Question

If its given For an isolated system∆U = 0

the n fine out ∆S?

Answer 1

$\implies \huge{\mathfrak{Answer :≈}}$

$\implies$ The Changes in internal energy (∆U) for an isolated system is zero

So,

$\implies$it does not exchange any energy in any form with the surroundings.

$\implies$But , The entropy tends to increase in case of spontaneous reaction.

•°• , ∆S > 0 or positive.

Answer 2

ANSWER :

Δ S > 0

Δ S will be positive .

Details :

We know that :

Δ S = p∆V /T

T Δ S = p Δ V

p Δ V is more than 0 .

Hence the Δ S has to be positive.

.

Logical :

Δ S has to be positive because the entrophy increases from reactant to product.

This justifies the answer !