Calculating entropy for a simple scenario

[zrek]

I'd like to create a simple model that demonstrates the basic values of thermodinamics of an ideal gas. I begin with two rooms, several molecules in them. Every data of every individual molecule is given (position, mass, speed, etc), so I can easily calculate the total energy, pressure, temperature. Now I open up the small door between the two rooms and I can easily model how the molecules mixing, causing the two rooms finally balancing their temperature and pressure.
But I have problems with calculating the entropy. Can you please help me?
Is there a formula for this? (If it is necessary, we can say that the available positions of the molecules are limited)
Thanks in advance.

 

[Stephen Tashi]

But I have problems with calculating the entropy.

The concept of entropy can be related to probability, which in turn, applies to probability spaces defined over the "possible" microstates of a system. If you are simulating a deterministic system to the level of detail of simulating the position and velocity of each particle, you haven't introduced any idea of probability. At a given time in the simulation, the system would be in one particular microstate, so this does not account for any concept of it having a probability for being in each of several "possible" microstates.

You need to introduce the idea of probability. You might be able to do this by running many simulations that begin with some random variation in the initial conditions. Then at time T, the system might be in a different microstates on different runs. You could estimate the probability that the system was in a certain microstate by the frequency that it appeared in that microstate on different runs of the simulation.

 

[zrek]

The concept of entropy can be related to probability (...). You might be able to do this by running many simulations that begin with some random variation in the initial conditions.

Thank you for your answer.
As far as I know, the probability calculations are (in nutshell) about the ratio of the all possible states and the desired result. This can be done even in deterministic scenarios: I don't have to run many simulations if I can calculate all of the possible ones. This is why I mentioned some limit for the molecule positions (and other property values). If we assume that it is possible to calculate for a scenario, can you suggest a formula that gives a concrete number for entropy?

 

[Stephen Tashi]

As far as I know, the probability calculations are (in nutshell) about the ratio of the all possible states and the desired result.

That only works if all possible states have the same probability.

You describe doing a deterministic simulation down to the level of individual gas molecules and apparently you want to compute entropy as a function of time. That is not a typical textbook problem in thermodynamics (as far as I know). The textbook problem would only ask us to find the change in entropy between the initial state and the state where the gases have reached equilibrium.

You are describing a system that is not in equilibrium. If the Wikipedia can be believed (
https://en.wikipedia.org/wiki/Non-equilibrium_thermodynamics ) :

Another fundamental and very important difference is the difficulty or impossibility in defining entropy at an instant of time in macroscopic terms for systems not in thermodynamic equilibrium.[1][2]

So we may be out of luck. Perhaps we can define entropy in microscopic terms if we introduce probability into the model. I'll have to think about it.

 

[Chestermiller]

Let me understand your question. You have two chambers, with a partition between them. You have the same gas in each chamber. You know the volumes of the two chambers. And you know the initial temperatures and pressures of the of the gas in each of two chambers. Now you remove the partition, and let the system re-equilibrate adiabatically. You want to know the change in entropy from the initial state of the system to the final equilibrated state of the system. Is this correct? Are you allowed to use classical thermodynamics?

 

[malemdk]

Net increase in entropy will be zero , since the intermixing of molecules are done in adiabatic condition , and all the position, velocity etc are known,and sum of the heat transfer will be zero , the change in entropy will be zero

 

[Chestermiller]

Net increase in entropy will be zero , since the intermixing of molecules are done in adiabatic condition , and all the position, velocity etc are known,and sum of the heat transfer will be zero , the change in entropy will be zero

Is this supposed to be the answer to the problem that I posed in post #6? If so, it is totally incorrect. Here is a link to a Physics Forums Insights article I wrote on how to correctly determine the entropy change for a process imposed on a closed system: https://www.physicsforums.com/insights/grandpa-chets-entropy-recipe/ See if you can figure out how to apply the recipe to this problem.

 

[zrek]

Let me understand your question. You have two chambers, with a partition between them. ... You want to know the change in entropy from the initial state of the system to the final equilibrated state of the system. Is this correct? Are you allowed to use classical thermodynamics?

