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I am not sure about the correct $pV$ diagram for the isobaric process of gas expanding inside a sealed cylinder covered with a piston when it is heated with candle. The best I can come up is

The zig-zag lines represent repeated changes in pressure in gas while it expands. When heat is added to the gas, its temperature and pressure changes. Higher gas pressure will cause it to expand in volume. While it expands, its pressure will return to its original state. This process will be repeated many times until the gas stops heating. Is that right?

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  • $\begingroup$ I mean, the context here is important. Usually in these situations, we're only thinking about how to model the process without having to think too carefully about the microscopic (or mesoscopic) details. In that case, you should just have a horizontal line (you did say isobaric, after all!). This is the answer, then, unless you really are wondering about the more complicated details of what's really going on (or are asked to think about that). In my opinion, the very slow heating due to a candle will result in a horizontal line, easily, within experimental precision. $\endgroup$
    – march
    Commented Feb 1, 2023 at 18:09
  • $\begingroup$ Yes, I am interested in microscopic details when the gas expands. The zig-zag lines are exaggerated on purpose. $\endgroup$
    – Sammy Lam
    Commented Feb 1, 2023 at 18:20
  • $\begingroup$ Why do you think that the heat is absorbed in discrete portions ("quantized")? $\endgroup$
    – Roger V.
    Commented Feb 1, 2023 at 19:29
  • $\begingroup$ @RogerVadim Not really, these discrete portions represent infinitesimal changes in pressure. As I said before, my drawing is exaggerated. $\endgroup$
    – Sammy Lam
    Commented Feb 1, 2023 at 21:01
  • $\begingroup$ There's nothing wrong with pressure growing continuously. You probably have some microscopic model in your head,suggesting that it is somehow in small steps... but it is really not needed here. $\endgroup$
    – Roger V.
    Commented Feb 1, 2023 at 21:05

2 Answers 2

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In real life, we might expect a notched appearance from friction, as static friction might cause the piston to stick and then slip, stick and then slip. However, it doesn't look like this aspect is included in your model.

Instead, it looks like you're assuming that the nature of molecular collisions causes the pressure to increase a certain amount, at which point the (well-lubricated) piston moves to equalize the pressure, and then the pressure increases again, and so on. This isn't what happens; if the piston has a constant weight and the bore has a constant area, then the gas pressure is constant.

But—you might ask—the pressure arises from discrete collisions, doesn't it? Can't we magnify the line to show these? Yes, but it would still not look notched. It would vary around the constant-pressure line as a random walk whose amplitude deviations depend on the sampling time. With a sufficiently short sampling time, the concept of pressure loses meaning, as the momentum transfer from individual molecular interactions during that interval goes to zero.

A similar argument applies to the heating: There is no start–stop mechanism that would produce notches with a characteristic frequency. Heat transfer is a continuous process from the candle to various regions of the gas, again arising from random molecular collisions. It's not as if (as occurs in some computer simulations, sometimes problematically) the system iterates forward a step, for example, checks the total heat input, and adjusts the pressure or temperature in discrete increments.

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  • $\begingroup$ I came up with my model in order to explain why the pressure is constant in the isobaric process. Do you have a better explanation for how can the gas pressure remain constant after it absorbs heat and increases its temperature? I believe the pressure increases with temperature because of the ideal gas law: $\Delta P=\frac{Nk\Delta T}{V}$. $\endgroup$
    – Sammy Lam
    Commented Feb 1, 2023 at 21:32
  • $\begingroup$ I used Newton’s second law: since the piston isn’t accelerating away, the system pressure equals atmospheric pressure plus the pressure from the piston’s weight. The system’s equations of state are irrelevant. $\endgroup$
    – Chemomechanics
    Commented Feb 1, 2023 at 22:50
  • $\begingroup$ @SammyLam Your statement of the ideal gas law has the volume as a constant which make it an isometric process not an isobaric process $\endgroup$
    – Stevan V. Saban
    Commented Feb 1, 2023 at 23:59
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The ideal gas law requires that all the work done by heat goes into PV work so that $$ nR\Delta T = \Delta (PV)$$

Any heat that does not provide work goes to change the internal energy of the system.

In an isobaric process P is a constant p so: $$ nR\Delta T = p\Delta V$$

An instantaneous change in temperature is followed by an instantaneous change in volume. There are no intermediate pressure changes so no zig zags since the flow of heat as a function of time is not included. These processes are assumed to be quasi-static and the application of heat is a slow process.

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  • $\begingroup$ Thank you, it seems like there is a relationship between isometric+isothermal process (illustrated by my zig-zag line) and the isobaric process. I wonder if isometric+isothermal process will become isobaric if changes in pressure are infinitesimally small? Or maybe they both refer to different physical situations. $\endgroup$
    – Sammy Lam
    Commented Feb 2, 2023 at 0:42
  • $\begingroup$ @SammyLam Yes, they both are different. The answers are very well written. Read these and imagine something by your own and then justifying if it's correct or not. $\endgroup$
    – Debanjan Biswas
    Commented Feb 6, 2023 at 9:36

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