Entropy and reversibility I'm very confused right now, I need to make sure of the following if they're correct or not:
A reversible path is a path in which if if you return back to your initial state, there would be no change in the system and the surroundings
Let's say I now increase the temperature of something by adding heat and then remove heat again to bring it back to its initial temperature, along the adding heat path, I should have increased the entropy of the system, if I go back to the initial state, I should have decreased the entropy again to its initial state and so the total change of the entropy of the system is zero.
Now from the above, atleast from my understanding which I'm not 100 % convinced about, I just concluded that the change in entropy of the system is zero for a reversible process, even though the process was not adiabatic, but the equation for entropy states that $\Delta S = \int_i^f{\frac{dQ}{T}}$ for a reversible path, so its not always equal to zero, it needs to be adiabatic, what is wrong in my statements? I hope you clear everything up for me please, thank you.
 A: Reversible just like you said doesn't mean that there is no exchange, it just mean that if you go back there would be no change. In the formula you are using everything depends on your final state. If the final state is the same as the initial state (f = i) then you would find that $\Delta S = 0$. But if you do not go back then it is expected that the entropy has changed. 
If you separate the path in two part the first part will have $\delta Q > 0$ for instance and when you go back you will have $\delta Q < 0$ so the integral will be null.
A: An example of a reversible path that is not adiabatic is a reversible isothermal process. During a reversible isothermal expansion of a gas, for example, heat is added. Since an isothermal process is constant temperature, the temperature comes out of the integral and the change in entropy is $\frac {+Q}{T}$ for the system and $\frac {-Q}{T}$ for the surroundings where Q is the heat transferred from the surroundings to the system undergoing expansion. The key for reversibility is the temperature difference between the system and the surroundings must be infinitesimally small and the expansion carried out very slowly (Quasi-statically). 
Now reverse the process and compress the gas to its original state. The entropy change of the system is $\frac{-Q}{T}$ and the surroundings is $\frac{+Q}{T}$. The key again is an infinitesimally small difference between the system and surroundings and a slow (Quasi-static) process. 
The total entropy change for the combination of the expansion and compression is zero for both the system and surroundings. 
Hope this helps.
