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#1 Mark

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Posted 18 August 2012 - 07:39 AM

reading a book on thermodynamics.

the expansion of an ideal gas is considered to be a process in which entropy is supposed to increase. right?

so if we have two equal volume connected containers and a 'removable partition' between them, one of the containers is filled with a mole of ideal gas, and the other one empty.
the partition is removed and the ideal gas equally fills both containers. yes.
the total energy of the ideal gas through the process is not changed. (the removing of the partition does not take kinetic energy from the molecules)
there is no work done on the system.
and there is no heat entering the system.
the temperature is (presumably) non-zero.
and the entropy increases...

gibbs equation:

dU = TdS - PdV

A contradiction?


dX = change in X
U = total internal energy
T = temperature
S = entropy
P = pressure
V = volume
note that PdV = work by the system.

Edited by Mark, 18 August 2012 - 07:41 AM.


#2 Egann

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Posted 18 August 2012 - 11:59 AM

Uhh, not exactly. I don't understand entropy too well, but I think you're making a mistake. As that the one container has no gas in it, it's initial entropy is theoretically zero and is an embodiment of perfect order. Consequently, releasing the mole of gas to fill both volumes averages the entropy across both containers and, if you consider both containers to be a single system, the entropy rises. The entropy does decrease in the one half, though, because it's only half of the system.

Correct me if I'm horribly misunderstanding what you mean, though. I don't do differential equations.

#3 arunma

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Posted 18 August 2012 - 12:48 PM

I think you may be forgetting that the gas does work by expanding, so negative work is done on the gas. Also, the temperature doesn't stay the same because the gas has to expand into the second container, and an expanding gas cools. I haven't done the calculation in awhile, but I'm pretty sure what will happen is that the gas will do work at the expense of cooling, and increase in entropy.

Btw Egann, actually entropy isn't defined for a system of zero particles. So before removing the partition it doesn't really make sense to talk about the entropy of whatever's inside the second container (because there's nothing to talk about).

#4 Mark

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Posted 19 August 2012 - 05:00 AM

I think you may be forgetting that the gas does work by expanding, so negative work is done on the gas. Also, the temperature doesn't stay the same because the gas has to expand into the second container, and an expanding gas cools. I haven't done the calculation in awhile, but I'm pretty sure what will happen is that the gas will do work at the expense of cooling, and increase in entropy.

Btw Egann, actually entropy isn't defined for a system of zero particles. So before removing the partition it doesn't really make sense to talk about the entropy of whatever's inside the second container (because there's nothing to talk about).


Ok, so firstly - I believe that there is no work done on the gas. I hardly see how the gas is pushing against anything if it is expanding into a vacuum.
we could (in our minds) construct the walls of both containers out of infinitely rigid infinitely thin material, hence no deformation (and no work on the containers) and no heat transfer from the containers (as they have neg. mass) and we will -for argument sake- assume that the removal of the partition itself does no significant work on the gas.
and particularly as it is an ideal gas, the forces between the atoms will be zero, so there is no kinetic to potential energy conversion in expansion.

perhaps the temperature does change, I dont see how this should be relevant - unless of course it is zero.

However it is my belief that the internal energy of the gas should not change on expansion (and here I may be mistaken), If you imagine the atoms of the ideal gas bouncing around (if even passing through each other) with no loss of kinetic energy in one chamber, if the partition is removed, I fail to see how this fact alone should slow the particles down.



EDIT: ok, I now know....... because the gas exerts a pressure, and there is a change in volume there is "work" done.... <_< thermodynamics feels like terrible science! there is no work done, but there is "work" done....

Edited by Mark, 19 August 2012 - 05:05 AM.


#5 arunma

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Posted 19 August 2012 - 04:37 PM


I think you may be forgetting that the gas does work by expanding, so negative work is done on the gas. Also, the temperature doesn't stay the same because the gas has to expand into the second container, and an expanding gas cools. I haven't done the calculation in awhile, but I'm pretty sure what will happen is that the gas will do work at the expense of cooling, and increase in entropy.

