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I was wondering what causes the difference between adiabatic expansion and Joules expansion? Because to me both seems to same. In a sense that in case joules expansion we open the valve in order to let gas expand on its own. In this case after a while the pressure inside the chamber change, volume change but not the temperature.

If I imagine the molecules seems to travel at same speed even if we let the valve open which causes the velocity of the particles to remain same.

So I want to know why cant I imagine same thing in case of adiabatic expansion ?

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The opposite of a conserved, adiabatic process, is just a non-conserved diabatic process. I have modeled the universe in my own approaches using diabatic processes.

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The opposite of a conserved, adiabatic process, is just a non-conserved diabatic process. I have modeled the universe in my own approaches using diabatic processes.

Does it make any sense to you? I wanted to know, why they are different,If you can help?

Does it have anything to do with the velocity of expansion being different than the velocity of wall?

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I was wondering what causes the difference between adiabatic expansion and Joules expansion? Because to me both seems to same. In a sense that in case joules expansion we open the valve in order to let gas expand on its own. In this case after a while the pressure inside the chamber change, volume change but not the temperature.

If I imagine the molecules seems to travel at same speed even if we let the valve open which causes the velocity of the particles to remain same.

So I want to know why cant I imagine same thing in case of adiabatic expansion ?

The difference is that, in an adiabatic expansion, work is done by the expanding gas: the system is thermally isolated from its surroundings, but work can be extracted. Whereas in Joule expansion, no work is done when the expansion occurs, because the expansion is into a vacuum.

Since no work is done, the mean kinetic energy of the molecules is the same at the end of the process as at the start. And it is mean kinetic energy that determines temperature.

In the adiabatic case, the gas molecules exert a pressure on some surface that moves in response, thereby exerting a force through a distance: F x d = Work. This is reflected in the loss of some of their momentum and thus energy, as they are rebounding from a moving surface. And so the temperature drops.

Edited by exchemist
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The difference is that, in an adiabatic expansion, work is done by the expanding gas: the system is thermally isolated from its surroundings, but work can be extracted. Whereas in Joule expansion, no work is done when the expansion occurs, because the expansion is into a vacuum.

Since no work is done, the mean kinetic energy of the molecules is the same at the end of the process as at the start. And it is mean kinetic energy that determines temperature.

In the adiabatic case, the gas molecules exert a pressure on some surface that moves in response, thereby exerting a force through a distance: F x d = Work. This is reflected in the loss of some of their momentum and thus energy, as they are rebounding from a moving surface. And so the temperature drops

Isn't it that in normal case in adiabatic expansion the work done by the gas must be induced which is stored as the potential energy.

for example in NTP an ideal gas inside a fully insulated container can be expanded by reducing the pressure inside the wall . How can we achieve it ? By pulling the piston .

In this case how would the gas inside the container knows that it is doing work ? Why can't we just consider adiabatic  expansion as the sum of small Joules expansion /

I know I am missing some key points right? Can you clarify what?

And how does it lose momentum. Isn't it redistributed to produce different energies since the velocity of the wall is supposed to constant?

I think it somehow arrange the momentum differently in case of adiabatic and Joules expansion.

And can we get different parameters of p,t,v when any one of them is predetermined .

Edited by Nishan
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Isn't it that in normal case in adiabatic expansion the work done by the gas must be induced which is stored as the potential energy.

for example in NTP an ideal gas inside a fully insulated container can be expanded by reducing the pressure inside the wall . How can we achieve it ? By pulling the piston .

In this case how would the gas inside the container knows that it is doing work ? Why can't we just consider adiabatic  expansion as the sum of small Joules expansion /

I know I am missing some key points right? Can you clarify what?

And how does it lose momentum. Isn't it redistributed to produce different energies since the velocity of the wall is supposed to constant?

I think it somehow arrange the momentum differently in case of adiabatic and Joules expansion.

And can we get different parameters of p,t,v when any one of them is predetermined .

I think you may need to slow down a bit and try to be more exact with your English. I am finding it difficult to follow what you are now asking.

The gas does not need to "know" it is doing work. But, as I said, if there is a moving piston, a molecule rebounding elastically from it will rebound at a velocity equal and opposite to the velocity at which it hits it, but relative to the piston surface, which is in motion. So from the point of view of the static gas, the velocity of the rebounding molecules has been slightly reduced. The piston surface resists the expansion, forcing the molecules to rebound off it. In other words the molecules push, with the force of pressure x piston area, against this resistance and thus do work, as soon as the piston is allowed to move in response to the pressure.

(If a molecule moving at v rebounds elastically from a surface moving in the same direction at δv, its velocity with respect to the moving surface will be v- δv. It will rebound from the surface with the same velocity in the opposite direction i.e. at v-δv relative to the moving surface. But from the point of view of the static gas, this will be v-2δv. So the molecule has lost kinetic energy, which it has given up to the moving surface).

In the case of Joule expansion, there is no rebounding at all as there is no piston: the molecules just shoot through the opening with nothing to resist them. No work is done.

You can turn the adiabatic case into the Joule case only if you have an infinitely light piston offering no resistance so no work is done by moving it. That would mean it moves out of the way, at infinite speed, in response to the first molecule hitting it and without causing any change to the trajectory of the molecule. In other words, it would mean it is not really there!

Edited by exchemist
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I think you may need to slow down a bit and try to be more exact with your English. I am finding it difficult to follow what you are now asking.

The gas does not need to "know" it is doing work. But, as I said, if there is a moving piston, a molecule rebounding elastically from it will rebound at a velocity equal and opposite to the velocity at which it hits it, but relative to the piston surface, which is in motion. So from the point of view of the static gas, the velocity of the rebounding molecules has been slightly reduced. The piston surface resists the expansion, forcing the molecules to rebound off it. In other words the molecules push, with the force of pressure x piston area, against this resistance and thus do work, as soon as the piston is allowed to move in response to the pressure.

(If a molecule moving at v rebounds elastically from a surface moving in the same direction at δv, its velocity with respect to the moving surface will be v- δv. It will rebound from the surface with the same velocity in the opposite direction i.e. at v-δv relative to the moving surface. But from the point of view of the static gas, this will be v-2δv. So the molecule has lost kinetic energy, which it has given up to the moving surface).

In the case of Joule expansion, there is no rebounding at all as there is no piston: the molecules just shoot through the opening with nothing to resist them. No work is done.

You can turn the adiabatic case into the Joule case only if you have an infinitely light piston offering no resistance so no work is done by moving it. That would mean it moves out of the way, at infinite speed, in response to the first molecule hitting it and without causing any change to the trajectory of the molecule. In other words, it would

Thank you for the answer. Now I get it . And sorry for the bad english

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Thank you for the answer. Now I get it . And sorry for the bad english

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