Yes, you described correctly what I want to do. I'd like to create a starting state with no equilibrium, different temperatures and pressures in the chambers, and watch what happens after I open the door. During the process while the termperatures balancing, I'd like to draw a graph with the values of the key properties: pressure, temperature ... and entropy. As far as I know, the entropy should increase. Since I can calculate the temperature and the pressure directly form the data of the particles, I'd like to do the same with the entropy. But I have no idea how to calculate it. Thank you for your help!

 

[zrek]

So we may be out of luck. Perhaps we can define entropy in microscopic terms if we introduce probability into the model. I'll have to think about it.

Then it is not possible to calculate the entropy during a process in a simulated environment? There is no concept at all for problems like this?

I'll have to think about it.

Thank you! I'd be happy if you have a good idea. A concrete modell with a simulation would make anyone understand the entropy better, I think.

 

[Chestermiller]

Are you willing to go to a continuum model? If so, you can get the transition of the entropy.

So you are able to get the entropy in the final state, correct?

 

[Stephen Tashi]

Then it is not possible to calculate the entropy during a process in a simulated environment?

For the sake of your other advisors, let's be clear that you want to

1) Model the gas as a finite number of particles
2) Compute entropy as a function of time - i.e. not merely compute the 2 numbers that give the entropy before the gases mix and the entropy after the gases have mixed and reached an equilibrium state.

For now, I leave the difficulties of 1) to you. (They including finding algorithms to determine when particles collide in the simulation.)

There is no concept at all for problems like this?

As far as I know, current textbooks do not provide any standard definition for the entropy of a system that applies to times when the system is going through states of non-equilibrium. It wouldn't surprise me if people have written papers proposing various definitions in that case. We'd have to search for such papers.

So, yes, the first difficulty of 2) is to define entropy for your simulation.

There are two basic approaches to defining entropy. One approach is "Shannon entropy", also known as "Information Entropy". This approach requires having a probability distribution. In thermodynamics, the entropy of a system can defined in terms of Shannon entropy by using a probability distribution for the system being in various possible microstates. However, at a given time t, your simulation is a single microstate - namely the microstate that specifies the positions and momenta of each of the particles. So you don't have a non-trivial probability distribution for the system being in several different microstates.The other approach to defining entropy is thermodyamic entropy, which defines it by a differential equation.
e.g.
$t$ time
$U(t)$: total energy of the system at time t
$T(t)$: temperature of the system at time t
$P(t)$: pressure of the system at time t
$V(t)$: volume of the system at time t

The defining equation for entropy $S(t)$ is $U'(t) = T(t) S'(t) - P(t) V'(t)$

However, there are difficulties in defining some of these quantities for non-equilibrium situations. Consider the concept of "Volume of the system at time t". Take an extreme case where the system consists of 3 particles. Suppose they are within cube that is 1 meter on a side.

What is the "Volume of the system"? At a given time, you could fit the 3 particles in a triangular shaped thin box with each particle in one corner of the box. Is the "Volume of the system at time t" equal to the volume of this thin box? - Or is the "Volume of the system" equal to the volume of the 1 meter cube?

The concept that a 1 meter cube is the "Volume of the system" makes sense if the system consists of particles uniformly distributed throughout the cube. So "Volume of the system" becomes unambiguous when we are considering an equilibrium situation.

 

[malemdk]

Consider the two room filled with few particles, let's 10, them how we will entropy?

 

[zrek]

Are you willing to go to a continuum model? If so, you can get the transition of the entropy.

What do you mean on "continuum model" exactly? The Holtsmark-continuum model for example? Or you mean the way of continuum mechanics? I guess I'd prefer a way to calculate directly from the data of the given, discrete particles. But if there is an easy and correct way to map the states of particles to a continuum model, then I'd accept it also, if it leads to a numeric representation of entropy.

 

[zrek]

1) Model the gas as a finite number of particles
2) Compute entropy as a function of time - i.e. not merely compute the 2 numbers that give the entropy before the gases mix and the entropy after the gases have mixed and reached an equilibrium state.

