Thermodynamics Basics: Questions Answered

[Urmi Roy]

Hi!

I just need some basic things cleared and I would really appreciate if someone helped me out.

Here are a few questions that put forward my doubts:

1. Thermodynamic equilibrium requires mechanical equilibrium--please explain.

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

3. In order for Boyle's law to be applicable to a process,does the process need to be quasi static and (or) isothermal?(Referring to this,the main thing is--is an isothermal process always quasi static?)

4.What kind of factors does continuum volume (the minimum volume needed to obtain a continuum in the system) depend upon?


5. why is density a thermodynamic quantity?(It says so in my book).

6. What is the significance of classifying systems into 'open','closed' and 'isolated' systems?(I mean in real life,do we consider them while performing adiabatic/isothermal/isochoric processes?)

7. Lastly,what is the difference between the Helmoltz equation and gibb's equation(just basic idea required).

Thanks.
(Btw,there are a lot of questions,but I guess they're short ones!)

 

[mikelepore]

5. why is density a thermodynamic quantity?(It says so in my book).

Notice that the ideal gas law PV=nRT can also be written in the form PM=dRT, where M is molar mass, d is density. Derivation: PV=nRT, substitute n=m/M (number of moles = sample mass / molar mass), PV=(m/M)RT, rewrite as PM=(m/V)RT, density d = m/V, PM=dRT.

 

[Urmi Roy]

Does that mean that density is a thermodynamic quantity because the pressure or temperature changes cause an overall density change of the gas?

 

[mikelepore]

I would say yes to your question. Volume is a thermodynamic variable, and when the volume changes the density changes, therefore density is a thermodynamic variable also.

 

[Urmi Roy]

Thanks!
Could you answer some of my other questions too,please?

 

[mikelepore]

question 3: Boyle's law just requires that you are comparing two instants in time when the temperature T and the amount of substance n are the same in both cases: P1 V1 = nRT = some constant, P2 V2 = nRT = the same constant, P1 V1 = P2 V2. There is no dependency on the kind of process. The main problems that cause Boyle's Law to fail are when the pressure or density get too high (because electrical intermolecular forces may become significant) or when the temperature gets too low (because a real gas can liquify). Otherwise, the process doesn't affect the correctness of the equation.

 

[mikelepore]

Your other questions, I'm afraid I might answer them incorrectly. There may be someone else here who has more certainty.

 

[my_wan]

1. Thermodynamic equilibrium requires mechanical equilibrium--please explain.

Consider a pressurized tank of air (gas). Now this tank of gas is the same temperature as outside air, so if it was in equilibrium it wouldn't leak under pressure, unless it's not only in thermal equilibrium, but also in mechanical equilibrium. At a fundamental level there's not much difference. Open the valve till all the air leaks out, then heat it and more air leaks out under pressure.

The pressurized tank is not in equilibrium because the gas is denser, due to more mechanical impacts of the molecules. Once in mechanical equilibrium, then heating the tank adds velocity to the gas particles. Now the gas molecules are not in equilibrium, not because there is more, but because they are hitting harder and faster than the molecules outside.

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

That depends on exactly what you want to specify. Pressure, volume, and temperature is enough for an ideal gas. Temperature is basically the average kinetic energy, 1/2Mv^2, of the molecules. Changing density also changes volume, and temperature due to the kinetic energy of the particles being closer together, thus more energy per volume.

3. In order for Boyle's law to be applicable to a process,does the process need to be quasi static and (or) isothermal?(Referring to this,the main thing is--is an isothermal process always quasi static?)

Yes, the temperature is what's defined to stay static in Boyle's law, but the pressure and volume are inversely related. So Boyle's law does not require constant pressure and volume, only temperature, but the ratio of pressure and volume will stay constant though.

4.What kind of factors does continuum volume (the minimum volume needed to obtain a continuum in the system) depend upon?

This is a more difficult question. It really doesn't make much sense to talk about the temperature of a single particle, though you can define its kinetic energy. The volume doesn't really matter either, but the fewer particles there are the bigger the volume you need to average over to make sense. So primarily it depends on an area big enough so that averaging the kinetic energy of the molecules in it will give you a true average. It's like asking what the average shoe size is. You can't learn that from 1 shoe.

5. why is density a thermodynamic quantity?(It says so in my book).

For the same reason I described in question 1. If you compress more gas closer together, then the average kinetic energy per unit volume increases, because the kinetic energy of the molecules are closer together. Increasing temperature also increases the average kinetic energy per unit volume, not by more particles with kinetic energy closer together, but by adding more kinetic energy per particle.

