# Why is the total change of entropy of a reversible process equal to zero

Why is it that, in order to attain reversibility, the temperature difference between the system and the surroundings must be infinitesimally small, I understand that it has to be quasi-static, but why can't there be a difference in temperature?

When I say the total entropy I mean the combined entropy of the system + surroundings, and I mean at every point on the reversible path, not only the initial state.

Just to be clear, my question is Why do reversible processes be like this? I'm not asking why if there is no temperature difference that there would be no change in the entropy of the universe, that is understood, but why the reversible process itself has to keep a difference in temperature between the system and surroundings that tends to zero.

• if there is a finite temperatute difference then you cannot control the speed of the heat transfer, making the process non quasistatic
– user65081
Commented Nov 24, 2018 at 20:34
• the entropy generated by transferring $\delta q$ over a temperature gradient is $dS_{irrev}=\delta q (\frac{1}{T'}-\frac{1}{T})$ and it is always positive. Commented Nov 24, 2018 at 20:35
• Would you like for me to describe for you an adiabatic process that is reversible for both the system and the surroundings? Commented Nov 24, 2018 at 22:43

Why is it that, in order to attain reversibility, the temperature difference between the system and the surroundings must be infinitesimally small, I understand that it has to be quasi-static, but why can't there be a difference in temperature?

Let there be two bodies, A and B, which are so massive that a transfer of heat between the two does not change their temperatures. For example, the ocean (body A) and atmospheric air (body B). Let the temperature of body A (the ocean) be $$T_H$$ (higher temperature) and body B (the atmosphere) be $$T_L$$ (lower temperature). Let there be a transfer of heat $$Q$$ occur between body A and B. Then the entropy changes are:

Body A

$$\Delta S=\frac{-Q}{T_H}$$

Body B

$$\Delta S=\frac{+Q}{T_L}$$

The total entropy change is

$$\Delta S_{Tot}=\frac{-Q}{T_H}+\frac{+Q}{T_L}$$

You can see that for all $$T_{H}>T_{L}$$, $$\Delta S_{Tot}>0$$

The only way for the total entropy change to be zero is if the temperature difference between the two bodies is zero. But then, of course, there would be no heat transfer. Bottom line, all real processes are irreversible. A reversible process is an ideal but not attainable, process.

Hope this helps.

• But this means that any reversible process is isothermal? [which should not be correct] Commented Nov 24, 2018 at 20:42
• @khaled014z No. an adiabatic process can also be reversible. In the case of a reversible adiabatic process, the difference in pressure between the system and surroundings must approach zero so that the process is carried out quasi-statically. Commented Nov 24, 2018 at 20:59
• In general we could say, in order to have a quasistatic process, the internal energy of both the system and surroundings must approach each other? [I got the idea regardless, thank you] Commented Nov 24, 2018 at 21:04