Entropy is a measure of energy availiable for work ????

Rap

"Entropy is a measure of energy availiable for work". Can someone explain this to me? Give some examples that show in what sense it is true. It has to come with a lot of caveats, proviso's etc. because its simply not true on its face.

I mean, if I have a container of gas at some temperature above 0 K, then I can extract all of its internal energy as work, just let it quasistatically expand to infinity.

Studiot

Such a container has internal pressure P.

Expanding from P into a vacuum does no work.
Expanding from P against an external pressure P' < P does work, but as this happens P diminishes until P = P' when the system is in equilibrium.

How would you proceed from this equilibrium to infinity, where P = 0 ?

Rap

Yes, you would need initial pressure (P) greater than zero, and ambient pressure (P') equal to zero, i.e. in outer space. But the point remains, the statement that "entropy is a measure of energy unavailable for work" is contradicted by this example.

Studiot

This is fundamental, pushing against something that offers no resistance does no work.

russ_watters

Where did you get that quote? It is wrong: it is missing the word "not"!

http://en.wikipedia.org/wiki/Entropy

Studiot

Agreed - I think that was a typo -, but we also need to dispel the misconception in the proposed counterexample that follows.

Once that is done it is easy to explain the correct reasoning.

Rap

Yes, sorry, I misquoted it. It should be something like "Entropy is a measure of the energy NOT available for useful work". I also see the point that the example I gave is not good. In order for the expansion to be slow, there has to be an opposing force almost equal to the pressure force, and that force has to diminish as the volume increases (and pressure decreases). This opposing force would be the mechanism by which work was done on the environment. Something like a mass in a gravitational field, in which the mass is slowly being reduced by removal.

I found a web site http://web.mit.edu/16.unified/www/SP...es/node48.html which seems to give an explanation, I will have to look at it closely.

Also the Wikipedia quote says it is "energy per unit temperature" not available for work, which I cannot immediately decipher.

Rap

Looking at the above site, it seems to me what it is saying is that the amount of energy unavailable for useful work in a Carnot cycle is equal to the entropy extracted from the hot reservoir (which is equal to the entropy deposited in the cold reservoir) times the temperature of the cold reservoir. How you get from this to the idea that "Entropy is a measure of the energy unavailable for work" still eludes me.

Entropy of what? I tried assuming that the hot reservoir was actually a hot system with a finite amount of energy and entropy. Again, you can prove that if you extract entropy ∆S from the hot body, the amount of unavailable energy is Tc ∆S where Tc is the temperature of the cold reservoir. But you can only extract so much entropy from the hot body using a working body at the cold reservoir temperature. If their temperatures are very close, you can extract very little entropy and energy. The great majority of the internal energy of the hot body is unavailable for work in this case.

If you set the cold reservoir to zero degrees, then the amount of energy unavailable for work is zero. You could extract all of the internal energy of the hot body as work. (right?).

I still don't get it.

Andrew Mason

We have hashed this over before. The problem with "entropy is a measure of the energy unavailable for work" is that standing alone it is not clear and open to a number of interpretations. It requires a special definition of "energy unavailable for work" as the explanation shows. Without that explanation it can easily lead to misinterpretation.

For example, between two temperatures, Th and Tc, the heat flow Q is capable of producing an amount of work, W = Q(1-Tc/Th) ie. with a Carnot engine. So there is energy E = QTc/Th energy that is, in that sense, "unavailable" for doing work. Yet, as we know, ΔS = 0 for a Carnot engine. So one could well ask, how is 0 a measure of QTc/Th? The answer is: "well, that is not what we mean by 'energy unavailable to do work'. We really mean 'lost work' which is the potential work that could be extracted minus the work that was actually extracted, or the amount of additional work required to restore the system and surroundings to their original state if you saved the output work and used it to drive the process in reverse." Hence the confusion.

So, as I have said before, this particular statement should not be used to introduce the concept of entropy. By itself it explains nothing and leads to great confusion.

AM

Rap

Well, I kind of thought that was case, but there was also the possibility I was missing something. Thanks for the clarification.

Studiot

The time for hand waving is over, here is some mathematics.

Consider a universe consisting of a system contained in a heat bath or reservoir at uniform constant temperature T.

Consider changes in the function Z = entropy of bath plus entropy of the system = entropy of this universe.

dZ = dS_b + dS_s.................1

Where b refers to the bath and s refers to the system.

