1]Suppose a system of ideal gas is enclosed adiabatically and with a piston roof. if the atmospheric pressure is suddenly reduced, it will expand, cool down and do some work(i.e transfer energy) to the surrounding. then how would that work be stored,i.e. in what form will the transferred energy be if the piston is weightless and frictionless and only force acting on it was atm pressure and gas pressure.
2]in the last phase of carnot cycle, we contract till it reaches the temperature of its heat reservoir, then how can we assume that the temp will be reached where the corresponding pressure and volume will be same as our initial ones,i.e. we will return to exact same state as before? and is the fact that perpetual motion or in this case continuos heat transfer from cold to hot body not possible, a consequence of second law of thermodynamics or is it vice versa??
> if the piston is weightless and frictionless
Please understand how work is defined -- it is the DISTANCE an object is moved against a FORCE. No movement or no force, no work is done.*
A weightless piston has no force on it. There is no movement AGAINST a force. Thus, there is no work done in this situation. It is no different from placing a balloon inside a vacuum chamber and then reducing the pressure in the chamber. The balloon will expand, but the air inside the balloon will do no work.
Recall that a Carnot Cycle Engine is an IDEALIZATION -- like a incompressible fluid with no viscosity (which has been called "completely dry water"). Carnot was able to show that there exists an upper limit on the efficiency of engines, and that the efficiency of an engine can be increased by using a higher temperature during the expansion sequence.
These two hyper-link lectures will explain things far better than I can. The first shows how various concepts relate to each other
allowing to study concepts in a proper order. The second
looks specifically at a Carnot Cycle.
* More precisely, it is the dot product of the distance vector and the force vector. If the movement is completely at right angles to the force, there is also no work done.