what rules, energy or entropy??

Andrew Mason

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

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.

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

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

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

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

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

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

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.

?? 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.

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

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

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.

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

Andrew Mason

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

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