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Intro:

Thank you for reading.

I know there are lots of questions on Stack Exchange about entropy. The following is the one that most closely addressed my confusion:

Entropy as a state property

However, I'm yet a beginner, the answer went way over my head, and sadly, I'm still quite confused.

I appreciate all and any help!

Thank you.


How my Confusion Arose:

I've been watching Shankar's lectures on YouTube, in which he shows that in the isothermal expansion and contraction of a gas in a Carnot Cycle, the entropy change of the gas in the isothermal expansion and contraction cancel out.

I'm referring to the following lecture:

https://www.youtube.com/watch?v=ouSLRgkPzbI&list=PLFE3074A4CB751B2B&index=24

enter image description here

That is, $\frac{\Delta Q}{T_H}=-\frac{\Delta Q}{T_C}$. The entropy change in the upper isotherm cancels out with the entropy change in the lower isotherm.

Conceptually, that part kind of makes sense to me, since at any fixed temperature, the work that we must do to the gas (and therefor the heat the gas gives off) for a little compression is inversely proportional to the temperature (since the pressure of the gas is inversely proportional to the temperature).

Therefor, the ratio of small changes in heat to temperature, $\frac{\partial Q}{T}$, will cancel out in the upper isotherm with those in the lower isotherm.

However, Shankar from the above somehow concludes that the entropy change in the entire Carnot cycle must've been zero.

Then, he claims that entropy must be a state function, dependent only on the point at which the gas is on the $pV$ diagram.

I don't understand either of these claims.

  1. I don't understand how Shankar got to the conclusion that in the closed-loop of the Carnot cycle, the entropy change must've been zero. He showed that the entropy change in the lower isotherm cancelled out with the entropy change in the upper isotherm. However, he’s forgetting to pay attention to the entropy changes in the adiabatic expansions and contractions. From his explanation, I could’ve just as easily come to the conclusion that points $d$ and $a$ have the same entropy, and points $b$ and $c$ have the same entropy, and then entropy wouldn’t be a state function.

Edit - The above question (1) was answered...changes in entropy in an adiabatic process are zero. However, the following questions still hold:

  1. Additionally, after this, I don't understand how he gets to the conclusion that in any closed loop the entropy change is zero. I understand that in the Carnot Cycle, it was zero. But why in any closed loop?

  2. Finally, why does the fact that in any closed-loop the entropy change is zero imply that entropy is a state function, and we can assign each point on the diagram a unique entropy, even if we have not yet discovered an equation for it (at that point in time, we only seemed to have an equation for entropy changes). Perhaps I can accept this so long as we DO move in a specific path on the $pV$ diagram. But...we calcualate entropy changes for processes that can't even be drawn on the $pV$ diagram, like the free expansion of a gas into a vacuum. How come we're allowed to do this -(yes, I know, because its a state function...but, why is it a state function)?

I feel like if I understood the above, I could follow the rest of his lecture, and understand entropy quite a bit better! But...I'm super confused 😔.

Thank you.

(As for what "the rest of his lecture" is, after he deduces that entropy is a state-function, he calculates the entropy change of a gas expanding into a vacuum by assuming it instead expanded along an isotherm and claiming that it wouldn't make a difference in terms of entropy change as long as the starting point and endpoint were the same, and shows that processes that happen spontaneously in the universe correspond to entropy increases of the universe, and from there arises the second law of thermodynamics).


Edit - Directly Addressing the Question Linked In The Intro:

Addressing the question linked in the introduction to this post...I love the answer which @joshphysics gives...but, its going way over my head. I'm hoping that pherhaps, in answering this question, people could elaborate on his answer there.

How can we show that Entropy is a state Function, if it seems that in the process of its discovery as a good way to characterize a gas in a certain state, they didn't even initially have a definite function for it, but only for its changes?

That is, referencing the question linked above, why is it that (and maybe its all obvious to some...but I don't really understand it):

Fact. (Physics) $\int_\gamma \frac{\delta Q}{T} = 0$ for any closed path γ in thermodynamic state space.

Why does that imply that...

Claim 1. $\delta Q/T$ is conservative, namely $\int_{\gamma_1} \delta Q/T = \int_{\gamma_2} \delta Q/T$ for any two path segments $\gamma_1$ and $\gamma_2$ with the same endpoints.

Why in the world does...

Claim 2. A $1$-form (this is just a fancy term for the kind of mathematical object $\delta Q/T$ is) is conservative if and only if it is exact (exact means it can be written as the differential of a scalar function)

And how the heck does all that mean...

Desired Result. There exists a scalar function $S$ such that $dS = \delta Q/T$.

Thank you!


