Absence of dissipative forces in a reversible process 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?
 A: 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.
