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PV work: differential vs instantaneous

  1. Jan 29, 2014 #1


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    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.
  2. jcsd
  3. Jan 29, 2014 #2
    If the latch is removed and the gas expands irreversibly, the pressure is not even uniform spatially within the system, so what pressure do you use to calculate the work done on the surroundings? What you use is the pressure at the interface with the surroundings, pI. If the force per unit area at the interface is held constant in the irreversible expansion (say by expansion against a large volume of air), then the work is pIΔV.

    If the process is carried out reversibly, then the pressure within the system is essentially perfectly uniform, and matches the pressure of the surroundings at the interface. This is the case that is approximated when the weights are removed incrementally.

    For more details, see my Blog at my PF personal page.

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