# Equilibrium=homogeneity of the intensive parameters?

1. Jan 10, 2006

### mma

Equilibrium = homogeneity of the intensive parameters?

It is obvious, that if all intensive parameters of an isolated thermodynamic system are homogeneous then the system is in equilibrium. But, what is the situation with the reverse statement? Is it true, that if an isolated thermodynamic system is in equilibrium, then all intensive parameters of it must be homogeneous?
I say a counter-example: a grape in a glass water. If the grape is strong enough that does not crack, then the pressure inside the grape will be greater than in the surrounding water.
So, it seems, that an isolated thermodynamic system can be in equilibrium, even the pressure isn't homogeneous inside.
But, what about the other intensive parameters, e.g. with the temperature? Can be an isolated system in thermodynamic equilibrium if the temperature isn't homogeneous inside?

Last edited: Jan 10, 2006
2. Jan 23, 2006

### Vey2000

I'm just a student, so I may be wrong. However, I don't think the problem you have posed is actually in equilibrium.

The grape will eventually decompose. Not only that, but water is most likely slowly seeping into the grape through the skin. There might be many more ways in which the system is currently progressing towards equilibrium, however slowly.

It won't reach equilibrium until the grape completely dissolves within the water. At which time all intensive parameters will be homegenous.

I suppose one might consider the grape to be in a metastable condition, but in that case I'm out of my league, so I'll just stop here.

3. Jan 23, 2006

### mma

I think, the notion of equilibrium depends on the time scale. For example, a glass of hot water reaches an equilibrium, when it cools down to the temperature of its environment.
The next equilibrium state is when the water completely evaporates.
Next equilibrium state, when the glass flows down on the table (the glass is really a very viscous fluid).
Next equilibrium is when the glass evaporates.
....
Final equilibrium is when all protons of its material decays.
So, I think that a grape berry in a glass of water has a kind of equilibrium.
---------------------------
But, I can tell a better example. Let's take an ideal gas column in homogeneous gravitational field, separated from the environment. Clearly, it can reach an equilibrium state. In this equilibrium state, the pressure will be greater at the bottom, as on the top. But the temperature will be homogeneous all over the column.
If P can be inhomogeneous in equilibrium, then T why not?

4. Jan 23, 2006

### pseudovector

If the outward pressure of the grapes is bigger than the inward pressure of the water, wouldn't the grape expand?

5. Jan 23, 2006

### vanesch

Staff Emeritus
Take a simpler example: some ice cubes in water of 0 degrees C, in an adiabatically closed container. There will be an equilibrium between both, nevertheless you will keep two phases.

6. Jan 23, 2006

### mma

OK, but in this case all intensives will be homogenious. Intensive means pressure, temperature and chemical potentials. What isn't homogeneous, that is density. But density isn't an intensive parameter.

7. Jan 23, 2006

### vanesch

Staff Emeritus
You're right! Sorry.

8. Jan 24, 2006

### mma

Yes, of course. And it seems that this is the point. The skin of the expanded grape berry exerts with an elastic force inward, but not outward. So the force of the skin of the berry isn't isotropic in our system! Just like the gravity force in our gas column. This is an anisotropic force too. The question is, how can we describe these forces in thermodynamics.

9. Jan 29, 2006

### mma

Well, let's take the most simple example. We can complicate it later on.

Ideal gas in an adiabatically closed rectangular box. The box has rigid and impermeable walls. Let be the axises of our coordinate system parallel to the edges of the box. The box is divided into two parts by a rigid, adiabatic, and in impermeable wall in the yz plane. If we change the x position of the wall, then the pressure of the gas on the left and right part will change according the ideal gas law. If the wall can move freely in the x direction, parallel with itself, then there will be a definite x0 equilibrium position of the wall, independently of its starting position.

The question is, what thermodynamic laws or postulates assure, that the pressure of the two parts will be equal in the equilibrium position?

Last edited: Jan 29, 2006
10. Jan 29, 2006

### Clausius2

Pressure would be equal due to mechanical equilibrium, not due to thermodynamics.

In general I think you're right. I think that gradients of intensive variables cause nonequilibrium. For example, gradients of concentration enable diffusivity, which is characteristic of non equilibrium flows. Gradients of temperature enable heat conduction. In general, I would say that any property transport due to diffussion mechanism corresponds to non equilibrium flow. But we usually assumme local thermodynamic equilibrium for our computations, because the relaxation time of this processes are much smaller than the typical residence time of fluid particle.

11. Jan 29, 2006

### mma

Do you mean that we cannot describe even this very simple system within the framework of thermodynamics? The equality of the pressures in equilibrium cannot be derived from the first and second laws of thermodynamics plus the ideal gas law? Or at least from Onsager's theory?
This would be very surprising!

