A thermally insulated cylinder is closed at both ends and is fitted with a frictionless, heat-conducting piston that divides the cylinder into two parts. Initially the piston is clamped in the center with V = 1 liter of air at T0 = 300◦ K and P0 = 105 Pa on one side, and V = 1 liter of air at T0 = 300◦ K and pressure 2P0 on the other. After the piston is released the system comes to equilibrium with the piston at a new position.

I have already solved the problem and obtained the final pressure, temperature and volumes, but how can I describe a process like this? Obviously the process is isothermal, the net change in energy is zero and entropy increases, but is this process quasi-static or not? Any help or explanations will be appreciated.


1 Answer 1


The process is not quasi-static. The gas in the high pressure chamber experiences an irreversible expansion, and the gas in the low pressure chamber experiences an irreversible compression. This all occurs spontaneously after you release the piston. The piston will at first overshoot the equilibrium position and will experience a damped oscillation, but will eventually come to rest, such that its kinetic energy is dissipated.

The process can be made closer to quasi static by using a more massive piston, but this will just require more time for the kinetic energy of the piston to be dissipated by viscous stresses. In the end, the net result of the irreversibility will be the same...i.e., the same final state and the same increase in entropy.


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