Answer to Question #93900 in Molecular Physics | Thermodynamics for Amit Rajaram Shinde

Question

we know that entropy is a point function i.e. it is independent on path followed by the system. Let system changes its state from 1 to 2 by a reversible path, then system returns to its original state by following process 2 to 1. If process 2 to 1 is a reversible process then we can write
[1-2]∫(δQ/T)+[2-1]∫(δQ/T)=0
and if process 2 to 1 is irreversible then we can write
[1-2]∫(δQ/T)+[2-1]∫(δQ/T)<0
entropy is a point function it is not dependent on path followed, but when we changed process 2 to 1 from reversible to irreversible then equation changed, how it is possible?
Note: "[1-2]∫" indicates integration from limit 1 to 2.

Answer 1

Revercible process is the process without energy loss. So released by the gas heat during the compression process 1-2 (it's for example) will be equal to absorbed by the gas heat during expansion process 2-1. Entropy change in the process 2-1 will be equal to the entropy change in the process 1-2, but with the opposite sign. So in the summ we receive zero.

In the case of irrevercible process, due to the energy loss in the process 1-2, the amount of heat, that we need to give to the gas to return it to its original state, should be bigger, then the heat, released by the gas. So in this case entropy change will be not equal to zero, because entropy change in the process 2-1 will be bigger, than the entropy change in the process 1-2.