What rules, energy or entropy?

[sugeet]

what rules, energy or entropy??

any physical system goes to a state of maximum entropy or minimum energy,

they don`t imply each other, by my understanding

1>third law
s->0 as e->0
2>system tends to be in a state of least energy

how??

 

[Andrew Mason]

any physical system goes to a state of maximum entropy

This is not really true, since there is no "maximum" to a system's entropy. The statement "any physical system experiences an increase in entropy in any thermodynamic process" would be true only for an isolated physical system. The entropy of the universe is always increasing but little parts of the universe, like the inside of a refrigerator, can experience decreases in entropy.

or minimum energy,

Again, this is true only of a system that is not isolated. If the system is losing energy, the energy has to be accounted for somewhere else.

they don`t imply each other, by my understanding

1>third law
s->0 as e->0
2>system tends to be in a state of least energy

how??

I am having difficulty understanding your question.
AM

 

[Cleonis]

what rules, energy or entropy??

The text of your message is very unclear, so I limit myself to the content of the subject line.

I remember the following classroom demonstration: a glass was put on the table and with some chemistry Iodine gas was produced. Iodine molecules are very heavy. Pure Iodine gas is much heavier than air. It also has a distinct brownish color. A piece of paper covered the glass, so that the Iodine gas wouldn't escape. A second glass was placed on top, inverted, so that when the piece of paper was pulled away the two compartments became one compartment.

Then we watched the brownish color climb up, and gradually it developed to an even distribution.

(I think it was Iodine gas, but maybe I don't remember correctly. Anyway, that detail is not important for the point I'm making.)

That was a demonstration of entropy overpowering energy. The gas mixture tends to develop towards the most probable state, and a uniform distribution is more probable. Iodine molecules migrated upward, against the gravitational potential.

The gain in gravitational potential energy is at the expense of thermal energy. If perfecly insulated from the environment the gas mixture will drop in temperature.


There are also circumnstances where energy overrules entropy. Example, spinning a tube containing blood in a centrifuge results in separation of constituent parts of the blood. Under normal gravity the proteins in the blood remain in suspension (entropy stronger than energy)
The centrifuge is pulling G's, far more than normal gravity. Now inside the blood containing tube there is a far steeper gradient in gravitational potential. Now energy overpowers entropy, and the proteins in the blood go out of suspension.



So, energy and entropy are independent, and there are lots of circumstances where the two have opposing influences. Depending on the precise circumstances one or the other is strongest.

 

[sugeet]

Yes I understand, that I did not put the question well...
I think last night I have got the concept as I was pondering.
Let me try to put it in a proper way,
any system has max prob to be in a state of maximized entropy(considering sys + surr), that is in a state having maximum number of microstates..., does this state always correspond to a state of least energy?... it this the state which is most probable, even energetically.

I have confused a bit, but I think, u all will understand.

 

[sugeet]

Andrew thanks for addressing my question, yah! I got it, I am talking in one place about a closed system, and then about an open system, energy is always conserved, so there is nothing like least energy also. But I want to understand the equivalence, in the fact(I think, it must be so, just a feeling) that any process, described energetically, that is this happened , because, the system wants to be in a state of lesser energy, is also the most probable state, If I am still wrong, please tell me!

 

[Andrew Mason]

But I want to understand the equivalence, in the fact(I think, it must be so, just a feeling) that any process, described energetically, that is this happened , because, the system wants to be in a state of lesser energy, is also the most probable state, If I am still wrong, please tell me!

It depends on what is happening with the system. If the system is the Sun, would you say this?

The first and second laws say that 1. energy will always be conserved in some form but 2. it will not flow on its own from a state of high entropy to a state of low entropy. 2. since entropy is defined as dQ_rev/T, this is just another way of saying that heat flow will not occur on its own from a colder to a hotter body.

You can answer your questions by simply applying those 2 principles. You have to be careful to distinguish between closed and open systems. You cannot make general statements about open systems, since it depends on how it is interacting with its surroundings.

For closed systems, entropy cannot decrease but energy will always be conserved. How much entropy will increase or how rapidly it will increase depends on what the system consists of.

AM

 

[Whovian]

Let's first determine what you mean by a closed system tending to "lower energy." By The First Law of Thermodynamics (aka Conservation of Energy), this energy can't be destroyed. So what you probably mean is the amount of usable energy is decreasing in a closed system. There are a few ways to think of Entropy. One's a well-defined measure of disorder. Another's a measure of the amount of information in a system. But another's a measure of the amount of unusable energy in a system.

So when the amount of usable energy is decreasing, entropy is necessarily increasing, and so they're equivalent.

