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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

`Delta S_("system") gt 0` only

B

`Delta S_("surroundings") gt 0` only

C

`Delta S_("system") + Delta S_("surroundings") gt 0`

D

`Delta S_("system") - Delta S_("surroundings") gt 0`

Text Solution

Verified by Experts

The correct Answer is:
C

`Delta S_("system") + Delta S_("surroundings") gt O` (for spontaneity)
Because of `Delta S= R "ln " (V_(2))/(V_(1))`
Here the volume of gas increases from `V_(1)" to "V_(2)` at constant temperature T. The total increase in entropy of the system and its surrounding during the spontaneous process of expansion considered above is, thus R ln `((V_(2))/(V_(1)))` Since `V_(2) gt V_(1)`, it is obvious that the spontaneous (irreversible) isothermal expansion of a gas is accompanied by an increase in the entropy of the system and its surrounding considered together.
`Delta S_("system") + Delta S_("surrounding") = -22.13+ 22.05 = -0.08`
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