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Work and expansion of a gas

  1. Aug 13, 2011 #1
    There's something I'm not understanding about work done on a system with constant velocity. The total net work is zero right? Say two friends are pushing a block on opposite sides. One friend manages to dominate the other so they go in one direction. The floor is frictionless. The total work done on the block has to be zero right? The two boys are pushing in opposite directions with equal magnitudes of force. What about raising a book with constant velocity against gravity. The total work is zero again but the book is gaining potential energy. Does the energy come from the kinetic energy that the book has while it's being raised up?

    This leads me to adiabatic gas expansion. An example I see quite often is the case of a cylinder filled with gas with a piston loaded with weight. The system is in equilibrium. However when calculations are done to determine the work done by the gas on the surroundings during an expansion...they use the force exerted by the surroundings on the cylinder system. For me that calculation only makes sense for a reversible reaction. The reason I bring it up is because when all of the weights are taken off at once (irreversible expansion). The work that it can do is at a minimum and it is calculated by also using the external pressure of the surroundings. This doesn't make sense to me. It's not a reversible expansion so shouldn't the pressure in the work energy formula be the internal pressure of the gas? The gas itself is doing work on the surroundings.

    Thank you.
     
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  3. Aug 14, 2011 #2

    Mapes

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    Hi Marwyn, welcome to PF.

    No way! Not if there's a resisting force. If you lift a book at constant speed, you're absolutely doing work on the system, which gains potential energy.
     
  4. Aug 14, 2011 #3

    atyy

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    In the case of gravity, the total work done by gravity and you is zero. Gravity is a conservative force, so if it were the only force acting on the book, energy would be conserved: if the book had an initial upward velocity, it would lose kinetic energy as it flies up, gaining potential energy equal to the loss in kinetic energy. However, when you raise the book at constant velocity, the book does not lose kinetic energy. The book gained energy because you did work against gravity.

    If a process is quasi-static, then the internal pressure is defined at all times, and you can use that to calculate work done. If it is not quasi-static, then the internal pressure isn't well-defined, and you have to calculate work done by some other means.
     
    Last edited: Aug 14, 2011
  5. Aug 14, 2011 #4
    oooh ok. So my only confusion has been that I assumed energy would be conserved as I raised the book? I assumed that the potential energy gained by the work done to raise the book should be offset by the negative work done by gravity.

    That's what I was confused about with irreversible expansions. They simply used the external pressure to calculate the work done by the system on the surroundings as it expands.
     
  6. Aug 14, 2011 #5
    Ya, but that resisting force is balanced by you pushing the book up isn't it?
     
  7. Aug 16, 2011 #6

    Andrew Mason

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    What if the external pressure was zero? Would the expanding gas do any work at all? Would any of the internal energy of the gas be transferred to the surroundings?

    AM
     
  8. Aug 16, 2011 #7

    cjl

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    The problem with the external pressure being zero is that it requires one of two things to be true. Either it requires the internal pressure to also be zero (which means you either have no gas, or it is at absolute zero), which is obviously not very useful, or it requires the problem to be a transient condition (basically free expansion of a gas). Since most of basic thermodynamics is founded on the assumption that processes are effectively reversible and quasi-steady state, this situation would basically force you to throw out the standard equations and use a far more complicated analysis.

    (My thermo is a bit rusty though, so I'm not promising that this is 100% correct)
     
  9. Aug 18, 2011 #8

    Andrew Mason

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    So, does an ideal gas do any work in an adiabatic free expansion? If not, what does the first law tell you about the change in internal energy and, hence, temperature?

    I am not sure what you mean. It is simply a matter of applying the first law. The first law is not based on any assumption that the process in moving from one state to another is effectively reversible or quasi-static.

    AM
     
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