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Work done by gas under constant pressure

  1. Nov 25, 2011 #1
    Hi, just wanted to ask a simple question:

    In the following equation:

    Work done by gas = Pressure x Change in volume

    does pressure refer to the pressure of the gas (that is doing the work), or does it refer to the external pressure (i.e. pressure of the surrounding gas)?

    I've heard that it is actually external pressure, and I can't for the life of me figure out why...

    Thanks in advance.
  2. jcsd
  3. Nov 25, 2011 #2
    Consider how this formula was derived:

    work = force X distance (Eqn. 1)

    force = pressure X area (cross section) (Eqn. 2)


    work = pressure X area X distance (Eqn. 3)


    volume = area X distance (Eqn. 4)


    work = pressure X volume (Eqn. 5).

    As you can see from Eqn. 2, this is the force (and hence pressure) of the gas doing the work.
  4. Nov 25, 2011 #3


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    For the work done by the gas, use the gas pressure. For the work done on the surroundings, use the external pressure. These are sometimes equal to each other.
  5. Nov 25, 2011 #4
    Thanks for your replies. They agree with what I originally (and still) thought. What I heard must be wrong then... thanks again :)
  6. Nov 25, 2011 #5

    Andrew Mason

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    It is not quite as simple as that. A free expansion of a gas (expansion against 0 external pressure) does no work, so you would not use the internal gas pressure to determine the work done by the gas (work done=0). If the expansion does dynamic work - in moving a piston with non-negligible mass vertically for example, you would not use external pressure to determine the work done on the surroundings (work done > ∫Pext dV)

    Generally, if the process is not quasistatic, you cannot use ∫PdV to accurately determine the work done by the gas (where P is either internal or external pressure) as this does not take into account dynamic work.

    However, if the dynamic work is much less than the static work, ∫PexternaldV would be a good approximation of the work done by the gas in an expansion and -∫PinternaldV would be a good approximation of the work done on the gas in a compression.

    Last edited: Nov 25, 2011
  7. Nov 26, 2011 #6
    Great! Thanks for your insightful points Andrew :)

    I would like to ask about some of them in more detail though:

    1. For a free expansion of a gas, could I consider it to be a special case where by w.d. = p(delta)V does not apply? My rationale is that since this equation is derived from w.d. = Fd, where F is the force exerted by the gas on ANOTHER body, it would not apply for free expansion since the gas does not exert a force on another body (the gas molecules may exert a force on each other but that still does not cause it to do any work as a whole)? To put it simply, this equation only applies to where there is an interaction of the gas with another body.

    2. Totally agree with your point about not using external pressure to determine dynamic work. But I'm still not sure why can't I use ∫PinternaldV to determine the work done by the gas in a non-quasistatic process. For example, if I have a high-pressure gas trapped in a chamber with a heavy piston, with external atmospheric pressure, the force exerted by the gas on the piston (pinternalA) does work on the atmosphere, and on the piston (giving it K.E.). Thus, work done by gas, ∫PinternaldV = Work done on atmosphere ∫PexternaldV + Increase in K.E. of piston. Am I missing anything?

    Thanks again.

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