# Few Fundamental Thermodynamics Questions

1. Aug 1, 2012

### Red_CCF

Hi

I had a few thermodynamics questions on a thread that was locked. The link is here and my questions are directed at the last post.

1. Does the author mean that all quasistatic irreversble processes are reversible or if for a quasistatic irreversible process that goes from state 1 to 2, there is actually another way to get from state 1 to 2 reversibly? Is there some kind of proof that this generalization is true as I can't find it in any books?

2. With regards to the proof for work for non-quasistatic systems, does the proof apply to quasistatic irreversible process as well? On page 17 of Anderson's modern compressible flow, he states that only for reversible process does W = integral of pdV. In addition, I thought that equation (5) that T dS >= dQ used to prove W < integral pdv also applies for any irreversible process quasistatic or not?

3. I'm a bit confused here because I always thought that internal energy was a measure of all the microscopic energies which includes kinetic energy (and that kinetic energy directly correlates to the temperature of gas)?

4. I was also wondering for a real gas (intermolecular forces important) or a chemically reacting system, why is internal energy also a function of specific volume and enthalpy a function of pressure while this isn't the case for ideal gas and non-chemical reacting system?

5. How are irreversibilities accounted for in terms of energy balance for closed or open systems? For a closed and insulated piston cylinder and W < integral PdV, does energy balance become something like:

W = U2 - U1 = integral PdV - loss​

So for closed systems do losses get accounted for in the internal energy of the final state and for steady state open systems does losses get accounted for in the enthalpy terms (assuming adiabatic processes)?

6. Is there an example that shows how heat is process dependent, in all the books I looked through they only give examples for work but just simply state this for heat without examples?

7. What is the difference between equilibrium and steady state in a general thermodynamics sense?

Thanks very much

2. Aug 5, 2012

### Simon Bridge

1. Quasistatic irreversible processes can always be realized in a completely reversible way.
... that would be: that there is a reversible process to get you between any two states. For proof - just see if you can find two states that can only be moved between irreversibly. Put simply: there is more than one way to skin a cat.

2. I think you need to be more careful about what people are saying when they write stuff in text books. Work is not always pV work.

3. it's badly explained - in incremental free expansion described the gas also gains a bulk KE. It's the randomized KE that relates to temperature.

4. because the ideal gas is a simplified model and the real gasses have to obey real physics.

5. the irreversable processes are accounted in all cases by the statistics of the setup.

6. The questions does not mean anything: heat, for example, is another name for energy. Similarly: work is not process dependent - thermodynamic work is a process. Work is not always path dependent either. eg. for a reversible adiabatic process.

7. same as in any sense - one can oscillate (for eg.) about an equilibrium but must remain at a steady state.

3. Aug 11, 2012

### Red_CCF

Hi

Thanks very much for the response.

With regards to work I should have said thermodynamic work to make things more clear. So with regards to my original question, does W = ∫pdV only apply for quasi-static reversible processes only and not for quasi-static irreversible as the original author had stated since TdS > δQ?

I'm not really familiar with gas interactions at a molecular level, is there a simple explanation on what real world effects that pressure and specific volume cause or reflect which in turn affect internal energy?

So was my original reasoning correct that all irreversibility and loses are accounted for by one of the variables in the energy balance (either ΔQ, ΔU, ΔH etc.)?

With regards to the path dependent nature of thermodynamic work and heat (in this case I'm more referring to the heat as a process of adding energy to a system or δQ as in the first law) between two states, since thermodynamic work is not path dependent for reversible adiabatic process, does this mean that heat is also not path dependent for a reversible process with no work input?

Is the path dependent nature of thermodynamic work and heat between the two states solely because there are two "source" of energy (work and heat for closed system) and one can play with different ratios of the two to get the same energy change of the system?

Thanks very much

4. Aug 11, 2012

### Simon Bridge

The process of adding energy to a system is called "doing work on the system".

Work is change of energy.

Ah - you are thinking of the process of adding heat - called "heating" (or "cooling") the system - as in an isochoric process.

But if no pV work is done isn't the path fixed?

Now I'm with you - you can put a container on a hot source, or you can do mechanical work on it, both will increase the internal energy.

5. Aug 12, 2012

### Simon Bridge

... talks in a simpler way about different irreversible processes in thermodynamics.
It should help with the earlier question - I'm sorry, it's 30mins of dull talk, however the lecturer has some good analogies and it's those that I think will help. He also deals with the processes in a very general way that makes many of the connections you are trying for.

