# Gibbs energy and compression ratio

I came across a very interesting Gibbs energy-based interpretation about the importance of compressing the fuel before combustion in an engine (Kennesaw State University, [1]). However, while the conclusions are sound, it seems to me that the reasoning is somehow flawed. But my own approach leads to incoherent results. So I wanted to check if my criticisms are correct, and where they fail.

The reasoning in [1] goes as follows. We want to evaluate the amount of work available when combusting some compressed fuel.

1. We ignore temperature effects: let's assume everything takes place at the same temperature T.
2. Gibbs energy is defined as $$\Delta G = \Delta H - T \Delta S$$ and commented as follows "the Gibbs equation implies that if we want a lot of work, we look for a high negative delta H and a decrease in entropy, or as small an increase as possible."
3. Combusting the fuel (octane) increases the number of moles of gas in the cylinder, so the entropy (and the pressure) of the system increases upon combustion: "$$\Delta$$ S is positive, so the Free Energy (G) available to do work is reduced by the value of T$$\Delta$$S."
4. Compare 2 different compression ration :
• Case 1: the fuel is compressed from ambient pressure to pressure $$p_1$$, then combustion takes place and increases entropy (as discussed above) by an amount $$\Delta S_1$$.

• Case 2: idem, but compression to pressure $$p_2>p_1$$.

The reference [1] argues that $$\Delta S_2$$ <$$\Delta S_1$$: "the actual increase in entropy is less because there is less disorder at [$$p_2$$] than at [$$p_1$$]. The molecules have less freedom, the increase of T delta S is less than for the first case."

1. Conclusion "The value of $$\Delta G$$ is more negative [in case 2 as compared to case 1], the system can do more work."

My criticisms are the following :

A1. Gibbs energy gives the maximal available work for transformations at constant T and p. In the situation considered here, the transformation occurs at fixed volume, and varying pressure. The maximal available work is thus given by Helmoltz energy (which doesn't make a big difference here, but still)

A2. If $$\Delta G$$ is to be large and negative to have a lot of work, then we look for an increase of entropy $$\Delta S$$. The interpretation by Atkins is as follows: if the entropy decreases, it means that the system has given some heat away, and this heat is remove from the available energy.

B. same criticism as A2. The available work is -$$\Delta G$$, so reducing $$\Delta G$$ with a positive $$\Delta S$$ actually increases $$-\Delta G$$, and thus the amount of available work.

C. I agree with statement 3. Using the Sackur–Tetrode formula for an ideal gas, considering the transformation $$(N_i,\, p_i,\, T,\, V)\rightarrow (N_f,\, p_f=p_i N_f/N_i,\, T,\, V)$$, I get $$\frac{\partial \Delta S}{\partial p_i} = -\frac{k_B (N_f - N_i)}{p_i}<0$$ The entropy variation is indeed smaller when the transformation occurs at a higher pressure.

D. This is where incoherence kicks in. According to B and C, $$0 < \Delta S_2 < \Delta S_1$$ so $$0>\Delta G_2 > \Delta G_1$$ and $$0<-\Delta G_2 < -\Delta G_1$$. So it seems that more work would be available in case 1 than in case 2, which is obviously wrong.

• The quantity discussed in ref. [1] is not the amount of work provided by the engine over a cycle, but the amount of work which can be recovered directly from the combustion reaction ($$T \Delta S - \Delta H$$ instead of $$-\Delta H$$). In the usual (air standard) analysis of combustion engine, the energy provided by combustion is treated as pure heat input, and the standard entropy of combustion is simply neglected as compared to other entropy variations throughout the cycle.