# Thermodynamics basics.

1. Jul 9, 2010

### 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!!)

Last edited: Jul 9, 2010
2. Jul 9, 2010

### mikelepore

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.

3. Jul 9, 2010

### 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?

4. Jul 10, 2010

### 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.

5. Jul 10, 2010

### Urmi Roy

Thanks!

6. Jul 10, 2010

### 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.

7. Jul 10, 2010

### mikelepore

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

8. Jul 10, 2010

### my_wan

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.

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.

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.

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.

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.

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.

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.

9. Jul 10, 2010

### 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.

10. Jul 10, 2010

### Urmi Roy

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.

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?

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.

11. Jul 10, 2010

### Urmi Roy

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!

12. Jul 10, 2010

### Andy Resnick

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 equilbirium (different temperatures= heat flow). Heat is not a mechanical phenomenon.

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.

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.

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

density of what? mass? energy?

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.

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.

13. Jul 12, 2010

### Urmi Roy

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

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?

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

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!

14. Jul 12, 2010

### Andy Resnick

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

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

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.

15. Jul 13, 2010

### 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.

16. Jul 13, 2010

### Andy Resnick

17. Jul 17, 2010

### 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..

18. Jul 17, 2010

### 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.

19. Jul 17, 2010

### Urmi Roy

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?

20. Jul 17, 2010

### Andy Resnick

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.

Last edited: Jul 17, 2010
21. Jul 17, 2010

### the_house

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.

Last edited: Jul 17, 2010
22. Jul 18, 2010

### Andy Resnick

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

23. Jul 18, 2010

### the_house

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:

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).

24. Jul 18, 2010

### 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)?

25. Jul 18, 2010

### Studiot

Hello spy and welcome.