Thermodynamics (work done by unrestrained gas expansion)

[albertov123]

https://www.physicsforums.com/attachment.php?attachmentid=20388&stc=1&d=1252066499When the partition removed, gas flows into the vacuum until all system has a uniform pressure. Now, is it possible to calculate the work?

Let's say gas is not an ideal gas.(Freon 12)
We know the volumes of both sides. Also the mass and the initial pressure of gas.

I think we can't use W=PdV because pressure is not constant. Is there a way to do it?

Edit: By the way the temperature will change but in the end returns to its initial value. So heat transfer will occur through the boundaries.

 

[MarcusAgrippa]

https://www.physicsforums.com/attachment.php?attachmentid=20388&stc=1&d=1252066499When the partition removed, gas flows into the vacuum until all system has a uniform pressure. Now, is it possible to calculate the work?

Let's say gas is not an ideal gas.(Freon 12)
We know the volumes of both sides. Also the mass and the initial pressure of gas.

I think we can't use W=PdV because pressure is not constant. Is there a way to do it?

Edit: By the way the temperature will change but in the end returns to its initial value. So heat transfer will occur through the boundaries.

Since the chamber into which the gas expands is a vacuum, the gas pushes against nothing at all. So, if the outer walls are rigid, the gas does no work. Not so?

The expansion is rapid, much faster than the thermal adjustment time, so the expansion process involves zero heat intake. The internal energy will thus remain constant in the expansion process. If the gas is ideal, this means that the temperature remains the same. If the gas is non ideal, then the kinetic energy of the molecules will suffer as a result of the expansion, leading to a change in temperature. If the forces between the molecules are attractive, then kinetic energy will drain into potential energy, since the average separation of the molecules increases due to the expansion, leading to a drop in temperature.

The reason you cannot use dW = p dV is because this expression for the work done applies only to reversible processes, which are necessarily quasi static. Free expansion is very rapid and definitely not reversible - apart from which pressure waves will travel through the chambers until the pressure equilibrates, which means that between the initial equilibrium state and the final one there is no uniquely defined single value of the pressure in the chamber - the pressure needs to be specified by a pressure field, until it reaches its final constant equilibrium value.

Am I being too naive? Did I misunderstand your question?

 

[albertov123]

Thank you, i understand why there is no work done in the process.