Question

In a particular spontaneous process the entropy of the system decreases. What can you conclude about the sign and magnitude of δssurr?

Answer 1

Answer:

▲Ssurroundings is +ve and greater than ▲Ssystem

Explanation:

The Second Law of

Thermodynamics

States that, a spontaneous process leads to an increase in the entropy of the universe. That is for a spontaneous process ∆Suniv > 0

or ▲Ssystem + ▲Ssurroundings > 0

Hence according to the Second Law of Thermodynamics when

▲Suniverse > 0, means that the process is spontaneous

▲Suniverse = 0 the process will not take place under ordinary condition

▲Suniverse < 0, The backward, opposite or alternate process occurs

spontaneously

Therefore since the process is spontaneous

∆Suniv = ▲Ssystem + ▲Ssurroundings > 0

However ▲Ssystem < 0

Therefore ▲Ssurroundings - ▲Ssystem

> 0 or

▲Ssurroundings > ▲Ssystem

Hence ▲Ssurroundings is +ve and greater than ▲Ssystem