Quote from Feynman Lectures. Lecture 44. Ch.44-6 Entropy:

$$ S_{b}-S_{a}=\int _{a}^{b}{\frac {dQ}{T}}$$ The question is, does the entropy difference depend upon the path taken? There is more than one way to go from a to b. Remember that in the Carnot cycle we could go from a to c in Fig. 44–6 by first expanding isothermally and then adiabatically; or we could first expand adiabatically and then isothermally. So the question is whether the entropy change which occurs when we go from a to b in Fig. 44–10 is the same on one route as it is on another. It must be the same, because if we went all the way around the cycle, going forward on one path and backward on another, we would have a reversible engine, and there would be no loss of heat to the reservoir at unit temperature. In a totally reversible cycle, no heat must be taken from the reservoir at the unit temperature, so the entropy needed to go from a to b is the same over one path as it is over another. It is independent of path, and depends only on the endpoints. We can, therefore, say that there is a certain function, which we call the entropy of the substance, that depends only on the condition, i.e., only on the volume and temperature.

In the fig. 44.10 we see that in the path a-b the substance increases its temperature and volume. So in path a-b the substance does work and increases its internal energy. The substance must not liberate the heat, but absorb according to the 1st law of thermodynamics $\Delta U = Q + W$ , where $W$ is negative, $Q$ is positive and $\Delta U$ is positive. In the path b-a in fig. 44-11 the volume is decreased, so the work is done on the substance. Total work is the area between the curves. Therefore, as there is the net work done on, the substance must absorb some heat from colder reservoir and liberate to hot reservoir (it is the heat engine switched backwards). I don't understand where are the borders of that engine, where are the hot and cold reservoirs? It seems that heat is taken from 1° reservoir, but Feynman said there is no heat to or from 1° reservoir.

I assume that little reservoirs are the part of the engine, because Feynman brings all the little reservoirs back to their original condition (maybe to be able to make a new cycle).


2 Answers 2


I assume that in figure 44-10 the engine, Feynman says about, consists of the next parts:
1) the substance;
2) the little reservoirs on a-b path (6 in figure, but infinite number is implied);
3) the little engines on a-b path (6 in figure, but infinite number is implied);
4) the little reservoirs on b-a path (not shown);
5) the little engines on b-a path (not shown)

To make a process reversible the temperature difference must be infinitesimal. So each little reservoir on a-b path must have temperature of substance at that moment $+\Delta T$. As temperature of the substance increases, we must have infinite number of the little reservoirs (otherwise temperature difference will be too big).

In figure 44-10 the direction of heat flows from little reservoirs through little engines to 1° is either incorrect or the quantities $dQ, dW, dS$ are negative.

On the a-b path we see that temperature of the substance increases, so the internal energy increases too. Also we see the volume increases, so the substance does the work. The only way it could be is: the substance absorbs some heat from the little reservoirs.

According to fig. 44-11 we see that the net work is done on the substance, because the area under b-a curve is larger than area under a-b curve (where the work is done by the substance). On a-b path some work done on the little engines, on b-a path some work done by the little engines. If on the b-a path the little engines liberate heat back to 1° reservoir, then total work must be zero.


I have not read the chapter in detail but if we consider the system and the small reversible engines, we are dealing with a bigger system. Its path is a reversible cycle in contact with the single (1°) reservoir. in this case, the total heat exchanged (and the total work exchanged) must be zero because otherwise, in one of the two possible directions, one could extract heat from a single source (1°) by a cyclic process?

Sorry for my poor english !


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