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When we have an exothermic reaction as well as a decrease of the entropy in the system after the reaction, why does the decrease in the entropy of the system diminishes the amount of useful work we can extract from the system?

In my opinion, if we have at the final state less accessible microstates than at the initial state, the kinetic energy (at STP most of them as translational energy) of the system must be more concentrated and therefore have a higher capability of doing useful work. Thus, it gains reversible heat from its system, "stores" it as kinetic energy and could do work.

Where is my mistake? Could you please help me, maybe with a depiction?

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The way to think about microstates is as alternative configurations of a system which satisfy constraints consistent with the observed macroscopic properties of the system. The collection of such microstates forms an ensemble.

The misconception lies in the idea of "concentrating" kinetic energy, and in thinking that reducing the number of available microstates changes the available energy. It doesn't necessarily do so. It is necessary to be careful, since the idea that "concentrating kinetic energy reduces the entropy" is not generally correct. In general you should consider all available degrees of freedom. An ideal gas only has translational degrees of freedom (kinetic energy) so in this case this restriction to kinetic energy applies.

In addition, during an isothermal process performed on an ideal gas, the internal energy does not change, and neither does the average kinetic energy ($K=\frac32 N \mathrm{k_B}T$), but the number of available microstates may change (for instance with an expansion), as reflected in the dependence of entropy on volume. The change in entropy reflects a change in the probability distribution over the ensemble.

To illustrate what is meant with "dispersion" (and by extension entropy) envision a maze in which the system wanders. As the maze gets larger the system gets more lost in that maze, it has more choices where to go.

If you use a particle in a box model, assuming quantum like behavior, then the gaps between energy levels become smaller as you increase the size of the container, that is, with expansion. This results in spreading probability of occupation (dispersion) over more states. Now, irrespective of how the isothermal expansion is performed - reversibly or irreversibly (assume a free expansion) - the entropy cost for the system is the same. However, in the case of the spontaneous expansion the surroundings remains at constant entropy, whereas its entropy is reduced in the reversible case. In addition (and this is a key point), no work was performed in the free expansion.

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  • $\begingroup$ I appreciate your text. For your example we have constant internal energy and the same average kinetic energy, respectively, throughout the expansion. The system has now more accessible microstates and can therefore disperses its energy more ergo (in my opinion) "looses" some energy per Kelvin. How can this work in terms of the molecular view. The more spread out thermal energy is less useful now, but why $\endgroup$ Jan 14 '20 at 8:29
  • $\begingroup$ I think the idea of dispersion is causing confusion. Hope my edits help. $\endgroup$
    – Buck Thorn
    Jan 15 '20 at 8:06
  • $\begingroup$ I like that maze comparison. Now where the system has more choices what happens to the energy quanta. I always read that the entropy increases with every irreversible heat transfer and becomes less useful to us due to its reduced concentration though the quantity stays the same. Where is the connection to the changes probability distribution. $\endgroup$ Jan 18 '20 at 11:52
  • $\begingroup$ What energy quanta? And in the second sentence, which quantity stays the same? $\endgroup$
    – Buck Thorn
    Jan 18 '20 at 19:20
  • $\begingroup$ I have meant with "quantity" the amount of energy and energy quanta, respectively. And to your first question I have learned that the internal energy is quantized and is therefore the sum of energy quanta which are distributed over the microstates. Do you know what i mean. I apologize for my english. I am not a native speaker. $\endgroup$ Jan 19 '20 at 12:23

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