Btw Egann, actually entropy isn't defined for a system of zero particles. So before removing the partition it doesn't really make sense to talk about the entropy of whatever's inside the second container (because there's nothing to talk about).


Ok, so firstly - I believe that there is no work done on the gas. I hardly see how the gas is pushing against anything if it is expanding into a vacuum.
we could (in our minds) construct the walls of both containers out of infinitely rigid infinitely thin material, hence no deformation (and no work on the containers) and no heat transfer from the containers (as they have neg. mass) and we will -for argument sake- assume that the removal of the partition itself does no significant work on the gas.
and particularly as it is an ideal gas, the forces between the atoms will be zero, so there is no kinetic to potential energy conversion in expansion.

perhaps the temperature does change, I dont see how this should be relevant - unless of course it is zero.

However it is my belief that the internal energy of the gas should not change on expansion (and here I may be mistaken), If you imagine the atoms of the ideal gas bouncing around (if even passing through each other) with no loss of kinetic energy in one chamber, if the partition is removed, I fail to see how this fact alone should slow the particles down.



EDIT: ok, I now know....... because the gas exerts a pressure, and there is a change in volume there is "work" done.... <_< thermodynamics feels like terrible science! there is no work done, but there is "work" done....


Oh, and I had a big long explanation planned out in my mind before I saw the edit...

I forget how exactly internal energy is defined, but if it doesn't account for potential energy then it absolutely makes sense that the internal energy decreases here. We know that the change in internal energy is the head added to the gas minus the work done by the gas. If gas expands then it's doing work. It cools, but there's no heat being added or removed (remember that temperature and heat aren't the same thing). So the internal energy would decrease. It's pretty closely analogous to compressing a spring and then letting it expand back to its equilibrium state. If you only start looking at the system after you compress it, you see the spring evolving without any external force; it decreases in internal energy, and you go "huh?" But if you realize that someone had to compress the spring (or in this case gas) to begin with, then it makes sense.

#6 Mark

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Posted 20 August 2012 - 08:07 AM



I think you may be forgetting that the gas does work by expanding, so negative work is done on the gas. Also, the temperature doesn't stay the same because the gas has to expand into the second container, and an expanding gas cools. I haven't done the calculation in awhile, but I'm pretty sure what will happen is that the gas will do work at the expense of cooling, and increase in entropy.

Btw Egann, actually entropy isn't defined for a system of zero particles. So before removing the partition it doesn't really make sense to talk about the entropy of whatever's inside the second container (because there's nothing to talk about).


Ok, so firstly - I believe that there is no work done on the gas. I hardly see how the gas is pushing against anything if it is expanding into a vacuum.
we could (in our minds) construct the walls of both containers out of infinitely rigid infinitely thin material, hence no deformation (and no work on the containers) and no heat transfer from the containers (as they have neg. mass) and we will -for argument sake- assume that the removal of the partition itself does no significant work on the gas.
and particularly as it is an ideal gas, the forces between the atoms will be zero, so there is no kinetic to potential energy conversion in expansion.

perhaps the temperature does change, I dont see how this should be relevant - unless of course it is zero.

However it is my belief that the internal energy of the gas should not change on expansion (and here I may be mistaken), If you imagine the atoms of the ideal gas bouncing around (if even passing through each other) with no loss of kinetic energy in one chamber, if the partition is removed, I fail to see how this fact alone should slow the particles down.



EDIT: ok, I now know....... because the gas exerts a pressure, and there is a change in volume there is "work" done.... <_< thermodynamics feels like terrible science! there is no work done, but there is "work" done....


Oh, and I had a big long explanation planned out in my mind before I saw the edit...

