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Thermodynamics

Question
CBSEENCH11006249

Prove that in a reversible process:
(system) + ∆S(surroundings) = 0
Or
Prove that there is no net change in entropy in a reversible process.

Solution

Suppose heat is absorbed by the system reversible and the heat is lost by the surroundings also reversibly (process occurs under complete reversible condition).
If qrev is the heat absorbed by the system reversibly, then the heat lost by the surroundings will also be qrev. If the process takes place isothermally at T kelvin, then Entropy change of the system
straight S subscript system space equals space straight q subscript rev over straight T space space space space... left parenthesis 1 right parenthesis 
Entropy change of the surroundings
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Hence the total entropy change for the combined system and surroundings will be:
increment straight S subscript sytem space plus space increment straight S subscript surroundings space equals space straight q subscript rev over straight T space minus space straight q subscript rev over straight T space equals space 0 space space space space... left parenthesis 3 right parenthesis
Hence in a reversible process, the net entropy change for the combined system and the surroundings is zero i.e. there is no net change in entropy.