Paradox with free expansion of ideal gas - where is the mistake? Suppose we have an isolated chamber of volume $V$ with a wall in the middle of the chamber and an ideal gas on one side of the wall. In a quasi-static process we expand the gas to the entire volume $V$.
My professor writes in his lecture notes, that since the energy of an ideal gas doesn't depend on it's volume we have
$$dU=0 \rightarrow dT=0$$
during the process. Also there's no change in particle number. So from the fundamental thermodynamic equation 
$$dU=TdS+pdV$$ we get
$$dS=\frac{p}{T}dV=\frac{nR}{V}dV \rightarrow \Delta S=nR\log(2) $$
Now I will present a different line of reasoning: Since the system is isolated there is no heat flow in the system 
$$\delta Q=0$$
Using
$$dS=\frac{\delta Q}{T}=0 \rightarrow \Delta S=0 \quad (*)$$
I think this paradox comes from using $(*)$ in a non-reversible process. But it's surprising to me that the fundamental thermodynamic relation yields the right answer, since it's basically derived by using $(*)$ and the first law.
$$dU=\delta Q+\delta W\stackrel{(*)}{=}TdS-pdV$$
Can someone shed light on this?
 A: The free  expansion is an irreversible process so that there are no equilibrium states joining the initial and the final one. For this reason the notation $dU$ and similar ones  is inappropriate: no infinitesimal changes exist here. 
During the free  expansion the work of (and on) the gas is evidently $0$ and  the net heat received by the system is $0$ since the system is isolated. From the first principle you find $\Delta U=0$ (notice $\Delta$ not $d$). Using the expression of $U$ of an ideal gas, you immediately conclude the the final  temperature coincides with the initial one (no intermediate temperatures can be defined during the expansion). 
The variation of entropy can be computed out of the known formula for an ideal gas as a state function of temperature and volume, again using only the initial and the final equilibrium states, finding your first result (your derivation is hower wrong if, as it seems, you integrate the infinitesimal variations along the actual transformation). Your last argument is untenable as you are improperly using the def of entropy as you also finally declare.
A: You are right in that * is not valid for irreversible processes, however there is no paradox here. The reason is that entropy is a function of state (it does not depend on how the system reaches a given state, only on the state itself), so you can calculate the change in entropy of an irreversible process by using a reversible process that has the same initial and final states.
A: 
In a quasi-static process we expand the gas to the entire volume V.

In response to my question as to how you would expand the gas quasi-statically, you responded:

Think of there being multiple walls. Each one with very small distance
  to the other one. We start pulling out wall after wall. Since the
  walls are very close to each other in each step we are close to
  equilibrium.

I would submit to you that your scheme would not result in a quasi-static (reversible) process. The reason is for a process to be quasi-static, thermal and mechanical disequilibrium needs to be minimized, i.e., approach zero. The procedure you describe doesn't, in my view, achieve this.
If I wish to compress a gas quasi-statically, the pressure of the surroundings must be infinitesimally greater than the pressure of the gas. Likewise, if I wish to expand a gas quasi-statically, the pressure of the gas must be infinitesimally greater than the pressure of the surroundings. In the scheme you propose you are making the displacements infinitely small, but no matter how small you make the volume of the evacuated space between the removable walls, it does not reduce the pressure differential between the gas and the vacuum when a wall is removed. 
Bottom line: When you remove the wall  no matter how small the volume, the pressure differential remains finite, not infinitesimally small. Therefore, in my view, the process is not quasi-static. 
Hope this helps.
A: I think that the reason is: In the 1st law, P in PdV is the pressure in the gas, but not the external pressure which is zero It was proven in my paper:I. A. Stepanov, Determination of the isobaric heat capacity of gases heated by compression using the Clément-Desormes method,  Journal of Chemical, Biological and Physical Sciences. Section C.  10(3), (2020), 108  116,. Free online. https://DOI.org10.24214/jcbsc.C.10.1.10816
A: Indeed the entropy of the outside (the environment) has not changed because the walls are adiabatic. This is what your second computation (*) showed, as you have referred to heat exchanged with the environment.
However, the entropy is increasing? Why?
In a (very metaphorical, not physical!) way, there is heat transfer (although a subtle one [and one that I would not call heat in a very proper sense]) between the two halves of the container as you remove the separation.  Basically, as there is the expansion, you might imagine there being a flow of moving molecules (i.e. kinetic energy) and thus a flow of heat (i.e. kinetic energy) between the two halves.
So there is entropy production inside the container, despite no heat coming from the outside.
That "hidden" heat is hard to define for an irreversible process. That is why we refer to that of a reversible one.
edit: as pointed out in comments, I do not intend to say that there actually is heat flow, but rather to give a "picture" of what is physically happening. 
The proper way to "see" where the entropy is coming from would be to refer to other definitions of entropy (the $S=klogW$ with $W$ number of microstates being the best one) or to compute it using a reversible path (isothermal+adiabatic). However that does not really explain why the entropy increases without any heat flowing in. 
