Thermodynamics Basics: Questions Answered

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    Basics Thermodynamics
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SUMMARY

This discussion addresses fundamental concepts in thermodynamics, including the relationship between thermodynamic and mechanical equilibrium, the necessary state variables to define a thermodynamic system, and the applicability of Boyle's Law. It confirms that thermodynamic equilibrium does not always require mechanical equilibrium, and that pressure, volume, and temperature are sufficient to specify an ideal gas. The conversation also clarifies that Boyle's Law does not necessitate quasi-static or isothermal conditions, and discusses the significance of density as a thermodynamic quantity. Additionally, it differentiates between Helmholtz and Gibbs free energy equations based on system constraints.

PREREQUISITES
  • Understanding of basic thermodynamic concepts such as equilibrium, state variables, and gas laws.
  • Familiarity with the Ideal Gas Law (PV=nRT) and its implications.
  • Knowledge of thermodynamic potentials, specifically Helmholtz and Gibbs free energy.
  • Basic grasp of continuum mechanics and the significance of density in thermodynamics.
NEXT STEPS
  • Study the derivation and applications of the Ideal Gas Law in various thermodynamic processes.
  • Explore the differences between open, closed, and isolated systems in thermodynamics.
  • Learn about the implications of quasi-static and isothermal processes in real-world applications.
  • Investigate the conditions under which Helmholtz and Gibbs free energy are utilized in thermodynamic analysis.
USEFUL FOR

Students and professionals in physics and engineering, particularly those focusing on thermodynamics, mechanical engineering, and chemical engineering. This discussion is beneficial for anyone seeking to deepen their understanding of thermodynamic principles and their applications in real-world scenarios.

  • #31
Andy Resnick said:
I have never heard this- can you provide a reference?

The way I stated this was way too strong, but for a system where there's enough energy for inelastic collisions (so particle types change and 2<->4, etc., processes also change particle number), it is unlikely for there to be any notion of conservation of particle number. It's certainly not a symmetry of our fundamental theories. There will still be associated chemical potentials for any other relevant conserved quantity though--electric charge, baryon number, etc.



Andy Resnick said:
Ah, ok. Oftentimes, thermodynamics is presented as some sort of 'average' statistical mechanics. That is, thermodynamics is based on a *mechanical* theory.

Since mechanical theories have a symplectic geometry while thermodynamics has a contact geometry, this cannot be the case.

Thanks, I think I'm starting to see how this fits together (especially after George's post, and looking at the blog post he linked to). If I understand correctly, the total number of state variables is (5+2n). I.e., there are always conjugate pairs plus one extra and this will always be odd. This is important for determining that the number of independent variables needed to specify the state is 2+n (note that my n--the number of conserved quantities--is the same as (n-2) in George's post). Note that this number of independent variables is not necessarily odd. I think there has been some confusion about this earlier in the thread.

This is made clear in the blog posting (and actually in your arXiv reference as well--see paragraph 4 on page 4). In the case of n=0, there are 5 variables (taken as U,T,S,P,V in these references), but only 2 of them are independent. Adding, e.g., a conserved particle number will add a conjugate pair N and mu (chemical potential) to the total number of variables, but only one of them is independent.


George Jones said:
The first link is to a blog posting by the author of the paper you cited. This posting works through some standard examples.

Thanks for that link. It was helpful.
 
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  • #32
Andy Resnick said:
I don't know much about that expression, but it's only valid for an ideal gas. And we already know that an ideal gas can be discussed using extremely simplified models. But ideal gases don't display every behavior possible for real gases, let alone more complicated substances like liquids and solids.

Edit: and in any case, your 'counterexample' still has an odd number of state variables, which is the real point.

Let me count: T, one, V, two. Thats an even number.
Btw, similar equations for U and S are available for non-ideal gasses, e.g. the van der Waals gas:
http://en.wikipedia.org/wiki/Van_der_Waals_equation. The equations for U and S only depend on V and T, too, as in the case of the ideal gas.
 
  • #33
Hi everyone!

