Entropy change for an irreversible isolated system is
Entropy change for an irreversible isolated system is Correct Answer > 0
From second law of thermodynamics$$\eqalign{ & TdS \geqslant \delta Q \cr & \Rightarrow TdS \geqslant dU + \delta W \cr} $$
For an irreversible process $$S - dU - \delta W > 0.$$ So, system undergoing irreversible change under constant internal energy and constant volume will have entropy change greater than zero $$dS>0.$$
And for an reversible process $$TdS - dU - \delta W = 0.$$
