Why is there a special connection between entropy and heat?

Question

As I understand it: $dS = \frac{1}{T}dU + \frac{p}{T}dV$

(for a thermodynamic system where $dN=0$) and since for an ideal gas $pV=Nk_BT$ and $U=C_VT$ we can say

$dS = \frac{C_V}{U}dU + \frac{Nk_B}{V}dV$

so ultimately entropy changes are caused by changes in proportional changes internal energy $U$ (proportional to the total internal energy already) and changes in volume (proportional to the total volume already).

This makes sense to me, especially the volume part since there will be more available microstates for a larger volume (and the proportionality bits are there to make entropy an extensive quantity).

What I would like to know is whether $dS=\frac{dQ_{rev}}{T}$ is just essentially a 'backward engineered formula' which accounts for both ways entropy can change since $dQ_{rev}=dU+pdV$, or whether it suggests there is some 'special' link between entropy and reversible heat transfer.

At the moment, I don't think there is a special connection since for a Joule expansion there is no heat transfer but there is an entropy increase due to the volume increase, and even though we can model this expansion as an equivalent reversible isothermal expansion with a certain %$Q_{rev}$, in 'reality' that isn't what is happening.

Am I correct to think of thermodynamic entropy as fundamentally to do with proportional changes $dU$ and $dV$ or should I be thinking more fundamentally in terms of heat instead?

Answer 1

Let the work done on the system be $\delta W$ while its internal energy change be $dU$, assume that the system may also exchange energy with a reservoir that is at temperature $T_r$. Then for an arbitrary process the entropy change $dS$ of the system satisfies $dS \ge \frac{dU-\delta W}{T_r}$. The equality sign holds for a reversible process.

When the process is reversible then the system's internal temperature is equal to that of its surroundings, here $T=T_r$. By the 1st law the heat exchanged with the surroundings is, of course, $\delta Q = dU-\delta W$, so you have Clausius's formula that $dS \ge \frac{\delta Q}{T_r}$ and also $dS = \frac{\delta Q_{rev}}{T_r}=\frac{\delta Q_{rev}}{T}$ (Note that "$T_r$" or "$T$" is in the denominator and not "$dT$".)

Of course one can always say that $\delta S_r = \frac{\delta Q}{T_r}=\frac{dU-\delta W}{T_r}$ is the entropy transferred from the reservoir to the system but unless the process is reversible it is not true that $\delta S_r$ is all the entropy change $dS=\frac{dU}{T}-\frac{\delta W}{T}$ in the system; in fact $dS = \delta S_r +\sigma$ where $\sigma \ge 0$ is the internally generated entropy due to dissipation and is never negative.

Answer 2

What I would like to know is whether ${\rm d}S=\frac{\delta Q_{rev}}{T}$ is just essentially a 'backward engineered formula'

In some sense it is. Dividing by temperature is what turns $\delta Q_{rev}$ into the exact differential ${\rm d}S.$ It is what Clausius did (in 1858 I think) when he found that there was such a state quantity, which he called entropy.

Answer 3

There is a special relationship between entropy and heat because when heat passes from $A$ to $B$ then entropy comes along for the ride, and this is unavoidable. The entropy of $B$ will go up. The entropy of $A$ may go down or up or stay fixed, but if the process is reversible then it will go down.

The only way for $B$ to avoid this entropy increase upon receiving heat from $A$ is by passing some heat on to some third party, or by giving up some of its own substance (an open system), or by going to the limit of high temperature.

In short, to transmit heat $dQ$ is to transmit entropy, such that the system receiving the heat gets an entropy increase of at least $dQ/T$ (assuming no further processes of exchange of heat or substance are going on).

Answer 4

Caveat: I have not done statistical mechanics. All my knowledge of this subject is based on classical thermodynamics. However, I tried to keep my answer factual by only referencing already well-accepted ideas on the topic while providing references.

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What I would like to know is whether $dS=\frac{dQ_{rev}}{dT}$ is just essentially a 'backward engineered formula' which accounts for both ways entropy can change since $dQ_{rev}=dU+pdV$, or whether it suggests there is some 'special' link between entropy and reversible heat transfer.

The first expression which you have written is wrong. The true expression for entropy in a reversible process is given by $ dS = \frac{dQ_{rev}}{T}$. I'm not quite sure how you were introduced to entropy, but when I studied thermodynamics, the lecturer integrated the quantity $ \frac{dQ_{rev} }{T}$ for a few different reversible cycles for an ideal gas and found that the line integral over the whole cycle was zero(A). After this, it was motivated to call that $ dS = \frac{dQ_{rev}}{T}$ as a state function since any line integral of this over a reversible path came out as zero (B).

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There are a few alternate explanations for the concept which I have seen over the time that I have studied this subject and I will list them out below:

1. Temperature can be thought of as a 'generalized force' and entropy can be thought of as a 'generalized displacement'. The product of these two conjugate thermodynamic parameters has dimensions of energy and hence we can think of entropy as some abstract quantity that is transferred (C).
2. The most popular interpretation of entropy is as a measure of disorder, people say that entropy is a measure of how 'spread out' energy is. As we spread out energy more and more the amount of useful energy reduces. (D)
3. Statistical mechanics: This is personally one of the most interesting interpretations (For me), in this subject, we think of entropy is a measure of the number of microstates of the system. (E)

Some other say the origin of entropy is in the idea that it is the integrating factor of the first law which can be multiplied so that $dq = dU + dW$ so that we get an exact differential but I'm not sure why this is a particular function which turns it exact. I have personally tried to derive that it is using some mathematical methods but I was unsuccessful. You can see my attempt here but if you still wish for a discussion of it, see this reddit post.

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On the topic of irreversible changes, we must use a different entropy expression and involve a term known as 'entropy generation' to account for the entropy generated by internal processes inside the substance. See the wonderful answer by Chet Miller (here)

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Ref: A: See the gradient theorem here, it is equivalent to say that the line integral of a differential has a potential function if it's integral over any loop is zero

B: This lecture around 4:00

C: see the answer by Chemomechanics here

D: See the video by Steve Mould here

E: https://en.wikipedia.org/wiki/Entropy#:~:text=In%20statistical%20mechanics%2C%20entropy%20is,volume%2C%20pressure%20and%20temperature).

Answer 5

At the moment, I don't think there is a special connection since for a Joule expansion there is no heat transfer but there is an entropy increase

In the "Joule expansion" the gas cools as it uses its thermal energy to accelerate itself. That is a reversible process. Then the mechanical energy of the gas heats the gas, which is an irreversible process and involves heating something.

Let's say we have hydrogen atoms on one side of a container, and deuterium atoms on the other side of the container. When the wall at the middle of the container is removed, the hydrogen gas cools as it accelerates itself, same is true for the deuterium gas.

Then the mechanical energy of the hydrogen gas heats the hydrogen gas, which is an irreversible process and involves heating something. Deuterium gas does the same thing.