In thermodynamics, given a piston containing a fixed quantity of gas, when the gas is heated the piston expands and does work on the environment. Using the convention that

$$\Delta U_{in} = Q_{in} - W_{out}$$

the work done is positive and is measured by

$$W_{out} = \int P \cdot dV$$

If the same gas is cooled to below atmospheric pressure, then the atmosphere will do work on the gas (nominally equal to the amount of heat that is withdrawn from it). In this case, the $dV$ will be negative, so the direction of the work sign is negative.

My confusion is this: it is still possible to extract work from a piston that is being compressed by atmospheric pressure (e.g. a Newcomen-style engine ). But then seems as though the work gets "double-counted"- done ON the piston system but also extracted BY the mechanism. On the one hand, I can say that the work done by the system is negative, since the area inside the P-V curve doesn't depend on atmospheric pressure, and this work is outside the P-V curve, and the sign of the heat transfer is negative, etc. On the other hand, the work done on the real world is definitely positive. How do I reconcile this?

Thanks in advance.

  • 1
    $\begingroup$ You are not counting twice the same work. The atmosphere actually does work simultaneously on two systems (and thus diminishing its "internal" energy), one on the gas and one on the weight that is lifted. But I am not sure if I interpreted correctly your doubt. $\endgroup$ – user126422 Oct 9 '16 at 4:39
  • $\begingroup$ I think the tricky part is that, although the system is mechanistically "responsible for" the work getting done, the system is not actually doing the work, and the work done by the atmosphere on the system is irrespective of the work done by the atmosphere on the world system. $\endgroup$ – Brandon Kuczenski Oct 10 '16 at 21:37

In a typical treatment of introductory thermodynamics the effects of atmospheric pressure are ignored for the sake of simplicity, and because it has, to good approximation, no net effect over a whole cycle. In situations where it cannot be ignored, you need to consider the system to be constructed of two separate bodies of gas - the gas in the cylinder and the pressure/temperature bath presented by the atmosphere. In the case where the pressure inside the cylinder is below atmospheric, work was done on the atmosphere to get it that way, and the atmosphere will happily do work to return it to a pressure equilibrium.

  • $\begingroup$ Right, and there are two systems expand/contracting and doing work that have to be considered - the atmosphere and the gas inside the cylinder. $\endgroup$ – Sean E. Lake Oct 8 '16 at 23:32
  • $\begingroup$ The net work done by the atmosphere is $0$, so it doesn't affect the global analysis of the cycle, but it does answer the question asked of how it affects the processes locally. $\endgroup$ – Sean E. Lake Oct 8 '16 at 23:39
  • $\begingroup$ @AlbertAspect I'm assuming you were the downvote on this answer- can you provide another answer? $\endgroup$ – Brandon Kuczenski Oct 9 '16 at 2:19
  • $\begingroup$ @SeanLake "the net work done by the atmosphere is 0" - I assume you mean when both expansion and contraction processes are considered- I agree that the atmosphere can be viewed as a second reservoir, but this doesn't help resolve my confusion about where the extracted work "comes from"- how can the atmosphere reservoir both do work on the enclosed system and also provide work to the outside world? $\endgroup$ – Brandon Kuczenski Oct 9 '16 at 2:22
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    $\begingroup$ It is mathematically no different from if the piston is vertical and you were to add to your considerations the work done by gravity on a piston with non-trivial weight. In the case of the piston with atmospheric pressure, you and the gas in the piston do work on the atmosphere to expand the piston past equilibrium. On the return stroke, the atmosphere does work on both the atmosphere and you the entire time. $\endgroup$ – Sean E. Lake Oct 9 '16 at 2:41

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