By Sunil Bhardwaj

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Suppose in a reversible expansion of an ideal gas, q J of heat energy is absorbed by the system from the surrounding at temperature T. Then the entropy change is given by, $${ \left( \Delta S \right) }_{ system } = \frac { q }{ T } \qquad { \left( \Delta S \right) }_{ surrounding } = \frac { -q }{ T }$$ Therefore total entropy change is $${ \Delta S }_{ 1 } = { \left( \Delta S \right) }_{ system } + { \left( \Delta S \right) }_{ surrounding }$$ $$= \frac { q }{ T } + \frac { -q }{ T } = 0\qquad ....(1)$$ Similarly for the reverse process when heat is absorbed by surrounding from system, $${ \left( \Delta S \right) }_{ system } = \frac { -q }{ T } \qquad { \left( \Delta S \right) }_{ surrounding } = \frac { q }{ T }$$ Therefore total entropy change is $${ \Delta S }_{ 2 } = { \left( \Delta S \right) }_{ system } + { \left( \Delta S \right) }_{ surrounding }$$ $$= \frac { -q }{ T } + \frac { q }{ T } = 0\qquad ....(2)$$ Or total entropy change is $${ \Delta S } = { \Delta S }_{ 1 } + { \Delta S }_{ 2 } = 0 + 0 = 0\qquad ....(3)$$ Therefore all reversible process takes place with constant entropy.

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