From your words I assume that you know the way how to calculate "the 2 numbers that give the entropy" form the given finite number of particles. If this is the true, I also would be happy to know this.

For the sake of your other advisors, let's be clear that you want to

$t$ time
$U(t)$: total energy of the system at time t
$T(t)$: temperature of the system at time t
$P(t)$: pressure of the system at time t
$V(t)$: volume of the system at time t

The defining equation for entropy $S(t)$ is $U'(t) = T(t) S'(t) - P(t) V'(t)$

I like your approach, the simple formula may work. I think, in case of ideal gases, the $U(t)$ is constant. $V(t)$ also constant, gases do not occupy a fixed volume, but expand to fill whatever space they can, don't they? However from the formula $U'(t) = T(t) S'(t) - P(t) V'(t)$ I can't get a concrete value for S(t), only its rate of change, right?

For the sake of your other advisors, let's be clear that you want to
... "Shannon entropy", also known as "Information Entropy". This approach requires having a probability distribution. In thermodynamics, the entropy of a system can defined in terms of Shannon entropy by using a probability distribution for the system being in various possible microstates. However, at a given time t, your simulation is a single microstate - namely the microstate that specifies the positions and momenta of each of the particles. So you don't have a non-trivial probability distribution for the system being in several different microstates.

The probability is easy to calculate from the following concept:
With closed doors, there are several particles in the first chamber and there are others in the second. Their macrostate is represented by their all possible microstate in this situation, knowing that their exact position in the actual chamber is irrelevant. Now if we open the door, we have to count all of the possible positions of the particles, and compare this number to the number of states in which the same number of particles are in the very same chambers. This gives the probability distribution of microstates for a given macrostate. But how to calculate a concrete number for entropy from this probability (ratio)?

 

[Chestermiller]

What do you mean on "continuum model" exactly? The Holtsmark-continuum model for example? Or you mean the way of continuum mechanics?

The latter.

I guess I'd prefer a way to calculate directly from the data of the given, discrete particles. But if there is an easy and correct way to map the states of particles to a continuum model, then I'd accept it also, if it leads to a numeric representation of entropy.

So you are talking about using molecular dynamics, right?

In your original post, I got the feeling you were looking for a simple example of a problem that could illustrate the evolution of the entropy vs time for a system experiencing an irreversible change. Is this correct? If so, would a simpler system be acceptable, or does it have to be the ideal gas example.

 

[Stephen Tashi]

From your words I assume that you know the way how to calculate "the 2 numbers that give the entropy" form the given finite number of particles.

No. I'm referrring to the 2 entropies that can be calculated from macroscopic parameters. Similar to Example 3 of https://www.physicsforums.com/insights/grandpa-chets-entropy-recipe/

I like your approach, the simple formula may work. I think, in case of ideal gases, the $U(t)$ is constant. $V(t)$ also constant, gases do not occupy a fixed volume, but expand to fill whatever space they can, don't they?

They don't expand to fill the space instantaneously. If you want to compute entropy as a function of time in the problem you described, you are dealing with times before the gases have had "filled whatever space they can".

The probability is easy to calculate from the following concept:
With closed doors, there are several particles in the first chamber and there are others in the second. Their macrostate is represented by their all possible microstate in this situation, knowing that their exact position in the actual chamber is irrelevant. Now if we open the door, we have to count all of the possible positions of the particles, and compare this number to the number of states in which the same number of particles are in the very same chambers. This gives the probability distribution of microstates for a given macrostate.

No. Counting the possibilities is unrelated to computing probability unless you can define the microstates so each microstate has the same probability of being occupied. How do you intend to define the microstates?

 

[Useful nucleus]

I'd like to create a simple model that demonstrates the basic values of thermodinamics of an ideal gas. I begin with two rooms, several molecules in them. Every data of every individual molecule is given (position, mass, speed, etc), so I can easily calculate the total energy, pressure, temperature. Now I open up the small door between the two rooms and I can easily model how the molecules mixing, causing the two rooms finally balancing their temperature and pressure.
But I have problems with calculating the entropy. Can you please help me?
Is there a formula for this? (If it is necessary, we can say that the available positions of the molecules are limited)
Thanks in advance.

