Consider a container with Volume V. The container is seperated by a non-movable, impearable wall, which allows heat flow. Both subvolumes are filled with an ideal gas. We can obtain informations about the final state by using the principal of maximum entropy. Let's say the total internal energy of the system is $U$, and the energy of the subsystems are $U_1$ and $U_2$:
$$S = S_1(U_1) + S_2(U_2) = S_1(U_1)+S_2(U-U_1) \rightarrow max$$
Because heat flow is allowed, $U_1$ can change:
$$0 = \frac{\partial S}{\partial U_1} = \frac{\partial S_1}{\partial U_1} - \frac{\partial S_2}{\partial U_1} = \frac{1}{T_1} - \frac{1}{T_2} \Rightarrow T_1 = T_2$$
Because volume and particle number of the two subsystems are fixed, we have obtained complete information about the subsystems.
More or less the same thing can be done if we use a movable, impearable wall (with heat flow), with the following result:
$$p_1 = p_2, T_1 = T_2$$
If you use a imovable wall, which allows particle exchange and heat flow one recieves:
$$T_1 = T_2, \mu_1 = \mu_2$$
In all of these 3 cases the principle of maximum entropy tells us what the final state of the two subsystems is.
Let's now consider an impearable, movable wall with no heatflow. We can also try to use the principle of maximum entropy:
$$S = S_1(V_1) + S_2(V_2) = S_1(V_1)+S_2(V-V_1) \rightarrow max$$
But now we encounter a problem: Because no heat flow is allowed ($0 = dQ = (1/T_1)dS_1 = -(1/T_2)dS_2$), the total entropy S won't change, no matter what value we choose for $V_1/V_2$.
In this case, the principal of maximum entropy won't tell us what the final state is.
If i imagine this scenario in real life it seems obvious to me that there must be a final state which the system occupies.
I had the idea to minimize one of the thermodynamical potentials, but since the minimum of the thermodynamical potentials follows from the principle of maximum entropy, that seems pointless to me.
How do i figure out the final state of the system?
