# Homework Help: Expansion of ideal gas into a vacuum

1. Jan 13, 2007

### erty

I have a box divided by a partition. On one side I have n mole ideal gas, on the other side there's vacuum.
If I remove the partition the ideal gas will expand into the vacuum. Since the box is adiabatic and no work is applied to or by the gas, $$\Delta U = 0$$.

But what about the work used when I remove the partition? Is this so small that we neglect it?

2. Jan 13, 2007

### Staff: Mentor

How big/heavy is the partition?
What is the work being done on?
What work is the problem asking for?

3. Jan 13, 2007

### Crosson

Ignoring the work done to remove the partition is a rephrasing of the frictionless assumption: instead of removing the partition imagine the gas expanding against a frictionless piston.

4. Jan 13, 2007

### Staff: Mentor

Well, you'd also need to assume a massless piston...

Really, the problem is fine on its own.

5. Jan 14, 2007

### Andrew Mason

The gas does not do work to remove the partition so ignore it.

You are right that after the gas reaches equilibrium, it should have the same internal energy. This is provided the walls do not absorb the energy of the gas. So you would have to add here a condition that the walls are absolutely rigid. (incidentally, it is the process that is adiabatic - no heat flow in or out - not the container. the container is insulated).

The problem here is that the gas actually does do work. It does work on itself, giving its centre of mass velocity (ie. kinetic energy). It ceases being in equilibrium while it is expanding into the vacuum. When it is in this dynamic state its temperature is not really defined. The blast of freely expanding gas carrying this mechanical energy impacts the end wall of the container and the gas bounces back. Provided the walls do not move (ie. do not aquire mechanical energy) the mechanical energy of the gas will eventually go into heat, returning the gas back to its original temperature when it reaches equilibruim.

AM

6. Jan 14, 2007

### erty

This is Joule's experiment. He found out that the internal energy U of an ideal gas only depends on the temperature.

7. Jan 14, 2007

### erty

Well, that's the problem. I consider the box with the gas to be my system and everything else to be the surroundings/universe.
But what if I turn it the other way around? If I consider everything but the box to be my system, and I pull up the partition, then I've applied work on the surroundings (by removing the partition).
What is the problem? When I consider the initial system, that is the box with ideal gas, $$\Delta S > 0$$. But I do not account for the work used to lift the partition! I could assume the partition/pistion to be massless, but then I would turn the whole thing into a thought experiment (which it already resembles because of the ideal gas-thing - I never quite understood the concept to its fully :p).

A perfectly rigid wall? That must be a idealistic borderline case...

Could you explain that a little further? I don't think I understand.

This must be the real gas vs. ideal gas scenario, right? Do you talk about the properties of a real gas, or does this apply to all kinds of gas, real and ideal?

Last edited: Jan 14, 2007
8. Jan 15, 2007

### loom91

Andrew has given a nice reply here. erty, what exactly is your question? What do you want to know? Whether or not you consider the objects to be ideal depends on the level of complexity you are prepared to accept in your work. However the ideal gas assumption is not really required here, the properties of the gas are irrelevant in this particular scenario (I can't call it a problem as nothing has been asked).

But if you are concerned with the evolving state of the system rather than simply the equilibrium thermodynamics of the initial and the final state, then you must take into account the properties of the gas. For example, an ideal gas does no work against its intermolecular forces when expanding, but a real gas (modelled by say Fermi-Dirac statistics) does. But this does not affect the equilibrium thermodynamics, only the intermediate non-equilibrium states.

Molu

9. Nov 3, 2008

### Simulana

There is no inherent problem here. If you consider the gas to be your system, then the system has not done work to remove the box lid. Thus, the system (the gas) is adiabatic. However, if you consider the system to be everything else, then yes, you did work on the box. But you did not do any work on the gas (in a frictionless assumption, of course).

If you consider the system to be box + gas, then yes, you did work on the system, but the outcome was an increase in gravitational potential energy for the box lid (assuming you lifted it up), and thus no heat was transferred. But really, the whole point of this exercise is that it is the *gas* that is moving adiabatically.

It's true...a perfectly rigid wall would be completely insulating - which implies we could reach absolute zero with no problems. Obviously not true. However, the amount of heat lost over time by molecules bouncing off the other 5 sides of the box is miniscule over the time scale of adiabatic expansion.