# Insights Understanding Entropy and the 2nd Law of Thermodynamics - comments

1. Dec 10, 2017

### Tahira Firdous

You have taken internal pressure times change in volume in work equation with a positive convention. I am learning from YouTube lectures they have taken work as external pressure times change in volume with a negative convention, in this way work done on the system becomes positive but J M Smith takes work done by the system positive.So can you please explain me some logical reason behind these conventions and also about internal and external pressures in work equation.
Thanks

2. Dec 10, 2017

### fox26

In "dq/T", does "q" stand for, as it did in my statistical mechanics courses, the thermal energy of the system in question, call it "SYS"? If so, then dq/T is dS, the change in entropy of SYS, so in, for example, a process such as slow compression of a gas in a cylinder by a piston, which we will call "SYS", which would be a reversible process, the environment external to SYS, which supplies the mechanical energy for compression, can have zero entropy change, so dq and so dS must be zero (assuming T>0), otherwise the entropy change of SYS together with the environment would be non-zero, so positive--it could hardly be negative--so the process wouldn't be reversible. The point of this is that you said that in all reversible paths between the initial and final equilibrium states of a system give exactly the same (maximum) value of the integral of dq/T, which is the total entropy change of SYS together with the environment in our example, so all other paths between the initial and final states must give less than or equal to zero entropy change, and a change less than zero would violate the second law, so all paths must give zero entropy change, so no paths which increase the entropy can exist, which is obviously false. Was there a typo in both 3. and 4. in your article, and it should have been (minimum)? The only other likely possibility that I can think of right now is that your dq = my -dq.

Last edited by a moderator: Dec 10, 2017
3. Dec 10, 2017

### Staff: Mentor

Some people take work done on the system by the surroundings as positive and some people take work done by the system on the surroundings as positive. Of course, this results in different signs for the work term in the expression of the first law. In engineering, we take work done by the system on the surroundings as positive. Chemists often take work done on the system by the surroundings as positive

4. Dec 10, 2017

### Staff: Mentor

I stated very clearly in the article that q represents the heat flowing into the system across its boundary, from the surroundings to the system. The T in the equation is the temperature at this boundary.

What you are calling the thermal energy of the system, I would refer to as its internal energy. But, the internal energy is not what appears in the definition of the entropy change.

I don't understand what you are trying to say here. Maybe it would help to give a specific focus problem to illustrate what you are asking.

Last edited: Dec 10, 2017
5. Dec 10, 2017

### fox26

Chet,
I didn't read your article before entering my post, just read your 5 listed points in reply to Khashishi, but a little while ago I did look at its first part, and found, of course, that what I called "thermal energy", q, is not, despite what you said, what you called the "internal energy", which according to your introduction includes both what I called thermal (heat) energy, but also mechanical energy, as it usually is meant to include. Also I found that you defined "equilibrium" so that even the atmosphere of earth at a uniform temperature and completely still would not be in equilibrium, because of the pressure variation with altitude. The situation that I was concerned with is stated in the first sentence of my previous post. For my last statement of what your 5 points implied, instead of "entropy", I should have had "the integral of dq/T". The main point where I disagreed with you, apparently, is in the definition of "equilibrium", and so of "state" of the system. About the only thing you would consider to be a system in equilibrium is a sealed container with a gas, absolutely still, at uniform pressure and temperature, floating in space in free fall, with the state of the system, for a given composition and amount of gas, specified completely by its temperature and pressure, as DrDu said. Then its entropy, given the gas, is determined by that temperature and pressure, and I am willing to believe that Clausius did, by calculating many examples, almost show, except for paths involving such things as mechanical shock excitation of the gas, what you claimed he did show, for such a system.

However, defining entropy from just changes in entropy isn't possible; a starting point whose entropy is known is necessary. Can this be, say, empty space, with zero entropy (classically)? Also, if the second law, that the entropy of a closed system (by this I, and most other people, mean what you mean by an "isolated system") never (except extremely rarely, for macroscopic systems) decreases, is to have universal applicability, it must be possible to define "the entropy" of any system, even ones far from equilibrium, in your or more general senses of "equilibrium". How can this be done? In particular, why the entropy of the entire universe, or very large essentially closed portions of it, always increases, or at least never decreases, is now a fairly hot topic. Do you think this is a meaningful question?