From a pure classical thermodynamic point of view, there is no answer to your question because entropy is defined for equilibrium states only. The situation you describe is a succession of non-equilibrium states and classical thermodynamics cannot address it.

Statistical mechanics can allow you to calculate the difference in free energy (and hence also the difference in entropy) between the initial and final states even if the path that leads between them is a non-equilibrium one. The mixing process you describe is a non-equilibrium path. Using many molecular dynamics simulations and Jarzynski equation, you can compute the difference in free energy and consequently entropy. But you need to run many simulations to get a good ensemble average.

In order to get a time-dependent evolution of entropy, one needs a non-equilibrium definition of entropy. I think you can find such definitions in literature, but I'm not familiar with them and I'm not sure how useful they can be.

 

[zrek]

They don't expand to fill the space instantaneously. If you want to compute entropy as a function of time in the problem you described, you are dealing with times before the gases have had "filled whatever space they can".

I don't mind if the expansion is not instantaneous. It is perfect also if it is not instantaneous, but much faster then that my time resolution scale requires. In a computer simulation I can't give exact numbers for the other values too for every single moment, but I can calculate them with approximations, for example even with simple linear interpolation.
For example if there are several particles, they sometimes hit the wall of the chamber. I can't calculate the pressure value for every moment (or step) as the particles moving, but I can count continously how many particle hits a given area of the surface of the chamber for the last time section, so in a less exact time scale with a linear interpolation I can get pretty good result for the pressure. By this method I can state that the movement of the particles is so quick, that for the next analysed step of time the gas clearly fills the space, and this is a very good approximation.

No. Counting the possibilities is unrelated to computing probability unless you can define the microstates so each microstate has the same probability of being occupied. How do you intend to define the microstates?

I can map any actual configuration of the microstates of the particles to a macrostate. I can also define what and how many microstate configuration is equal for any macrostate. Isn't this enough? For example I can count how many particles are in one chamber and how many are in the other. I can also calculate how many possible microstate configuration results the same macrostate. I can clearly give a probability percentage for this current state compared to the all possible microstate configurations. For example I can calculate that "the current state have 35% chance to exist", and for sure I'll get the a maximum percentage if there are same number of particles in the two chambers. Isn't this percentage convertable to entropy somehow?

 

[zrek]

... you were looking for a simple example of a problem that could illustrate the evolution of the entropy vs time for a system experiencing an irreversible change. Is this correct? If so, would a simpler system be acceptable, or does it have to be the ideal gas example.

This is correct, the simplest system is acceptable as far as it can be modeled from deterministic microstates, since I'd like to run my modell in a computer. I'd like to see a number for entropy just like for the temperature and pressure, during the process. I haven't realized, but you got it right, it is my intension to create a spectacular demonstration of an irreversible change. (Thank you for making my intension clearer even for me :-) )

 

[zrek]

... Using many molecular dynamics simulations and Jarzynski equation, you can compute the difference in free energy and consequently entropy. But you need to run many simulations to get a good ensemble average.

This sounds interesting. Could you give me a very simple example for the connection between the free energy and the entropy?

 

[Stephen Tashi]

I don't mind if the expansion is not instantaneous. It is perfect also if it is not instantaneous, but much faster then that my time resolution scale requires. In a computer simulation I can't give exact numbers for the other values too for every single moment, but I can calculate them with approximations, for example even with simple linear interpolation.

You're completely ignoring the point that the "volume of a gas" is not defined unless the gas is at equilibrium.

I can map any actual configuration of the microstates of the particles to a macrostate.

You don't understand the definition of a "microstate". One microstate of a gas specifies the position and momentum of each molecule in the gas. If you are simulating the particles of a gas then at a given time, the gas is in a single microstate. It isn't one particle that occupies a microstate. It is the whole gas.

 

[zrek]

You're completely ignoring the point that the "volume of a gas" is not defined unless the gas is at equilibrium.

