I can't understand why dissipative forces must be absent during a reversible transformation. Aren't they a way of exchanging heat with ambient? Since the system is allowed to exchange heat with the ambient during these processes, why can't we consider a reversible transformation with dissipative forces?
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$\begingroup$ Moving a mass backwards on a rough surface does not extract heat from its surroundings. It always creates heat, in a reversed process or not. Here friction is the dissipative force. $\endgroup$– WarreGCommented Jun 12, 2019 at 8:31
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$\begingroup$ But isn't heat a way of transfering energy and not something that is created/owned? So shouldn't I consider that heat as a transfer of energy from the kinetic energy of the mass to the internal Energy of the ambient? So isn't this a legit process in a reversible transformation? $\endgroup$– StefanCommented Jun 12, 2019 at 9:27
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$\begingroup$ You could indeed see it as a transfer of energy, but the transfer will always be in one direction. Never from the ambient to the mass. $\endgroup$– WarreGCommented Jun 12, 2019 at 9:32
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$\begingroup$ Ok, I got it now. Just one more thing, why can I say the transfer will be in only one direction? For the definition of dissipative force? $\endgroup$– StefanCommented Jun 12, 2019 at 9:42
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$\begingroup$ If it could go in two directions, the mass has to extract heat from the environment and convert it into work to move itself. This is in contradiction with the second law of thermodynamics (entropy must increase in a isolated system) and also is intuitively just impossible. Look here : physics.stackexchange.com/questions/226826/…. $\endgroup$– WarreGCommented Jun 12, 2019 at 9:50
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I can't understand why dissipative forces must be absent during a reversible transformation. Aren't they a way of exchanging heat with ambient?
Yes they are a way of exchanging heat between the system and ambient (surroundings), but when friction is present more heat must be transferred to the surroundings than without friction in order to bring the system back to its original state. When more heat is transferred to the surroundings, more entropy is transferred to the surroundings such that the total change in entropy will be greater than zero with friction than without friction. That makes the process irreversible.
Since the system is allowed to exchange heat with the ambient during these processes, why can't we consider a reversible transformation with dissipative forces?
To illustrate why we can't consider a process to be reversible with dissipative forces, consider first the quasi-static isothermal expansion and compression of an ideal gas not involving friction in a piston cylinder.
During the reversible expansion heat $Q$ is transferred from the surroundings to the system to perform work $W$ by the system on the surroundings such that $\Delta U=Q-W=0$. We now do a reversible isothermal compression to bring the system back to its original state. This requires the same heat $Q$ to be transferred from the system back to the surroundings. If the temperature difference between the system and surroundings is infinitely small, then the change in entropy of the system and the change in entropy of the surroundings will both equal zero. The process is considered reversible.
Now consider the same process but with friction between the piston and cylinder walls. During the expansion friction work has to be done by the system to overcome friction. The temperature of the piston and cylinder walls will increase thereby increasing the internal energy of the system because of the friction. The same thing will happen during the compression. This generates entropy in the system In order to bring the system back to its original state (no overall change in internal energy and entropy of the system) it is necessary to transfer additional energy out of the system in the form of heat, which we will call $Q_f$ (friction heat) to the surroundings, in addition to the heat $Q$.
The change in entropy of the surroundings with friction during the compression becomes
$$\frac{Q+Q_{f}}{T_{surr}}$$
The change in entropy of the surroundings during the isothermal expansion was
$$\frac{-Q}{T_{surr}}$$
That makes the total entropy change of the surroundings due to friction in the system being
$$\frac{Q+Q_{f}}{T_{surr}}+\frac{-Q}{T_{surr}}=\frac{Q_f}{T_{surr}}>0$$
Since the total entropy change of the surroundings with system friction during the expansion and compression is greater than zero, the process is irreversible.
Hope this helps.
