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Entropy is a measure of energy availiable for work ? 
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#1
Dec1812, 12:44 AM

P: 789

"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. 


#2
Dec1812, 02:14 AM

P: 5,462

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 ? 


#3
Dec1812, 04:48 AM

P: 789




#4
Dec1812, 05:13 AM

P: 5,462

Entropy is a measure of energy availiable for work ?



#5
Dec1812, 05:49 AM

Mentor
P: 22,243




#6
Dec1812, 05:56 AM

P: 5,462

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


#7
Dec1812, 10:47 AM

P: 789

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. 


#8
Dec1812, 11:00 PM

P: 789

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. 


#9
Dec1912, 07:26 AM

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P: 6,654

For example, between two temperatures, Th and Tc, the heat flow Q is capable of producing an amount of work, W = Q(1Tc/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 


#10
Dec1912, 09:45 AM

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#11
Dec1912, 10:54 AM

P: 5,462

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 rearranging 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. 


#12
Dec1912, 12:52 PM

P: 741

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 monoatomic 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] (PPf) 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. 


#13
Dec2012, 04:30 AM

P: 5,462




#14
Dec2012, 10:16 AM

P: 741

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. 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. 


#15
Dec2012, 11:35 AM

P: 5,462

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] 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. 


#16
Dec2012, 01:26 PM

Sci Advisor
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P: 6,654

[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 quasistatic expansion, in which case T is constant for an ideal gas. AM 


#17
Dec2012, 02:17 PM

P: 789

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. 


#18
Dec2012, 04:30 PM

P: 741

Oops. My typo. I meant, The quantity of the ideal gas in the closed container is constant. The number of atoms per molecule is constant. For an isothermal expansion, the temperature of the ideal gas is constant. Therefore, the internal energy of the ideal gas is constant for an isothermal expansion. The internal energy never changes during the entire expansion, even in the limit of infinite volume. However, work is being done for the entire time. Therefore, the work can't come from the internal energy. An isothermal expansion is actually the most inefficient way to use the internal energy of the ideal gas. None of the internal energy of the ideal gas becomes work in an isothermal expansion. All heat energy absorbed by the ideal gas instantly turns into work on the surroundings. In an isothermal expansion, not a single picojoule of work comes from the internal energy of the gas. It all comes from the heat reservoir connected to the ideal gas. One should note that the total amount of work done by the gas in the isothermal expansion is infinite. The work done by the gas increases with the logarithm of volume. So an infinite volume means that an infinite amount of work is performed. Do the calculations for yourself. Obviously, the infinite energy that is going to become work is not in the internal energy of the gas. In fact, the internal energy of the gas remains the same even in the extreme limit of infinite work. The energy for work is being supplied by the heat reservoir, not the ideal gas. In order to maintain a constant temperature, the container has to be in thermal contact with If you want to extract all the internal energy of the ideal gas to work on the surroundings, then you have to decrease the temperature to absolute zero. This can be done in an adiabatic expansion. It can't be done in an isothermal expansion. Entropy can change only two ways. It can move or it can be created. In an adiabatic process, entropy can't move. The temperature changes instead. In an isothermal process, the entropy moves in such a way as to keep the temperature constant. 


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