If the system absorbs heat dq the same amount of heat is lost by the bath so the entropy change

dS_b =-dq/T...................2

substituting this into equation 1

dZ = dS_s - dq/T..............3

Now consider a change of state of the system from state A to state B

By the first law

dU_s = dq - dw ..................4

Combining equations 3 and 4 and rearranging

dS_s = (dU_s+dw)/T + dZ

re-arranging

dw = TdS_s - dU_s -TdZ

Since TdZ is always a positive quantity or zero

dw ≤ TdS_s - dU_s

Where dw is the work done by the system

This is the principle of maximum work and calculates the maximum work that can be obtained from the system. As you can see it has two components vis from the entropy created and from the change in internal energy and these act in opposite directions so in this sense the TdS_s term reduces the amount of work obtainable from the internal energy of the system and accounts for unavailable energy since TdS_s has the dimensions of energy.

Note the usual caveat
The inequality refers to irreversible processes, the equality to reversible ones.

Darwin123

What you said is only possible if the gas expands both adiabatically and reversibly. In an adiabatic and reversible expansion, the change in entropy of the gas is zero. Under that condition, one could turn all the internal energy into work.

Any deviation from the conditions of adiabatic and reversible would result in some internal energy not being turned to work.


First, I prove that one can extract all the internal energy from a monotonic ideal gas using an expansion that is BOTH adiabatic and reversible.

Suppose one were to take an ideal gas in a closed chamber and expand it both adiabatically and slowly, so that it is in a state near thermal equilibrium at all times. No entropy goes in or out of the chamber.

At the end of the expansion, even in the limit of infinite volume, you would end up with a gas of finite temperature.

The ideal gas law is:
1) PV=nRT
where P is the pressure of the gas, V is the volume of the chamber, n is the molarity of the gas, R is the gas constant and T is the temperature.

The internal energy of the ideal gas is:
2) U=(3/2)nRT
where U is the internal energy of the gas and everything else is the same.


Substituting equation 2 into equation 1:
3)U=(3/2)PV

Before the gas starts expanding, let P=P0, V=V0, T=T0, and U=U0. The chamber is closed, so "n" is constant the entire time. There are three degrees of freedom for each atom in a mono atomic gas. Therefore, for a mono atomic gas:
3) U0=1.5 P0 V0

The adiabatic expansion of an mono-atomic gas is:
4) P0 V0 ^(5/3)= P V^(1.5)

Therefore,
5) P = P0 (V0/V)^(5/3)

The work, W, done by the gas is
6) W = ∫[V0→∞] P dV

Substituting equation 5 into equation 6:
7) W = (P0 V0^(5/3)))∫[V0→∞] V^(-5/3) dV

Evaluating the integral in equation 6:
8) W = (3/2)(P0 V0^5/3)V0^(-2/3)

9) W=(1.5) P0 V0

The expression for W in equation 9 is the same as the expression for U0 in equation 3. Therefore, the internal energy has been taken out completely.

I’ll solve the problem later (a week or so) for an expansion with sliding friction. There, the increase in entropy characterizes the amount of internal energy not turned into work. However, I will set up the problem. I will show the two equations that makes the case with sliding friction different from the reversible condition.

One remove the reversibility condition by adding sliding friction,
10) Wf = ∫[V0→Vf] (P-Pf) dV
where Pf is the pressure due to sliding friction and Vf is the volume when the P=Pf. The chamber stops expanding when P=Pf.


However, another expression is necessary to describe how much entropy is created.
11) dQ = Pf dV

Equation merely says that the energy used up by the sliding friction causes entropy to be created. The heat energy, dQ, is the energy used up by friction.


I don’t have time now, so I leave it as an exercise. Honest, moderator, I promise to get back to it. However, he wants an example where the creation of entropy limits the work that can be extracted. This is a good one.


Spoiler

Wf<W. Not all the internal energy is turned into work with sliding friction included. Let Q be the work done by the sliding force alone. The increase in entropy is enough to explain why the internal energy is not being turned into work.

Studiot

Do you not agree that the maximum possible work is extracted in a reversible isothermal expansion?

Are you sure you mean this: you have different values of gamma on each side?

Darwin123

In an isothermal expansion, energy is entering the gas from a hot reservoir. Therefore, one can't say that one is extracting the energy from the ideal gas. Most of the energy is being extracted from the hot reservoir, not from the ideal gas.

In the corresponding isothermal expansion, the ideal gas is acting like a conduit for energy and entropy. The ideal gas is not acting as a storage matrix for the energy.