I forget how exactly internal energy is defined, but if it doesn't account for potential energy then it absolutely makes sense that the internal energy decreases here. We know that the change in internal energy is the head added to the gas minus the work done by the gas. If gas expands then it's doing work. It cools, but there's no heat being added or removed (remember that temperature and heat aren't the same thing). So the internal energy would decrease. It's pretty closely analogous to compressing a spring and then letting it expand back to its equilibrium state. If you only start looking at the system after you compress it, you see the spring evolving without any external force; it decreases in internal energy, and you go "huh?" But if you realize that someone had to compress the spring (or in this case gas) to begin with, then it makes sense.


ok so first things first, it is an ideal gas therefore it dosn't have potential energy. only kinetic energy, the sum of which I believe is the internal energy. I am led to believe that ideal gas molecules colide briefly and rarely.

If gas expands then it's doing work. It cools, but there's no heat being added or removed ... So the internal energy would decrease

I disagree, firstly because I cant imagine that the ideal gas molecules would spontaneously slow down because of the removal of the partition, but also because my text says otherwise, so I quote:

chapter 4.2 of "Basic Thermodynamnics" by Gerald Carrington

A straitforward illustration of entropy creation is given by the free expansion of an ideal gas .... blah blah.... consists of two volumes .... partition suddenly removed .... gas will fill larger space ... We can now calculate the change in entropy due to the change from initial to final equilibrium states. Since no work is done, and no heat is exchanged with the surroundings, Q=0, W=0, and therefore dU=0. hence Ta=Tb (final temp = initial temp) and: Sb-Sa = NRln(Vb/Va),


I believe that the temperature does not change in this case is because the process is not "Quasi-static" (ie. that the gas is not in equalibrium throughout the process), if it were quasi-static, then the change in volume would happen continuously (not suddenly), and the only way this can happen is if there is a wall (or something) gradually moving such as to expand the volume.
in this case it is true that the gas molecule collisions with the gradually moving wall will involve them loosing kinetic energy to it, hence the temperature and internal energy of the gas will decrease as you say.
there is actual mechanical work done, the gas molecules exert a force on the wall and it moves in the direction of the force.

however in the case in which a partition is removed there is a sudden change, and there is no movement of anything except the gas molecules themself (except of course the momentary removal of the partition, and assuming that the containers are perfectly rigid). what do you propose that the gas molecules succesfully do work on? themself? - remember no potential energy.

I think that I might be right on this one, I will share what I have learnt.
I didnt pay significant enough attention to this line.

Chapter 4.1:
...
dU = TdS - PdV
...
(for reason unstated) But remember, we can interpret the two terms on the right hand side of the Gibbs equation as Heat and Work only when the change of state occurs quasi-statically.

In this case, I believe that there is actually - and genuinely - no actual mechanical work performed at all, yet there is a pressure which the gas exerts and there is a change of volume, thus there is PdV..... - and no, there isnt any work.... lol

Edited by Mark, 20 August 2012 - 08:13 AM.


#7 J-Roc

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Posted 29 August 2012 - 08:39 AM

However it is my belief that the internal energy of the gas should not change on expansion (and here I may be mistaken), If you imagine the atoms of the ideal gas bouncing around (if even passing through each other) with no loss of kinetic energy in one chamber, if the partition is removed, I fail to see how this fact alone should slow the particles down.


I thought the fact that your gas is dissipating would slow the particles down because they aren't colliding and creating as much energy anymore.

#8 Veteran

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Posted 29 August 2012 - 06:09 PM

gibbs equation:

dU = TdS - PdV

A contradiction?


dX = change in X
U = total internal energy
T = temperature
S = entropy
P = pressure
V = volume
note that PdV = work by the system.


Forgive me if I've missed something obvious, but I don't see the contradiction. By removing the partition you increase the volume and so the entropy also increases. The equation is balanced.

I always considered entropy to be the bastard child of physics and chemistry. Too many uses of the word 'work' for my liking.

#9 Mark

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Posted 30 August 2012 - 02:03 AM


However it is my belief that the internal energy of the gas should not change on expansion (and here I may be mistaken), If you imagine the atoms of the ideal gas bouncing around (if even passing through each other) with no loss of kinetic energy in one chamber, if the partition is removed, I fail to see how this fact alone should slow the particles down.


