Gas diffusion into a vacuum is an irreversible process, and its entropy increases even if no heat transfer is required (as for an ideal gas). However, the entropies of the initial and final states can be calculated from equilibrium thermodynamic theory. For the ideal gas, S = RlnV/N, so the rise in entropy is Rln(V2/V1) where V1,V2 are the initial and final volumes. Anyway, saying "that the second law of thermodynamics cannot be calculated" doesn't make sense; you can't calculate a law but you can use it to make calculations which might test its validity. You did make such a calculation, but it is incorrect.

1. Put the calculation results from other places here, and the problem will be solved?

2. Physics pursues universality and consistency, where a theory cannot be applied at one moment and cannot be applied at another.

3 S (t1) does not exist, which means that thermodynamic entropy does not have temporal continuity.

Look, you simply have written down the wrong law. dS is not just dQ/T but dQrev/T, i.e. the reversible heat change divided by T. Generally, dQ≤dQrev. So David Aitchisons Smith's answer was right; to calculate the entropy change for an irreversible process, you have to find a reversible replacement process between the same initial and final states to obtain dQrev. You can find those elementary considerations in each decent textbook on thermodynamics. I suggest Callen, Thermodynamics and an Introduction to Thermostatistics.

Enrico Pier Giorgio Cadeddu

Yes, he is entirely wrong, precisely because dQrev isn't the dQ=0 of the process considered. The entropy change due to the process considered is not dQ/T, to calculate it you have to calculate dQrev/T. For the entropy change all that matters is the difference between the final and initial states, not the process by which that state change is achieved. Entropy is a state function, not a process function.

Of course, in books, you will often just see dQ/T, for the simple reason that only equilibrium processes are considered in the corresponding chapter of the book, which may be only about equilibrium thermodynamics. But as soon as you consider nonequilibrium processes, you have to make the distinction. And as soon as you distinguish correctly between dQ and dQrev, you have to account for the fact that the differential of entropy is given by dQrev/T, which is not zero for the state change considered. The calculation to be made here has actually been outlined above by David Aitchisons Smith.

Enrico Pier Giorgio Cadeddu

it is not appropriate to write dQ when the heat is not exchanged reversibly (quasi-static transformation), it is not an exact differential so you should write 𝛿Q. As Kassner said dQ stays for the exact differential in a reversible process, dQ = 𝛿Qrev .

"Ds=dQ/T" is defined as a reversible process that can be used, but an irreversible process that cannot be used. This violates the universality and consistency of natural science.

The formula change in S = dQ/T refers to a particular case, that of simple thermal heating. The expansion of an ideal gas into a vacuum involves no change in energy, so this formula does not apply to that case.