5251 Views
5997 Views
5386 Views
6302 Views
4600 Views
5926 Views
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.
4916 Views
8179 Views
4830 Views
3750 Views
11068 Views
Apply to join the passionate instructors who share their expertise and knowledge with the world. You'll collaborate with some of the industry's best producers, directors, and editors so that your content is presented in the best possible light..