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Simple Thermodynamic extremum problem

  1. Apr 23, 2013 #1

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    Suppose I have a cylinder with a movable partition inside, separating it into two subsystems. The partition is movable, but will not transmit heat or matter. The same type of gas is contained in each subsystem, but the pressures and temperatures are different. The same amount of mass (M) is in each subsystem. The question is, how to find the equilibrium situation, given P1, T1, V1, P2, T2, V2, the initial pressures, temperatures, and volumes of the two subsystems, assuming the motion of the partition is quasistatically slow.

    I have solved it by invoking conservation of total mass, energy, volume and entropy (no entropy production for a quasistatic process, i.e. reversible), and requiring the pressures on either side to be equal.

    I would like to solve it by finding the minimum or maximum of something, but I don't know what that something is. It can't be total energy U, that's constant. It can't be total enthalpy (H) because, for an ideal gas H=(5/3)U and U is conserved. It can't be total entropy, that's constant. That leaves Helmholtz free energy, or Gibbs free energy or something else? If so, why?

    Thanks
     
  2. jcsd
  3. Apr 24, 2013 #2
    nice approach, same problem here
     
  4. Apr 24, 2013 #3

    DrClaude

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    Why do you say that?
     
  5. Apr 24, 2013 #4

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    Because quasistatic compression or expansion is reversible, creating no entropy. I see the problem, though - If the "skids are well greased" on the movable partition, it will move fast and the only thing slowing it will be viscosity - creating entropy. If not greased, there will be friction - creating entropy. If well greased, and an outside force is applied to the partition to make sure it moves quasistatically slowly, then entropy is conserved, but energy is not conserved.

    So, lets suppose an outside force is applied to make it move quasistatically, , so that the force on partition is essentially zero, conserving entropy but not energy. I think that's a well-posed problem. What quantity gets maximized or minimized? To make the piston move slowly, the force applied must be opposite to the motion, so energy is extracted from the system. I guess the energy is minimized?
     
  6. Apr 24, 2013 #5

    DrClaude

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    No. Reversible implies quasistatic, but a quasistatic process doesn't have to be reversible.
     
  7. Apr 24, 2013 #6

    DrDu

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    If you want the process to be quasi-statically slow you have either to excert external force to make up for the Delta p force on the membrane, in which case you will provide external ##\int \Delta P dV ## work or this holding force is brought up by very high friction of the gliding piston. In that case, there is heat generated. If the frictional force is proportional to velocity, the integral ##\int T dS=\int F(t) dt \sim \int v dt ## will not vanish.
     
  8. Apr 24, 2013 #7
    The thing that's extremized is the amount of work done.
     
  9. Apr 24, 2013 #8

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    Yes, right. I should have said the process was reversible. The case for "ungreased skids" could be quasistatic, but not reversible, and I did not mean to consider that case.

    In this particular case, is there a reason why the minimum energy principle is wrong? It seems to me that, in this case, they are both valid approaches.
     
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