Maybe you are right if we take this rigorously, but there is no gas in a real environment in which the gas is in perfect equilibrium. For example in my very room the air flows because of my breathe, causing turbulence, maybe there are even several microliters of vacuum in it. Still we can say that it is in general in a given volume, since at average it fills up the space. I wanted to say that when you open a (tiny) door between the chambers, then the slow (relatively to the movement of the particles) process of balancing have no rigorous effect on determining the volume, since the gas in the two chambers in average have specific, slowly changing macroparameters (pressure and entropy) that can be calculated. Maybe I'm wrong.

You don't understand the definition of a "microstate". One microstate of a gas specifies the position and momentum of each molecule in the gas. If you are simulating the particles of a gas then at a given time, the gas is in a single microstate. It isn't one particle that occupies a microstate. It is the whole gas.

No, I understand this. I just mean that I can not only compute the actual microstate, but every time I can calculate every microstate and I can classify them. I can compare the current, single microstate to all of the possible ones, and draw conclusions, measure the probability. This method is similar to that if I run lots of simulations with random initial parameters, but better: I can map every possible states, and not only many of them.

 

[Stephen Tashi]

Maybe you are right if we take this rigorously, but there is no gas in a real environment in which the gas is in perfect equilibrium.

I agree. However, the thermodynamic laws that refer to "Volume" are based on a volume enclosing a gas in equilibrium. So for such laws to apply, the gas must at least be in an approximate equilibrium. For example, you can't claim $U'(t) = T S'(t) - P V'(t)$ if $V(t)$ is the volume of a container enclosing a gas that is not in equilibrium. In the scenario you describe, a partition is removed between two gases at different pressures. So time elapses before the gas mix and reach equilibrium. If you are interested in what happens during the time the gases are mixing, you are trying to deal with a (very) non-equilibrium state of the gases.

No, I understand this. I just mean that I can not only compute the actual microstate, but every time I can calculate every microstate and I can classify them. I can compare the current, single microstate to all of the possible ones

It's unlikely that the gas in one run of your simulation will visit each possible microstate - if "possible" refers to those microstates that might be visited by gases that began in slightly different initial conditions. (There are lots of initial conditions that are consistent with the verbal description that two gases, "each at equilibrium" are in chambers separated by a partition.)

Those microstates that the gas does occupy during the non-equilibrium phase of the simulation will likely be visited only once. The idea of "microstate" is that it is a very fine division of the intervals of position and momentum.

Can you give an example of how you intend to define the microstates for a 3 particle system? Can you list four of the possible microstates?

Furthermore, at a given time (e.g. 2 seconds) the gas will be in a single microstate. If the gas is not in equilibrium then to simulate the a probability distribution for what microstate the gas is in at t = 2 seconds you would have run multiple repetitions of the simulation with different initial conditions in order to explore the possible microstates at t = 2 seconds.

The microstate the gas is in at t = 100 seconds is not valid data for the probability distribution for the microstates of the gas at t = 2 seconds unless the gas is in equilibrium from t = 2 seconds to t = 100 seconds.

(Which illustrates why a very useful consequence of equilibrium is the "ergodic" property . For example if the gas is in equilibrium from t = 2 to t = 100 seconds then data at any time in that interval does provide valid information about the probability distribution for the gas visiting various microstates at t = 2 seconds.)

 

[zrek]

I agree. However, the thermodynamic laws that refer to "Volume" are based on a volume enclosing a gas in equilibrium. So for such laws to apply, the gas must at least be in an approximate equilibrium. For example, you can't claim $U'(t) = T S'(t) - P V'(t)$ if $V(t)$ is the volume of a container enclosing a gas that is not in equilibrium. In the scenario you describe, a partition is removed between two gases at different pressures. So time elapses before the gas mix and reach equilibrium. If you are interested in what happens during the time the gases are mixing, you are trying to deal with a (very) non-equilibrium state of the gases.

