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.
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.