Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 1 / 35D (Statistical Physics)

Macroscopically, an adiabatic process exchanges no heat with its surroundings: $\delta Q=0$. A reversible adiabatic process is therefore isentropic, since $dS=\delta Q_{\mathrm{rev}}/T=0$. Microscopically, the system is thermally isolated, so changes in its energy arise from work done by changing external parameters rather than random energy exchange with a heat bath; in a sufficiently slow change, the populations of the corresponding microscopic states are carried along without producing entropy.

For repeated operation the working substance must execute a thermodynamic cycle. Reversible isothermal contact at $T_2$ absorbs heat $Q_2$, a reversible adiabat changes the temperature to $T_1$, reversible isothermal contact at $T_1$ rejects heat $Q_1>0$, and a second adiabat returns to the initial state. This is the Carnot cycle. Reversibility and zero net entropy change giveThe thermal efficiency is the net work output divided by the absorbed heat:ConsequentlyThe Carnot theorem shows that any irreversible cycle has smaller efficiency, so the two isotherms joined by two adiabats are optimal.

At fixed volume, entropy from heat capacity obeys $dS=C(T)dT/T$. The Third law of thermodynamics fixes $S(0)=0$ when the ground state is nondegenerate, so below the transitionAbove the transition,For a transition with no latent heat, entropy is continuous, and comparison with $S=\beta \log T+\gamma$ gives the directly testable relationThus calorimetric measurements of $C(T)$ and $T_c$ verify $\gamma$. More generally, if a measurable latent heat $L$ is present, $S(T_c^{+})-S(T_c^{-})=L/T_c$, and the formula becomes