I went through this a while back.

If we had 2 air tanks, say 1 full, 1 empty, connected together via hose to each tanks valve, valves closed, then we open the valves till the pressure levels out in the 2 tanks from the full one.

Now, if we were to measure the amount of energy that 1 full tank could provide to say an air driven motor from full till empty, and compared that to measuring the output of the air motor running on the 2 half full tanks, measuring the total for 1 tank and then the other and adding, do we get the same amount of energy from our air motor using the full tank till empty, and from the total of the 2 half empty tanks?

Did we lose half of the energy stored in the full tank, when we connected the empty tank to the full tank and leveled their pressures?

And when we level out the 2 tanks, is there a loss in the system because of heat produced by the transfer in the system?

The parallel between the two capacitors paradox and the two tanks problem is interesting and there are real similarities nevertheless we are not exactly in the same conditions.

Firstly you must give the way according which you fill one tank with the other: is it adiabatic or not? You may have noticed that I used the term "adiabatically" to qualify the charge of the capacitor from a current source. This meant that we "filled" the capacitor step by step, electron by electron, without creating a wide potential difference. In this case we avoid heat and entropy increase, and this is the interest of the above mentioned method. For pressure, it's the same. If your filling is not adiabatic, as the charge of a capacitor from a voltage source like from another capacitor, you will produce heat and so, loss energy.

The second point is that you suppose that your experiment is made in an environment which is neither at the absolute zero nor at a zero pressure. You must firstly obtain the pressure difference with a work exerted from a start point including the ambient temperature and pressure. The initial energy is this work. At the end of your experiment, the two tanks would be at the same pressure, but the possibility to extract energy from this state depends on the difference from the ambient pressure unlike the case of the capacitors that can be fully discharged in any case whatever the ambient potential because they are dipoles while your tank is like a monopole.

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If we had 2 large caps, 1 full and the other empty, and we put a light bulb in line, we would get light and heat, and the caps will still be half the original voltage, but we made use of the transfer. Or a motor. Or a JT.

That's correct. When we charge a capacitor from a voltage source we could use half the energy that otherwise would be wasted. It's an other option than the goal of the proposed method, which is to move without loss the energy from a capacitor to another one.

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Now, it is said, that if it were all supper conducting, that the caps will be more than half of the original voltage. But I dont know.

The caps wouldn't be more than half of the original voltage (see the arXiv papers, especially this one involving thermodynamics).

If you led the experiment with surperconductors, there would be an infinite current, incompatible with an experimental setup. So what happens in reality?

In fact in this case, if the conductors don't fuse and the strong magnetic field doesn't collapse the superconducting state, then you have to take into account the inductance of the circuit, because it becomes no more negligible in comparison with the circuit resistance. In the real life, all circuits have an inductance. The time constant due to the inductance will limit the current, working exactly as the circuit with the coil, the coil being replace by the natural inductance of the circuit.