12. Jan 29, 2006

### Clausius2

Well, I think that the concept of mechanical equilibrium is present everywhere. Kinetic theory, in my opinion, does not escape from this law. Maybe you can get around claiming something about the minimum entropy of the system, but I am pretty sure the law you are going to use is built just at the bottom by means of some kind of primary ideas of mechanics. Why do you think that thermodynamics is decoupled of mechanics? In fact, in many times is the mechanics of the thermodynamic system which ultimately establishes its non equilibrium. One only has to take a look at Navier Stokes equation and notice for example that the Internal Energy Equation is not decoupled of the mechanic field.

Last edited: Jan 29, 2006
13. Jan 30, 2006

### mma

I think only that thermodynamics is a discipline which can describe a wide range of phenomena with the same way, independently of the actual mechanical background. For example, if we simplify our system still a little bit, so that we take heat conducting wall instead of the adiabatic one, then the entropy maximum principle yields the equality of the pressures directly.

If the wall is heat conducting, then the temperature of our ideal gas in the two parts of the box will be the same (after a while), when we fix the wall in any definite x position. After reaching this common temperature, let the wall move freely to the equilibrium x0 position. Clearly, the energy of each part of gas will remain during this, because the following reasons.

1. The energy of the whole system remains, because it is closed.
2. The temperature of the system will also remain because the energy of the ideal gas depends only on its temperature.
3. Due to the same reason ,both part of the gas will keep its energy (because it keeps its temperature).

So, in this case we have to find the maximum of the $$S(x) = S(E_1,V_1(x),N_1) + S(E_2,V_2(x),N_2)$$ function. We know the S(E,V,N) function of ideal gas from thermodynamics textbooks, $$V_1(x)=Ax,V_2(x) = A(L-x)$$, so the problem can be solved.

Last edited: Jan 30, 2006
14. Jan 30, 2006

### Vey2000

Sorry for not answering earlier, but I've been (and am) busy with exams.

Looking back on my previous response, I realize now that I fell for your trick question. Equilibrium does not necessarily mean that the intensive parameters are homogenous. Equilibrium means that the entropy of a system is maximum. So take dS=0 and solve for the equilibrium conditions given the contraints of the system.

So for example, if we assume the grape skin is diathermal and rigid then in equilibrium the temperature will be the same, but the pressure won't necessarily be the same.

You're right, the decomposition happens on a larger time scale, so for all intents and purposes it can be treated as equilibrium. I was just trying to think of some reason to explain why the system could not be in equilibrium as you had defined it: homogeneity of all intensive parameters.

As for the example of the column of gas in a homogenous gravity field, I had a look at Landau's Statistical Physics which has a section on the topic of equilibrium in external fields.

Essentially, you have to consider the column as made of infinitesimal systems layered one upon the other. Clearly, they are separated by diathermal walls as heat can be transfered across them (since there aren't actually any walls), so in equilibrium the temperature will be equal across all the subsystems.

Since the gas molecules can obviously move across layers, the walls are considered permeable, and hence the chemical potential must be the same for all the subsystems.

Here's the kicker: the chemical potential also depends on the height by including a term for the gravitational potential.

Unfortunately, other than saying that the pressure will be inhomogenous it doesn't really explain why. I think it's because since each infinitesimal subsystem is defined exclusively by its height, in essence its walls are rigid, since any particle moving up or down will be transfered to another system rather than pushing that system's walls up or down. Hence, the pressure does not factor at all into the derived equilibrium conditions.

15. Jan 31, 2006

### mma

I also looked into Landau's book. He explains here the homogeneity of temperature simply by the fact of equilibrium. But I don't know why is it enough. It's clear that gravity changes the spatial distribution of molecules, so it affects density and pressure. But gravity also changes the speed of molecules, which is in close connection with temperature. However, if we consider the Maxwell-Boltzmann distribution, it can be shown that this distribution remains the same at any height in the gas column. At least for non-interacting, or centrally colliding point-like particles. But this is a statistical mechanical consideration only for a special case. I'm sure, that this can be done for more general cases too. But I conjecture, that there are general thermodynamical considerations also, what yield this result generally.

I think that we will come closer to this problem, if we study the behavior of intensive parameters in different situations. That's why I started to analyze a very simple system in my previous post. Perhaps we can find the proper treatment, or maybe somebody knows it already and shares his knowledge with us.

