Why Is Reversibility Essential for Using dS=δQ/T?

[Proust]

Homework Statement

In my book it says, "only if the transformation is reversible can we use dS=\frac{\delta Q}{T}", but I don't know why? I think even if it's not reversible we also use \delta Q=TdS to solve the thermo problems, don't we?

I'm looking forward to your ideas, thanks.

 

[rude man]

Entropy is a state variable. That means that, no matter how we get from state A to state B, the change in S is the same. Doesn't matter if the change is reversible or not.

The point is that in order to CALCULATE the change in S, you must integrate along a reversible path.

In fact, the integral TdS from state A to B is always less for an irreversible change than for a reversible one. E. Fermi proves this in his Dover book "Thermodynamics' but I never followed the proof. Since the states are undefined for an irreversible change I don't lose too much sleep over it. But Fermi actually invokes that fact later in the book on a different topic.

If you want to get totally snowed, that is the book for you!