Prove that in a reversible process:
∆(system) + ∆S(surroundings) = 0
Or
Prove that there is no net change in entropy in a reversible process.
Suppose heat is absorbed by the system reversible and the heat is lost by the surroundings also reversibly (process occurs under complete reversible condition).
If qrev is the heat absorbed by the system reversibly, then the heat lost by the surroundings will also be qrev. If the process takes place isothermally at T kelvin, then Entropy change of the system
Entropy change of the surroundings
Hence the total entropy change for the combined system and surroundings will be:
Hence in a reversible process, the net entropy change for the combined system and the surroundings is zero i.e. there is no net change in entropy.