Oh, and double-posting isn't ideal, there's an edit button there for a reason ...

 

[Andrew Mason]

Let's first determine what you mean by a closed system tending to "lower energy."

I am not sure whose post you are referring to. The quote button is there for a reason ...

In any event, a closed system does not and cannot tend to lower energy. Total energy is always conserved: first law.

By The First Law of Thermodynamics (aka Conservation of Energy), this energy can't be destroyed. So what you probably mean is the amount of usable energy is decreasing in a closed system. There are a few ways to think of Entropy. One's a well-defined measure of disorder. Another's a measure of the amount of information in a system. But another's a measure of the amount of unusable energy in a system.

How is it a measure of the unuseable energy in a system?

First of all, the entropy of a closed system is inversely related to the amount of useful work that the system is capable of doing.

Furthermore, entropy is not directly related to the capacity to do useful work. A system with high entropy can easily have more capacity to do useful work than a system with much lower entropy. For example a system consisting of a tank of steam and a block of ice has much more entropy than a system consisting of just a block of ice. But the former can do a lot more work than the latter. What is material is the change in entropy, and that depends on how the energy in the system is distributed and in what form it is in.

What you can say about entropy is that the universe tends toward ever increasing entropy and that a closed system will not experience a decrease in entropy.

So when the amount of usable energy is decreasing, entropy is necessarily increasing, and so they're equivalent.

They are not equivalent unless the decrease in energy is equal to the increase in entropy. Are they?

Oh, and double-posting isn't ideal, there's an edit button there for a reason ...

Not sure who this is directed to.

AM

 

[Studiot]

Do you guys not think this reference to a closed system is a tad disingenuous?

How is energy preserved within a closed system?
By definition a closed system is one in which energy can pass across the system boundary between the surroundings and the system.
A thermometer is a closed system. Place one in a refrigerator and you remove energy.

With respect to the original question:

A system is said to be in equilibrium when no further spontaneous change occurs. The entropy of an isolated system tends to increase until no further spontaneous change can occur.

So in an isolated system in equilibrium at constant energy and volume the entropy is a maximum.

S → max at const E (U) and V

If instead we have constant entropy and volume then the equilibrium criterion is that the energy is a minimum.

S → min at const S and V


So the drive to max S and min E compete and only if E is held constant can S achieve its max or if S is held constant can E achieve its min.
In order to calculate what happens in the intermediate stages you need to consider one of the free energy or work functions.

does this help?

 

[Andrew Mason]

Do you guys not think this reference to a closed system is a tad disingenuous?

How is energy preserved within a closed system?

Technically, you are correct. We are using "closed system" to mean "isolated system", one that is, by definition, thermodynamically isolated from the rest of the universe ie: it cannot exchange matter or energy with the rest of the universe. Energy is preserved in an isolated system.

With respect to the original question:

A system is said to be in equilibrium when no further spontaneous change occurs. The entropy of an isolated system tends to increase until no further spontaneous change can occur.

So in an isolated system in equilibrium at constant energy and volume the entropy is a maximum.

How about an isolated system consisting of a Carnot engine, operating between hot and cold finite reservoirs, compressing a ratcheted spring. Can any further spontaneous change occur? When the temperature difference becomes arbitrarily small, the engine cannot do any more work and the spring remains compressed. Has entropy increased? Is it at a maximum?

S → max at const E (U) and V

If instead we have constant entropy and volume then the equilibrium criterion is that the energy is a minimum.

If it is an isolated system, where does the energy go?

AM

 

[Studiot]

How about an isolated system consisting of a Carnot engine, operating between hot and cold finite reservoirs, compressing a ratcheted spring. Can any further spontaneous change occur? When the temperature difference becomes arbitrarily small, the engine cannot do any more work and the spring remains compressed. Has entropy increased? Is it at a maximum?

How is the spring (gaseous?) compressed at constant volume in an isolated system?

If it is an isolated system, where does the energy go?

Good question to a poorly worded statement.
In a system at constant entropy there can be no heat change so the only energy change is mechanical work. So this is the statement of ordinary mechanical least action.

 

[Andrew Mason]

How is the spring (gaseous?) compressed at constant volume in an isolated system?

It doesn't really matter. Let's just say that the engine through a system of gears compresses the spring or lifts a weight - just something that stores mechanical energy indefinitely.

In a system at constant entropy there can be no heat change so the only energy change is mechanical work. So this is the statement of ordinary mechanical least action.

Since the Carnot engine is doing work, heat is flowing out of the hotter reservoir and into the colder reservoir. So there is a conversion of heat energy into mechanical work and all of this is kept within the isolated system. But there is no increase in entropy.

