InoOP Chem

Let a system undergoes a change of state reversibly at a constant temperature T. The heat absorbed by the system from the suuroundings may be represented by qrev . The increase in entropy of the system (ΔS_system) will then be given by

ΔS_system = qrev/T …(i)

Since the surroundings have lost heat equal to qrev , reversibly at the same temperature T , the entropy change of surroundings ΔS_surroundings will given by

ΔS_surroundings = – qrev/T …(ii)

(since the surroundings have lost heat , therefore ,it is given a negative sign)

Therefore , total change in entropy of the system and its surroundings during reversible process is given by

ΔS_system + ΔS_surroundings= qrev/T – qrev/T = 0

It may, therefore , be concluded that in a reversible process, the net entropy change of the system and the surroundings is zero .In other words , there is no net entropy change in a reversible process.