Entropy change for an irreversible isolated system is

Solution (By Examveda Team)

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 $$TdS - dU - \delta W = 0.$$