![]() Phase space volume is something like a counter of possible states of the system (in the most simple case, the positions and momenta of all particles). ![]() It is synonymous to a certain phase space volume $\Omega$ which we have to fix somehow. All allowable microstates have been made accessible and equally probable! Once the system has reached complete thermodynamic equilibrium no more entropy increase in possible. Think of the system spreading in the phase space. Thus, when you equilibrate more microstates become accessible. Since the total number of microstates you can access is small, entropy is lesser than what it could be if you remove that thermally insulating wall letting the system equilibrate. But you can calculate entropy by summing up entropy of different local equilibrium parts (e.g., entropy of the hot part and cold part separately). Technically, there is no unique global thermodyanmic state for the whole system and you cannot define entropy. Hence the system is restricted to fewer microstates. In fact if the system is stuck in non-equilibrium ( e.g., a hot part and cold part in the box separated by a thermally insulating wall) it cannot access some microstates at all. If the system is in non-equilibrium the system doesn't have equal probability of being in every microstate. In such a state there is no more increase in entropy possible. Therefore the each micostate has equal probability and $S=k \ln\Omega$. If the system is in equilibrium, all these microstates are equally probable and the system visits each of these microstates over the course of time (also known as ergodic hypothesis). What does that mean in terms of microstates? ![]() So is the entropy constant? Yes, if the system is in equilibrium. Isolated system: Since the matter, energy, and momentum is fixed, the total number of microstates available that satisfy these constraints is fixed/constant. ![]()
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