Past exam of the mathematics course of the University of Cambridge / 2022 / ii / Paper 4 / 35A (Statistical Physics)

The Carnot theorem states that no engine between two fixed heat reservoirs can be more efficient than a reversible engine, and that all reversible engines between those reservoirs have the same efficiency. Thus the heat ratio of a reversible engine depends only on the two reservoirs. Its multiplicative consistency for three reservoirs permits a state variable $T$, unique up to an overall scale, such thatFixing one reference value defines the thermodynamic temperature scale, and the reversible efficiency is

At fixed volume, a reversible infinitesimal heating supplies $\delta Q_{\mathrm{rev}}=C_{V}dT$. Since entropy satisfies $dS=\delta Q_{\mathrm{rev}}/T$,

Let the common final temperature be $T_f$. A reversible isolated process has zero total entropy production, so part (i) givesHencefor both bodies. The decrease in their total internal energy is extracted as work by the reversible engine mediating the transfer.

Direct contact permits no work extraction, so conservation of internal energy givesThe total entropy change isThe arithmetic-geometric mean inequality makes the last expression positive when $T_1\ne T_2$, with equality only when the bodies were already at the same temperature. This is the entropy production of irreversible heat flow.

The fundamental thermodynamic relation isBecause $E(S,V,N)$ is extensive, the Euler theorem for homogeneous functions givesSubstitution into the Gibbs free energy $G=E+pV-TS$ therefore yieldsFor a one-component equilibrium phase, the intensive chemical potential depends on the intensive state variables $T$ and $p$.

A first-order phase transition occurs where two phases have equal Gibbs free energy but a first derivative of that free energy is discontinuous. Typical jumps occur inA nonzero entropy jump gives a nonzero latent heat $T\Delta S$.

At coexistence, $G_{\mathrm{I}}(T_0,p,N)=G_{\mathrm{II}}(T_0,p,N)$. Stability saysThus $G_{\mathrm{I}}-G_{\mathrm{II}}$ crosses zero with negative slope as $T$ increases through $T_0$. On the two smooth phase branches,ThereforeThe transition from phase II to phase I consequently absorbs positive latent heat $T_0(S_{\mathrm{I}}-S_{\mathrm{II}})$.