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Adiabatic system, work done

  1. Dec 27, 2010 #1
    I'm not sure if this should be in the physics section or chemistry, but I think it's covered in both. I apologise if this has been asked numerous times before!
    I am interested in the following adiabatic system, I think it counts as an isolated system:


    Here is my attempt to explain what is happening:
    The internal energy of this system decreases when I remove one of the weights (gold) on the lid (red). I was wondering why this is so. The change in the internal energy is a result of either heat being trasnferred or introduced, or work being done by or to the system. In this case, heat is not a factor, so it must be solely due to work. I can see that work is being by the system. But I am interested in the 'factors' involved.
    The lid pushes down on the system due to gravity (green arrow). The gas pushes back with a net force that is due to the particles each exerting a force on the lid by bumping in to it. The lid does not go down or up as the two forces are equal. I assume that the kinteic energy of the system is constant because the energy a particle transfers to the lid is 'recouped' by the lids collision with the particle?! I ask this, becasue it ties in with the next bit.
    If I remove a gold weight, the force exerted by the lid is reduced, and the particles push against the lid. They transfer kinetic energy to the lid (and gold weight), this kinetic energy is then transformed in to gravitational potential energy as the lid moves up. This happens until the particles loose enough kinetic energy in moving the lid (by transferring their kinetic energy to the lid) AND because the volume is larger that the particles collide with the lid less often (reducing the force they exert), that the force exerted by the lids new weight due to gravity is equal to the new kinetic energy in the system.
    Is this the correct interpretation?
    Any help appreciated.
  2. jcsd
  3. Dec 27, 2010 #2


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    It is correct. Can you complete it though, by figuring out what what happens to the three macroscopic thermodynamic variables, namely pressure, volume and temperature? We know the volume increases, so what happens to the other two and why?
  4. Dec 27, 2010 #3
    Thanks for the response kuruman.
    The pressure goes down as there are less collisions. I guess this is due to both an increase in volume and a decrease in kinetic energy of the particles.
    The temperature also goes down, as the tempertaure is a measure of the kinetic energy in the system, and since kinetic energy has been trasnferred, the temperature goes down. Which is interesting becasue if it's correct it means that the temperature of an abiabatic system can change, even though it is not transferring heat or acquiring heat.
    Are these correct statements?
    Many thanks!
  5. Dec 27, 2010 #4


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    These are correct statements, but I was hoping you would consider and interpret equations as a means to support your arguments. Equations often allow you to see clearly in cases where you doubt your judgment.

    For example, pressure = Force/Area of piston.
    Since the piston is at equilibrium in each of the two cases,
    pressure = Weight on piston/Area of piston.
    Since the area of the piston does not change, it takes less pressure to support less weight.

    The First Law of Thermodynamics says
    Change in internal energy = Heat added - Work done by the gas
    Here, Heat added = zero and Work done by the gas is positive because the gas expands. Therefore,
    Change in internal energy is negative, therefore the temperature decreases.

    I am glad you realized that the temperature can decrease even though no heat leaves the system. Some people believe that if heat is not allowed in or out the gas, the temperature has to remain constant, which is not necessarily true because the gas may do positive or negative work at the same time.
  6. Jan 2, 2011 #5
    Thanks Kuruman.
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