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.

pressure vs volume graphical representation of thermodynamic cycle

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

$$\Delta G = - T\,\Delta S.\tag{2}$$

$$ \begin{align} \Delta G_\mathrm{A \to B} &= \pu{-0.25 J K^-1}\times\pu{1000 K} = \pu{-250 J}\tag{3}\\ \Delta G_\mathrm{D \to C} &= \pu{0.25 J K^-1}\times\pu{300 K} = \pu{75 J}\tag{4} \end{align} $$

  • $\begingroup$ From what I've learnt in high school, the formula for entropy change is $\int\frac{dQ}{T}$. Here T is constant in the two processes asked for, so it simply becomes $\frac{\Delta Q}{T}$. I've never done entropy calculations with Gibbs free energy so far... that's probably just my inexperience, but then IChO is meant for high school students $\endgroup$
    – TRC
    Commented Aug 25, 2021 at 5:22
  • 2
    $\begingroup$ $\Delta G=0$ for process at equilibrium NOT for reversible process. $\endgroup$
    – Jay
    Commented Aug 25, 2021 at 5:34
  • $\begingroup$ @Jay The working fluid experiences a closed cycle, so of course $\Delta G=0$, since G is a function of state. $\endgroup$ Commented Aug 25, 2021 at 11:06
  • $\begingroup$ The reason the calculation is wrong is that the heat removed at the cold reservoir is not 250 J. It is something less than this. $\endgroup$ Commented Aug 25, 2021 at 11:09

1 Answer 1


The cycle shown is a classic reversible Carnot cycle. An intuitive reason for the change in Gibbs free energy not being equal to zero for such a step is that the pressure is not constant: the internal and external pressure are throughout in equilibrium, but since $p_\mathrm{ext} = p = nRT/V = \mathrm{constant}/V$, $p$ is not constant. A condition for applying $\Delta G$ as a criterion of spontaneity (reversibility) is that the pressure is constant.

In fact you can show (by extension of the calculations show below) that the free energy change is equal to the expansion or compression work done in that step: $$\Delta G_\mathrm{isothermal} = w$$ as follows.

Computing the entropy changes for the isothermal steps from the heats is straightforward using the first law of thermodynamics, since

  • for an ideal gas undergoing an isothermal expansion or compression $\Delta U=0$ so that $w=-q$
  • for a reversible isothermal process $q=T \Delta S$

such that we can compute

  • for the first step, $\Delta S = \frac{\pu{250 J}}{\pu{1000 K}} = \pu{0.250 J K-1}$

For the second isothermal step ($D\rightarrow C$) we use the fact that the total entropy change for the system, given that it is a cyclic process, is $\Delta S = 0$, therefore

  • for the second isothermal step, $\Delta S_\mathrm{DC} = - \Delta S_\mathrm{AB} = \pu{-0.250 J K-1}$

Regarding now the free energy computation, you would use (sure enough) $$\Delta G_\mathrm{isothermal} = \Delta H_\mathrm{isothermal} - T \Delta S_\mathrm{isothermal} = - T \Delta S_\mathrm{isothermal}=-q=w$$

since $\Delta H_\mathrm{isothermal}=0$ for such a process.

  • 2
    $\begingroup$ It maybe of interest to note that, in this case, the efficiency of the engine is given as $\frac{\Delta G}{q_{in} }$ that is ratio of the gibbs free energy by the heat put into the sytem $\endgroup$
    – Babu
    Commented Aug 25, 2021 at 12:54

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