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The usual explanation of isothermal work, especially in the context of the Carnot cycle, is that the "heat" absorbed in the isothermal leg is converted into "work" done on the environment as the "heat" flows in and the "work" flows out. Of course, this argument completely hinges on the particular constitutive relationship $U=f(T) = c_v(T) T$ of the caloric equation defining the ideal gas. Because of this particular relationship $U=f(T)$ the internal energy does not change during isothermal expansion, $\Delta U =0$, and while $T=const$ the absorbed heat at every instant must, therefore, be equal the work done, $\Delta Q_ {|T} = \Delta W_ {|T}$ as claimed.

But what happens if $U \ne f(T)$? Then we cannot claim $\Delta Q_ {|T} = \Delta W_ {|T}$, in fact, in general, $\Delta Q_ {|T} \ne \Delta W_ {|T}$ and depending on the particular caloric equation of state the left side could be larger or smaller than the right side. Is there a microscopically accessible statistical explanation that would confirm the convertibility of isothermal heat transfer to work? So how does "heat" get converted into "work" during the isothermal absorption stage of the Carnot cycle? What happens to the "unused" absorbed "heat" inside the working substance; e.g., will it get converted later into something else, etc.?

Is there any experimental evidence beyond verbalized interpretation that heat absorbed isothermally is directly converted for work? The reason why this question is important because there is another completely consistent way of interpreting the same phenomenon, namely, starting with the statement that isothermal heat transfer does no work.

(The issue reminds me a bit of the question as to what is the true volumetric EM energy density or flux? Normally we say that it is $\delta w = \mathbf E d \mathbf D + \mathbf H d \mathbf B$ or $ \mathbf S = \mathbf E \times \mathbf H$ but experimentally this is unsupported because we only ever have access to their full volume integral and then verify energy conservation by relating said integral to the outside sources: Poynting's Theorem. There is clear advantage in assuming that the particular spatial distributions $\delta w$ or $\mathbf S$ are true, the assumption is a great simplifier, but I do not see any advantage in maintaining the interpretation that isothermal heat transfer can do work.)

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  • $\begingroup$ "Because of this particular relationship $U=f(T)$ the internal energy does not change during isothermal expansion, $\Delta U =0$,.." $\endgroup$
    – Bob D
    Commented May 11, 2022 at 17:27
  • $\begingroup$ I was considering posting this as an answer, but it seems more appropriate as a comment for discussion. How about considering that instead of heat being directly converted to work, between equilibrium states the energy used to perform work by the gas is taken from internal energy causing it to decrease and that heat transfer to the gas increases the internal energy by an equal amount. When equilibrium is reestablished the change in temperature and change in internal energy is zero. $\endgroup$
    – Bob D
    Commented May 11, 2022 at 17:28
  • $\begingroup$ I don't believe the application of the first law requires that a process be reversible. Even if the process is not reversible, as long as the initial and final equilibrium temperatures of an ideal gas are the same, $\Delta U=0$ and $Q=W$. Wouldn't this allow for the possibility of temperature and internal energy varying between equilibrium states? Your thoughts? $\endgroup$
    – Bob D
    Commented May 11, 2022 at 17:28
  • $\begingroup$ Once you consider that the internal energy is the source of isothermal work then the natural question arises as to which part of the internal energy $U = TS - pV + ...$ is converted to external work. This is what Bronsted's analysis is all about, just ignore the awkward and very unfortunate terminology that makes it sound like the old caloric theory but has absolutely nothing to do with it. $\endgroup$
    – hyportnex
    Commented May 11, 2022 at 21:11
  • $\begingroup$ I don't agree that TS and pV are parts of internal energy. Heat and work may be need to effect a change in energy but they are not stored as energy. That's my biggest objection to Bronsted's approach. He is treating heat and work as if they are components of internal energy. Internal energy is the sum of all the kinetic and potential energies at the microscopic level. As far as the energy used for work, its kinetic in the case of ideal gases. $\endgroup$
    – Bob D
    Commented May 11, 2022 at 21:22

2 Answers 2

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What happens to the "unused" absorbed "heat" inside the working substance; e.g., will it get converted later into something else, etc.?

The problem may lie in part with this statement, which is reminiscent of the debunked theory of caloric. Heat is not a "thing" that resides within a body. In fact, it may be best to not think of "heat" as a noun at all. The closest match to what you're saying may be that heating transfers entropy, whereas reversible work doesn't. (To operate in a cycle, therefore, we must dump this entropy, which we do by heating the so-called cold reservoir.)

As you know, heating is driven by a temperature difference; mechanical work is driven by a stress difference. In general, the two processes are uncoupled. Sometimes we implement the processes simultaneously (or model them as occurring simultaneously). Sometimes the temperature is controlled to remain constant (or happens to remain constant). But this is for practical or modeling expedience. If the energy transferred out through work is less (more) than the energy transferred in by heating, then the system temperature will rise (fall) or a phase change will occur.

Isothermal heating, on its own, does not necessarily do work. Work is obtained by letting the system expand. It could lead to confusion to literally conceptualize the outcome as "heat being converted to work," even if this shorthand description is often used. The nature of the two processes and their driving forces are completely different.

During isothermal expansion, energy is being transferred (through heating and work), and the transfer is complete in the special case of the ideal gas. (For other materials, the strain energy might be altered, or a phase change might occur, for example). In addition, entropy is being transferred into the system and not removed. Does this address your question?

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  • $\begingroup$ I completely agree with you. Not only your answer addresses the question, and at least partially solves it, but also you have just made Bronsted's argument when he says that isothermal heat transfer cannot do any work and denies coupling from the $TS$ content as a potential source of work, see physics.stackexchange.com/q/707654. $\endgroup$
    – hyportnex
    Commented May 11, 2022 at 18:58
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In the isothermal reversible expansion of a non-ideal gas, $dU=C_vdT+\xi dV$, where $\xi$ is a function of the non-ideal PVT behavior of the gas. For isothermal expansion, you would then have $$dU=\xi dV=dQ-PdV$$ If the temperature is constant, you could use this to solve for Q associated with the volume change.

Actually, the quantity $\xi$ is given by $$\xi=-P+T\left(\frac{\partial P}{\partial T}\right)_V$$So one would have $$dQ=T\left(\frac{\partial P}{\partial T}\right)_VdV$$ along with $$dW=PdV$$

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  • $\begingroup$ That is all fine but I do not understand how it answers my question as to what is the underlying physical process, if there is any, by which we can claim that the isothermally transferred and absorbed heat is the source of the work done. $\endgroup$
    – hyportnex
    Commented May 11, 2022 at 19:32
  • $\begingroup$ That is the case only in the limit of an ideal gas. Otherwise, no. $\endgroup$ Commented May 11, 2022 at 22:02
  • $\begingroup$ @ChetMiller Hi Chet. To get a better idea for where hyportnex is coming from, see his answer to the following post: physics.stackexchange.com/questions/707625/… The gist of the issue, as I understand it, is for an ideal gas isothermal process is heat directly converted to work. $\endgroup$
    – Bob D
    Commented May 11, 2022 at 22:39
  • $\begingroup$ I'd be interested in your take on the Bronsted's analysis that hyportnex is referring to. $\endgroup$
    – Bob D
    Commented May 11, 2022 at 22:41
  • $\begingroup$ @BobD I have no idea what he's saying. $\endgroup$ Commented May 12, 2022 at 11:00

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