In problem 6, part 6.5 of the 2020 IChO (PDF), a thermodynamic cycle is given for “one mole of monoatomic perfect gas”:
- $\mathrm{A \to B}$; isothermal reversible expansion receiving $\pu{250 J}$ by heat transfer $(q_H)$ at a temperature of $\pu{1000 K}$ $(T_H)$ from a hot source.
- $\mathrm{B \to D}$; reversible adiabatic expansion.
- $\mathrm{D \to C}$; isothermal reversible compression at a temperature of $\pu{300 K}$ $(T_C)$ releasing some amount of heat $(q_C)$ to a cold sink.
- $\mathrm{C \to A}$; reversible adiabatic compression.
6.5 Calculate the entropy change $(\Delta S)$ for $\mathrm{A \to B}$ and $\mathrm{D \to C}$ processes in the heat engine in terms of $\pu{J K^-1}.$
To my understanding, the answer should be $\pu{0 J}$ for both steps since $\Delta G = 0$ for any reversible process, which $\mathrm{A \to B}$ and $\mathrm{D \to C}$ are. But the solution below gives non-zero values of $\Delta G.$ $ΔS$ for both steps was calculated in previous parts, $\pu{+0.25 J}$ for $\mathrm{A \to B},$ $\pu{-0.25 J}$ for $\mathrm{D \to C}.$ What is wrong with my reasoning here?
$$\Delta G = \Delta H - T\,\Delta S,\tag{1}$$
but $\Delta H = 0$ for an isothermal process, so
$$\Delta G = - T\,\Delta S.\tag{2}$$
$$ \begin{align} \Delta G_\mathrm{A \to B} &= \pu{-0.25 J K^-1}\times\pu{1000 K} = \pu{-250 J}\tag{3}\\ \Delta G_\mathrm{D \to C} &= \pu{0.25 J K^-1}\times\pu{300 K} = \pu{75 J}\tag{4} \end{align} $$