Frenkel defects and the formation of defects with $n_d/N = \exp(-\Delta H_d/kT)$ on point defects and their energy cost; Arrhenius conductivity, MnCl₂ doping and the $\log\sigma$ plot on ionic conductivity and batteries.

Plain words, (a). $x = 1$ is the point where the entropy gain of opening a link (from the degeneracy $g$) balances the energy cost. Setting $g\,e^{-\varepsilon/kT} = 1$ and solving for $T$ gives $T_c$ on the DNA zipper page. The lecture’s plot shows a small zipper smears the change out.

Plain words, (b) and (c). A random sample of all configurations would waste almost every trial on high-energy ones that never matter. See the simulations page. In the Ising model, the net magnetisation $m$ is the sum of the spins, so random spins cancel.

Model answer.

(a) $x = 1$ gives $g\,e^{-\varepsilon/kT_c} = 1$, so $T_c = \varepsilon/(k\ln g)$. If $\varepsilon$ doubles, $T_c$ doubles. Sharp transitions only occur in the thermodynamic limit. For $N = 20$ the partition function and the fraction unzipped vary gradually with $T$, but as $N$ grows the curve steepens, and at $N = 10^5$ it is a step from 0 to 1 at $x = 1$.

(b) Pick a random particle, change its state randomly, and calculate the new energy. If the energy went down, accept the move. If it went up, accept with probability $\exp(-\beta\Delta E)$. Generating configurations with probability $e^{-\beta E}$ and weighting them equally (importance sampling) concentrates the effort on the configurations that matter at equilibrium.

(c) Far above $T_c$, $m \sim 0$, because the spins point randomly. Far below $T_c$, $\lvert m\rvert \sim 1$, because almost all spins point one way. Real iron has magnetic domains in which the spins align. If the domains are randomly oriented the net magnetisation is zero, and a field can align them.

Very good answer: a gold nanoparticle shows the same size effect as the short zipper. With few particles there is no longer a sharp melting temperature.

Where this is taught: $T_c = \varepsilon/(k\ln g)$, the $N$ dependence and why phase transitions need the thermodynamic limit on interactions and the DNA zipper; Metropolis, importance sampling, $m \sim 0$ and $\lvert m\rvert \sim 1$, and domains on simulations, Monte Carlo and the Ising model.

Plain words, (a). Put the numbers into the formula for $T_c$. Then read off the graph on the mean-field page: a steep line meets the curve once, a shallow line three times.

Plain words, (b) and (c). Molecules cannot overlap, so each one takes up some room. Molecules also pull on each other at medium range, and the equation averages that pull over the whole volume. The van der Waals page shows the loop and why the real path cuts across it.

Model answer.

(a) $T_c = Jd/k = 75 \times 4 = 300$ K. At 500 K ($T > T_c$) there is one solution, $m = 0$, the unmagnetised phase. At 100 K ($T < T_c$) there are three: $m = 0$ (a free-energy maximum) and two non-zero solutions, the spin-up and spin-down magnetic phases.

(b) $b$ accounts for the excluded volume, $Nb$, because the repulsion at very short range means molecules cannot overlap, so $V$ is replaced by $V - Nb$. $a$ accounts for the attraction between molecules, which acts as a (negative) pressure correction $-aN^2/V^2$ and is found by assuming the other molecules are evenly spread (the mean-field step). At low pressure the molecules are far apart, the interactions are weak and the gas becomes ideal.

(c) Below $T_c$, compression seems to make the pressure fall, which is unphysical. The true equilibrium is set by the Gibbs free energy (the common tangent), and the system goes straight from the gas to the liquid, an abrupt change in volume, with liquid and gas coexisting at the volumes between. Above $T_c$ there is only a supercritical fluid.

Very good answer: mean-field theory predicts the abrupt transition in both the magnet and the fluid, but it is an approximation, so its numerical predictions are not accurate.

Where this is taught: $\tanh$, the three solutions and $T_c = Jd/k$ on mean-field theory and the Ising magnet; $a$, $b$, the loop, the common tangent and the limits of the model on the van der Waals fluid and the liquid-gas transition.

Plain words. Chemical potential is the Gibbs energy per mole, like a height that substances run downhill on. The phase diagram page draws $\mu$ against $T$ for each phase and shows that the lowest curve wins. On a boundary two curves cross, so the two values are equal.

Model answer.

(a) The chemical potential $\mu_i$ is the Gibbs free energy per mole of substance $i$ (the molar Gibbs energy). The phase with the lowest $\mu$ is the more stable. Along a coexistence curve $\mu_\alpha(p,T) = \mu_\beta(p,T)$.

