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Simple doubt in work done by gas

  1. Feb 20, 2014 #1
    In First law of thermodynamics , work done by the gas is given by pΔV .

    I have few doubts regarding this-

    1)Is 'p' ,pressure of the gas or external pressure of the surroundings ?If gas is expanding then surely gas pressure should be more than that of the surroundings and vica versa .

    2)There are three things ,gas,container fitted with piston and surroundings .Now ,gas is exerting pressure on the piston i.e doing work on the piston .Piston is doing work on the surroundings. Then why is that considered as gas doing work on the surroundings ?

    3)Is the piston considered in the arrangement always massless or it may have some mass ? How does that affect our results ?

    I would be grateful if somebody could give his/her inputs .
    Last edited: Feb 21, 2014
  2. jcsd
  3. Feb 21, 2014 #2

    Simon Bridge

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    FLT - Fork lift trucks? Fermi Liquid Theory? First Law of Thermodynamics ... ah, probably that one :)
    I'll try addressing your concerns in a different order.

    ##W=P\Delta V## if the pressure is kept constant and the volume changes.

    The piston will always have mass - it makes no difference.
    In the model, the pressure is commonly supplied by the mass of the piston.

    In lifting the mass of the piston, the gas has done work against gravity.

    The pressure of the gas is treated at all times as the same as the pressure exerted on it - since there is no unbalanced force, this means that the piston is lifted at a constant velocity. It's an idealization - naturally there must have been an unbalanced force at some stage to start the piston moving. Think of the process as happening very slowwwly.
  4. Feb 21, 2014 #3


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    For #1

    If you had a container with a piston on the bottom ( ie no gas inside ) and you pulled on the piston, how much work do you have to do to move the piston a distance x. Of course you know that is just pressure of the atmosphere times the area of the piston times the distance the piston moved, W = Fx.

    Now, have a container with some air inside and pull the piston a distance x. Of course the work done on the surroundings is still the same as before, but since there is gas pushing on the other side of the piston to help you out this time, you have to do less work to pull the piston a distance x. You are interested in how much work the gas inside the container has done to help you out, so the PΔV of the gas inside the container is what is what you want to calculate.

    Extending that to a machine, it is the work of the gas inside the container that will give you usable work to extract from the machine.
  5. Feb 21, 2014 #4
    If the expansion is taking place irreversibly, then the pressure within the system is not uniform with respect to spatial position. So what pressure do you use to get the work done on the surroundings? You always use the pressure at the interface with the surroundings, p=pI. This is equal to the pressure of both the system and the surroundings at the interface, since pressure is continuous at the interface.

    If the process is reversible, the pressure of the system is essentially uniform throughout, such that p=psystem=pI, where psystem is the uniform pressure throughout the system.
    In the first law, we divide things into only two entities, system and surroundings. If there is a piston present, you have to decide whether you want the piston included with the system or with the surroundings. If you include the piston with the system, then the gas and piston are doing work on the surroundings. If you include the piston with the surroundings, then the gas (system) is doing work on the piston (which is lumped into the surroundings). If the system consists of the gas and the piston, then you need to include the potential energy change of the piston in determining the total energy change within the system in transitioning from the initial to the final equilibrium state. If the piston is included with the surroundings, then the pressure at the interface with the surroundings pI must be taken as the force per unit area at the inside piston face (associated with pushing the piston and whatever is behind the piston) in calculating the work.
    The piston can be considered to have mass, or to be massless. If it has negligible mass, then, by Newton's second law, the pressure on the inside face of the piston is always equal to the pressure on the outside face of the piston, and no work is done on the piston. It doesn't matter whether the piston is lumped with the system or with the surroundings. This situation is easier to conceptualize.

    If the piston has mass, this isn't much more difficult to work with. You just need to do what I described in the previous answer.

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

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