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Confusion about the work done by an ideal gas

  1. Aug 27, 2018 #1
    When an ideal gas,in a piston kind of system and whose equilibrium state is mentioned, is allowed to expand (piston is allowed to move and not gas leaking )against a constant external pressure very quickly, then, is the work done by gas zero or not zero ?

    The arguement for work being zero is that "the constituents take sometime to move towards the piston, as the piston moves away, and hence there's no pressure one the piston due to one side (inside ) and hence zero " I feel this argument is wrong because the pressure is 0 from the inside then the piston would be compressed by the external pressure.

    Thank you in advance.
     
  2. jcsd
  3. Aug 27, 2018 #2

    stockzahn

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    I think the question is if the expanding gas moves the piston (for example the piston ist locked the gas pressure is increased e.g. by isochoric heating and then the piston is unlocked) or if another force moves the piston and the gas is expanded by the increase of available room.

    Wouldn't that mean that the piston moves faster than the gas is expanded? I think theoretically that effect could maybe play a role when approaching the speed of sound. So I think that you are correct with your last statement, but where do you get the quoted statement from?
     
  4. Aug 27, 2018 #3

    Thats what my friends argued.
     
  5. Aug 28, 2018 #4

    Chandra Prayaga

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    1. If the external pressure is constant, and you want to move the piston quickly, then there must be an internal pressure which is greater than the external pressure. This gives rise to a non-quasistatic process, which blows the piston outwards. In general, it is not possible to calculate the work done by the gas.
    2. Zero work is done if the gas expands into a vacuum. This is also a non-quasistatic process.
     
  6. Aug 28, 2018 #5
    When you release the piston, since the gas has inertia, a dilatation region begins to form in the gas immediately adjacent to the piston face. Within this region, assuming a massless piston, the pressure is virtually equal to the outside pressure (since the piston has no mass, it can accelerate at finite speed with even the slightest differential in pressure between inside and outside). As time progresses, the size of the dilatation region gets larger, and its leading edge travels backwards at the speed of sound within the gas (along the cylinder axis). So the phenomenon is similar to that of a sound wave.

    In addition to this, there are viscous fluid stresses that develop within the gas that also contribute to the force per unit area exerted by the gas on the inside face of the piston. But, again, for a massless piston, Newton's third law tells us that the force exerted by the gas on the inside face of the piston must equal the constant external force exerted on the outside face of the piston.
     
  7. Aug 28, 2018 #6
    This is not correct. If the piston is massless, the work done by the gas is equal to the external pressure integrated over the change in volume. If the piston has mass, and before the piston comes to rest, the work done by the gas is equal to the kinetic energy of the piston plus the integral of the external pressure over the volume change (up to that time); if the piston has come to rest, so that the gas is at thermodynamic equilibrium in its final state, then the work done by the gas is equal to the integral of the external pressure over the volume change (up to the end of the process). This all follows from a straightforward force balance on the piston.

    The bottom line is that, if you can control the external pressure exerted on the piston, then you are automatically controlling the work done by the gas on the piston.
     
    Last edited: Aug 28, 2018
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