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I am currently confused as to how exactly a Carnot engine (ideal engine) produces work. I will explain my current understanding of how a Carnot engine works:

  1. A piston containing gas is placed into contact with a hot reservoir. To prevent the temperature of the gas from increasing, the applied pressure on the piston is decreased and the gas is allowed to expand. (I have seen flywheel, and rocks being removed that were resting the piston as a way to explain this).

  2. The reservoir is removed, and the pressure applied to the piston is decreased further, this causes the temperature of the gas to decrease.

  3. The piston is brought into contact with a cold reservoir, to prevent the temperature inside the piston from decreasing, the applied pressure on the piston is increased.

  4. The cold reservoir is removed, but the pressure applied to the piston is increased, this causes the temperature of the gas to increase.

The piston can then be brought into contact with the hot reservoir again...

My issue is that I do not see work being done by the gas on the piston, at any point in this cycle. It seems intuitive that this would take place during step 1, because the piston is in contact with a hot reservoir, but for the gas to start moving this piston the pressure and temperature would have to increase while the volume is fixed, but this isn't supposed to happen at this stage in the Carnot cycle, as this stage is completely isothermal.

I thought that the whole point of the Carnot cycle was that the pressure applied to the piston is manually controlled, specifically so that the temperature of the gas remains constant during step 1.

Please don't close this question, if there are similar questions elsewhere. I have looked and can't find one that directly answers my question.

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    $\begingroup$ A review of expansion work $\int P\,dV$ would be useful. Work is done by a system any time it expands against a retreating boundary. It doesn’t matter if the temperature is changing or fixed. The work corresponds to the area under the pressure–volume curve. Work is done by the gas in steps 1 and 2, as confirmed by the positive areas under these curves. Please clarify what is unclear. $\endgroup$
    – Chemomechanics
    Commented Jul 31 at 22:33
  • $\begingroup$ ahh so are you saying work is done by the system, regardless of whether or not the force it exerts on the piston changes? $\endgroup$
    – cookiecainsy
    Commented Jul 31 at 22:36
  • $\begingroup$ Yes. Expansion work involves some nonzero (but not necessarily constant) pressure and a nonzero change in volume. In addition, no fine-tuning of the piston movement is required except that it be slow enough to readily allow heat transfer and minimize frictional dissipation. Being exposed to a heat reservoir is enough to ensure that the system tends to that temperature automatically. $\endgroup$
    – Chemomechanics
    Commented Jul 31 at 22:38
  • $\begingroup$ @Chemomechanics Thanks, that makes sense, however I don't really understand where the work is going. Let's say for example that the piston is connected to a flywheel, if the force exerted by the gas on the piston does not change, the flywheel will not gain energy through each completed cycle $\endgroup$
    – cookiecainsy
    Commented Aug 1 at 16:08
  • $\begingroup$ That’s more of a gearing/linkage question than a Carnot engine question. The piston moves in and out every cycle (with a varying force); the details of how to use this motion to run machinery, for example, is up to the user. $\endgroup$
    – Chemomechanics
    Commented Aug 1 at 17:00

1 Answer 1

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You seem to be confused as to how the isothermal expansion process works. The temperature and pressure of the gas doesn’t have to increase while the volume is constant to do expansion work.

The external pressure is first infinitesimally decreased. That results in an infinitesimal amount of expansion work and an infinitesimal decrease in temperature below the hot reservoir temperature. That, in turn, causes infinitesimal heat transfer from the hot reservoir to the gas raising its temperature back to that of the hot reservoir until the next infinitesimal decrease in external pressure occurs. The sequence is then repeated.

The end result is expansion work is done by the gas while the temperature is constant equal to the reservoir temperature. Thus the pressure drops while the volume increases in such a way that, for an ideal gas, $PV$=constant.

Hope this helps.

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  • $\begingroup$ Thanks for your answer, however I'm still not really sure where the work is going. Let's say for example that the piston is connected to a flywheel, I was told that the flywheel would gain energy through each completed Carnot cycle, but if the force exerted by the gas on the piston does not change, the flywheel will not gain energy through each completed cycle? $\endgroup$
    – cookiecainsy
    Commented Aug 1 at 16:14
  • $\begingroup$ is there any particular reason you have removed the link to your other answer? I was reading it and trying to make sense of it, and thought it was very good. Have you decided it is no longer relevant? $\endgroup$
    – cookiecainsy
    Commented Aug 2 at 0:22
  • $\begingroup$ @cookiecainsy I removed it because it didn’t apply to flywheels. It applied to work done in slowly lifting a variable weight $\endgroup$
    – Bob D
    Commented Aug 2 at 0:30
  • $\begingroup$ while I was reading it, I thought it could be applied to flywheels perfectly well? With the only difference being instead of incremental weights being moved, you consider incremental movements of the piston due to the flywheels momentum $\endgroup$
    – cookiecainsy
    Commented Aug 2 at 0:44
  • $\begingroup$ @cookiecainsy perhaps it could. But I haven’t given it sufficient thought. Anyway, here’s the link again: physics.stackexchange.com/questions/530018/… $\endgroup$
    – Bob D
    Commented Aug 2 at 0:46

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