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Fundamental doubt in Thermodynamics

  1. Mar 15, 2009 #1
    in Thermodynamics, in a book i have read that during expansion of a gas(system), W(on system by surroundings)=-ve..They have proved it using W=-force(external)*displacement. They are saying as displacement is positive and force is always considered as positive, W=-ve. But, how did they determine displacement of a gas as positive, even before introducing the concept of volume in the equation. I mean they should introduce the volume in the equation first, then say about the sign of W. I mean how can they say displacement of a gas. is there any such concept as displacement of a gas, consequently centre of mass of a gas, etc?? Plz help
     
    Last edited: Mar 15, 2009
  2. jcsd
  3. Mar 15, 2009 #2

    Doc Al

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    Which is it? You start by saying the work is positive, then end by saying it's negative. :wink:

    The work done on the gas will be negative, since the force on the gas is opposite to the displacement of that force.
    Think of how the force is moving. Is it doing positive or negative work? (Somehow you have no problem with the concept of "expansion" without worrying about the details of volume.)
     
  4. Mar 15, 2009 #3
    but even after thinking how the force moves, direction and all, it wont help me understand the significance of displacement in the equation...displacement of a gas is a concept hard to understand..
     
  5. Mar 16, 2009 #4

    Doc Al

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    Don't think in terms of some vague displacement "of a gas", whatever that might mean. Imagine a cylinder of gas expanding against a piston. The gas exerts a force on the piston (and vice versa). It's the displacement of the piston--the interface between "the system" and "the environment"--that we are concerned with.
     
  6. Mar 16, 2009 #5
    Even if there weren't a piston the expanding gas would do work by pushing against/lifting the atmosphere.
     
  7. Mar 16, 2009 #6
    Really? I'll agree that it transfers energy through the occasional interaction (and therefore does work on the atmosphere, but I'm not sure it "lifts or pushes against" the atmosphere. There's no boundary between the expanding gas and the atmosphere, so the particles will diffuse. As such a displacement doesn't make much sense in this case. And I don't think that it can really "push against" the atmosphere, it's more that it diffuses through.

    Perhaps you could talk about the average loss of energy by a particle as it travels and interacts with the atmosphere, but I'm not sure that this would be a simple dW = -F(x).dx, because the forces come from interactions between individual molecules of gas (Van der Waals, etc) and so you'd need density of the target atmosphere, a cross section of interaction, etc.
     
  8. Mar 16, 2009 #7
    But the gas will expand against atmospheric pressure and work will be done.it will be an approximately isobaric change.
     
  9. Mar 16, 2009 #8
    However, the work done in an isobaric process is related to the change in volume. How do you define the volume of such a gas? It will have no clear cut boundaries and will be mixed with the atmospheric gas. I'm not sure isobaric applies here.
     
  10. Mar 16, 2009 #9
    The pressure remains practically constant at atmospheric pressure and the work done equals this pressure times the volume change.Of course with thermodynamics we can only get approximate answers.Even Van Der Waals equation is an approximation.
     
  11. Mar 16, 2009 #10

    Yes, ths situation is similar to free expansion in vacuum. Thermodynamics does not apply during the irreversible change. But you can apply thermodynamics to the initial and the final state. Suppose the initial state is a big container of gas kept at constant volume V at some temperature T and pressure P. Inside this container, there is small gas container at volume V' and at temperature T' and pressure P'. Then the small container is opened, the gas in there leaks out.

    If the big container is thermally insulated, then all the work done by the gas in the small container stays in the system. So, no net work is done by the whole system and the internal energy of the final state is thus the same as the internal energy of the initial state. You can then compute the change in temperature from the known thermodynamic properties of the systems.

    If the big container was instead kept at constant pressure, then the total internal energy would change due to work done by the whole system. This is close to what happens if you let gas escape from a gas cylnder in your house. The air in your house is approximately in pressure equilibrium with the outside atmosphere (even if all windows are closed).

    So, if you let the gas from the (high pressure) cylinder escape, the total internal energy in the final state will be less than in the intitial state by an amount of P delta V, where P is the atmospheric pressure and Delta V is the amount by which the total volume of the gas in your house has expanded in the final state relative to the initial state. This means that some of the gas will have leaked away into the atmosphere. Now, you can deal with that to a good approximation by focussing on the part of the old volume of V - Delta V that will now occupy the whole house. This expands under atmospheric pressure.
     
  12. Mar 16, 2009 #11
    Also note that diffusion is a slow process. This is due to the fact that the mean free path of molecules is small under normal conditions. This small mean free path allows you to pretend as if there is a hard bundary exerting pressure on the gas. So, in practice, it is possible to look at a time scale in which the gas expands by pushing away the surrounding gas and thus performing work. You can then look at the much slower process of the expanded gas diffusing into the surrounding gas.
     
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