Last edited: Jan 31, 2006
16. Feb 2, 2006

### mma

More precisely, we don't need S(E,V,N) for it.
The solution:

$$\frac{dS}{dx} = \left( \frac{\partial S}{\partial V} \right)_{E_1,N_1} \frac{dV_1}{dx} + \left( \frac{\partial S}{\partial V} \right)_{E_2,N_2} \frac{dV_2}{dx} = \frac{p_1}{T_1} A + \frac{p_2}{T_2}(-A)$$

Taking into account that $$T_1 = T_2$$, this means, that the derivative vanishes when $$p_1=p_2$$. So the equality of the pressures is a necessary condition for the maximality of the entropy.

Last edited: Feb 2, 2006
17. Feb 3, 2006

### mma

I forgot to explain why.
Well, let's take the wall fixed in any x position.
Then, because our system is closed, the first law reads:
$$dU=T_1dS_1 + T_2dS_2 = 0$$
Introducing the notation $$Q=T_1dS_1$$
$$dS_1=\frac{Q}{T_1}, dS_2=-\frac{Q}{T_2}$$

The second law :
$$0 \le dS = dS_1 + dS_2 = \frac{Q}{T_1} - \frac{Q}{T_2} = Q (\frac{1}{T_1} - \frac{1}{T_2})$$.

In equilibrium dS=0, i.e $$\frac{1}{T_1} = \frac{1}{T_2}$$.

Last edited: Feb 3, 2006
18. Feb 4, 2006

### mma

By the way, this solution is wrong. The result is good, but the derivation of it is bad.

The error is, that I took the internal energy of the parts, $$E_1$$ and $$E_2$$, constant. The justification of this was the following: The system is closed, hence the total energy of it remains. The total energy consists of the energy of the parts of the gases. In equilibrium, the temperature of the parts is equal. The energy is an injective function of the temperature, so this common temperature must be the original temperature. But, if the temperature of the parts doesn't change, then the energies of them remain also unchanged.

But, surprisingly, this isn't true in general. Namely, a $$dx$$ change of the position of the wall causes a $$dE=(p_1-p_2) Adx$$ change in the total internal energy of the gas. In the case of $$p_1\not=p_2$$ this quantity doesn't vanish. That is, the internal energy of the gas isn't unchanged when changing the position of the wall. If the system is really cosed, then the difference will be given to or getting from the kinetic energy of the wall. In a correct solution we cannot neglect this.

Our fortune was that we got just $$p_1=p_2$$, the only case when $$dE=0$$ really.

Last edited: Feb 4, 2006
19. Feb 6, 2006

### mma

Indeed. We cannot apply our previous considerations to the original problem, i.e., when the wall is adiathermal, because in this case the entropy of our closed system does not depend on the position of the wall. But mechanical analysis shows, that our system is in equilibrium, when the pressure is equal on the left and right side, i.e, when it is homogeneous all over the system. This example suggests, that homogeneity of the intensive parameters is a more general condition for equilibrium than the maximality of entropy. On the other hand, there are those puzzling counterexamples....

Last edited: Feb 7, 2006
20. Feb 27, 2006

### Clausius2

Intensive parameter gradients do definitely cause non equilibrium in the system. At least they cause traslational non-equilibrium. I'm not going to demonstrate it here mathematically, you can find it on your own only looking for the equations I am going to mention here.

We understand non equilibrium as a departure from the maxwellian distribution (let's call it $$f_o$$). The transport equation of $$f$$ is the Boltzmann equation:

$$\frac{D(nf)}{Dt}=\frac{\partial (nf)}{\partial t}\right]_{collision}$$

where the term on the right accounts about the transport of f due to individual collision dynamics. In his model, Krook approximates this term as an small departure from equilibrium:

$$\frac{\partial (nf)}{\partial t}\right]_{collision}\sim n\nu(f-f_o)$$

where $$\nu$$ is the collision frequency (very large in flows near equilibrium). In this way, one may accomplish the Chapman-Enskog expansion of the Boltzmann equation using the Krook Model. Doing that, one is able to see that, in first order:

$$f=f_o\left[1+\Phi\left(\nabla T, \nabla u\right)\right]$$

where $$u$$ is the bulk velocity. Therefore, the departures from translational equilibrium are enhanced by macroscopic gradients of temperature and velocity at first order. You will see this equation leads directly to the Navier-Stokes equations, Fourier Law (heat transfer) and Poisson's Law (viscous stress), which are by the way the result of the first order correction to the maxwellian distribution.

A remarkable point is that pressure is not anywhere in the correction. One may have an hydrostatic field and be in translational equilibrium. I ignore if this effect has some contribution in subsequent corrections of $$f_o$$. All the assumptions made for obtaining the Chapman-Enskog expansion of the Krook model are: translational non equilibrium and monoatomic gas.

See you.