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    $\begingroup$ Your "short version" is a series of questions about the answer that was given in the link. If you have questions concerning that answer, you should direct it to the person who gave it. $\endgroup$
    – Bob D
    Commented Feb 11, 2020 at 21:36
  • $\begingroup$ @BobD thank you. You're right, I'm going to ask him in a comment on his question right now...however, I also wanted more than just his input, which is why I asked a separate question. $\endgroup$ Commented Feb 11, 2020 at 22:43
  • $\begingroup$ Ok. I am posting an answer related to the "long version". Hope it helps. $\endgroup$
    – Bob D
    Commented Feb 11, 2020 at 22:48
  • $\begingroup$ @BobD 😃😃😃😃 I'm looking forward to it!!! I edited my question and messaged him in a comment. I also edited this question so that it wasn't primarily a questioning of his answer there, but moved that part to the end (since I do think that if people could go in depth into his answer, it would help me a lot). Your advice was good, thanks again! $\endgroup$ Commented Feb 11, 2020 at 22:50
  • $\begingroup$ en.wikipedia.org/wiki/Entropy#Classical_thermodynamics $\endgroup$
    – user65081
    Commented Feb 12, 2020 at 0:10

2 Answers 2

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This is more a mathematical problem than a physical one.

As I understand, you accept that $ \oint_\gamma \frac{\delta Q}{T} = 0$ if $\gamma$ is a Carnot cycle.

Now, any cycle in the thermodynamic state space can be approximated with fragments of Carnot cycles with an aritrary precision. It can be shown that for such cycles you will also have $ \oint_\gamma \frac{\delta Q}{T} = 0$ because the region surrounded by the graph can be divided into many subregions that are surrounded by Carnot cycles, and $\oint_\gamma \frac{\delta Q}{T} = \sum_i \oint_{\gamma_i} \frac{\delta Q}{T}$.

Since the precision of approximation can be as good as we want, by continuity we can prove that $ \oint_\gamma \frac{\delta Q}{T} = 0$ for any cycle.

Now, consider two curves (processes) going from point A of the state space to the point B, let's call them $\gamma_1$ and $\gamma_2$. If you follow one of the curves and then backtrack the other, you get a cycle $\gamma$. You have $$ 0 = \oint_\gamma \frac{\delta Q}{T} =\int_{\gamma_1} \frac{\delta Q}{T} -\int_{\gamma_2} \frac{\delta Q}{T} $$ that is $$ \int_{\gamma_1} \frac{\delta Q}{T} =\int_{\gamma_2} \frac{\delta Q}{T} $$

The curves $\gamma_1$ and $\gamma_2$ were chosen arbitrarily, which means that the integral $\int_\gamma \frac{\delta Q}{T}$ over a curve (process) $\gamma$ linking two states $A$ i $B$ does not depend on the specific of the process; it can only depend on the starting and ending points.

Let us choose an arbitrary state $O$. For $A$ being another state, let us define a function $$ S(A) = \int_O^A \frac{\delta Q}{T}$$ where the integration is performed over any curve linking state O to state A (as w ehave proven, it doesn't matter which one). This is what we called entropy. By construction, it's a function of state.

It can also be proven that $$ \Delta S_{AB} = S(B)-S(A) = \int_A^B \frac{\delta Q}{T}$$ where the integration is over any process from state A to state B.

There is some arbitrariness in the definition of the entropy function (choosing state $O$) but choosing a different state $O'$ would only change the entropy function by a constant.

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  • $\begingroup$ Adam, thank you. I feel like in your response lies the answer to my question, but...I yet do not know enough to understand it. I'm going to have to tackle it piece by piece. Could you perhaps point me to somewhere where... "Now, any cycle in the thermodynamic state space can be approximated with fragments of Carnot cycles with an arbitrary precision." ...is explained more in depth? Then, I'll continue reading, and keep you updated as I go. $\endgroup$ Commented Feb 14, 2020 at 4:03
  • $\begingroup$ @JoshuaRonis It is easier to look at the graphs of the processes on the T-S plane, rather than on the p-V plane, because on the T-S plane the Carnot cycles are rectangles. You can approximate a region bounded by a general curve with a set of rectangles, and the more rectangles you can use, the better approximation. The fact that $\oint_\gamma = \sum_i \oint \gamma_i $ is known as decomposition lemma, and appears in several places, for example in the proof of the Green's theorm, you may look it up. $\endgroup$ Commented Feb 14, 2020 at 9:02
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Regarding your "short version" you really should pose your questions to the person who gave that answer in your reference link. I will respond to your "long version". In particular, there are the following key points you are missing.

However, he’s forgetting to pay attention to the entropy changes in the adiabatic expansions and contractions.

