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What is the mechanism and the reasons of this heat release? I know that the molecules of oxygen are solvated and the hydrogen bonds are rearranged.

But for what reasons are heat released and what does that concretely mean? Does it mean that the mean kinetic energy of the molecules is a bit increased?

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There are two things that drive the mixing of components: entropy and energy.

  • Entropy always promotes mixing because components have more microscopic configurations in the mixture than they have in the pure state. It is the reason that ideal gases mix, even though there is no interaction of any kind between molecules.

  • Energy can work in either direction. If unlike molecules "like" each other more than they like themselves, i.e., the solution has lower energy than the pure components, then energy promotes mixing. In this case the system must pass the extra energy to a bath as heat. In the opposite case where interactions between unlike molecules is unfavorable, the solution has higher energy (endothermic mixing) and we must supply the heat to produce the mixture.

The general qualitative answer your question is that water, being a good solvent for many molecules, interacts with these molecules in such way that the solvated solute plus water has lower energy than the pure components, i.e., it is more stable, and it must pass, therefore, the extra energy to the surroundings as heat.

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    $\begingroup$ “Entropy always promotes mixing” Entropy can also work in either direction, like enthalpy. Oil and water separate because water forms ordered cages around dispersed oil molecules, which decreases entropy; see this discussion. $\endgroup$
    – Chemomechanics
    May 13 at 13:59
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    $\begingroup$ @Chemomechanics I need to think about this. While I don't reject it off-hand, my intuition tells me otherwise. Obviously, if $\Delta S_\text{mix}<0$ we must have a strong enthalpic effect to make $\Delta G_\text{mix}$ negative in order to explain the mutual solubility between water and oil, which is small but not zero. Thanks for the link. $\endgroup$
    – Themis
    May 13 at 14:14
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    $\begingroup$ It doesn't look like it's addressed in the article, but I agree, the behavior must be different for extremely dilute solutions. The entropy increase from a single oil molecule wandering around a cosmos of water solvent must outweigh any entropy decrease from local water ordering around it. The full story must be more nuanced. My point was only that "always promotes mixing" may not be unequivocally true for real solutions. $\endgroup$
    – Chemomechanics
    May 13 at 15:48
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The anser is yes. Temperature always is the spread in the Maxwell distribution of velocities. Binding forces, potentials etc do not enter into the expectation value of v^2 ~ T.

The second part of the answer as always: If the gas dilutes, then it does because the diluted state has a lower potential energy.

By the physical conditions of such an experiment: Pressure constant by an open surface to the atmosphere, constant volume and isolated boundary, the kinetic energy has to go up.

And if the system paramters T, p change, things may go into the reverse direction.

Nice example: dilution and recondensation of H2O in air, our wheather.

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In the hope of providing a not-so-scientific answer that might be a bit more accessible to OP:

The molecules (or atoms, in case of a noble gas) are "zipping around" at a very high speed, so each of them has a comparably high kinetic energy. While the molecules of the fluid, in your case water, are moving much slower at the same temperature.

If the gas molecules get diluted in the water, they can't keep their kinetic energy, they collide with the much slower water molecules all the time, speed those up a bit, and lose their own speed. We can measure the "speeding up" of the water molecules as the water temperature rising.

This is basically the same mechanism that's behind heat of condensation.

This does not yet take any chemical bonds into consideration - forming chemical bonds will, in many cases, release additional energy. But this does not seem to be what your question is about.

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    $\begingroup$ "While the molecules of the fluid, in your case water, are moving much slower at the same temperature" This is wrong. At the same temperature the velocity distribution is the same in the liquid as in the gas. That's what Maxwell-Boltzmann says. Your explanation of the heat of condensation is also wrong: it is due to molecular interactions that are present in the liquid but not in the gas. $\endgroup$
    – Themis
    May 14 at 10:23

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