I've been trying to find a treatment of an isenthalpic process as being particulaly relevant to your questions ... this had dQ=0 but TdS>0, it's irrversible, and iirc it follows an isotherm on the PV diagram [dh=0=cpdT]... which means that W≠∫pdV in this case. You have to analyze it with a T-s diagram.

For the kind of things that real-gas models have to take into account, wikipedia has a nice summary. From there you should be able to see how the extra terms you ask about get to be included.

The question you have asked are a tad general for quickie pat answers - you have more reading to do. The above should point you in fruitful directions.

Last edited by a moderator: Sep 25, 2014
6. Aug 22, 2012

### Red_CCF

Hi

Thank you very much for the link, hopefully it'll be able to answer some of my questions.

In the meantime, I was reviewing Moran and Shapiro's fundamental of thermodynamics, and had a question about the state principle for simple systems. I attached the segment of the book that talked about it.

I was just confused about their explanation of the state principle. To me it seems to say that if I have a rigid tank and I start heating the system from state 1 to state 2, only one state property is required to define all other state properties as no work is being done. However, I am confused as if for example the container has ideal gas, the ideal gas equation still dictates that two state variables be known to find another regardless of the system?

Also, does this principle apply to specific internal energy and enthalpy of gases given that if a gas is calorically or thermally perfect h and u are functions of temperature only, and also in Anderson's Modern Compressible Flow, these two properties are said to depend on chemical reactions and intermolecular forces, with no mention of the state principle? Does h and u depend on more than two variables for a non-simple system and what if the gas is thermally or calorically perfect?

Returning to my question about equilibrium, in Anderson's Modern Compressible Flow, he state that

"equilibrium is evidenced by no gradients in velocity, pressure, temperature, and chemical concentration throughout the system i.e. the system has uniform properties."​

To me this statement implies that some rigid insulated tank that contains Gas A would be considered in equilibrium, but if I have another insulated rigid tank of gas (Gas B) at a different temperature and pressure and I put the two tanks side by side and I consider both tanks as the system, the system is now not at equilibrium even though the components of the system is at equilibrium?

Thanks very much

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7. Aug 22, 2012

### Simon Bridge

This is correct - if the container is rigid.
This is also correct - but recall that, in this case, the container is rigid. This means that the volume cannot change.
Note that the state principle described is for simple systems ... you are describing complicated systems so what do you think the answer is?

Also - consider the stress on "independence" in the description of the state principle.
No. If the two tanks are thermodynamically isolated, then each is a separate thermodynamic system. It is an error to consider them as a combined system in this context.

8. Aug 22, 2012

### Studiot

That is one view, but I see no reason for not considering them jointly as a single thermodynamic system.

In that event RED is correct the system is not in stable equilibrium.

Here is another example from chemistry.

Some paper is sitting on my desk in an atmosphere containing oxygen.

Is it in equilibrium?

No because if I apply a match it will catch fire and continue to burn spontaneously after removal of my flame.

That is because the final burnt state has lower energy and higher entropy than the initial state.

However I need to add the activation energy before anything happens.

It is the same with the two tanks of gas - there is an energy and entropy available from their mixing (assuming no chemical reaction), its just that the activation energy of breaching the tank walls is required as a starter.

Last edited: Aug 22, 2012
9. Aug 22, 2012

### Simon Bridge

The fact they cannot interact in any way does not deter you? Interesting...
But in this example, the paper and air are not thermodynamically isolated.

Yes - you have to bring them into contact.

OP is talking about the definition of "equilibrium" in thermodynamics. In the real world - nothing is in stable equilibrium ... ever. But that does not affect the definition of the equilibrium.

OPs example cites a "system" where each component is in thermodynamic equilibrium and wants to say that the differences (temperature say) means that the system is, by definition, not in equilibrium.

Since the systems are not changing, they are not affecting each other. Ergo, the system is in equilibrium.

In terms of the text book extract provided, this would be understood as two separate thermodynamic systems in the abstract sense.

10. Aug 22, 2012

### Studiot

No it doesn't there exists a thermodynamic potential for the process to take place which is not altered by the fact that there is a (temporary) barrier in place. The isolation between the tanks is mechanical, not thermodynamic.

In thermodynamics there are several types of equilibrium, as with any other branch of physics.

Note I said the tanks or paper are not in stable equilibrium.

They are technically in metastable equilibrium, but I'm sure you appreciate this.

