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First law of thermodynamics of gas

  1. Jan 27, 2014 #1
    The first law can be stated as ΔU = Q + W and we adopt a sign convention as follows, if the system under study does work then W is negative and if work is done on the system W is positive.
    Now consider the example of a gas expanding against a constant external pressure p. W would be negative as the gas does work, and from a classical mechanics approach this can be explained by imagining the external force from pressure moving through a displacement that is in an opposite direction to the force, so the work done ON the system is negative. But picture this; what if while the gas expands, i pull the piston so my force is in the same direction as the force by the gas, does that mean work done on the system is positive although the gas is clearly doing work? Would that violate the first law of thermodynamics?
     
  2. jcsd
  3. Jan 27, 2014 #2
    What you do outside of the system is your own work and not that of the system. If the system is expanding it is doing work, and if it is being compressed than work is being done on it.
    Your scenario is the same as a piston with weights on it. With more weight, with the system expanding, it is doing more more than if there is less weight on the piston. By your pulling on the piston, that would be the same as less weight on the piston.
     
  4. Jan 27, 2014 #3

    jbriggs444

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    You would have to arrange for the gas in the cylinder to have a negative pressure. Otherwise both you and the gas would be doing positive work on the piston. If the piston is massless (as is implied), that would violate conservation of momentum along with the first law of thermodynamics.

    If the piston has mass, it will merely be accelerated, thereby balancing the books for both energy and momentum.
     
  5. Jan 27, 2014 #4
    jbriggs444, how would this violate the first law of thermodynamics and the conservation of momentum?
     
  6. Jan 27, 2014 #5
    From the force balance on a frictionless, massless piston, the tension you are pulling with is equal to the atmospheric pressure on your side of the piston times the area of the piston, minus the gas pressure within the cylinder on the other side of the piston times the area of the piston. The work you are doing on the piston is TAd, where T is the tension, and A is the piston area, where d is the displacement of the piston. The work being done on the gas on the other side of the piston is (T-pa)Ad, where pa is atmospheric pressure. This is minus he work done by the system on its surroundings, which, in this case, occurs at the inside face of the piston. So part of the work you are doing to pull the piston out is to push back the atmosphere on your side of the piston. But this is not the work you are doing on the gas on the other side of the piston or the work that the gas is doing on its surroundings.

    Chet
     
  7. Jan 28, 2014 #6
    Chet,
    Good post
    Much like I said but way more explanatory.
    Cheers
     
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