6. Dec 10, 2017

### Staff: Mentor

Not so. Internal energy is a physical property of a material (independent of process path, heat and work), and heat and work depend on process path.
Not correct. The atmosphere at a uniform temperature and completely still would be in equilibrium even with pressure variation. The form of the first law equation I gave, for simplicity, omitted the change in potential energy of the system.

I still don't understand what you are asking or saying. Why don't you define a specific problem that we can both solve together? Define a problem that you believe would illustrate what you are asking. Otherwise, I don't think I can help you, and we will just have to agree to disagree.

7. Dec 11, 2017

### fox26

Is this not possible (classically, ignoring the internal energy of atoms and molecules and the relativistic rest-mass equivalent E = mc^2 energy): Total internal energy E of a closed (in my sense) system, in the center of mass frame = mechanical (macroscopic, including macroscopic kinetic and internal potential energy) energy + thermal (microscopic kinetic) energy? That is what I meant and, when I wrote my first comment, thought you meant, by "mechanical" and "thermal" energy. (I used "heat", non-precisely, in parenthesis after "thermal" in my second comment to try to indicate the meaning of "thermal" just because you seemed, in your reply to my first comment, to think my "thermal energy" meant the total internal energy of the system, which of course it didn't.) My comment that the atmosphere of the earth would not be a system in equilibrium under my stated conditions, according to your definition of "thermodynamic equilibrium state", follows from your definition of that in the first sentence after the second bold subheading "First Law of Thermodynamics" in your article. I agreed, in the last sentence of the first paragraph of my second comment (this is my third comment) with your statements 3 and 4, of 5 total, in your reply to Khashishi ("and" in 3 should be "which"). Two other specific problems are stated in the second and last paragraph of my second comment.

8. Dec 11, 2017

### Staff: Mentor

Huh??? From what you have written, I don't even really know whether we are disagreeing about anything. Are we?

By a specific problem, what I was asking for was not something general, such as systems you have only alluded to, but for a problem with actual numbers for temperatures, pressures, masses, volumes, forces, stresses, strains, etc. Do you think you can do that? If not, then we're done here. I'm on the verge of closing this thread.

9. Dec 12, 2017

### fox26

I asked general questions because those were what I was interested in, not a specific calculation. You mostly made general statements, instead of specific calculations, in your article and answers to replies, which often were themselves general. However, if you won't answer general questions from me, here's a specific one, even though it's a particular case of the first general question in the last paragraph of my last previous reply:

Suppose a closed (in your sense) system SYS in state S1 consists of a gas of one kilogram of hydrogen molecules in equilibrium at 400 degrees kelvin in a cubical container one meter on a side; I leave it to you to calculate its internal pressure approximately, if you wish, using the ideal gas law. How can its entropy be calculated? Integrating dq/T over the path of a reversible process going from some other state S2 of SYS to S1 can give the change of entropy Δentropy(S2,S1) caused by the process, and entropy(S1) = entropy(S2) + Δentropy(S2,S1), but what is entropy(S2), and how can that be determined by thermodynamic considerations alone, without invoking statistical mechanical ones? You did state somewhere that some important person in thermodynamics, I don't remember who, so call him "X" (maybe it was Clausius), had determined that the entropy of any system consisting of matter (in equilibrium) at absolute zero would be zero, so letting S2 = SYS at absolute zero, we would have entropy(S2) = 0, so the problem would be solved, except for the question of how X had determined that entropy(S2), or any other system at absolute zero, = 0, using only thermodynamic considerations. You wrote "determined", so I assume he didn't do this just by taking entropy(any system at absolute zero) = 0 as an additional law of thermodynamics, or part of the thermodynamic definition of "entropy", but instead calculated it. How? It can be done by statistical mechanical considerations (for the SM idea of entropy), but you presumably would want to do it by thermodynamics alone.

10. Dec 13, 2017

### Staff: Mentor

Wow. Thank you for finally clarifying your question.

You are asking how the absolute entropy of a system can be determined. This is covered by the 3rd Law of Thermodynamics. I never mentioned the 3rd Law of Thermodynamics in my article. You indicated that, in my article, I said that "some important person in thermodynamics, I don't remember who, so call him "X" (maybe it was Clausius), had determined that the entropy of any system consisting of matter (in equilibrium) at absolute zero would be zero, so letting S2 = SYS at absolute zero, we would have entropy(S2) = 0." I never said this in my article or in any of my comments. If you think so, please point out where. My article only deals with relative changes in entropy from one thermodynamic equilibrium state to another.