Last edited: Aug 22, 2012
11. Aug 22, 2012

### Simon Bridge

Ahhh I see - but you were responding to my statement: If the two tanks are thermodynamically isolated, then each is a separate thermodynamic system.
... I wrote that as an attempt to make a clear distinction between thermodynamically ideal situations used in the models and real-world situations. Notice the "if".
In that statement I was specifying "thermodynamically isolated" as a model - not as an approximation or anything. I'll concede that there is no such absolute in the real world.

In the real world there is no such thing as a permanently stable equilibrium that could apply to the description ... things will always fall apart eventually. Do you really think this is what OP had in mind when he asked the question?

While there are, indeed, many different types of equilibrium, OP was asking about a particular definition of equilibrium that had been found. It was to this definition that I offered my reply. It is the kind of thing that occurs near the start of a thermodynamics course alongside other idealizations like "diathermal wall" end suchlike.

So we are agreeing in general - I'm replying from an idealized model, you from real life.
Lets allow OP to tell us which was relevant to the question?

12. Aug 22, 2012

### Studiot

Simon, if you are familiar with chemical kinetics you will understand that thermodynamics tells us the answer to the question

'Can something happen spontaneously?'

It does not usually answer the questions 'will it happen spontaneously?' or 'How fast does it happen?'

The description of equilibrium as the condition of no change is of little value in applying the classical equations of thermodynamics since they rely heavily on reversible processes, being those that are in equilibrium.

Taken literally such a description would imply that no process can ever take place!

A better description of equilibrium is the idea of what happen to an infinitesimal displacement from equilibrium.

In stable equilibrium the system restores itself whatever the direction of the displacement.

In metastable equilibrium restoration depends upon the direction.

In the attached sketches a small push to the left or right return the ball in stable equilibrium, but only to the right in metastable equilibrium.

This allows us to introduce the idea of a quasi static or reversible process.

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13. Aug 22, 2012

### Simon Bridge

Yes it is.

In fact this allows for the idea to be extended easily to a dynamical equilibrium.
(Note: your examples are also in the lecture I linked to earlier.)

The book is using poor pedagogy.
Just reading through the few bits shown to us here you can feel your brains congeal.
We must be able to point OP at an introductory lecture course or something better than that book.

14. Aug 24, 2012

### Red_CCF

Hello

Thanks very much for the responses. When I posed the question about equilibrium, I was not thinking about the real world cases Studiot had pointed out but I can see how they are important. However, I don't think that I'm at a stage where I can understand these complexities; I'm more focused on getting the fundamentals from an idealized perspective first before considering real-world cases.

The aim of my original question was just trying to understand what is equilibrium and the difference between equilibrium and steady state. Based on the definition in Anderson, it seems that if I have a turbine whereby there would be a gradient of at least one state property, the turbine would not be considered at equilibrium but at steady state? Also, by this definition a system in equilibrium would have to have no gradients in entropy as well so the inlet and exit of a control volume must have the same entropy?

If these cases are true, I'm now confused about what constitutes a thermodynamic state. In the Moran and Shapiro book I am reading they introduce a thermodynamic state as a condition where the system is at equilibrium, such as the end states from pushing a piston from state 1 to 2 or the intermediate states of a quasistatic process. However, for a slightly more complex system like a turbine at steady state which will probably have a gradient of some sorts, can the system even have a state as constituted by the term steady state?

With regards to my question about specific internal energy and enthalpy, I'm still confused about whether the state principle dictates that they must be functions of 2 state variables and why they are only functions of one for certain idealizations?

Lastly, the insulated rigid tank mentioned in my earlier post, although volume is constant, can the specific volume change? For instance, if I add in more gas to this tank or have another identical tank with more gas in mass , would these constitute the same system or be considered different systems?

Thanks very much

15. Aug 24, 2012

### Studiot

'Steady State' usually refers to what are known as the flow equations, for example in your turbine.

So fluid is constantly entering and leaving the turbine, carrying with it physical quantities such as energy, entropy, mass, momentum etc.

We say that the system is in a steady state when there is no change to these fluxes, ie the difference between (X) in and (X) out is a constant.

However the process undergone within the turbine may be reversible or irreversible.

So a constant mass of fluid may be flowing through the turbine and undergoing irreversible ( or irreversible) expansion, loosing some heat in the process.
The rate of work output from the turbine and the rate of heat lost will also be constant if the turbine is in a steady state.

However the process is only an equilibrium one if the process is reversible.