6. What is the significance of classifying systems into 'open','closed' and 'isolated' systems?(I mean in real life,do we consider them while performing adiabatic/isothermal/isochoric processes?)

An open and isolated system are opposites. A closed system is a bit different. An isolated system cannot exchange any parts or energy with anything outside that system. A heated gas that can't interact with anything outside that gas can never heat up or cool down, if the volume stays the same. Insulation is an attempt to isolate the air system in a house. It's never perfect under any circumstances. An open system is the opposite and allows parts and energy in and out of the system.

Now a closed system is when no parts are allowed in or out of the system, but energy is. The freon in an air conditioner is 'enclosed', but is what is used to transfer heat out of your house. That's what the big coils outside the house is for, as a heat exchanger.

7. Lastly,what is the difference between the Helmoltz equation and gibb's equation(just basic idea required).

The is very rough, but basically the Helmholtz equation separates two variables to each side of an equation such that both side are equal to some constant. This allows you to separate those variables into two separate equations, both equal to that constant. It's useful for defining angular frequency in linear trig functions, Fourier transforms, etc., depending on which form you use. I assume, when you say Gibbs equation you mean the Gibbs–Helmholtz equation. It simply defines the change in Gibbs energy as temperature changes.

These last concepts need a better understanding of the previous questions. It would be best to learn how and why the equation of state of an ideal gas are manipulated the way they are. With 3 variables you can hold one of them constant to see how the other two work together. Then choose another variable to hold constant and learn how another pair of variables work together. When you learn how any combination of two variables work together, you can start treating all 3 as variables. That's how you learn about the answers to these questions you are asking.

 

[johng23]

The difference in Helmholtz and Gibbs energy:

In general, the equilibrium state of a system is the state of maximum entropy. Since entropy is not a quantity that can be easily understood or measured in real situations, we can define other thermodynamic potentials that are relevant to the conditions at hand. Gibbs energy and Helmholtz energy are two such potentials. You can show that, in the right conditons, the entropy maximum condition for equilibrium is equivalent to either the minimum Gibbs free energy or the minimum Helmholtz free energy.

In chemical reactions that occur commonly on earth, the pressure is fixed at atmospheric pressure, and the temperature is fixed by the surroundings. Then P and T are the controlled variables. In this situation, you can show that maximum entropy is equivalent to minimum Gibbs free energy. Thus, it is the Gibbs energy that is generally used to analyze the system.

In a pressure cooker, the situation is different. In this case, the volume is fixed by the container. Temperature and volume are the controlled variables, and the system will spontaneously evolve until the Helmholtz energy is minimized.

In general, it's all about maximizing entropy, but how that manifests itself depends on the physical constraints of the system.

 

[Urmi Roy]

Consider a pressurized tank of air ...At a fundamental level there's not much difference. Open the valve till all the air leaks out, then heat it and more air leaks out under pressure.

This is exactly I was cofused about initially. It seems that due to the fact that both pressure (hence mechanical equilibrium) and temperature (hence thermal equilibrium) both vary similarly on heating or cooling,there isn't much difference between mechanical equilibrium and thermal equilibrium . I read up on this later,and I found that pressure (hence
mechanical equilibrium ) can be affected in many ither ways also,hence it is different from thermal equilibrium.

Yes, the temperature is what's defined to stay static in Boyle's law.

Thanks for the clarification! I also figured out that the process to which Boyle's law is applied has to be quasi-static also,else we can't trace the progress of the process and plot it,right?

So primarily it depends on an area big enough so that averaging the kinetic energy of the molecules in it will give you a true average. It's like asking what the average shoe size is. You can't learn that from 1 shoe.

Thanks,I found your explanation very helpful!

An open and isolated system are opposites. A closed system is a bit different...

I now understand exactly what these terms mean. Just going a bit further,would it be right to say that for most of our experiments on adiabatic processes,isobaric processes and isochoric processes are conducted with isolated systems? On the other hand,for isothermal process,we need a closed system.


The other answers were also very helpful. Thanks,my_wan!

 

[Urmi Roy]

The difference in Helmholtz and Gibbs energy:
In general, it's all about maximizing entropy, but how that manifests itself depends on the physical constraints of the system.

Right,so that means both these equations are used to check the feasibilty of a process without directly taking recourse to entropy,but in two different kinds of situation (Gibb's free energy for P-T invariant processes and V-T invariant processes use Helmoltz free energy).

