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Irreversible Compression and Expansion

  1. Oct 10, 2012 #1
    Say I have a gas enclosed in a bottle with a weight on top exerting pressure P0. Let the gas pressure be P. If P0 is not equal to P the compression/expansion will be irreversible.

    If P0>P and the gas is being compressed:
    Work done on the gas is: -PdV
    Work done on surroundings by the gas is: P0dV

    If P0<P and the gas is expanding:
    Work done on the gas is: -PdV
    Work done on surroundings by the gas is: P0dV

    Is this correct? My textbook says that work done on the gas during compression should instead be -P0dV but it seems to me that the internal gas pressure is the one in contact with the gas.

    My second question is about how the first law of thermo works. I know that engineering textbooks often define their work as work done by the gas while my chemistry textbook defines work as work done on the gas. It would seem though from the list I made above that the work done on the gas is not equal to the work done by the gas during irreversible compression/expansion and so one would get different results depending on which version of the first law was used.

    Thanks!
     
  2. jcsd
  3. Oct 10, 2012 #2
    In an irreversible compression or expansion, the pressure within the gas is not uniform, and varies with spatial position. At the boundary with the surroundings, the gas pressure must match the pressure of the surroundings. However, away from the boundary, the gas pressure can be higher or lower, depending on whether expansion or compression is occurring. But the work done at the boundary by the gas on the surroundings is determined by the pressure at the boundary, which is equal to the pressure of the surroundings.
     
  4. Oct 11, 2012 #3
    Simply put,
    engineers define work as being done by the gas because what you ultimately pay for in dollars and cents is the cost of energy, and the less energy input for maximum work output costs less.

    Chemists are interested in whether a reaction is exothermic or endothermic, so they are interested from an energy balance point of view.
     
  5. Oct 11, 2012 #4
    From what I said above, it follows that the work done by the gas on the surroundings is exactly equal to minus the work done by the surroundings on the gas. The pressure is continuous at the interface between the gas and the surroundings, even though, in an irreversible process, the pressure varies with position within the gas.
     
  6. Oct 11, 2012 #5
    Thanks for the replies.

    So would the weight be considered part of the surroundings or part of the gas? Based on what Chestermiller has said I think this is the point on which I am confused. Because the pressure on the top of the weight is different from that on the bottom during an irreversible compression/expansion this leads to my getting different expressions for the work done on the gas vs work done by the gas.

    Also can you verify the expressions I wrote for the 4 cases?
     
    Last edited: Oct 11, 2012
  7. Oct 11, 2012 #6
    From the problem statement, the weight is supposed to supplying all the pressure of the surroundings. Any pressure on the top of the weight is supposed to be negligible, or, effectively zero (vacuum).

    This problem description is not as well posed as it could be. If the weight is supposed to be supplying the pressure P0 during the entire expansion or contraction, then this goal will not be accomplished. As soon as the weight starts to accelerate, the pressure at its base will differ from what it was when the weight was static. To accomplish what the problem description desires, one could flush mount a rapid response pressure transducer on the base of the weight, and manually impose whatever motion (of the weight) is necessary to maintain the pressure at the base constant.
     
    Last edited: Oct 11, 2012
  8. Oct 11, 2012 #7
    There is only ONE first law of thermodynamics, and that holds for both engineering or chemistry. The only difference is whether work put into the system or extracted is positive or negative.

    With that in mind, there are not 4 cases to calculate the work, but only one.

    If you consider yor system to be a cylinder with a piston, the force F on the piston is what keeps the gas inside at the pressure P. The force could be a mechanical linkage to a shaft, a set of weights, or an external pressure from the surounding gas. In either case, W=Fdx where dx would be the distance the piston has moved. F could be a variable force as a function of x ( everyone knows how much harder it is to push a bicycle pump the farther the gas is compressed ) or possibly a constant.

    so you can calculate the work performed by the system for a compression or expansion by using the external force if you know the equation on how F is related to x.

    You can also calculate the work from the change of state of the gas within the piston cylinder volume. Here, W= PdV. P is can also be constant or a variable function of V and it is up to you to determine that function whether the process is isothermal, adiabatic, etc.

    work is a path function meaning that it not dependant upon the initial state and the final state but on how it got from initial to final ie isothermal, adiabatic etc.

    You do realize that the piston is only a rigid barrier between the internal system and the external surroundings and being massless the internal pressure P = the external pressure Po, or external force F/A ( A= area of the piston ).
     
  9. Oct 12, 2012 #8
    You cannot always calculate the work this way, since, under irreversible conditions, the pressure P will be a function of position within the cylinder, so what pressure do you use? The correct pressure to use is the gas pressure in contact with the piston, if you can calculate this from gas dynamics, or measure it.
     
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