AM

 

[Studiot]

How does this machine do work without expansion, since constant volume means no expansion.

Whether you raise your weight by placing it on a piston or by heating a gas or raising steam and turning a turbine or whatever you are not operating at constant volume.

 

[Andrew Mason]

How does this machine do work without expansion, since constant volume means no expansion.

Whether you raise your weight by placing it on a piston or by heating a gas or raising steam and turning a turbine or whatever you are not operating at constant volume.

It does work over many cycles and returns to its original volume. It is a Carnot engine operating between two finite reservoirs using a working volume of a gas that goes through many, many cycles.

 

[sugeet]

How about an isolated system consisting of a Carnot engine, operating between hot and cold finite reservoirs, compressing a ratcheted spring. Can any further spontaneous change occur? When the temperature difference becomes arbitrarily small, the engine cannot do any more work and the spring remains compressed. Has entropy increased? Is it at a maximum?


AM

AM, I guess it is the maximum, consider two metal containers of water at diff temp, in contact
isolated, from everything else, at equilibrium, both have same temp , and yes, this is a state of maximised entropy!
but i cannot explain it in energetics...

 

[Andrew Mason]

AM, I guess it is the maximum, consider two metal containers of water at diff temp, in contact
isolated, from everything else, at equilibrium, both have same temp , and yes, this is a state of maximised entropy!
but i cannot explain it in energetics...

But what if you operate a Carnot engine between them and extract work that you store in a mechanical or electrical device (eg. a spring or battery). There is no increase in entropy but the two reservoirs end up at the same temperature. Can you say that the entropy is maximized?

 

[Studiot]

So in an isolated system in equilibrium at constant energy and volume the entropy is a maximum.

I'm sorry I still don't see how your engine meets the stated criteria.

 

[Andrew Mason]

I'm sorry I still don't see how your engine meets the stated criteria.

The engine keeps running until the two reservoirs are arbitrarily close to the same temperature. Think of the engine and reservoirs being in a insulated container of fixed volume. The total energy is the same, it is just distributed over both reservoirs and in the spring or battery. The total entropy has not changed because the engine operated on a Carnot cycle.

AM

 

[Studiot]

Oh dear, I fear this thread is wandering off topic and sugeet is still interested in it.

Never mind I think that discussing Andrew’s system is bringing out some related fundamental issues.

For someone making their way in Thermodynamics there are a lot of formulae and definitions to find their way round. It is easy to apply the wrong one or the right one in the wrong way. So easy that even the most experienced sometimes slip up and do it.

The key to a good analysis is the proper definition of the system.

To be complete a proper system definition must define

1) The system components
2) The system boundary
3) The system process.

In each case the definition should specify what is included and what is not included. This applies both to the physical participants such as 'bucket of water' and to the mathematical considerations such as 'quantity of heat', boundary conditions, location etc. By changing one or more of these we can convert between isolated, open and closed systems.

Andrew, I see nothing in your system contrary to what I said. However your system initially is not in equilibrium so does not conform to the conditions of my statement until the reservoirs have equilibrated. At this time the heat flow has ceased and, as you say the entropy is constant, since no further change occurs.

I do have two reservations, however.
Firstly you have yet to describe a mechanism for doing work at constant volume.
Any change to mechanical energy at constant volume involves the product VdP rather than PdV. Unlike PdV, VdP is not work.

Secondly you are misapplying the first law. The first law connects a property of the system components (internal energy) to a property of the system boundary the (energy flows across it). By including both the carnot engine and the mechanism within your system and making it isolated (as specified) you exclude applying the first law across the boundary. In order to do this you must divide the system into subsystems.

 

[Andrew Mason]

I do have two reservations, however.
Firstly you have yet to describe a mechanism for doing work at constant volume.
Any change to mechanical energy at constant volume involves the product VdP rather than PdV. Unlike PdV, VdP is not work.

It is an ENGINE!. The whole point of a thermodynamic engine is to do ∫ PdV work, of course. The engine does work using a complete thermodynamic cycle. The gas expands doing work and is then compressed (but using less work to compress than was performed during the expansion). The gas ends up back in its initial state after having done net work.

Secondly you are misapplying the first law. The first law connects a property of the system components (internal energy) to a property of the system boundary the (energy flows across it). By including both the carnot engine and the mechanism within your system and making it isolated (as specified) you exclude applying the first law across the boundary. In order to do this you must divide the system into subsystems.