I'm not sure if this is appropiate,especially with the debate that's going on about one of my questions regarding the number of state variables needed to define the system (I'm waiting for the debate to come to an end so that I get my final answer), however,my teacher taught something really interesting and I'm really curious to know about it.

Here's what it is:
Well,the Steady Flow Energy Equation (SFEE) basically says that when heat is supplied to to a thermodynamic sysytem or work is done on it,the energy so gained may be stored in the form of internal energy or pdv work( completely thermodynamic properties) or in the form of macroscopic enrgies(like kinetic energy of the fluid as a whole or its potential energy)...now whenever heat is provided to a system,it is supposed to supply kinetic energy to the molecules/atoms thereby increasing the temperature of the fluid.

However,according to the SFEE,supplying heat to the fluid in this way could also give kinetic energy to the fluid as a whole too! How is this possible?

Similarly,I can't see how the heat supplied to a system could provide potential energy to the entire fluid...I mean heat and flow work(pdV) are supposed to affect the molecules/atoms within the system,not the fluid as a whole!

This might have some relation with macroscopic and microscopic internal energies,but I'm not sure.

Also,another quick question: When we measure Cp(specific heat at constant pressure),is the process isothermal as well as isobaric?
 
  • #34
When you heat something up it expands.

This expansion can do work and/or result in an increase in potential energy.

But you can't use all the input heat this way.

I'm sure you can think of lots of ways this might happen.
 
  • #35
Studiot said:
I'm sure you can think of lots of ways this might happen.

No,I mean suppose we heat something up,suppose water...I could easily visualise the heat making the water boil...this is a result of increasing the internal kinetic energy...but I couldn't imagine the water starting to flow,like in a pipeway because of the heat!

P.S I found something called the two property rule which says that the state of a thermodynamic system in equilibrium can be completely defined by 2 variables.
 
  • #36
Urmi Roy said:
2. How many state variables (Voume,pressure,temperature etc.) are needed to specify a thermodynamic system?
Andy Resnick said:
Please see, for example,

http://arxiv.org/abs/math-ph/0703061

For a single phase, the 5 variables are pressure/volume, temperature/entropy, and U.
DrDu said:
Andy, one counterexample may be sufficient. For an ideal gas, U is a linear function of T only and S can be expressed in terms of T and V (Sackur Tetrode equation). So e.g. T and V are sufficient to specify the state in all respects.
Urmi Roy said:
I found something called the two property rule which says that the state of a thermodynamic system in equilibrium can be completely defined by 2 variables.

Urmi Roy, there is an ambiguity in question 2. from your first post. Did you mean:

2. How many state variables (Voume,pressure,temperature etc.) are needed to specify an arbitrary (possibly non-equilibrium) state (at a fixed time) of a thermodynamic system?

If you meant this, then I think the answer is 5, as Andy has said.

Or did you mean

2. How many state variables (Voume,pressure,temperature etc.) are needed to specify an equilibrium state of a thermodynamic system?

If you meant this, then I think the answer is 2, as DrDu and the_house have said. So, I think Andy and DrDu have been answering different questions.

Things are still fuzzy for me, but here is what I think is going on. It takes 5 variables, pressure, volume, temperature, entropy, and U, (in the simple, but general systems we are considering) to pin down an arbitrary state. Since Q = TdS for quasi-static processes, the (hyper)surface on which dU = TdS - PdV holds is the set of all equilibrium states. I think it can be shown that this surface (in our case) is 2-dimensional. To find a specific 2-dimensional manifold of equilibrium states, an equation of state is used. The equation of state together with dU = TdS - PdV is used to generate two Maxwell(-like) relations. The equation of state together with the two Maxwell relations act as three equations of constraint that restrict the 5-dimensional manifold of arbitrary states to the 2-dimensional manifold of equilibrium states.
 
  • #37
George Jones said:
there is an ambiguity in question 2...
did you mean

2. How many state variables (Voume,pressure,temperature etc.) are needed to specify an equilibrium state of a thermodynamic system?

Oops! Sorry for the ambiguity...I didn't realize the gravity of my mistake.

