why is the entropy for irreversible adiabatic process not zero? Provide a mathematical proof.
Answers
There are two ways to understand this question.one is theoretical another is mathematical . Firstly we have to understand about irreversible process .Irreversible processes are one directional processes ,because of some external losses .Suppose any piston and container assembly , now if any gas is expands adiabatically in the container it will do some work on the piston to move it in the upward direction and some amount of its energy is taken by the frictional losses .Now if we consider initial state having most arranged molecular arrangement and we want to go back to it’s initial state we have to do some work on it .Because this is a one directional process even we are doing some work on it to restore its initial position we cant get same molecular arrangement as before .so that means randomness or say number of micro-states of a system increases. so entropy will be more than before. If we look mathematically ds = (dQ/T) + ds,generation now if process is adiabatic than dQ=0 that means change in entropy is only depends on entropy generation.And for irreversible process entropy generation is positive due to external losses. So, ds > 0