What is the source of isothermal work?

Question

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.)

Answer 1

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?

Answer 2

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 $$dW=PdV$$