Adiabatic Expansion: Final Temperature?

pdw
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Just a question that will hopefully improve my lack of understanding.

A thought experiment.
I have two cylinders both with frictionless pistons constrained to move in the z direction (against gravity) the same distance in both. So, the intial volume and final volume will be the same for both cylinders. One piston is heavier than the other, it will require more work to move to the limit of travel. Each piston is held at the lower limit of travel until it is released and then can move to the upper limit. Each cylinder is filled with the same mass of gas at the same temperature. The pistons are released and both travel to the upper limit.
The process is adiabatic.

What will be the final temperature of the gas in both cases?

I would think different - the gas that does more work will be cooler. But if both start from the same point (P,V,T) both are constrained to follow an adiabat, so then both will end up at the same temperature (due to having the same final volume). Have I missed the meaning of an adiabat? Have I shown, in my thinking, a complete ineptness in understanding Physics?

Regards
pdw
 
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It seems that those two cylinders are 'not' at the same pressure since one supports heavier piston than the other.
 
But the gas is not supporting the piston until the piston is released. The piston froms a rigid wall of the cylinder until release. Have I missed something obvious?
 
pdw said:
I would think different - the gas that does more work will be cooler. But if both start from the same point (P,V,T) both are constrained to follow an adiabat, so then both will end up at the same temperature (due to having the same final volume).

Yes, the gas that does more work will be cooler. The contradiction appears because your process is non-quasistatic: the pistons accelerate, then strike the second stop. You can't say what path the system is taking. In contrast, the adiabatic path you're comparing it to is quasistatic.

The standard procedure is to mimic a non-quasistatic process through multiple quasistatic alternative processes. You can take the cylinders to their final state this way by assuming that a reversible machine resists expansion in both cylinders and keeps the system quasistatic while the gas is expanding. Then, at the end of the expansion the energy obtained by the machine is dumped back in. Since the machine played less of a part in resisting expansion in the cylinder with a heavier piston (and therefore obtained less energy from work), that cylinder ends up cooler. Does this make sense?
 
Excellent, thank you.
 
To solve this, I first used the units to work out that a= m* a/m, i.e. t=z/λ. This would allow you to determine the time duration within an interval section by section and then add this to the previous ones to obtain the age of the respective layer. However, this would require a constant thickness per year for each interval. However, since this is most likely not the case, my next consideration was that the age must be the integral of a 1/λ(z) function, which I cannot model.
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