Considering the first step of the Carnot process, heat is transferred from a bath to the system with both at the same temperature. But how does this process start? Why should the system spontaneously absorb heat and expand as a result. And even if I pull on the piston an infinitesimal bit, why should the process continue and do not stay at the new equilibrium state?


It is as you say. On the isothermes the system is at the same temperature as the bath. Therefore no energy is exchanged at first. Now reduce the pressure on the system ("pull the piston"). The system reacts by increasing its volume to go back to mechanical equilibrium. It does work on you! It does work because a pressure difference implies a net force along which the piston moves.

If it were not for the bath providing energy in the form of heat, the temperature would decrease.

Since you are pulling very slowly, it will also stay in thermal equilibrium with the bath. So temperature keeps constant over the process. Should the internal energy $U$ depend on temperature only, as is the case for an ideal gas, we have $dU=0$.

But because energy leaves the system as work, it must be compensated for by a heat flow into the system, since $dU = \delta Q + \delta W$. That heat is absorbed from the bath, although they are at the same temperature.

The trick is to successively disturb the system infinitesimally out of mechanical equilibrium, giving it the chance to effectively stay in thermal equilibrium with the heat bath.

  • $\begingroup$ But if I reduce the pressure (or "pull the piston"), why does the system then work on me? What is my misconception here? $\endgroup$ – McLawrence Feb 2 '17 at 11:43
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    $\begingroup$ If you have a bunch of tiny masses sitting on top of the piston, and you slide one of the tiny masses off the piston at the existing elevation, the piston will now be slightly out of equilibrium, and the gas will do work on the rest of the masses to raise them a small amount. Then you remove another tine mass from the piston (at the new, slightly higher elevation), and allow the gas to do work on the rest of the masses again. Meanwhile, you are adding heat to the gas to keep its temperature constant. If you don't add the heat, its temperature will drop since it is doing work. $\endgroup$ – Chet Miller Feb 2 '17 at 13:21

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