Thermodynamics: Reversible adiabatic process is always isentropic but viceversa is not true why?

This question has produced a lot of interest. If we really want to understand the question we need to first understand what entropy is and what heat is. Entropy is a concept that is well recognized for its difficulty. My own belief is that neither concept can be well understood unless we study thermal energy at the molecular level. Imagine an ideal gas expanding in a perfectly insulated piston-cylinder device. The molecules have kinetic energy and as the expansion takes place they impart this by doing work on the piston. No heat is transferred to/from the molecules from a source/sink. This is an adiabatic process. The temperature of the gas is directly related to the average kinetic energy of the molecules by the Boltzmann constant and is only defined when the gas is at equilibrium. However we have a process which is not an equilibrium situation so we have to idealize the process by imagining it passing through a series of equilibrium points - we call the process quasi-static. In this way it can the be regarded as reversible and therefore isentropic, i.e. the gas has constant entropy. If friction occurs during the expansion process, as in the case of a turbine expansion, where high gas velocity is present, the process deviates from the ideal isentropic process and is not reversible and therefor not isentropic, even though it may be adiabatic. *quoted from https://www.researchgate.net/profile/William_Dartnall/  msg

Yes.This is where the understanding of entropy generation helps. dS(gen)>=0 That is in a reversible process the entropy generated is zero. That is, dS=integral of dQ/T (only for reversible process) Then for an irreversible process the equation is written as dS>0 and integral of dQ/T<0 Then, dS=dQ/T + dS (gen) for an irreversible process. Now coming to your question, For an reversible adiabatic process the system and surroundings should be capable of reverting back to their original states. Consider a hot fluid expanding on a turbine, the entropy of system decreases due to heat loss to surroundings, which could be compensated by an increase entropy due to irreversibilities such as friction. Thus an increase of entropy due friction compensates the decrease causing entropy to reduce to zero. Hope I have answered your question

If we assume that all process are quasi-static - meaning that the process is done through a dense succession of equilibrium states-, then is true. With this hypothesis we can identify the heat transfer with TdSTdSTdS, δQ=TdSδQ=TdS\delta Q=TdS (keep in mind that T>0T>0T>0 always) then is clear that: if δQ=0⇒TdS=0⇒dS=0⇒S=constantδQ=0⇒TdS=0⇒dS=0⇒S=constant\delta Q=0\Rightarrow TdS=0\Rightarrow dS=0\Rightarrow S=constant. An adiabatic procces implies an isontropic one. if  S=constant⇒dS=0⇒TdS=0⇒δQ=0 S=constant⇒dS=0⇒TdS=0⇒δQ=0\ S=constant\Rightarrow dS=0\Rightarrow TdS=0\Rightarrow \delta Q=0. An isontropic procces implies an adiabatic one.

well said thank you but my doubt is here, is it cylinder refers to a insulated surface?if so then heat is transfering from cylinder to fins also...thenm how it will be adiabatic ?