Thanks,johng23!

 

[Andy Resnick]

1. Thermodynamic equilibrium requires mechanical equilibrium--please explain.

Not always- a freely-falling body is not in mechanical equilibrium, but can have a well-defined temperature (thermodynamic equilibrium). Two objects may be at mechanical equilibrium (say, at rest toughing each other) but not in thermal equilibrium (different temperatures= heat flow). Heat is not a mechanical phenomenon.

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

The state space has 5 dimensions- volume, pressure, temperature, entropy and enthalpy. That the state space has an odd number of dimensions, whereas the mechanical state space has an even number, is significant.

3. In order for Boyle's law to be applicable to a process,does the process need to be quasi static and (or) isothermal?(Referring to this,the main thing is--is an isothermal process always quasi static?)

Boyle's law is an approximate law which gives accurate predictions for dilute cold gases. AFAIK, the process must simply be isothermal, and does not have to be quasi-static.

4.What kind of factors does continuum volume (the minimum volume needed to obtain a continuum in the system) depend upon?

The spacing between atoms (mean free path). The volume element must be much larger than this distance.

5. why is density a thermodynamic quantity?(It says so in my book).

density of what? mass? energy?

6. What is the significance of classifying systems into 'open','closed' and 'isolated' systems?(I mean in real life,do we consider them while performing adiabatic/isothermal/isochoric processes?)

The significance is simply convenience: each type of system has certain constraints (e.g. conservation of energy/mass for a closed system) that simplify the model.

7. Lastly,what is the difference between the Helmoltz equation and gibb's equation(just basic idea required).

The Helmholtz free energy is good for systems at constant V and P, the Gibbs free energy is preferred for systems at constant P and T.

 

[Urmi Roy]

Not always- a freely-falling body is not in mechanical equilibrium, but can have a well-defined temperature (thermodynamic equilibrium). ...

Hmm... the concept of mechanical energy when introduced into thermodynamics seems to be very ambiguous!

The state space has 5 dimensions- volume, pressure, temperature, entropy and enthalpy. That the state space has an odd number of dimensions, whereas the mechanical state space has an even number, is significant..

What I originally asked was how many state veriables completely define the thermodynamic state...entropy and enthalpy are determined by the other three,so I guess the answer to my original question,as my wan said, is 3.

Btw,why is it that "state space has an odd number of dimensions, whereas the mechanical state space has an even number" is important to note?

Boyle's law is an approximate law which gives accurate predictions for dilute cold gases. AFAIK, the process must simply be isothermal, and does not have to be quasi-static..

But if it isn't quasi-static,we can't even plot the resulting P-V diagram,as we can't define the Pressure for non-equilibrium states!

density of what? mass? energy?

Density of mass. my wan gave a good answer to this question.

The significance is simply convenience: each type of system has certain constraints (e.g. conservation of energy/mass for a closed system) that simplify the model.

Okay,so would it be right to say that for most of our experiments on adiabatic processes,isobaric processes and isochoric processes are conducted with isolated systems wheras for isothermal process,we need a closed system?

Thanks for the help,Andy Resnick!

 

[Andy Resnick]

Hmm... the concept of mechanical energy when introduced into thermodynamics seems to be very ambiguous!

Not really- the work (mechanical energy) is just the PV term, oftentimes considered as P dV.

What I originally asked was how many state veriables completely define the thermodynamic state...entropy and enthalpy are determined by the other three,so I guess the answer to my original question,as my wan said, is 3.

Btw,why is it that "state space has an odd number of dimensions, whereas the mechanical state space has an even number" is important to note?

Not quite- entropy and enthalpy also require the heat (Q), they aren't completely specified by specifying P, V, and T.

As to the difference in even- versus odd- dimensioned phase space, I can only give a partial answer: in mechanical theories, phase states with an even number of dimensions have a symplectic structure: there are conjugate pairs of variables (i.e. position and momentum, q and p). Odd-dimensioned spaces have a contact structure:

http://en.wikipedia.org/wiki/Symplectic_vector_space
http://en.wikipedia.org/wiki/Contact_geometry

But if it isn't quasi-static,we can't even plot the resulting P-V diagram,as we can't define the Pressure for non-equilibrium states!

You are raising an interesting question, regarding the appropriateness of thermodynamic variables in the context of non-equilibrium conditions. Thermodynamics is a continuum theory- it doesn't really apply to dilute gases.