(b) $\dfrac{dp}{dT} = \dfrac{\Delta S_{m,\text{tr}}}{\Delta V_{m,\text{tr}}}$. For melting, $\Delta V_{m,\text{melt}}$ is small, so $dp/dT$ is large and the curve is steep. Vaporisation has a larger $\Delta V_{m,\text{vap}}$, so the liquid-gas line is more sensitive to pressure (more curved).

(c) Beyond the critical point liquid and gas cannot be told apart: there is only a supercritical fluid. Supercritical CO₂ is easily reached and can be completely removed by lowering the pressure, so it is used as a solvent (for example to decaffeinate coffee).

Where this is taught: $\mu$ as molar Gibbs energy, phase stability, the coexistence condition, Clausius-Clapeyron and the critical point on chemical potential and phase diagrams.

Plain words. Heat capacity says how hard something is to warm. A bigger object needs more heat, so the lecture quotes it per gram or per mole. The calorimetry page lists the specific heat of water and defines $C = dQ/dT$. Rearranged, $Q = C_s\,m\,\Delta T$.

Model answer.

(a) Heat capacity is the heat needed to raise an object’s temperature by one kelvin, $C = dQ/dT$, in J K⁻¹. It is extensive, so we quote the specific (per gram) or molar (per mole) value. $Q = C_s\,m\,\Delta T = 4186 \times 2.0 \times 5.0 = 41\,860$ J, about $4.2\times10^{4}$ J.

(b) A first-order transition involves a latent heat, and the first derivative of $G$ changes discontinuously, so $S$ and $V$ jump (boiling, freezing). A second-order transition has no latent heat, and the discontinuity is in the second derivative, so $S$ and $V$ are continuous but the heat capacity $C_p = T\,(d^2G/dT^2)_p$ is not (ferromagnetic and superconducting transitions).

(c) Heat flow divided by the temperature scan rate is the heat capacity: $\dfrac{dQ/dt}{dT/dt} = C$. DSC compares a sample with a reference, and the heaters keep both at the same temperature, so the extra heater power the sample needs is the signal.

Where this is taught: $C$, $C_s$, $C_m$ and water’s value, the Ehrenfest classification and the DSC relation on heat capacity, calorimetry and DSC.

Plain words, (a). The first term pays you for making solid. The second charges you for the surface. Small nuclei are all surface and lose, big ones are mostly volume and win, and the crossover is $r^*$. The nucleation page gives $r^*$ and $\Delta G^*$ in terms of $\Delta T$, so doubling $\Delta T$ is a ratio problem.

Plain words, (b) and (c). Slow cooling stays close to $T_m$, where nucleation is slow but growth is fast. Rapid cooling gets you into the range where nucleation is fast. A glass is what you get if the liquid is cooled so fast that the atoms never find the crystal. See glasses and metastable states.

Model answer.

(a) $r^* = 2\gamma/\Delta G_V$, found from $dG/dr = 0$. Nuclei smaller than $r^*$ shrink and those larger grow. Since $\Delta G_V \approx \Delta H_V\,\Delta T/T_m$, $r^* \propto 1/\Delta T$ and $\Delta G^* = 16\pi\gamma^3/(3\Delta G_V^2) \propto 1/\Delta T^2$. Doubling $\Delta T$ halves $r^*$ and cuts $\Delta G^*$ to a quarter.

(b) Slow cooling means minor undercooling, so growth is rapid and nucleation slow, which gives few coarse-grained crystals. Rapid cooling allows more undercooling, so nucleation is rapid and growth slower, which gives many fine-grained crystals. Very rapid cooling involves little or no nucleation or growth and produces a glass.

(c) A glass is an amorphous solid that exhibits a glass transition, the gradual and reversible change between the brittle glassy state and the liquid as temperature rises. It is not a transition between two equilibrium states: $T_g$ depends on the cooling rate (always below $T_m$), so the glass depends on its history and cannot be the equilibrium state. It is kinetic: the thermal energy becomes too small for atoms to cross the barrier to crystallise on any practical timescale.

Very good answer: a glass is a metastable, frozen-in liquid-like structure, so it can have a non-zero residual entropy at $T = 0$.

Where this is taught: $r^*$, $\Delta G^*$, the $\Delta T$ dependence and the cooling-rate picture on nucleation, growth and undercooling; the definition, $T_g$, kinetics and metastability on glasses and metastable states; residual entropy in glasses on heat flow, the third law and residual entropy.