Last edited: Aug 24, 2012
16. Aug 24, 2012

### Simon Bridge

To add to what Studiot said - with our usual difference in perspective ;)
You are not the only one and some authors do not make a clear distinction. For example, in the lecture I gave you (have you watched it yet) the lecturer describes the state of a ball rolling down a hill as an equilibrium.

Usually we would think of an equilibrium as being a state in which a system naturally remains. A weight hanging from the end of a string has an equilibrium point that is easy to picture for example: where the weight is directly below the hook.

A steady state of the weight-on-string thing would be when it is hanging at equilibrium - but also when it is swinging with a constant amplitude and period.

So to say that something is in a steady state has a bit of leeway.
If it were turning steadily - if the turning were fluctuating then it would neither be in steady state nor equilibrium.
I'll defer to studiots answer here ;)
It is any set of values that completely describes the system.
That would not agree with my understanding - a system may pass through thermodynamic states. But they may be thinking of the system being in thermodynamic equilibrium with itself just to be pedantic. After all, when you do something like push in a piston, not all the gas changes temperature at the same time - there would be an initial gradient that settles down in a while. Similarly if you heat something - you may join a heat source via a diathermal wall ... again, not all the gas under study will heat to the same temperature at the same time: there will be convection etc. For the entire bottle of gas (or whatever) to be considered to be in a well defined state, then there should be no internal fluctuations.
In the example you gave, there are only two parameters that vary - count them - ergo, knowing one, and the relationship between them, you know everything.

The simple model you are being treated to at this early stage only admits the three variables - and only works for specific types of systems - there are other models that would apply to the other examples. Try not to get ahead of yourself. No - that's not right - rather: try to be aware as you read that you are not being supplied with everything in one go: the authors are providing stepping-stones to more complicated methods.

You do risk overreaching yourself by trying for more complex models before you understand the core concepts. Make sure you can do the problems in the chapters you have read.

17. Aug 24, 2012

### Studiot

18. Aug 27, 2012

### Red_CCF

Hi

Thanks very much for all the help so far.

I get equilibrium from a mechanical sense (ball in a bowl, pendulum hanging) but I'm having trouble with equilibrium in a thermal sense. I'm wondering if my understanding right now is correct.

Steady state is just that the derivative of each property in the system with time is 0 and can apply to essentially any system. Would a stable thermal equilibrium would be like an non-insulated tank that is at the same temperature as the surroundings while a unstable thermal equilibrium would be this tank maintained at a higher temperature than surrounding by a constant heat addition?

With regards to a pendulum swinging periodically or ball rolling down a hill, how come these systems are considered at steady state since as their velocity and elevation (or kinetic and potential energies) are functions of time?

I attached the excerpt from the book just so you can take a look and make sure I did not mis-interpret their words. This was introduced early on and they did distinguish that quasistatic process is an idealization that real world applications can only approximate. I believe that they see a quasistatic process as being so slow that the gradients you mention take an infinitesimal amount of time to disappear. In a separate section they state that the end state of any process must be in equilibrium as well to be considered a "state" as is any intermediate state of the quasistatic process.

Lastly, does this mean that an open systems (not insulated so inlet and exit entropies are not the same) at steady state even in ideal scenarios (no irreversibility) would not be in equilibrium and thus does not have a state?

I was actually referring to any general simple system not the example I mentioned earlier. So h = h (T,p) and u = u (T, v) or any two state properties is only the case when the system is simple and real gas is considered? Why does the assumption of calorically or thermally perfect gas make h = h (T) and u = u (T)? I originally thought that p and v become purely functions of T but I remembered that the ideal gas law would prohibit this.

Are the three variables you mention temperature, pressure, and specific volume and why does the simple compressible model only "admit" these variables and not other state properties? In both books I referenced above, the entire book essentially assumes simple compressible model for all examples (as they define a state variable to be a function of at most 2 other state variables) which they use later in the book for complex systems such as turbines and engines (although they are broken down into individual stages). What kind of system would be classified as a non-simple compressible system such that more than 2 state variables is needed to define another variable? I'm imaging that a non-insulated turbine doing shaft and constant pressure expansion work to be an example?

Unfortunately most book problems are application based, where solving them is essentially defining the system, stating assumptions, and figuring out which equations to use which does not really help me with the theory aspects. I have already taken a thermodynamics course and did well in it but I have a significant gap in my understanding of basic concepts mentioned above. The video you recommended was somewhat confusing at times as I was never introduced the concepts of Gibbs free energy (which may be due to the fact that I took an engineering thermo course) so I was lost when he was describing steady state as a condition where free energy is dissipated (ball down a hill, heat transfer via temperature gradient).