 

[DrDu]

Andi, what do you say! To specify the equilibria of the simplest systems two out of the three variables p, V, T are sufficient. The heat is not needed to specify U uniquely. Of course there are more complicated systems which need more variables for a complete specification of the equilibrium state, so there may also be electric and magnetic field variables, stress and strain etc. Thermodynamics does not fix this number but one has to find them for the specific system. Hence one sometimes says that thermodynamics is a meta-theory.

 

[Andy Resnick]

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.

 

[alynfazlin]

want to ask a question..is it true the absolute pressure in a liquid of c0nstant density d0ubles when the depth is d0ubled?explain please..

 

[DrDu]

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]

want to ask a question..is it true the absolute pressure in a liquid of c0nstant density d0ubles when the depth is d0ubled?explain please..

I think it's true...this is because the pressure at any point in a liquid is given by dgh (where d is the density,g is the acceleration due to gravity and h is the height of the column of the water above the point in concern.

So if you double the depth,the 'h' doubles,so P(pressure) =2(dgh).

Intuitively,we could say that the molecules at the depth 2h have to move around with twice the amount of kinetic energy than the molecules at 'h' in order to support the liquid above it.
(This is my personal idea...let's see what the experts say!)

P.S According to my intuition (that I put forward in the last paragraph),the temperature at the greater depths should be higher,as temperature depends upon the kinetic energy of the molecules...someone told me that this assumption is correct but due to rapid convection currents,we can hardly notice the temperature difference...is that right?

 

[Andy Resnick]

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.

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.

 

[the_house]

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

Can you elaborate a bit on this?

If I can shake out the cobwebs and try to remember correctly, in thermodynamics as it's usually taught in undergraduate and graduate courses, it is stated that all information about a thermodynamic system is contained in the relation for a thermodynamic potential--e.g., U(S,V,N1,N2,...) or the various Legendre transforms of this relation (e.g., enthalpy, Gibbs free energy, Helmhotz free energy, etc.). So then the state can be completely specified by 2+n quantities (where n is the number of conserved charges--in textbooks this is usually particle number for n species of particles). Any other quantities can be derived from the thermodynamic potential. Even your arXiv reference appears to say this is true in "classical mechanics", at least for n=0.

What has to be true for this not to be the case, and what exactly does it change?

Edit: OK, having skimmed the introduction to that arXiv reference, it says it's trying to answer the question. "Is there such a thing as ‘quantum thermodynamics’ where pressure or volume are represented as operators?" This work seems well outside the purview of standard thermodynamics and I believe it needlessly complicates the thread (titled "Thermodynamic basics"). In answering the original question, it's probably best just to state the standard thermodynamics answer: only 2+n quantities must be known to completely specify a thermodynamic system. If I've misunderstood something, let me know.

 

[Andy Resnick]

Can you elaborate a bit on this?

Unfortunately, not too much- I'm still learning it myself.

Other than the wiki page on contact geometry, there's some discussion in an appendix of Arnold's "Mathematical Methods of Classical Mechanics". There's a lot available on google scholar, but I'm not able to separate wheat from chaff

http://scholar.google.com/scholar?q...odynamics&hl=en&as_sdt=0&as_vis=1&oi=scholart

 

[the_house]

Unfortunately, not too much- I'm still learning it myself.

Then I'll have to go with the assumption that all this business about quantizing contact geometries goes well beyond standard thermodynamics (if it's even related at all).

So I'll have to stick with my previous answer: in standard thermodynamics you can completely specify a thermodynamic state with 2+n variables, where n is the number of conserved quantities. As an explicit example, from earlier in the thread:

Not quite- entropy and enthalpy also require the heat (Q), they aren't completely specified by specifying P, V, and T.

Assuming particle number is not conserved, you actually only need to know 2 of the variables. If you know V and T, then the entropy can be found by differentiating the relation for the Helmholtz free energy F(T,V): -S(T,V) = [dF/dT]_V (Sorry for the notation. I'm having trouble getting LaTeX to work). The pressure is P(T,V) = [dF/dV]_T. You can then plug in the values of S and P to determine the enthalpy H(S,P). (The enthalpy is related to F by a Legendre transform with respect to both T and V: H = F + TS + PV.)

Alternately, if you start with T and P, you can use the Gibbs free energy relation G(T,P) to calculate the entropy -S(T,P) = [dG/dT]_P, which you can then plug in for the enthalpy H(S,P). (Again, the enthalpy is related to G by a Legendre transform: H = G + TS).