I thought the fact that your gas is dissipating would slow the particles down because they aren't colliding and creating as much energy anymore.


An 'ideal' gas can be visualised as a room full of pingpong balls, all flying around, under no significant gravity, each bouncing off each other without loosing energy in collisions.
immagine that there are two rooms connected by a door. and there are two pingpong balls flying around in room 1, the door is opened to room 2, and one of the balls enters room 2.
why would the ball that entered room 2 slow down because it passed thorugh the door?


gibbs equation:

dU = TdS - PdV

A contradiction?


dX = change in X
U = total internal energy
T = temperature
S = entropy
P = pressure
V = volume
note that PdV = work by the system.


Forgive me if I've missed something obvious, but I don't see the contradiction. By removing the partition you increase the volume and so the entropy also increases. The equation is balanced.

I always considered entropy to be the bastard child of physics and chemistry. Too many uses of the word 'work' for my liking.


lol, yes. I hate the term 'work' and 'entropy' too.
you are right, the equation is balanced. I didnt put the contradiction clearly.
the removal of the partition does no work - force times displacement - on the gas. but work is supposed to be Pressure times change in volume. and there is a pressure and a change in volume.

the equation is balanced, and there isnt a lasting contradiction. you are right.

#10 J-Roc

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Posted 30 August 2012 - 11:39 AM



However it is my belief that the internal energy of the gas should not change on expansion (and here I may be mistaken), If you imagine the atoms of the ideal gas bouncing around (if even passing through each other) with no loss of kinetic energy in one chamber, if the partition is removed, I fail to see how this fact alone should slow the particles down.


I thought the fact that your gas is dissipating would slow the particles down because they aren't colliding and creating as much energy anymore.


An 'ideal' gas can be visualised as a room full of pingpong balls, all flying around, under no significant gravity, each bouncing off each other without loosing energy in collisions.
immagine that there are two rooms connected by a door. and there are two pingpong balls flying around in room 1, the door is opened to room 2, and one of the balls enters room 2.
why would the ball that entered room 2 slow down because it passed thorugh the door?


Great question, Mark! Honestly, I thought it was because the collision were creating friction and heat. My understanding was that a gas in a compressed space would move faster because the particles are more likely to collide and thus are heated. Is that wrong?

#11 Mark

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Posted 31 August 2012 - 10:46 PM




However it is my belief that the internal energy of the gas should not change on expansion (and here I may be mistaken), If you imagine the atoms of the ideal gas bouncing around (if even passing through each other) with no loss of kinetic energy in one chamber, if the partition is removed, I fail to see how this fact alone should slow the particles down.


I thought the fact that your gas is dissipating would slow the particles down because they aren't colliding and creating as much energy anymore.


An 'ideal' gas can be visualised as a room full of pingpong balls, all flying around, under no significant gravity, each bouncing off each other without loosing energy in collisions.
immagine that there are two rooms connected by a door. and there are two pingpong balls flying around in room 1, the door is opened to room 2, and one of the balls enters room 2.
why would the ball that entered room 2 slow down because it passed thorugh the door?


Great question, Mark! Honestly, I thought it was because the collision were creating friction and heat. My understanding was that a gas in a compressed space would move faster because the particles are more likely to collide and thus are heated. Is that wrong?


fair enough.
In discussing 'thermodynamics' one of the models used of a substance is that of an 'idea' gas. which is an approximation of real gasses consisting of treating gas molecules like bouncing ping-pong balls.
and that these ping-pong balls bounce of each other 'elastically' - meaning no energy lost from the motion in a colision.
and in that case for energy concervation, no energy is moved from the motion to anywhere else.
the 'ideal' gas is an approximation. In real gasses I understand that there is indeed energy lost in a colision - to EM blackbody type radiation.
however this amount of energy isnt particularly big (or so I understand) hence the ideal model works as an approximation for most gas behavior.

#12 arunma

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Posted 01 September 2012 - 12:29 AM

Mark, you've got a point. Excuse me while I go back and dig out one of my old textbooks so I can see how exactly internal energy is defined.