1. The problem statement, all variables and given/known data A container of volume 2V is divided into two compartments of equal volume by an impenetrable wall. One of the compartments is filled with an ideal gas with N particles. The gas is in equilibrium and has a temperature T. How does the total energy, the entropy, the temperature, and the pressure change when the separating wall is suddenly removed? (There is no interaction with the surroundings). 2. Relevant equations I'm guessing there are several, but perhaps dE=TdS-PdV+μdN, dS/dV=P/T, and PV=NkT are the most important ones. 3. The attempt at a solution Here's what I think is the answer, but I want to be sure: Since there is no energy added to the system, and there is no reason energy would be removed from the system, the total energy should remain the same. Since no thermal energy has been added or removed, the temperature should also remain unchanged. Since the temperature is unchanged, no particles have been added or removed, and the volume has been doubled, we can use PV=NkT to deduce that the pressure has to be halved. The problem is that I haven't found a good way of showing how the entropy would increase. It's obvious that it should increase, but I don't know by how much. Help????