Reversible Adiabatic Process/Isentropic Process. (PART-I) 

Thank you for your appreciation. I am happy as I was able to make it easier for understanding.

@@AllAboutMechanicalEngineering I am the one to thank you! By the way, are you aware of any references in the literature about irreversible adiabatic processes in a similar way than yours?

@@lucascesargomessquillante1192 I appreciate your comment, it motivates me to keep going. In reference to your question, I would like to suggest you to go through the topic Entropy Generation from the Book "Fundamentals of Thermodynamics" By Borgnakke and Sonntag. In regard to your question, it is important to understand what is entropy generation presence of which makes the process irreversible. And I hope this book will help you in clarifying the doubt.

@@AllAboutMechanicalEngineering Thank you very much. I will definitely take a look at it!!!

Basically there are two reasons for change in entropy 1. Heat interaction between system and surrounding 2. Entropy generation. We are talking about reversible Reversible Adiabatic process When we say Reversible means there is no entropy generation Because we all know reversible process can be brought back to its inital conditions And this is only possible when entropy generation is zero because entropy cannot be destroyed while bringing the system back to initial. Now regarding temperature decrease As per first law of Thermodynamics Amount of heat supplied = change in internal energy + work done by the system. i.e. Q = (U2-U1) +W In Adiabatic process Q =0 So substituting Q=0 we get W = U1-U2 Where W = work delivered by the system U1 = initial internal energy before the process is started U2 = final internal energy when the process is completed That means the work is obtained at the cost of decrease in internal energy of the system.. Here NOTE THAT Internal Energy is the function of Temperature So if internal energy decreases that means there is decrease in temperature. But entropy is unchanged because system has neither recieved heat from surrounding nor lost any heat to the surrounding As well as there is no entropy generation (since it is reversible)

You are correct but practically not there is heat loss in process

@@AllAboutMechanicalEngineering well explanation.. god bless u