11. Dec 16, 2017

### fox26

Chet,
My statement about what you had said regarding 0 entropy at 0° Kelvin did not involve a direct
quote from you, using “ “, it involved an indirect quote, using the word “that”, and included a part
which I wasn’t attributing to you, the “some important person in thermodynamics, I don’t remember
who, so call him “X” (maybe it was Clausius)” --that was my comment about what you had said. I
admit it wasn't perfectly clear which parts were ones that I was saying that you had said, and which
were mine, but making such things completely unambiguous in the English language often, as with
what I intended to say in this case, requires overly long and awkward constructions. Also, I didn’t
say that you had made the 0 entropy at 0° K statement in your article; in fact, I thought that you
X in that quote was "Kelvin", not "Clausius". According to INFO-MAN, Kelvin had said that a pure
substance (mono-molecular?--fox26's question, not Kelvin’s) at absolute zero would have zero
entropy. Using "entropy" in the statistical mechanical sense, this statement attributed to Kelvin is
true (classically, not quantum mechanically).

Fine, but that brings up what may be a serious problem with the thermodynamics equation:
Δ(entropy) for a reversible process between equilibrium states A and B of a system SYS = the
integral of dq/T between A and B. If SYS is a pure gas in a closed container, and A is SYS at 0° K,
and the relation between dq and dT, which one must know to evaluate the integral, is either
dq = C(dT), which you've used in evaluating such integrals, with C = the (constant) heat capacity,
say at constant volume, of SYS, or dq = k(dT/2)x(the number of degrees of freedom of SYS), which
is implied by the Equipartition Theorem, then the integral of dq/T between A and B is [the integral,
between 0° K and the final temperature T1, of some non-zero constant P times dT/T] =
P[ln(T1) - ln(0)] = ∞ (infinity [for T1 > 0], but actually even then it might be better to regard the
integral as not defined). This problem isn’t solved by requiring the lower (starting) temperature
T0 to be non-zero, but allowing it to be anything above zero, because the integral between
T0 and any T1 > 0 can be made arbitrarily (finitely) large by making T0 some suitably small but
non-zero temperature. Thus, if (1), Kelvin’s sentence is true with “entropy” having the
thermodynamic as well as with it having the statistical mechanical meaning, (2), the Δ(entropy) =
∫dq/T law is true for thermodynamic as well as statistical mechanical entropy, and (3), a linear
relation between dq and dT holds
, then the thermodynamic entropy for any (non-empty)
system in equilibrium and at any temperature T1 above absolute zero can’t be finite, even though
the statistical mechanical entropy for such a (finite) system can be made arbitrarily small by taking
T1 to be some suitable temperature > 0° K. Surely the thermodynamic entropy can’t be so different
from the statistical mechanical entropy that the conclusion of the previous sentence is true. The
problem's solution might be that the heat capacity C varies at low temperatures in such a way, for
example C ∝ √T, that the integral is finite, or that the Equipartition Theorem breaks down at low
temperatures, but at least for systems which are a gas composed of classical point particles
interacting, elastically, only when they collide, which is an ideal gas (never mind that they would
almost never collide), the Equipartition Theorem leads to, maybe is equivalent to, the Ideal Gas Law,
which can be mathematically shown to be true for such a gas, even down to absolute zero, and
experimentally breaks down severely, at low temperatures with real gases, only because of their
departures, including their being quantum mechanical, from the stated conditions. What is the
solution of this problem? Must thermodynamics give up the Δ(entropy) = ∫dq/T law as an exact, and
for low temperatures as even a nearly exact, law?

12. Dec 17, 2017

### Staff: Mentor

For pure materials, the ideal gas is only a model for real gas behavior above the melting point and at low reduced pressures, $p/p_{critical}$. For real gases, the heat capacity is not constant, and varies with both temperature and pressure. So, the solution to your problem is, first of all, to take into account the temperature-dependence of the heat capacity (and pressure-dependence, if necessary). Secondly, real materials experience phase transitions, such as condensation, freezing, and changes in crystal structure (below the freezing point). So one needs to take into account the latent heat effects of these transitions in calculating the change in entropy. And, finally, before and after phase transitions, the heat capacity of the material can be very different (e.g., ice and liquid water).

13. Dec 18, 2017

### DrDu

No, the third law was formulated by Walter Nernst. He also did not find that the absolute entropy at T=0 was 0. Rather he found that the entropy of an ideal crystal becomes independent of all the other variables of the system (like p) in the limit T to 0. So entropy at T=0 is a constant and this constant can conveniently be chosen to be 0.

