10
$\begingroup$

I am reading up on entropy in a textbook and I got confused by this:

It says that to calculate the entropy for an irreversible process using heat flow, one must imagine the reversible process in which the initial and final states are the same as for the irreversible process. Why is this so and why do we need to think of it in the reversible way?

$\endgroup$
  • $\begingroup$ Short answer: you don't. Just calculate the entropy of the initial and final states and take the difference. If there's a heat bath you have to include the change in entropy of the heat bath as well, but that's fine. I've never understood why some textbooks teach this weird and complicated reversible vs. irreversible path stuff, it's just unnecessary. $\endgroup$ – Nathaniel Oct 1 '12 at 21:03
9
$\begingroup$

There are two parts to this answer. The first is that the entropy change is defined only for reversible processes. The second is that the entropy is a state function. Many, but not all, the functions in thermodynamics (including the internal energy) are state functions.

A state function is one that is independent of path. That is, if one goes from state $A$ to state $B$, the change in internal energy is independent of how the change takes place. The same is true for the entropy.

So the entropy change in going from state $A$ to state $B$ is independent of how the change takes place. But we can calculate the change ONLY for reversible changes.

So we mentally find a reversible path from state $A$ to state $B$ and calculate the entropy change. Since entropy is a state function, that must also be the change for a non-reversible path.

$\endgroup$
0
$\begingroup$

For a closed system, if you calculate the integral of $dq/T_I$ over all possible process paths between two thermodynamic equilibrium states of a system (where $T_I$ is the temperature at the portion of the system interface (boundary) with its surroundings at which the heat transfer takes place), the integral will differ from path to path. The maximum value of the integral over all possible process paths will be found to occur when the path is reversible. Over any reversible path, the system temperature T will be essentially uniform throughout, and the temperature at the interface $T_I$ will be essentially equal to the system temperature T. All reversible paths will give the same maximum value for the integral. This maximum value is the change in entropy of the system between the two thermodynamic equilibrium states. It is a function of state because the maximum value of the integral is determined only by the initial and final equilibrium states.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.