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Considering entropy (S) as a thermodynam...

Considering entropy `(S)` as a thermodynamics parameter, the criterion for the spontaneity of any process is

A

`DeltaS_("system") + DeltaS_("surrounding") gt 0 `

B

`DeltaS_("system") - DeltaS_("surrounding") gt0`

C

`DeltaS_("system") gt 0`

D

`DeltaS_("surrounding") gt 0`

Text Solution

Verified by Experts

The correct Answer is:
A
For spontaneous process, `DeltaS` must be positive .
In reversible process
`DeltaS_("system") + DeltaS_("surrounding")=0`
Hence, system is present in equilibrium . (i.e., it is not spontaeous process)
While in irreversible process
`DeltaS_("system") + DeltaS_("surrounding") gt 0`
Hence, in the process `DeltaS` is positive.
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