I am reading a book that asserts the following:

The second law arises from an equation of the form

$$dS = \mu\,dE + \mu\,dW \tag{1}$$

where $dS$ is the change in entropy, $dE$ the change in system energy, and $-dW$ is the work done externally on the system. Thus the second law expressed as $dS \ge 0$ simply asserts that the system cannot receive more in energy than the amount of external work supplied.

I am confused by this. I know we normally have something like

$$dU=T\,dS-dW \tag{2}$$

where $U$ is internal energy, $dW$ is work done by the system, $T$ is temperature. This makes sense to me. The gain of internal energy is equal to energy flowing in as heat ($T\,dS$ is positive if heat is flowing in) less the energy flowing out as work ($dW$ is positive if the system is doing work).

We can get Eqn 1 from Eqn 2, if we write $U \to E$, $\frac{1}{T} \to \mu$ and take the $dW$ to the other side. Furthermore, $dW$ in Eqn 2 is the work done by the system, so it is correct to say, with respect to Eqn 1, that $-dW$ is the work done on the system.

So Eqn 1 looks fine. The use of $dS \ge 0$ as a statement of the second law of thermodynamics also looks fine.

If we then call the work done on the system $dW'$ so that $dW=-dW'$, we can write Eqn 1 as

$$dS=\mu\,dE-\mu\,dW' \tag{3}$$

If we then assert that $dS \ge 0$, this implies that


which implies that $dE \ge dW'$, i.e. the system must receive more energy than the external work supplied. Not only is this contrary to the statement of the book, but it does not make sense.

So how do starting points that make sense lead to something that does not make sense?

I believe it is something to do with the fact that the $dS$ in Eqn 1 refers to the entropy of the system, while the $dS$ in $dS \ge 0$ refers to total entropy of system plus surroundings.

Am I correct that the book is confusing these and leaving out some steps in the argument, so that, as written, it doesn't make sense? Or is the book perfectly correct, and it is me that is missing something?

  • 1
    $\begingroup$ Just curious, in which book did you see this? I would like to check the context of the chapter. $\endgroup$ – Charlie Mar 21 at 0:39
  • $\begingroup$ Wilson, A.G. 1970. Entropy in urban and regional modelling. p. 12. $\endgroup$ – JaredM Mar 21 at 21:34

So the second law does say that the entropy of an isolated system must increase, but there is nothing saying that for a general system the entropy cannot decrease. It seems like you already know that, but I just wanted to point that out. So we are looking at when the entropy of a system increases, but this is not a condition we always have to impose on our system.

I am going to stick with the form I am more familiar with, along with a change of sign convention (I guess your primed values) $$\text dU=T\text dS+\text dW'$$

Certainly, if $\text dS\geq0$ then we have $\text dU\geq\text dW'$. All this means is that when the entropy of a system increases, any extra energy that comes of this must come from heat. i.e. the energy change cannot all come from work done on the system.

On the other hand, if $\text dS\leq0$ then we have $\text dU\leq\text dW'$. This means that the system's energy will change by less than the amount of work we put into it. Well where did that extra energy go? It left as heat, which is why the entropy of our system decreased.

You have to keep in mind that $\text d U$ already includes energy supplied by work. So there really isn't any issue in having $\text dU\geq\text d W'$. It just means that energy is coming from more than just work done on the system.

It might help to look at when there is no work being done, then $\text dW'=0$. Then for entropy increase $\text dU\geq0$ and for entropy decrease $\text dU\leq0$. Perhaps this might make more sense based on what you have seen?

In terms of thinking about what we mean by the energy here and the distinction between the system and what is external, see the answer by @BobD

  • $\begingroup$ Thanks. I agree. This clarifies my understanding of the physics. Nevertheless, I still have a problem with how to interpret the book so that it makes sense. For example, you say "if $dS \le 0$"...but the book relies on the principle that this is never true because the second law says $dS \ge 0$. The discrepancy is because you are talking about entropy of the system and the 2nd law is talking about total entropy. So how do we get from $dS \ge 0$ to "the system cannot receive more energy than external work supplied"? I've not had time to look at this today. I will work on it tomorrow. $\endgroup$ – JaredM Mar 21 at 21:49
  • $\begingroup$ @JaredM It seems like for that statement they are talking about the entropy of just the system. $\endgroup$ – Aaron Stevens Mar 22 at 1:49

First let me say I agree with @Aaron Stevens answer, particularly the last paragraph to help make sense of things. I think there are some ambiguities in the book’s statement, some alluded to by Aaron, which could contribute to the confusion you’ve expressed. I agree with @Charlie 10 comment that the context is needed in order to clear up the ambibuities.