Yes,I did mean the number of state variables needed to define an equilibrium state,as that's all we've been taught to think about yet!
So I guess that answer has been given by DrDu.However,I think I gained a lot in looking through their arguments...It's interesting to know that even a system in a non-equilibrium state can be defined by a fixed number of variables i.e 5.
 
  • #38
Urmi Roy said:
It's interesting to know that even a system in a non-equilibrium state can be defined by a fixed number of variables i.e 5.

Don't get too attached to that statement. I admit I don't know too much about non-equilibrium thermodynamics, but I don't think it's accurate. George himself admitted things are still "fuzzy" for him, so unless he wants to give a reference that describes the out-of-equilibrium part of this 5 dimensional state space, I would take it with a grain of salt. I'm not convinced it even has any meaning, let alone that it uniquely specifies a "thermodynamic system" (whatever that means when the system is far from equilibrium).

Anyway, it does seem we have agreement on the equilibrium case. There are only 2 independent variables that uniquely specify an equilibrium system (plus another for each conserved quantity, but in the simplest case we can ignore that.)
 
  • #39
the_house said:
Don't get too attached to that statement. I admit I don't know too much about non-equilibrium thermodynamics, but I don't think it's accurate. George himself admitted things are still "fuzzy" for him, so unless he wants to give a reference that describes the out-of-equilibrium part of this 5 dimensional state space, I would take it with a grain of salt. I'm not convinced it even has any meaning, let alone that it uniquely specifies a "thermodynamic system" (whatever that means when the system is far from equilibrium).

Anyway, it does seem we have agreement on the equilibrium case. There are only 2 independent variables that uniquely specify an equilibrium system (plus another for each conserved quantity, but in the simplest case we can ignore that.)

Yeah...things do still seem a bit fuzzy,but I got a concrete answer to what I was actually looking for!

Sorry if my questions seem never-ending,but could you people please look into what I put forward in the first post of this page too(and the rest of my brief converstaion with Studiot about it),please? It's something that has got me thinking!
(It's about the SFEE).
 
  • #40
the_house said:
Don't get too attached to that statement. I admit I don't know too much about non-equilibrium thermodynamics, but I don't think it's accurate. George himself admitted things are still "fuzzy" for him, so unless he wants to give a reference that describes the out-of-equilibrium part of this 5 dimensional state space, I would take it with a grain of salt. I'm not convinced it even has any meaning, let alone that it uniquely specifies a "thermodynamic system" (whatever that means when the system is far from equilibrium).

Yes, I see what you mean. I am struggling to find, in this context, a question that has answer "5." Clearly, such a question should exist. I'll continue on with my method of successive approximations :smile:, and modify what I wrote in my previous post.

How many state variables are needed to accommodate (in one space) all possible equilibrium states of all thermodynamic systems?

It takes 5 variables, pressure, volume, temperature, entropy, and U, to accommodate (in one space) all possible equilibrium states of all thermodynamic systems. The relations 0 =dU - TdS + PdV shows that set of equilibrium states for anyone thermodynamic system system forms a 2-dimension surface in the 5-dimensional space. To find a specific 2-dimensional manifold of equilibrium states, an equation of state for a specific system is used. The equation of state together with dU = TdS - PdV is used to generate two Maxwell(-like) relations. The system's equation of state together with the two Maxwell relations act as three equations of constraint that restrict the 5-dimensional manifold of equilibrium states for all arbitrary systems to the 2-dimensional manifold of equilibrium state for the specific system.
 
  • #41
George Jones said:
The relations 0 =dU - TdS + PdV shows that set of equilibrium states for anyone thermodynamic system system forms a 2-dimension surface in the 5-dimensional space. To find a specific 2-dimensional manifold of equilibrium states, an equation of state for a specific system is used. The equation of state together with dU = TdS - PdV is used to generate two Maxwell(-like) relations. The system's equation of state together with the two Maxwell relations act as three equations of constraint that restrict the 5-dimensional manifold of equilibrium states for all arbitrary systems to the 2-dimensional manifold of equilibrium state for the specific system.

Sounds good to me. I think we're making progress.