14. Dec 19, 2017

### DrDu

Even in phenomenological thermodynamics, the heat capacity C generically depends on temperature. The equipartition theorem is a theorem from classical mechanics. It is approximately applicable if the number of quanta in each degree of freedom is >>1. In solids, this leads to the well known rule of Dulong-Petit, stating that the heat capacity per atom in a solid is approximately $3k_\mathrm{B}$. At lower temperatures, the heat capacity decreases continuously as the degrees of freedom start to "freeze out", with the exception of the sound modes. This leads to the celebrated Debye expression for the heat capacity at low temperatures $C_V \approx T^3$.

15. Dec 28, 2017

### fox26

Chet,
Is this too long for a comment?

Thank you for explaining the solution, for real materials, to my infinite entropy change
problem--maybe; I, as indicated, suspected something of the kind might be the explanation. Do
you know, however, that taking into account both the variation of heat capacity with temperature
and pressure, and also phase transitions, always leads to a finite value of ∫dq/T when integrated
between 0°K and a higher temperature? Do you care? Maybe you are concerned only with
changes in entropy for processes operating between two non-zero temperatures. Did you use
entropy change calculations in your chemical engineering job? I know that they can be used in
some cases to indicate that a proposed process is impossible, by showing that it would involve
reduction in entropy of an isolated system, thus violating the second law. A well-known example
is the operation of a Carnot cycle heat engine with efficiency greater than that set by the
requirement that the reduction in entropy caused by the removal of thermal energy from the high
temperature heat bath must be accompanied by at least as great an increase in entropy caused
by the addition of thermal energy to the low temperature heat bath.

One might think that the Δ(entropy) = ∫dq/T law should always give a finite Δ(entropy) for pure
ideal gases as well as real materials, but as I (simply) demonstrated, it doesn’t do so for ideal
gases with the lower temperature equal to 0°K, even though ideal gases would not experience
any variation of heat capacity with temperature or pressure, or any phase transitions. I recently
obvious to me) solution, or at least an explanation, that actually favors the thermodynamic
definition of entropy change, which involves infinite, or arbitrarily large, entropy change for
process (involving ideal gases) with starting temperatures at, or arbitrarily near, absolute zero,
over the statistical mechanical view, which requires any finite system to have only finite absolute
entropy at any temperature, including absolute zero, so gives only finite entropy change for
processes, for finite systems, operating between any two states at any two finite temperatures,
even when one is absolute zero. This solution or explanation will require quite a number of lines
to state. I hope that it is not so obvious a one that I am wasting your, my, and any other persons
who are reading this posts’ time by going through it.

The basic reason that the statistical mechanical (SM) entropy (S) of a pure (classical) ideal gas
SYS in any equilibrium macrostate at 0°K or any temperature above that is zero or a finite
positive number, whereas its thermodynamic (THRM) entropy change between an equilibrium
macrostate of SYS at 0°K and one at any temperature above that is infinite, is that the SM
entropy of SYS in some equilibrium macrostate MAC is calculated using a discrete
approximation NA to the uncertainty of the exact state of SYS when in MAC--NA is the number
of microstates available to SYS when it is in MAC, with some mostly arbitrary definition of the
size in phase space of a microstate of SYS-- whereas the THRM entropy change between two
equilibrium macrostates of SYS is calculated using the (multi-dimensional) area or volume in
phase space of the set of microstates available to SYS when in those macrostates, which can
be any positive real number (and for a macrostate at 0°K is 0). The details follow:

