What they were calling "sulphurous acid" back then is not what we would call an acid today. It was anhydrous sulphur dioxide which has a boiling point of $-10^\circ$C.
When liquid sulphur dioxide was poured into the red-hot vessel, due to the Leidenfrost effect, it would form itself up into globules and float on a layer of its own vapour. In this state the temperature of the globules would be just below that of its boiling of $-10^\circ$C as it evaporates away at a now greatly reduced rate. Pouring in a small amount of water, which freezes at $0^\circ$C, while the sulphur dioxide is in this state results in it freezing within a few seconds. Once all the sulphur dioxide has evaporated off, the ice will quickly melt again before being brought up to just below its boiling point of $100^\circ$C as it assumes its spheroidal form due to the Leidenfrost effect. If one is quick, before all the liquid sulphur dioxide has disappeared one can thrown out a small lump of ice from a red-hot crucible!
Update with a Reference
A source confirming the liquid sulphurous acid was indeed (anhydrous) sulphur dioxide can be found on page 645 of volume 3 of Lehrbuch der Physik by O. D. Chwolson, a copy of which can be found here. To quote, on the sixth and seventh lines down from the top of the page Chwolson writes (in German)
Boutigny brachte flüssige SO$_2$, welche bei $-10^\circ$C siedet, in einen glühenden Platintiegel, $\ldots$
which in English translation roughly reads
Boutigny brought liquid SO$_2$ [sulphur dioxde], which boils at $-10^\circ$C, into a glowing platinum crucible, $\ldots$
You went wrong in interpreting the Kelvin-Plank statement of the second law. There is no violation. The statement is (the details of the statement may differ depending on the references)(italics done by me for emphasis):
"It is impossible to devise a cyclically operating heat engine, the effect of which is to absorb energy in the form of heat from a single thermal reservoir and to deliver an equivalent amount of work"
No work is done by the system in the free expansion. In this case no heat is absorbed by the system in the cycle, only rejected by the system to the surroundings as a result of the isothermal compression done by the surroundings needed to return the system to its original state. The system does not deliver work. The surroundings does work on the system. There is no violation.
You can look at this also from the perspective of the surroundings. Since the system is isolated from the surroundings during the free expansion of the system, the surroundings does not partake in that process. When the surroundings does work to compress the gas in the system, it performs work, But this work is not part of a cycle. It is simply a single process, namely, a reversible isothermal process. Net work done during a single process using heat from a single reservoir (as opposed to a sequence of processes comprising a cycle) does not violate the law.
I think there has been a misunderstanding. Summarizing my question is:
"Why free expansion is said to be irreversible although (at least in
my opinion) you can bring back the whole universe to its initial
state?
But the whole universe has not been brought back to its original state.
The system has been brought back to its original state (including original entropy) but the state of the surroundings has now been changed. Heat has been transferred to the surroundings increasing its entropy. Thus there is a positive total entropy change of the universe (system + surroundings) making the entire process irreversible.
The following is in response to your follow up questions:
I'm trying to find another variable of the system or of the enviroment
(except entropy) which has changed at the end of the cycle in order to
show the irreversibility.
If by the term another "variable" you mean another property, the short answer is there is no other property that I am aware of to account for a process being irreversible than the property of entropy. This is the reason the second law and its associated property, entropy, was needed. The first law (conservation of energy) is satisfied by a process even if the process is impossible, as long as energy is conserved. The simplest example is that of natural heat flow.
We know that heat only flows naturally, or spontaneously (without external influence), from a hot body to a cold body. The reverse has never been observed to occur spontaneously, even though the reverse process does not violate the first law of thermodynamics if the heat lost by the cold body equals the heat gained by the hot body.
The example of the free expansion is more subtle. But the idea of spontaneity applies. The gas spontaneously expands from its chamber to the evacuated chamber. We would never see all the gas that expanded into the evacuated chamber spontaneously return to its original chamber. To be more precise, the probability is essentially zero. But there is no violation of the first law.
But the system comes back exactly to its
initial state, I was thinking that if we consider the enviroment made by the cold
reservoir, since it exchanges heat with the system, if we not consider
its thermal capacity infinity then its temperature is increased. Could
be this the variable which generates irreversibility in your opinion?
Good question. But if the temperature of the environment increases, and only heat transfer to the environment is involved, that would mean the internal energy of the environment would also increase. But that would be impossible in this example.
The work done by the environment on the system to return it to its original state exactly equals the heat transferred from the system to the environment (being that compression of the gas is isothermal). That means the change in internal energy of the environment has to be zero. For the entire cycle the change in internal energy of both the system and the environment is zero. This can be so regardless of whether the cycle is reversible or irreversible. The property of entropy is needed to show irreversibility.
Hope this helps.
Best Answer
There is a huge difference with the usual Joules experiment. Here the system is not isolated during the gas expansion into the evacuated chamber. Two things enter in competition in principle when a setup as the one considered in the article you mention is considered:
Having an evacuated chamber deals with point 1 and tells us that the gas does not have to work (and hence loose energy) to expand in the chamber to gain extra room. The fact that the whole system is connected to a piston in contact with atmospheric pressure tells us that the system is a) not isolated and b) that the atmospheric pressure will "help" the gas expanding in the chamber.
Now, assuming an expansion that is quasi-static, the pressure on the piston has to be balanced at every time during the expansion process. Basically, the piston will work a little bit each time there is a pressure drop on the gas side (due to leaking into the chamber) so as to balance pressures again. If the process is adiabatic, the work provided by the atmosphere has nowhere to go and stays in the gas as an increase in temperature/kinetic energy.