Sorry, Urmi, I don't think I have anything intelligent to add about your other question, but don't let that stop you from asking more of them.
 
  • #42
I want to add two points. 1. even for an equilibrium system, there may be more variables to specify the state than two. Two is probably the minimum for most realistic systems.
2. I don't think that any non-equilibrium state can be specified completely by the five variables listed by Andy. Think of a heat driven turbulent flow. You need a whole fields of variables to define the state.
 
  • #43
DrDu said:
even for an equilibrium system, there may be more variables to specify the state than two. Two is probably the minimum for most realistic systems.

Exactly. At the risk of sounding repetitive, the number is 2+n, where n is the number of relevant conservation laws for the system in question. 2 is the minimum and the simplest case, but there are many systems that require the treatment of various chemical potentials.
 
  • #44
No,I mean suppose we heat something up,suppose water...I could easily visualise the heat making the water boil...this is a result of increasing the internal kinetic energy...but I couldn't imagine the water starting to flow,like in a pipeway because of the heat!

All that elasticity theory we did last term?

Let the expansion work against some sort of spring. Energy will be stored in the spring.

Let one side of the expanding object push against a fixed object (the ground?).
The expanding object will expand and push itself bodily upwards, gaining potential energy against gravity.
 
  • #45
DrDu said:
2. I don't think that any non-equilibrium state can be specified completely by the five variables listed by Andy. Think of a heat driven turbulent flow. You need a whole fields of variables to define the state.

I agree it's not '5'. However, the total number required is an *odd* number.
 
  • #46
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  • #47
Studiot said:
All that elasticity theory we did last term?

Let the expansion work against some sort of spring. Energy will be stored in the spring.

:blushing: I'm sorry if I'm missing something really obvious!

Okay,so I see what you mean...we have a container full of an ideal gas and one side of it is movable,and attached to a spring...now we heat the gas,and it makes the container's side move and compress the spring...in that way,we sort of see the gas flowing...but we calculate this work as the PdV work(flow work),don't we? Not as the (m(v squared))/2!.

...again,suppose we have a container in which,for the sake of simplicity,the top is open...so now when we heat the gas,it flows upward...we could say the gas molecules acquire potential energy from the heat we supplied it(actually it's because we normally assume the ideal gas molecules to be point masses,really light, that all my confusions arise).
 
  • #48
However,according to the SFEE,supplying heat to the fluid in this way could also give kinetic energy to the fluid as a whole too! How is this possible?

I was just trying to answer this question.

If you heat the water in the boiler of a steam locomotive some of that heat is used to impart velocity to the water as it travels along with the train.
 
  • #49
Studiot said:
I was just trying to answer this question.
If you heat the water in the boiler of a steam locomotive some of that heat is used to impart velocity to the water as it travels along with the train.

You mean as in overcoming the inertia? That's an interesting point! But then,wouldn't it be analogous to a train carrying a big box(or something of that kind) just as heavy as the water...
I think I still need an explanation to help me understand what's going on in this example :blushing:

EDIT: I think I've found another example in which heat energy could be converted to the m(vsquared)/2 i.e kinetic energy of the liquid as a whole...in convection!
 
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  • #50
Hi,
I've run into something else...
We have a heat engine that is operated between two energy reservoirs which have finite heat capacities,my book says the work would continue to be produced till the temperatures of the two reservoirs became the same...but the two reservoirs are not directly in contact with each other...then why do they wait for the entire heat engine to come to a particular temperature?
 
  • #51
This time I'm really really really confused...
Here's what it is..please help...

The clausius inequality says that for an irreversible process,(sum of Q/T)<del S...where S is the entropy change of a reversible process acting between the same two points on the T -S diagram...however,the (sum of Q/T)is basically the entropy change of the irreversible process(acting between two specific points on T-S diagram)...so shouldn't that be greater than the S (entropy change in the reversible process acting between the same two pints on the T-S diagram)?

On the other hand,entropy is a point function...so the (sum of Q/T) for irreversible process which is basically the entropy change for the irreversible process should be equal to the entropy change delS of the reversible process acting between the same two points on the T-S diagram...
 

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