Past exam of the thermal physics course of the University of Oxford / 2016 / 1

A state function is determined by the equilibrium state, rather than the path used to reach it. Its differential is exact and its change around a closed thermodynamic cycle is zero. Internal energy, entropy and enthalpy are examples; transferred heat and work are not state functions.

For slow expansion in contact with a large thermal reservoir, the fixed temperature of that reservoir maintains an isothermal process. For a fixed amount of ideal gas, internal energy and enthalpy depend only on temperature, soThe ideal gas law also gives $pV=\mathrm{constant}$. For a reversible expansion the absorbed heat equals the work done by the gas, while its entropy increases by $n_{\mathrm{mol}}R\ln (V_2/V_1)$; constant internal energy does not mean zero heat transfer.

With a thermally insulating boundary, $\delta Q=0$. If the slow piston motion is frictionless and reversible, it is an isentropic process:The first law of thermodynamics gives $dU=-pdV$, so expansion lowers the internal energy and temperature. For constant heat capacity ratio $\gamma$, $p{V}^{\gamma }$ and $T{V}^{\gamma -1}$ are constant combinations of state functions. Insulation alone does not guarantee constant entropy: an irreversible adiabatic process can produce entropy. The conclusion uses the intended reversible, quasistatic piston idealization.

For steady insulated flow through a porous plug, neglect changes of bulk kinetic energy and gravitational potential energy, and allow no shaft work. Each amount of gas brings flow work $p_1{V}_1$ upstream and delivers $p_2{V}_2$ downstream. The steady energy conservation balance isThus enthalpy is conserved in this Joule-Thomson effect. For an ideal gas, $H=H(T)$, so $T_1=T_2$ and $U_1=U_2$ as well. The entropy increases:The porous-plug process is irreversible despite being slow and insulated; it is not the reversible expansion of the preceding part.