In a particular spontaneous process, the entropy of the system decreases?

Question

What can you conclude about the sign and magnitude of Δsurr?

-For a reversible process, ΔSuniverse = 0.

-For a reversible process, ΔSuniverse > 0.

-For a reversible process, ΔSuniverse < 0.

-For a spontaneous process, ΔSuniverse < 0.

-For a spontaneous process, ΔSuniverse = 0.

-For a spontaneous process, ΔSuniverse > 0.

Answer 1

*I&#039;m assuming Δsurr=ΔSuniverse=the change in entropy of the universe

A spontaneous process is one that takes place on its own, which means no work is being done on it. If no work is being done, then the entropy must increase because of one of the laws of thermodynamics (I believe the third). To summarize it, the entropy of the universe will always increase because no external work can be done on it (because there&#039;s nothing external to the universe).

As for a reversible process, because it can be reversed, entropy can go up or down(reversible means opposite things will happen each way), and it can also stay the same (work in = work out, entropy stays the same).

Answer: -For a spontaneous process, ΔSuniverse &gt; 0.

Answer 2

for a spontaneous process,  the entropy of universe increases