PV work: differential vs instantaneous

mercredi 29 janvier 2014

Hello,



I am having a conceptual difficulty understanding the following scenario



Suppose you has a tank with a piston and some gas. In the first situation, there is some weight on the piston holding it down, and when you remove a differential amount of weight from the piston, you end up getting different PV work than if say the piston was latched, and then you suddenly release the latch. I suppose its due to the reversible vs. irreversible process,



In the case of the differential weight removal, the integral becomes



W = nRT ∫dV/V



Why could this not be done for the irreversible process? Is the nRT/V term supposed to indicate that the volume slowly changes, so the pressure is changing slowly, but for the irreversible process the pressure change is near instant?



In the latch case, I don't know when calculating the work done by the gas, if the work should be the initial pressure times change in volume, or the final pressure times change in volume, and why.





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