I try to understand what you mean and try to find a solution paralelly so I continue with questions if you don't mind, maybe you will have further ideas. (Thanks in advance)
So, in the initial conditions, with two separated chambers, I have clear volume values, each of the volumes of the chambers are 1 unit, total volume of the two separated gases are 2 units. (The particles have same properties, I use one type of gas molecules) Now I open the door between them. Say that the left chamber have higher pressure, so the gas flows from the left chamber to the right one. The total volume does not change. Isn't this information enough?
If not, then I assume you mean I have to calculate the two new volume values of the two gases. I'm sure that the sum of these values is 2 units. I have two options (at least):
1. As the time passes in tiny steps, I can simply make good approximation with regression, how the volume changes. During the process as much the left gas gain volume, the right gas loses the same amount, they "transfer" volumes. The pressure is known (I can count continously the hits on the wall of the chambers), by this I can say that if the pressure difference is big, then the volume "transfer" rate is higher, and later in the process the rate decreases, as the pressure balances.
2. I can easily follow every particle if necessary. As soon as I detect one particle stepping to the other chamber, I can say that particle shares the volume with those they were already there, with the ratio of the 1 to the number of the particles are already there. By this I can calculate the new volume values and next time when a new particle changes places, I can use it as base.

What do you think?

 

[Stephen Tashi]

I have clear volume values, each of the volumes of the chambers are 1 unit, total volume of the two separated gases are 2 units.

The volumes of the chambers are clear. But the volumes of the chambers don't make sense as the volumes of the gases after door is open and the gases begin to flow. Unless a gas uniformly fills a chamber, the "volume of the chamber" and the "volume of the gas" are not the same. In fact, if a gas does not uniformly fill a chamber, the "volume of the gas" has no standard definition.

2. I can easily follow every particle if necessary. As soon as I detect one particle stepping to the other chamber, I can say that particle shares the volume with those they were already there, with the ratio of the 1 to the number of the particles are already there.

I don't know what you mean by "shares the volume".

By this I can calculate the new volume values and next time when a new particle changes places, I can use it as base.

Perhaps if you give an example of that calculation, it will be clear what you mean.

The fact that you can do a calculation doesn't establish that the calculation has anything to do with the way "volume of a gas" is defined in thermodynamics. I agree that you can experiment with various calculations.

-----
You have not yet given an example of how you intend to define microstates.

 

[zrek]

In fact, if a gas does not uniformly fill a chamber, the "volume of the gas" has no standard definition.
...
The fact that you can do a calculation doesn't establish that the calculation has anything to do with the way "volume of a gas" is defined in thermodynamics.

If the "volume of the gas" have no standard definition for the situation, then you can't expect me to give one that is completely correct. This would be a scientific breaktrough :-)
However you are wrong if you state that I can't figure out a calculation that has anything to do with the way "volume of a gas" is defined in thermodynamics. In fact we know very much of it, we have clear limits and good way of approximations. Please do not simply ignore my logic. The upper limit is 2 units in the case I mentioned, since the volume of the gas can't be bigger than the volume of the chamber. Also, at the initial state we know for sure (because it fits perfectly to the standard definition you mentioned) that the volumes of each the gases are 1 unit. Also known that at the end of the process (since the molecules are the same) the volume of the two gases will be at the same ratio as the ratio of the number of their molecules. Also we know that the process is not instantaneous. During this process we can make good approximations for the volume changes for example by those two methods I mentioned.

I don't know what you mean by "shares the volume".
Perhaps if you give an example of that calculation, it will be clear what you mean.

If there is a gas A in a chamber and you put in gas B, then they will share the volume. For sure, if the A and B gases have different particle with different behavior, we are in trouble to determine in which ratio they share the space. But if they have the same type of molecules, then we can make a good approximation if we say that the two gases share the volume by the ratio of their number of particles. It is easy to calculate. Isn't this a good approximation?

-----
You have not yet given an example of how you intend to define microstates.

Currently I tried to answer in my other browser tab, but it takes more time, thank you for your patience.

 

[Stephen Tashi]

However you are wrong if you state that I can't figure out a calculation that has anything to do with the way "volume of a gas" is defined in thermodynamics. In fact we know very much of it, we have clear limits and good way of approximations. Please do not simply ignore my logic.

You will have to state the algorithm for the calculation before any logic can be evaluated.

