About Feynman's treatment of entropy

Question

Feynman said in this chapter that if a system absorbs (rejects) an amount of heat $d Q$ at a temperature $T$, then we say the entropy of the system increased (decreased) by an amount $dS=d Q /T$. And he previously showed that for a reversible engine operating between two temperatures, the entropy lost from the hotter reservoir is equal to the entropy gained by the cold reservoir $$Q_{h}/T_{h}=S=Q_{c}/T_{c}.$$

So the entropy here is some quantity related to changes of the system: heat flows out of our system at a given temperature, we lose some entropy. He then said that when a system goes from state $a$ with temperature $T_{a}$ and volume $V_{a}$ to state $b$ with temperature $T_b$ and volume $V_b$, we can write the change of entropy as $$S_a-S_b=\int_{a}^{b} \frac{dQ}{T}.$$

What does $S_{a}$, or $S_b$, represent in this case? In the first case entropy was something flowing in and out and was connected to the amounts of heat leaving or entering the system. Now, entropy is connected only to the volume and temperature of the system, regardless of heat flows. Aren't these two interpretations at odds with each other?

A heat $Q_1$ at temperature $T_1$ is “equivalent” to $Q_2$ at $T_2$ if $Q_1 / T_1 = Q_2 / T_2$, in the sense that as one is absorbed the other is delivered. This suggests that if we call $Q/T$ something, we can say: in a reversible process as much $Q/T$ is absorbed as is liberated; there is no gain or loss of $Q/T$. This $Q/T$ is called entropy.

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We can move around on a $pV$ diagram all over the place, and go from one condition to another. In other words, we could say the gas is in a certain condition $a$, and then it goes over to some other condition, $b$, and we will require that this transition, made from $a$ to $b$, be reversible. Now suppose that all along the path from $a$ to $b$ we have little reservoirs at different temperatures, so that the heat $dQ$ removed from the substance at each little step is delivered to each reservoir at the temperature corresponding to that point on the path. Then let us connect all these reservoirs, by reversible heat engines, to a single reservoir at the unit temperature. When we are finished carrying the substance from $a$ to $b$, we shall bring all the reservoirs back to their original condition. Any heat $dQ$ that has been absorbed from the substance at temperature $T$ has now been converted by a reversible machine, and a certain amount of entropy $dS$ has been delivered at the unit temperature as follows: $$dS=d Q /T.$$ Let us compute the total amount of entropy which has been delivered. The entropy difference, or the entropy needed to go from $a$ to $b$ by this particular reversible transformation, is the total entropy, the total of the entropy taken out of the little reservoirs, and delivered at the unit temperature: $$S_a-S_b=\int_{a}^{b} \frac{dQ}{T}.$$

Answer 1

Feynman said in this chapter that if a system absorbs (rejects) an amount of heat $d Q$ at a temperature $T$, then we say the entropy of the system increased (decreased) by an amount $dS=d Q /T$.

If you re-read your link, you will see that Feynman did not say that. What you left out of your description is that Feynman explicitly specified that his equation applied to a reversible path. [The difference between $Q$ and $Q_{rev}$ is not a minor typo. Rather, it is crucial to understanding the issues you raise in your question.]

I.e., here is what he is saying:

$$dS=d Q_{\textbf{rev}} /T$$

[You only mention the reversibility constraint as applying to his equations for the efficiency of heat engines.]

Likewise, I'm afraid this is also incorrect, because it again leaves out the fact that Feynmann added the key constraint of reversibility:

He then said that when a system goes from state $a$ with temperature $T_{a}$ and volume $V_{a}$ to state $b$ with temperature $T_b$ and volume $V_b$, we can write the change of entropy as $$S_a-S_b=\int_{a}^{b} \frac{dQ}{T}.$$

Instead, Feynman's message is that we can calculate the entropy change associated with a process if we can connect the initial and final states by some reversible path (of which there are an infinite number), and integrate over that path, i.e.:

$$S_a-S_b=\int_{a}^{b} \frac{dQ_\textbf{rev}}{T}.$$

Entropy is a property of the system. It is a state function, and the change in the entropy of a system depends only on the difference between the intial and final states. The change in entropy is thus independent of the path through which that change is made; it doesn't matter whether the change is made reversibly or irreversibly—for a given initial and final state, the change in entropy will always be the same.

Heat flow, by contrast, is not a state function—i.e., it is not a property of the system. Rather (like work), it is a property of the path that connects changes in the system. Thus, if we use two different paths to connect an initial and final state, the heat and work flows along these paths can be different, even though the initial states for both paths are the same, and the final states also are the same.

Having said that, if we constrain a path in a very special way—namely, if we constrain the path to be reversible—we can calculate the change in entropy by integrating $\frac{dQ_\textbf{rev}}{T}$ along that path.

Answer 2

One can define / calculate the amount of entropy present in a system under specified conditions - these are your $S_a$ and $S_b$. You can also measure the difference between them $\Delta S$ which you could use in your first equation. It's analogous to mass - you can say I have 3 kg of apples, I give you 2 kg of apples, and now I have 1 kg of apples. These 3 are all masses even though the middle one (2 kg of apples transferred) is a change in mass, not a state of the system.

Answer 3

With all due respect to Feynman, his description of entropy leaves a lot to be desired. Otherwise you would be able to follow it.

The best way to think about entropy (per unit mass) is that it is a physical property of the material under consideration, rather than a function of any specific process. Like internal energy and enthalpy, it depends only on the present state of the material, as characterized by the present temperature and specific volume (or, equivalently, pressure) of the material (assuming a material of constant chemical composition and phase). So, once you specify the temperature and specific volume, the entropy of the material is determined.

Like internal energy and enthalpy, it is possible to change the temperature and specific volume of a material, and thus its entropy, by subjecting the material to a physical process. The way we do this is by subjecting the material to heat flow and work at its interface with its surroundings. In the case of entropy, there are two mechanisms for entropy change in a specific process:

1. Heat transfer through the interface with the surroundings, determined by the integral of $dQ/T_I$, where $T_I$ is the temperature at the interface where the heat transfer is taking place

and

2. Entropy generation within the material (system) which is produced by dissipative transport processes such as heat conduction and viscous deformation in the material.

In a reversible process, only the first mechanism is significant, but in an irreversible process, involving dissipative transport in the system, the second mechanism is also significant. This second mechanism is very difficult (although not impossible) to quantify, and usually involves the solution of a complicated set of partial differential equations. That is why, in the case of an irreversible change between two states of a material, we usually devise an alternative reversible path between the same two end states to establish the entropy change for the process.