I don't really understand how to couple the idea of entropy as info with heat. Let's say that you have an isolated system. A cube with a a series of movable partitions parallel to a face of the curve. Let's say that half of the cube is filled with an ideal gas, and you can remove each partition with no energy or arbitrarily low energy, e.g. you can slide out a partition, if it has no friction due to the parallel placement, it costs no energy. You don't even need to remove it all the way, just bigger that the mean path of the gas molecules.
So each time you move a partition the gas expands, if you have enough partitions and move them slowly enough you can guarantee a continuous path in a space of equilibrium states. You can adiabatically go through this path and no energy will leave or enter the system, as no work is done, no heat is exchanged.
So $dU = 0$
But the entropy changes, and because there is no change in energy so the change at the end of the process when the volume has increased to the entire cube, the change in entropy is proportional to $\ln V_2/V_1$.
We can also write $dS=\partial S /\partial UdU + \partial S/\partial V dV = 1/Tdu - p/TdV = 0/T - p/TdV$.
This is an irreversible change, as each time we remove a partition the gas grows a little and we would need to push it back and thus spend some energy. The entropy is then greater than the change in heat(as it is suppossed to be in an irreversible process), which is zero, so so far everything is consistent.
Yet it is written $0 = dU = dQ - pdV = 0 - pdV$. Because there is no flow of energy from the outside, so $dQ$ = 0 but this cannot be right as the pressure is always positive and the volume changes, $pdV \neq 0$: $dU = \partial U/ \partial S dS + \partial U / \partial V dV = TdS - pdV = 0$.
So therefore $TdS \neq 0$ so $dQ \neq TdS$. $dU \neq dQ - pdV$.
So even for a continuous path in the space of equilibrium states $dU \neq dQ - pdV$. To solve this problem either we change of definition of work and heat, so that $dW = pdV$ can be work even if there is no opposing force, which is what we mean in work in mechanics.
And heat would be just $TdS$; or instead we keep work as mechanical work and heat as flow of energy in which case the equation for the first law either is wrong $dU \neq dQ - pdV$.
Or we write $dU = dQ - dW, dQ = 0, dW = 0$ which is true but useless as we cannot use it to get the change in entropy, we have to get the formula from entropy from statistical mechanics and the equation of state, in this case the ideal gas law. It makes the definition of heat weird my opinion as flow of energy that is not work (which I guess is easy to measure empirically) as opposed to $TdS$ which is just conceptually clear and comes from the physics and makes sense to me.
I am asking specifically because I have seen formulas of change in heat in terms of specific heat and other variables.
So given that heat is supposed to be an inexact differential is there a way to always get a formula for it based on the partition function and not only for ideals gases?