The upper limit is 2 units in the case I mentioned, since the volume of the gas can't be bigger than the volume of the chamber. Also, at the initial state we know for sure (because it fits perfectly to the standard definition you mentioned) that the volumes of each the gases are 1 unit. Also known that at the end of the process (since the molecules are the same) the volume of the two gases will be at the same ratio as the ratio of the number of their molecules. Also we know that the process is not instantaneous. During this process we can make good approximations for the volume changes for example by those two methods I mentioned.

All that shows is that at the beginning of the process, your calculation of the volumes of the gases agrees with the thermodynamic definition and at the end of the process, when the gases have reached equilibrium the total volume of the mixed gases agrees with the thermodynamic definition. Any type of calculation that interpolated between those values could be called "logical" in that sense. If you reveal your formula , mathematics may prove that your interpolation is correct at the two extremes. But the logic of your argument doesn't explain why your particular way of interpolating a volume in between the extremes is better than any other way of doing it.

If there is a gas A in a chamber and you put in gas B, then they will share the volume. For sure, if the A and B gases have different particle with different behavior, we are in trouble to determine in which ratio they share the space. But if they have the same type of molecules, then we can make a good approximation if we say that the two gases share the volume by the ratio of their number of particles. It is easy to calculate. Isn't this a good approximation?

How would we judge if it was "good" approximation? I agree that we can select criteria for "good" and test if the approximation is good by those criteria. (For example, we could test the gas law PV = KT and see if the law was upheld by that definition of volume V. ) Without specific criteria, there is no way to tell whether you method is a "good" approxmation.

Currently I tried to answer in my other browser tab, but it takes more time, thank you for your patience.

Ok. This is an interesting discussion. Before this discussion, I didn't pay attention to the fact that the "thermodynamics" taught in textbooks is properly "equilibrium thermodyamics".

 

[zrek]

This is an interesting discussion. Before this discussion, I didn't pay attention to the fact that the "thermodynamics" taught in textbooks is properly "equilibrium thermodyamics".

I think also that this is interesting and I can learn from you. Finally it would be great if I can find a solution for my problem :-) Please don't be upset if I repeat myself, I just try to understand the things and find an acceptable way.

Before I continue, I'd like to know what do you accept or agree from my statements. Please classify my points one by one. The situation is the same I already described: two chambers, same type of molecules (every particle have the same speed and mass), same volumes, only the number of paricles are different, in the left chamber there is a higher pressure.
1. With the door closed, the total volume of the gases is 1+1=2 units, they are in equilibrium, and this fits to the standard definition. (agree or disagree?)
2. If we open the door, the volumes does not change instantaneously (so at the beginning the volume of the gases are 1 and 1 units even if the door is open) (agree or disagree?)
3. At the end, when the two gases mixed and reach the equilibrium, together their volume is also 2 units, they are in equilibrium, and this fits to the standard definition. (agree or disagree?)
4. During the process, the volume of gas from the left chamber increases, while the volume of the other gas decreases. (a or d?)
5. During the process the total volume of the gases are always 2 units. (a or d?)

Thank you!

 

[Stephen Tashi]

1. With the door closed, the total volume of the gases is 1+1=2 units, they are in equilibrium, and this fits to the standard definition. (agree or disagree?)

Agree - assuming the gases are each at equilbrium.

2. If we open the door, the volumes does not change instantaneously (so at the beginning the volume of the gases are 1 and 1 units even if the door is open) (agree or disagree?)

Agree

3. At the end, when the two gases mixed and reach the equilibrium, together their volume is also 2 units, they are in equilibrium, and this fits to the standard definition. (agree or disagree?)

Agree - the volume of the single gas that results from the combination of the two gases is 2 units.

4. During the process, the volume of gas from the left chamber increases, while the volume of the other gas decreases. (a or d?)

Disagree - because how we define the volume of gas in a non-equilibrium situation had not been decided. Particles from each chamber can go into the other chamber. So if we identify each gas by the particles originally in it, particles from the gas in the chamber initially at lower pressure chamber might go into the chamber that initially had the higher pressure chamber.