Thanks very much

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19. Aug 28, 2012

### Studiot

You are right it would be worth you while to review some fundamentals.

I have three gold blocks A, B & C on my desk, sitting quietly side by side.

As such they are in mechanical equilibrium.
Since they are touching and no heat is transferred between them they are said to be in thermal equilibrium. (note I say thermal equilibrium, not thermodynamic equilibrium).
This exemplifies what is sometimes called the zeroth law of thermodynamics.

Two bodies in thermal contact are at the same temperature if no heat passes between them.
Further if B is at the same temperature as A and at the same temperature as C then A is at the same temperature as C.

Seems obvious, but it is surprising how often we use these.

Final point from this. One body is not in ‘thermal equilibrium’ – it takes at least two bodies.

Now let me put these blocks in my pocket take the lift (elevator) up to the 10th floor.

The blocks are no longer in mechanical equilibrium since they are being accelerated by the lift.

But they are still in thermal equilibrium since they are all still at the same temperature as each other.

Now look at the container of fluid.

Before I light the burner, the container and the fluid are in (thermal) equilibrium with the surroundings.
The temperature of the fluid is steady
No work is being done.
Neither the volume nor the pressure are changing. They are steady.

The system (fluid) is in a steady state.

When I light the burner and begin the add heat, the average temperature in the fluid rises and the average pressure increases.
However I can no longer talk of the temperature or the pressure of the fluid as both vary from top to bottom and the fluid churns over.

As the pressure increases this means that the average internal pressure exceeds the external pressure (plus the piston weight) so the piston rises and work is done against the external pressure.

Further since the container is not insulated, heat is lost to the surroundings.
Another way to say this is.
The fluid and its surroundings (remember I said you need two bodies) are at different temperatures and no longer in thermal equilibrium and , by definition heat is transferred to the surroundings.

So whilst the temperature and piston are rising the system is not in equilibrium.

Nor is the system in a steady state.

If I keep the burner low enough eventually a form of dynamic equilibrium is reached where the heat from the burner exactly equals the heat lost to the surroundings.

The system pressure and temperature stabilise at a single value and the piston stops rising so the volume stops changing and no work is now being done.

The system is again in a steady state.

But the system is still not in thermal equilibrium with its surroundings because its temperature is different from that of its surroundings.

There is much more to be learned from these simple examples if you like the story so far.

This would include the difference between dynamic equilibrium and steady state and how we can sidestep the reversibility/irreversibilty problem.

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20. Aug 28, 2012

### Simon Bridge

You are still over-thinking this: one state variable can be expressed as a function on just one other state variable when all other state variables are held constant.

In the example, volume is held constant. (Rigid walled container.) By the Idea gas law, then, P=nRT/V - since nr/T is a constant, P is a function of T alone. Similarly, T is a function of P alone.

I can also put a non-rigid container in a large heat bath and make only small/slow changes allowing the gas to reach thermal equilibrium so it maintains a constant temperature. In which case, the pressure is a function of volume alone.

By definition - it is a model - not reality. When we make models we can admit or exclude anything we like: they are just made up. It is common to make simplifications in our mathematical models to help us understand quite complicated processes in real life.

You'll meet those later - a reasonable one would be any system where a chemical reaction is possible, where a liquid or a phase change is involved, where material can be added or removed as part of the process.

Aimed at engineers then?
Then you did not do well at it - despite decent grades (education vs schooling) ;) but maybe the course emphasis was not in the direction of underlying theory?
There are earlier lectures in the series.

Gibbs Free Energy is part of a more complete model that allows for material to be move in and out of the "container". I gave you the lecture to give you a better idea about "equilibrium" concepts.

To get at the core concepts, you should start with a year 1 college physics course, you know: one aimed at physics students. You'll find a lot of it familiar and it will allow you to make links between the course you did well at and the physics concepts you want to know more about. That will put you on a more stable footing for further exploration.

If you still want to see the fundamental ideas that are thought to underlie all this, you want to read around "statistical mechanics". Personally I didn't get it until post-grad quantum mechanics - but I'm slow - so be patient: it is a big, deep, hard, subject.

The texts just want you to get a basic idea about thermal equilibrium so you can use it to help understand the zeroth law of thermodynamics, temperature, and the concept of a "thermodynamic state". Later, these concepts are used for other things as you have seen. The author is just trying to be careful with terminology. I don't think you need to worry about it so much - you'll "get it" in time.