The OP was saying that the work was being extracted from the internal energy that was initially embedded in the ideal gas. All work energy comes from the container of gas, not outside reservoirs. So the walls of the container have to be thermal insulators.

If you allow heat energy to conduct through the walls of the container, then the work may exceed the initial value of the internal energy. The hot reservoir can keep supplying energy long after the internal energy is used up.

So I stick to my guns with regards to the specificity of the OPs hypothesis. He was unconsciously assuming that the expansion is both adiabatic and reversible.

I diagree with those people who said that the gas would remain the same temperature during the expansion, and that not all the energy could be extracted from the internal energy. I showed that the internal energy could be entirely extracted for an ideal monotonic gas under adiabatic and reversible conditions.

The big problem that I have with the OP's question is with the word "quasistatic". I conjecture that the OP thought that "quasistatic" meant "both adiabatic and reversible".

A quasistatic process can be both nonadiabatic and irreversible. However, "quasistatic" is a useful hypothesis. The word "quasistatic" implies spatial uniformity. In this case, the word quasistatic implies that the temperature and the pressure of the ideal gas is uniform in the chamber.

Quasistatic implies that enough time has passed between steps that both temperature and pressure are effectively constant in space. Thus, temperature is not a function of position. Pressure is not a function of position.


I think this is correct. I didn't spend much time checking my work. If you see an arithmetic blunder, feel free to correct me.

Also, I specified a specific case to simplify the problem. So even if I did it correctly, the gamma value that I used was atypical. Next time, I will let gamma be a parameter of arbitrary value.

I specified a monotonic gas. The gas is comprise of individual atoms. There are no internal degrees of freedom in these atoms. Any correlation between coefficients may be due to my choice.

There are many ways an expansion can be irreversible. I think the one most people think of is where the expansion is not quasistatic. Suppose the gas is allowed to expand freely, so that temperature and pressure are not uniform. This is irreversible. However, the mathematics is way beyond my level of expertise.

The problem is tractable if the process is quasistatic. So, I think the best thing would be to show how it works with an quasistatic but irreversible. For instance, what happens if one turns on the sliding friction and the static friction in this expansion. That would result in a process where entropy is created. In other words, friction would result in an irreversible expansion even under quasistatic conditions.

I don't have time now. I will post a solution to that later. For now, just remember that not all quasistatic processes are reversible.

Studiot

I wanted to be more subtle and polite but


[tex]{P_1}{V_1}^\gamma = {P_2}{V_2}^\gamma [/tex]


Whereas you have


[tex]{P_0}{V_0}^{\frac{5}{3}} = {P_1}{V_1}^{\frac{3}{2}}[/tex]

Is this a rejection of Joule's experiment and the definition of an ideal gas?
If so you should make it clear that your view is not mainstream physics.

It should be noted that Joules experiment was both adiabatic and isothermal and has been repeated successfully many times.

You are correct in observing that during an isothermal expansion neither q nor w are zero.
However what makes you think the internal energy is the same at the beginning and end, in the light of your above statement?

One definition (or property derivable from an equivalent definition) of an ideal gas is that its internal energy is a function of temperature alone so if the temperature changes the internal energy changes. To remove all the internal energy you would have to remove all the kinetic energy of all its molecules.

Andrew Mason

I think it was a typo. Darwin did write: P = P0 (V0/V)^(5/3) a little farther down.


I am not sure that you are both talking about the same process. Darwin was referring to a quasi-static adiabatic expansion of an ideal gas. Temperature is given by the adiabatic condition:

[tex]TV^{(\gamma - 1)} = \text{constant}[/tex]

So, if volume changes this cannot be isothermal. Temperature has to change.

I think you (Studiot) may be talking about free expansion, not quasi-static expansion, in which case T is constant for an ideal gas.

AM

Rap

Studiot - I agree with your derivation but with regard to the OP, the correct statement would then be: "an infinitesimal change in entropy is a measure of the minimum infinitesimal amount of energy unavailable for work given a particular ambient temperature."

Darwin123 - Thank you for clarifying the muddled OP. Depending on conditions, all, some, or none of a system's internal energy can be converted to work, and so the statement "Entropy is a measure of the energy unavailable for work" is ambiguous at best, wrong at worst. You are right, I was assuming adiabatic and reversible. Adiabatic or else you are potentially using energy from somewhere else to do the work. Reversible, because it will give the miniumum amount of work unavailable. I should have said that instead of quasistatic. I take quasistatic to mean, by definition, a process can be described as a continuum of equilibrium states.