Who says that you cannot include the reservoirs in the system? I am defining the system as an isolated container composed of two reservoirs that are thermally isolated from each other that are initially at different temperatures plus a Carnot engine operating between them. Nothing is preventing you from applying the first law anywhere. It always applies between states of thermodynamic equilibrium. Since it is a Carnot engine, the working substance in the engine is in thermodynamic equilibrium with the reservoir that it is in thermal contact with at any given time.

AM

 

[Studiot]

Who says that you cannot include the reservoirs in the system? I am defining the system as an isolated container composed of two reservoirs that are thermally isolated from each other that are initially at different temperatures plus a Carnot engine operating between them. Nothing is preventing you from applying the first law anywhere. It always applies between states of thermodynamic equilibrium. Since it is a Carnot engine, the working substance in the engine is in thermodynamic equilibrium with the reservoir that it is in thermal contact with at any given time.

The first law says:

If you have a system, the internal energy of that system increases precisely by the amount of heat added and work done on it. Or if you like by the amount of energy transferred from its surroundings.

You can add more energy types into the balance if you wish.

But

The definition of an isolated system is

There is no heat added and no work done on it.

Consequently the internal energy of an isolated system is constant. (and boring).

Can we agree on this?

 

[Andrew Mason]

The first law says:

If you have a system, the internal energy of that system increases precisely by the amount of heat added and work done on it. Or if you like by the amount of energy transferred from its surroundings.You can add more energy types into the balance if you wish.

But

The definition of an isolated system is

There is no heat added and no work done on it.

Consequently the internal energy of an isolated system is constant. (and boring).

Can we agree on this?

Yes. The internal energy of the isolated system is constant. But the internal energy of the parts of the system can change. It can change because some of that internal (thermal) energy is converted into useful mechanical energy by doing work.

In this case, you have to factor in the internal energy of the spring or battery. Some of the internal energy of the hot reservoir has been converted into useful energy stored in the spring/battery. The total internal energy of the engine and reservoirs has decreased.

The point of this exercise is to show that it is not true that every system will maximize entropy. It depends on how it is configured. In this case maximum entropy would be attained by letting the hot and cold reservoirs connect so that they reach thermal equilibrium. But the system does not permit that to occur.

AM

 

[Studiot]

dE = q + w (with appropriate sign conventions)

I'm glad we can agree on this so we can move on.

If we can agree on the next two statements we can move on another step.

E refers to energy within the system. By itself it says nothing about energy that is added to the system from somewhere else (the surroundings).

Both q and w refer to energy transferred into the system from somewhere else (the surroundings). Neither say anything about energy within the system or can be used to calculate anything about its disposition within the system.

 

[Andrew Mason]

dE = q + w (with appropriate sign conventions)

I'm glad we can agree on this so we can move on.

If we can agree on the next two statements we can move on another step.

E refers to energy within the system. By itself it says nothing about energy that is added to the system from somewhere else (the surroundings).

Ok. We know that E, the total internal energy of what is in the isolated box is constant.

Both q and w refer to energy transferred into the system from somewhere else (the surroundings). Neither say anything about energy within the system or can be used to calculate anything about its disposition within the system.

Sure it can. You just have to define a subsystem within your first system. In this case, that could be the Carnot engine. You could then analyse Q and W. Q is the heat flow into the engine and W is the work output of the engine.

AM

 

[Studiot]

Sure it can. You just have to define a subsystem within your first system.

Thank you.

That is exactly what I said in post#19 .

So does the carnot engine becomes the no longer isolated (sub) system and the rest of your original isolated system becomes the surroundings?

Well actually we have to subdivide the carnot engine still further (you have already said this), since otherwise temperature is not a well defined system property.

So the system comprising the hot reservoir is not in equilibrium with the system comprising the cold reservoir, unless they are at the same temperature.

But I specified that the original system should be in equilibrium, which it will not be until both reservoirs are at the same temperature.

I am sorry to labour these points so much but they are fundamental and sometimes minute attention to detail is necessary.

 

[Andrew Mason]

Thank you.

That is exactly what I said in post#19 .

So does the carnot engine becomes the no longer isolated (sub) system and the rest of your original isolated system becomes the surroundings?

The Carnot engine cannot be isolated from the reservoirs. It has to be in thermal contact with the reservoirs for the Carnot cycle to operate.

Well actually we have to subdivide the carnot engine still further (you have already said this), since otherwise temperature is not a well defined system property.

?? A Carnot engine is always in equilibrium. It is in thermal equilibrium with one of the reservoirs undergoing isothermal quasi-static compression or expansion or undergoing quasi-static adiabatic expansion or compression.

So the system comprising the hot reservoir is not in equilibrium with the system comprising the cold reservoir, unless they are at the same temperature.

But I specified that the original system should be in equilibrium, which it will not be until both reservoirs are at the same temperature.

