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Work done by an expanding gass against a mass

  1. Sep 14, 2012 #1
    Now that I am making progress, I am ready to ask more complex questions :approve:
    Please correct me where my reasoning is wrong.

    Consider a cylindrical piston in which an ideal gas is sealed. The gas is initially at temperature T. The piston is well insulated, so that all thermodynamic processes are at least adiabatic. Assume that the piston is massless and frictionless. Atmospheric pressure is assumed to be negligible.

    Now the cross-section of the piston is A1. A block of mass M is placed on the piston and the cross-section of the block is A2. Immediately, due to the weight of the block, a downward force of Mg acts on the piston. This can equally be interpreted as a pressure of Mg/A2 (or is it Mg/A1 ? ).

    On the other hand, the gas exerts an outward pressure on the piston much greater than that of the block, so that the piston begins to expand. As the piston expands, the gas's volume increases proportional to the height of the piston, since the cross-section of the piston is a constant.

    Now as the gas's volume increases, its pressure must decrease. Eventually, its pressure must reduce to a value equal to the pressure exerted by the block. The piston is now in static equilibrium, since the gas pressure on it equally opposes the pressure due to the block's weight.

    My question is, over this process, as the gas expands, can the work it does on the block be calculated? What about the work that the block does on the gas? Which of these is positive, and why?

    Is the overall expansion process isothermal? One argument tells me that since the process is adiabatic (by assumption), nonzero work done must be accompanied by nonzero internal energy change (due to the first law), and thus nonzero temperature change (due to the relation between U and T). Thus the temperature must change and the process cannot be isothermal.

    Another argument tells me that it is mathematically impossible to calculate the work done by the gas when both temperature and volume change?

    So I appreciate all help in this helping understand this concept.

    BiP
     
  2. jcsd
  3. Sep 15, 2012 #2

    Andrew Mason

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    The pressure on the gas is the force per unit area of the block on the gas. The force from the block on the gas is distributed over what area?

    One is positive and the other is negative. Work done by the gas in an expansion is positive. Work done on the gas by the surroundings (in this case by the block) in an expansion is negative. The reason has to do with the definition of work as the dot product of a force and the displacement through which the force acts. W = Fd if the directions of force and displacement are the same. W=-Fd if they are in opposite directions. The force of the gas on the mass/piston is in the same direction as the displacement, so the work done by the gas is positive. The force of the Mass/piston on the gas is opposite to the direction of the displacement through which it acts, so the work done by the mass/piston is negative.

    And you would be correct.

    It is more difficult. But why is it impossible? It is a just calculus problem.

    AM
     
  4. Sep 15, 2012 #3
    OK thanks for your reply.
    So the work done by the gas is positive, I get that. It equals the integral of F dot d over the entire interval of displacement.

    So what would be the value of F? Would it be equal to the gas's internal pressure times the cross-section of the cylinder, or would it be equal to the weight of the block, times the cross section of the block?

    Also, would the work that the block does on the gas just be the negative of the above?

    Also, my book uses [tex] W = -\int^{V2}_{V1}\frac{nRT}{V}dV [/tex] to calculate the work done. It says that the temperature is constant, and takes it out of the integral.

    But I thought we established that the process was not isothermal, so why are we allowed to do this?

    BiP
     
  5. Sep 16, 2012 #4

    Andrew Mason

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    It depends on how much greater the gas pressure was than the pressure due to gravity. If the initial gas pressure P is just a little greater than Mg/Area so the piston moves slowly from the beginning, then the pressure in the gas is equal to (Mg + Ma)/Area where a is the acceleration of M. If F is much greater than Mg so that the piston accelerates very rapidly then the gas is not in equilibrium - pressure and temperature are not uniform within the gas so that are undefined for the whole gas.

    I think you can work that out. What is the force that the block exerts on the gas? What direction is the displacement?

    T is not constant in the example you gave of a quasi-static adiabatic expansion. This integral applies to isothermal processes only.

    AM
     
  6. Sep 16, 2012 #5

    morrobay

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    Then what would the integral be for Work in this case when T is not constant ?
     
  7. Sep 16, 2012 #6
    I suppose the textbook never really assumed that the process is adiabatic. That is the only way to resolve this issue without violating the first law of thermo and at the same time accept the textbook's mathematics.

    BiP
     
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