A process that is both reversible and adiabatic is also isentropic, meaning a constant entropy process. That means there is no entropy change in either the adiabatic expansion or compression. Since you already agree that the entropy changes in the isothermal processes cancel each other, then the conclusion has to be the entropy change of the system at the completion of the cycle is zero.

Additionally, after this, I don't understand how he got to the conclusion that in any closed loop the entropy change is zero (he just seems to jump to it) and that therefor, entropy is a state function, and we can assign each point on the diagram a unique entropy, even if we have not yet discovered an equation for it (at that point in time, we only seemed to have an equation for entropy changes).

For any cycle (Carnot or otherwise), the change in all properties of the system at the end of the cycle is by definition zero. So the real issue here is whether or not you accept the fact that entropy is a property of the system. If you do, then just like any other system property (Pressure, volume, temperature, internal energy, ...) it will return to its original value at the end of a complete cycle. The unique thing about entropy, however, is that the change in entropy of the system is always zero for a complete cycle, but the entropy change of the surroundings will only be zero if the cycle is reversible. For irreversible cycles the entropy change of the surroundings will be positive.

Then, he claims that entropy must be a state function, dependent only on the point at which the gas is on the $pV$ diagram.

I haven't got time to watch the video, so I don't know specifically what led him to conclude entropy is a state function. But like I said, a state function has a unique value at each equilibrium state of the system.

From there, he calculates the entropy change of a gas expanding into a vacuum by assuming it instead expanded along an isotherm and claiming that it wouldn't make a difference in terms of entropy change as long as the starting point and endpoint were the same, and shows that processes that happen spontaneously in the universe correspond to entropy increases of the universe...etc...

The free expansion of a gas into a vacuum is an irreversible process. The typical example is a joule expansion. There is no work done, no transfer of heat, and therefore per the first law no change in internal energy. Finally, if it is an ideal gas, then there is also no change in temperature. You know the process is irreversible as you would never expect the process to naturally reverse itself. What he is saying is you can calculated the increase in entropy by conducting a reversible isothermal compression of the gas returning it to its original state (pressure, volume, internal energy and entropy). The heat transfer out divided by the constant temperature then has to equal the increase in entropy that occurred during the expansion, since the overall entropy change is zero.

In conclusion, one of the reasons a system property called entropy was "discovered" and defined, as well as the second law of thermodynamics, is that the first law of thermodynamics did not prohibit certain processes that were found to be impossible.

The first law is a statement of conservation of energy. If I have a hot object and cold object placed in contact with one another we always observe that energy transfer, called heat, occurs from the hot object to the cold object. Heat $Q$ leaves the hot object and goes to the cold object. The change in energy of the hot object is $-Q$ and of the cold object is $+Q$ for an overall change of zero. Energy is conserved and the first law is satisfied. However, the first law would also be satisfied if $Q$ left the cold object and went to the hot object. But this is never observed to occur naturally, and necessitates external work to accomplish. Thus the need for the second law and the property of entropy that prohibit this from occurring naturally.

Hope this helps.

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  • $\begingroup$ Bob thank you for your answer. I understand that IF entropy is a state function, then the entropy change in a closed loop is zero. But...why can we say that entropy is a state function in the first place? You see, in the video, he seemed to go in the opposite direction - he first showed that for the Carnot cycle, the entropy change was zero in the closed loop, and from this ended up concluding that in ANY closed cycle, the entropy change is zero, and THEREFOR its a state function. But...I neither understand his jump to ANY closed loop, nor why that makes it a state function. Thanks again! $\endgroup$ Commented Feb 11, 2020 at 22:55
  • $\begingroup$ @JoshuaRonis I'm sorry I don't have the time to delve into the video so I can't follow logic from what you've said. I would start with the fact any property change of the system is zero for a closed cycle. The entropy change of the surroundings, for any cycle, will also be zero if the cycle is reversible. The unique thing about the Carnot cycle is it is the most efficient of all reversible cycles. But the reason why entropy is a state function is because all the other properties, particularly the internal energy, can not explain why certain processes are impossible. $\endgroup$
    – Bob D
    Commented Feb 11, 2020 at 23:01
  • $\begingroup$ That's alright. Lets ignore Shankar's specific process. I understand that IF we accept that entropy is a state function, then the second law of thermodynamics can explain why certain processes happen (even without delving into atomic theory and Boltzmann's explanations, which I'm told came way later). However, I don't understand how scientists deduced that entropy indeed IS a state-function! Even if they deduce that in ANY loop the entropy change is 0 (which I still don't understand HOW they show)...why can they then apply it to non-equillibrium processes not characterized by ANY path? Thanks! $\endgroup$ Commented Feb 11, 2020 at 23:07

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