  1. What is the author’s purpose of substituting $μ$ for =$\frac{1}{T}$?

  2. In order for the two equations to be the same, $E$ must explicitly refer to the internal energy of the system. Does the author say somewhere that $E$ means internal energy? The total change in system energy is the sum of changes in its internal (microscopic) and external (macroscopic) energies. The long form of the first law for a closed system is $\Delta E_{Tot}=\Delta U +\Delta KE+\Delta PE$. The change in potential and kinetic energies are changes in the external energy of the system as a whole, for example, due to changes in elevation or velocity of the system as a whole with respect to some external frame of reference. These are typically ignored for closed systems, thus leading to the short form.

  3. In both equations $dS$ is the differential change in entropy of the system only, since the equations apply to the system. The $dS$ in the inequality $dS≥0$ is, as you said, the differential change in entropy of the system plus surroundings because $dS$ of the system can be negative in a process. An example is the isothermal compression of an ideal gas.

Hope this helps.

  • $\begingroup$ Thanks. I guess the answer to 1 is that he wanted to isolate $dS$ because he is focusing on entropy. He doesn't explain the $\mu$. As you can see from my quote, he doesn't even mention it, which is poor form...he ought to say "where $\mu$ is a constant" or something. Nor does he define what he means by the energy of the system. He doesn't actually use $dW$. I simplified there. He has $\sum_{k} X_k\,dx_k$ where "$X_k$ is the external force corresponding to external coordinate $x_k$" but he then says that this is the work done on the system so it is extra detail that makes no difference. $\endgroup$ – JaredM Mar 21 at 22:03

To answer my own question, I think the quote I gave from the book is confused to the point of being nonsensical.

I will use the discussion from the Wikipedia article on the 2nd Law of Thermodynamics, under the heading of "Energy, available useful work". I am adapting and simplifying the argument. See Wikipedia for a fully rigorous treatment, including issues I will leave out (chemical energy, difference between work and useful work).

We imagine the system in contact with an infinite reservoir, whose pressure and temperature can be assumed to remain constant. Let $S$ be the entropy of the system, $S_R$ be the entropy of the reservoir, and $S_{TOT}$ be the total entropy. By the 2nd Law, we have $$dS_{TOT}=dS+dS_R \ge 0 \Rightarrow dS \ge -dS_R \tag{1}$$ By the 1st Law, we have for the system $$dU=dQ-dW \tag{2}$$ where $U$ is its internal energy, $dQ$ is the heat flowing into the system and $dW$ is the work done by the system. Now $$dQ=T(-dS_R) \le T dS$$ where $T$ is the temperature of the reservoir, and we use Eqn 1. Substituting for $dQ$ in Eqn 2 and rearranging, we get $$dW \le -dU + TdS \tag{3}$$ Now we define the 'exergy' of the system (I think this is energy available for doing useful work) by $$E=U - T\,S \Rightarrow dE=dU-TdS$$ Eqn 3 can therefore be rewritten as $$dW \le -dE$$ $$\Rightarrow dW+dE \le 0 \tag{4}$$ Now if we consider work done on the system, $dW'$, we have $dW=-dW'$, so Eqn 4 becomes $$-dW'+dE \le 0$$ $$\Rightarrow dE \le dW'$$ or in other words, "the system cannot receive more in energy [available for work] than the amount of external work supplied". QED.

This is the path by which we go from "$dS \ge 0$" (meaning $dS_{TOT} \ge 0$) to this statement, which was my original question.

Having gone through this, I think the confusions of the book are as follows:

  1. Uses $S$ to represent both total entropy and system entropy, and does not distinguish between them.
  2. Writes the equality $dS = \mu\,dE + \mu\,dW$, which should really be an inequality $dS \ge \mu\,dE + \mu\,dW$ (a rearrangement of Eqn 3 above, with $\mu=1/T$, and the symbol $E$ in place of $U$).
  3. Uses $E$ to represent internal energy, instead of the usual $U$, and fails to make clear that, in the statement about "energy" received being less than work applied, this really refers to exergy or available energy, and is not the same as the $E$ in the equation.

Edited to add...

In fact, I now see a simpler way of looking at this. The basic problem is that $\mu$ in the original equation is negative, and this is concealed.

Imagine there is no heat transfer, so it is just a question of work done. If, as usual, $dW$ represents work done by the system. We have, by the 1st Law, $dU=-dW$. With the definition of exergy, $E$, above, $dE=dU-T\,dS$, this becomes $dE+T\,dS=-dW$ or, rearranging, $T\,dS=-dE-dW$, which is the same as the author's equation, with $\mu=-1/T$. Now if we have work done on the system, $dW=-dW'$, we can write $T\,dS=-dE+dW'$, and if $dS \ge 0$, this becomes $dW'-dE \ge 0$, which means $dW' \ge dE$, i.e. the useful energy received is less than the work done on the system, QED. I think this must be basically what the author had in mind. So maybe he didn't confuse $U$ and $E$, but it seems he did obscure some issues to do with the signs, and this is why his equation, taken at face value, produced the opposite result to what he claimed.


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