5. During the process the total volume of the gases are always 2 units. (a or d?)

Disagree - again because there is no established definition for the volume of a gas that is not in equilibrium.

Suppose we have chamber_A with a volume of 1 unit containing a gas and door that separates it from chamber_B that has a volume of 1,000,000 units and is empty. When the door is opened an a few molecules of gas go into chamber_B , are we to say the combined volume of the gases in each chamber is 1,000,001 units? The first few molecules of the gas that go into chamber_B might be nowhere near a wall of that chamber. At that time, what difference does it make whether chamber_B has a volume of 1,000,000 units or 20,000,000 units?

Consider a cloud (of water vapor) in the sky. How do you define its volume? Do you define it as the volume of the entire atmosphere because the atmosphere "contains" the cloud? Why should the volume of a chamber always have a direct relation to the volume of something within the chamber ?

 

[zrek]

OK, then first I have to try to convince you about the disagreed ones before I can step forward.

4. During the process, the volume of gas from the left chamber increases, while the volume of the other gas decreases. (a or d?)

Disagree - because how we define the volume of gas in a non-equilibrium situation had not been decided. Particles from each chamber can go into the other chamber. So if we identify each gas by the particles originally in it, particles from the gas in the chamber initially at lower pressure chamber might go into the chamber that initially had the higher pressure chamber.

You mentioned 2 things against it:
4.a: The volume of gas had not been decided in a non-equilibrium situation.
4.b: Particles from lower pressure chamber may go into the higher pressured one.

4.b is not a problem. It is clear that if we run a simulation, we will see that in average more particles will go from the higher pressured place than from the lower pressured one. In the point 4. I ment the volume increases and decreases in average.
Lets refine my statement, please answer (please don't forget that initially the only difference between the contents of the chambers is the number of the particles):
4.2: During the process more particles will go from the higher pressured place to the lower one than to the opposite direction. (a or d?)

4.a is not a real problem.
Please note that you agreed 3. and 2.
That means:
At the end of the process, the particles of the two gases mixed, occupy the available space with the same rights. Every single particle occupies the same amount volume partition in average (sorry for my bad english, I hope you understand what I mean). And since the number of the particles from the left chamber is higher, they occupy more space from the available 2 units. Since there is no other difference, also we can say that the ratio between the volume of the two gases is the same as the ratio between their number of particles. (For example if the left chamber contained 2000 particles and the right camber only 1000, then after the mixing, the total available 2 units of volume also shared between the gas molecules with 2:1 ratio)
At the beginig of the process the volume of the particles was exactly 1 unit from each of the chamber, 1:1 ratio (with opened door too).

I refine my statement here too (please note it is based on the point 2 and 3 that you already accepted)
4.1: At the beginning the gas from the left chamber had 1 unit volume, and at the end it had more. Also, at the beginning the gas from the right chamber had 1 unit of volume, but at the end it had less. This volume change was not happened instantaneously, but during the process. (a or d?)

Suppose we have chamber_A with a volume of 1 unit containing a gas and door that separates it from chamber_B that has a volume of 1,000,000 units and is empty. When the door is opened an a few molecules of gas go into chamber_B , are we to say the combined volume of the gases in each chamber is 1,000,001 units? The first few molecules of the gas that go into chamber_B might be nowhere near a wall of that chamber. At that time, what difference does it make whether chamber_B has a volume of 1,000,000 units or 20,000,000 units?

Earlier I also was thinking about this situation, but simply ignored it, as it is an edge problem like the division by zero. But now you forced me to give an explanation.
If the first molecule comes out from the chamber, the main problem is not with the volume, but with the definition of gas. One molecule is not gas. For example one molecule of water can not be solid, fluid or gas. Also problematic to measure its temperature I think you also know why. But in our case it is not a problem. As I mentioned in my earlier post, we can wait until these parameters have meaning, when we can measure these. For example when the first few molecules reach the walls of the chamber_B, I can calculate the pressure in it. Our measuring time scale must fit to the processes of the changes of microstates: for the calculation of the macro values like temperature, pressure, we have to wait a lot compared to the quick movements of the particles. Can you accept this?