The two reservoirs are in thermal equilibrium with each other only when the Carnot engine stops. At this point the reservoirs are at the same temperature and the spring/battery has maximum potential energy. Total change in entropy of the system is (arbitrarily close to) zero. The change in entropy of the universe outside the box is 0. Total change in entropy = 0

I am not sure what you are getting at here. It is a simple point. In the insulated box you have finite reservoirs at different temperatures with a Carnot engine operating between them. You have a device to store energy from the work done by the engine.

AM

 

[Studiot]

Rather than going backwards and forwards again and again perhaps yoy would post your definition of an isloated system that is in equlibrium?

Since interaction with its surroundings are not simply zero but disallowed this does not and cannot refer to equilibrium with its surroundings.

 

[Andrew Mason]

Rather than going backwards and forwards again and again perhaps yoy would post your definition of an isloated system that is in equlibrium?

Since interaction with its surroundings are not simply zero but disallowed this does not and cannot refer to equilibrium with its surroundings.

Who says the system has to be in total equilibrum?

Perhaps you could reread my posts and tell me what the difficulty is. The question is whether a system will always tend toward a state of maximum entropy. So I gave an example of a simple system that does not maximize entropy as it reaches thermal equilibrium.

AM

 

[Studiot]

Who says the system has to be in total equilibrum?

Well Andrew it seems that we have been posting at cross purposes in this discussion since my first post (#9) and what I thought was your reply (#10) where I stated conditions for an isolated system in equilibrium.

I have already noted that post#9 was poorly worded (although strictly correct)
Unfortunately the last line was the most important from the OP point of view and it seems to have been generally overlooked.

In order to calculate what happens in the intermediate stages you need to consider one of the free energy or work functions.

sugeet if you are still with us do you wish to pursue this?
I, or I'm sure Andrew, will happily explain.

 

[Andrew Mason]

Well Andrew it seems that we have been posting at cross purposes in this discussion since my first post (#9) and what I thought was your reply (#10) where I stated conditions for an isolated system in equilibrium.

I have already noted that post#9 was poorly worded (although strictly correct)
Unfortunately the last line was the most important from the OP point of view and it seems to have been generally overlooked.

I am not sure what you mean. In post #9 you said:

"A system is said to be in equilibrium when no further spontaneous change occurs. The entropy of an isolated system tends to increase until no further spontaneous change can occur.

So in an isolated system in equilibrium at constant energy and volume the entropy is a maximum."

to which I replied in post #10:

"How about an isolated system consisting of a Carnot engine, operating between hot and cold finite reservoirs, compressing a ratcheted spring. Can any further spontaneous change occur? When the temperature difference becomes arbitrarily small, the engine cannot do any more work and the spring remains compressed. Has entropy increased? Is it at a maximum?"


I simply pointed out that this is a system that is at constant volume and constant energy that is initially not all at the same temperature but which does not maximize entropy as it moves to a state of thermal equilibrium.

AM

 

[Studiot]

One thing you cannot describe a Carnot cycle engine as is a constant volume process. None of the four legs on an indicator diagram occur at constant volume. I do not know of a mechanism (if that is the right word) that is capable of extracting work with the working fluid at constant volume in a carnot engine.

There is a type of heat engine known as constant volume ( or Rochas) cycle engine that has two of the legs vertical (const vol)and two with the adiabatic gamma law expansion.

Another type with a (single) constant volume leg and a single adiabatic leg is the Lenoir gas engine.

However work is only only during the non constant volume parts of each cycle.

 

[Andrew Mason]

One thing you cannot describe a Carnot cycle engine as is a constant volume process. None of the four legs on an indicator diagram occur at constant volume. I do not know of a mechanism (if that is the right word) that is capable of extracting work with the working fluid at constant volume in a carnot engine.

There is a type of heat engine known as constant volume ( or Rochas) cycle engine that has two of the legs vertical (const vol)and two with the adiabatic gamma law expansion.

Another type with a (single) constant volume leg and a single adiabatic leg is the Lenoir gas engine.

However work is only only during the non constant volume parts of each cycle.

We are really going in circles now. Your concerns have nothing to do with whether an isolated system which is not internally in thermodynamic equilibrium will tend toward maximum entropy. I gave you a system that is not in thermodynamic equilibrium and showed you that it will end up in a final state of thermal equilibrium with zero entropy increase.

Why does the heat flow have to occur in a constant volume process? If heat flow occurs in a constant volume process, there will always be a net increase in entropy. It is only if heat flow occurs isothermally that there can be no increase in entropy. That is what occurs in a Carnot engine.

AM