Reversibility related to entropy

[Vanessa Nicoll 3H]

Reversible process ∆S(total) = 0 and irreversible ∆S(surr) = 0. Can someone explain why that is?

[Miyu Tang Dis3I]

The total entropy change of a system and its surroundings is given by:

∆S(total) = ∆S(system) + ∆S(surroundings)

A reversible process is one where the system and its surroundings can be returned to their initial states after the process, without any change in entropy. This means that the entropy change of the system and its surroundings are equal and opposite:

∆S(system) = -∆S(surroundings)

Substituting this into the first equation gives:

∆S(total) = ∆S(system) + ∆S(surroundings) = 0

Therefore, in a reversible process, the total entropy change of the system and its surroundings is zero.

In an irreversible process, the system and its surroundings cannot be returned to their initial states after the process, and there is always some increase in entropy. However, it is still possible for the entropy change of the surroundings to be zero if the process occurs adiabatically (i.e., with no heat exchange between the system and surroundings):

Q = 0, so ∆S(surroundings) = 0

Substituting this into the first equation gives:

∆S(total) = ∆S(system) + ∆S(surroundings) > 0

Therefore, in an irreversible process where there is no heat exchange with the surroundings, the total entropy change of the system and its surroundings is still greater than zero, but the entropy change of the surroundings is zero.

[Damin Rawlins 3B]

For a reversible process, the total entropy change is zero as there is no net transfer of heat or energy between the system and surroundings. In an irreversible process, the total entropy change is still zero, but the change in entropy of the surroundings is negative and equal in magnitude to the change in entropy of the system. This is because the total entropy change of the universe must always be positive. ΔS(total) = ΔS(system) + ΔS(surroundings) = 0 for both reversible and irreversible processes.