The state of an ideal gas SYS composed of N point-particles each of mass m which interact
only by elastic collision is specified by a point Ps in 6N-dimensional phase space, 3N
coordinates of them for position and 3N of them for momentum. If the gas is in equilibrium,
confined to a cube 1 unit on a side, and has a thermal energy of E, SM and THRM both consider Ps
to be equally likely to be anywhere on the energy surface ES determined by E, which is the set of all
points corresponding to SYS having a thermal energy of E, and the probability density of Ps
being at any point x is the same positive constant for each x ∈ ES, and 0 elsewhere . Since E is
purely (random) kinetic energy, E = Σ1Npi2/2m, where pi is the ith particle's momentum,
so this energy surface is the set of all points with position coordinates within the unit cube in the
position part of the phase space for SYS, and whose momentum coordinates are on the 3N-1
dimensional sphere MS in momentum space centered at the origin with radius r = √(2mE). The
area (or volume) where Ps might be is proportional to the area A of MS, and A ∝ r3N-1. The entropy
S of SYS is proportional to ln(the area of phase space where Ps might be), S ∝ ln(A), therefore
S = const1.+ const2. x ln(E), and since E ∝ T by the equipartition theorem, S = const1.+ const2. x
[const3. + ln(T)]. Thus dS/dE ∝ dS /dT = const2. x 1/T, so dS/dE = const4. x 1/T, and choosing const4.
to be 1, dS = dE/T. This shows the origin of your THRM dS law, for ideal gases (with dE = dq), which
you probably knew. SM approximates this law, adequately for high T and so large A, by dividing
phase space up into boxes with more-or-less arbitrary dimensions of position and momentum, and
replacing A by the number NA of boxes which contain at least one point of ES. This makes S a
function of T which is not even continuous, let alone differentiable, but for large T the jumps in NA,
and so in S, as a function of T are small enough compared to S to ignore, and the SM entropy
can approximately also follow the dS = dE/T law, and be about equal to the THRM entropy, for
suitable box dimensions. However, as T approaches 0°K, the divergence of the SM entropy from the
THRM entropy using these box dimensions becomes severe. As T decreases in steps by factors
of, say, D, the THRM entropy S decreases by some constant amount ln(D) per step, becoming
arbitrarily negative for low enough T, but with T never quite reaching 0°K by this process. For
T = 0°K, A = 0, so S = (some positive) const. x ln(A) = const. x ln(0) = minus infinity. Since the
energy surface ES must intersect at least one box of the SM partition of phase space, NA can never
go below 1, no matter how small T and so A become. Thus the SM entropy S can never go below
const. x ln(1) = 0. The THRM absolute entropy can be finite, except at T = 0, because, although Δ(S)
from a state of SYS whose T is arbitrarily close to 0°K to a state at a higher T can be arbitrarily
large (positive), S at the starting state can be negative enough that the resulting S for the state at
the higher temperature is some constant finite number, regardless of how near 0°K the starting
state is. For the SM entropy, a similar situation is not the case, since although the SM Δ(S) is
about as large as the THRM Δ(S), the SM S at the starting state can never be less than 0. The
temperature at which the SM entropy S gets stuck at 0, not being able to go lower for a lower T, is
not a basic feature of the laws of the universe. Making SYS bigger or making the boxes of the
SM partition of phase space smaller would result in the sticking temperature being lower, and
of course making SYS smaller or the boxes larger would raise the sticking temperature.

I have read somewhere (of course, maybe it was written by a low-temperature physicist) that the
amount of interesting phenomena for a system within a range of temperatures is proportional to
the ratio of the highest to the lowest temperature of that range, not to their difference. If so,
there would be as large an amount of such phenomena between .001°K and .01°K as between
100°K and 1000°K, but the usual SM entropy measure would show no entropy difference
between any two states of a very small system in the lower temperature range, but a non-zero
difference between different states in the upper range, so would be of no help in analyzing
processes in the lower range, even though of some help in the upper range (or if not, for a given
system, for these two temperature ranges, it would be so for some other two temperature ranges
each with a 10 to 1 temperature ratio). On the other hand, the THRM entropy measure would show
as much entropy difference (which would be non-zero) between states at the bottom and at the top
of the lower range as between states at the bottom and at the top of the upper range.

16. Dec 28, 2017

### Staff: Mentor

Actually, as an chemical engineer who worked on processes substantially above absolute zero, I have no interest in this whatsoever.
The concept of entropy figures substantially in the practical application of thermodynamics to chemical process engineering, but not in the qualitative way that you describe. Entropy is part of the definition of Gibbs free energy which is essential to quantifying chemical reaction and phase equilibrium behavior in the design and operation of processes involving distillation, gas absorption, ion exchange, crystallization, liquid extraction, chemical reactors, etc.
I personally have no interest in this, but other members might. Still, I caution you that Physics Forums encourages discussion only of mainstream theories, and specifically prohibits discussing personal theories. I ask you to start a new thread with what you want to cover (which seems tangential to the main focus of the present thread), possibly in the Beyond the Standard Model forum. You can then see whether anyone else has interest in this or whether it is just deemed a personal theory. I'm hoping that @DrDu and @DrClaude might help out with this.

For now, I think that the present thread has run its course, and I'm hereby closing it.