I think I figured it out. Since the process is isothermal, we know that ΔQ=ΔW. We also know that W=NkT⋅ln(V2/V1).
Since ΔS=ΔQ/T, it follows that ΔS=Nk⋅ln(V2/V1)), which in out case yields ΔS=Nk⋅ln2
Right?
But there is no interaction with the surroundings as stated in the original problem. I don't think comparing this to a reversible process is of much help. What I need help with is finding a relation I can use to explain how much the entropy increases.
I've tried using dS/dN=-μ/T , dS/dV=P/T ...
Moving the wall to one side of the container through a quasistatic process (i.e. "infinitely slowly"). But this still doesn't help me find how much the entropy increases.
EDIT: Wait, since entropy does not increase in a reversible process, is the entropy the same? But hasn't the number of...
Homework Statement
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...
But that doesn't work.
Take the case I described in my attempt at a solution: If you simply count by hand, you will find 10 ways to get 3 units of energy out of a system of three particles each in its own quantum well with four energy levels (Ranging from E0=0 to E3=3). But nCr(3,3) ≠ 10, nor...
Homework Statement
"A model system consists of N identical ”boxes” (e.g. quantum wells, atoms), each box with only two quantum levels, energies E0 = 0 and E1 = ε What is the number of microstates corresponding to the macrostate with total energy Mε?"
The Attempt at a Solution
I've done...