 

[SpY]]

I was wondering about linear thermal expansion:If a rectangular plate with a hole in the middle is heated and expands, will the hole get bigger or smaller?
Does the whole thing just scale up (bigger hole) or swell like a donut (smaller)?

 

[Studiot]

Hello spy and welcome.

You should really start your own thread to ask separate questions, rather than adding to existing ones.

It is a good question which probably merits some discussion since there are several possible answers.

The short answer is that if the plate is free to expand around its outer edges the hole will also get bigger.

 

[SpY]]

Thanks for the tip, I'm just new new here. And thanks for a quick reply :)

 

[Andy Resnick]

Assuming particle number is not conserved, you actually only need to know 2 of the variables.

I only know of one system with non-conservation of particles: photons. Otherwise, exempting chemical reactions, particle number is pretty well conserved.

I'm not really sure what your point is- when you write H = F + TS + PV or H = G +TS, how can you claim that thermodynamics does not have an odd number of state variables?

 

[the_house]

I only know of one system with non-conservation of particles: photons. Otherwise, exempting chemical reactions, particle number is pretty well conserved.

Any relativistic system will not conserve particle number. Even a non-relativistic system with fixed particle number will not have any associated chemical potentials. Mostly I just wanted to simplify things so there weren't a bunch of chemical potentials or conserved charges running around. Also, the only variables that were mentioned were T, P, V, S, and H, so I presumed there were no chemical potentials in this proposed system. In any case, the extension is straightforward. Instead of 2 variables to completely describe the system it takes 2+n.

I'm not really sure what your point is- when you write H = F + TS + PV or H = G +TS,

This is a http://en.wikipedia.org/wiki/Legendre_transform" . Incidentally, it is also the relationship between the Lagrangian and the Hamiltonian in classical mechanics.

how can you claim that thermodynamics does not have an odd number of state variables?

I don't think I made that claim, although the relevance is unclear to me. It's certainly not something that was ever mentioned in any thermodynamics course I've taken. Everything I have actually stated is definitely standard thermodynamics.

 

[Andy Resnick]

Any relativistic system will not conserve particle number.

I have never heard this- can you provide a reference?

I don't think I made that claim, although the relevance is unclear to me.

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.

 

[George Jones]

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.

I have been doing some reading, and, while things are still fuzzy for me, I'm starting to see an overview. These variables are five possible coordinates for 5-dimensional thermodynamic phase space. The possible states of a system (in equilibrium) is a 2-dimensional (Legendre) submanifold of 5-dimensional thermodynamic phase space. In general, a thermodynamic system that has n degrees of freedom is an n-dimensional submanifold of the (2n + 1)-dimensional contact manifold that is thermodynamic phase space. n can be even or odd, but, trivially, 2n + 1 is always odd.

From the paper:

Consider a material in an enclosure of volume V, pressure P, temperature T, entropy S and internal energy U. ... In our example, the dimension of a Legendre submanifold is two, which therefore is the number of independent thermodynamic degrees of freedom.

Other references:

http://sgrajeev.com/geometry-of-thermodynamics/

http://www.sci.sdsu.edu/~salamon/MathThermoStates.pdf.

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

 

[the_house]

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.

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.

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.

 

[DrDu]

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.

 

[Urmi Roy]

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?

 

[Studiot]

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.

 

[Urmi Roy]

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.

 

[George Jones]

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

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.

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.

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.

 

[Urmi Roy]

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.

 

[the_house]

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.)

 

[Urmi Roy]

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).

 

[George Jones]

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 , 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.

 

[the_house]

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.

 

[DrDu]

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.

 

[the_house]

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.

 

[Studiot]

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.

 

[Andy Resnick]

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.

 

[Andy Resnick]

a reference that describes the out-of-equilibrium part of this 5 dimensional state space

Here's a few:

http://iopscience.iop.org/0305-4470/36/17/301

http://ieeexplore.ieee.org/Xplore/l...256250.pdf?arnumber=5256250&authDecision=-203

http://jmp.aip.org/jmapaq/v39/i1/p329_s1?isAuthorized=no

Edit: Upon reading my posts, I think they may come across as a little 'gruff'. Honestly, I'm just excited to be learning something new.

 

[Urmi Roy]

All that elasticity theory we did last term?

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

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).

 

[Studiot]

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.

 

[Urmi Roy]

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

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!

 

[Urmi Roy]

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?