3
$\begingroup$

I came across this paper:

Baker, B. (1999). An easy to perform but often counterintuitive demonstration of gas expansion. American Journal of Physics, 67(8), 712-713. http://scitation.aip.org/content/aapt/journal/ajp/67/8/10.1119/1.19357

And I didn't fully understand the explanation. I can't see how this is any different from the Joule free expansion experiment with no temperature change.

Essentially the author of the article says that the atmosphere does work on the gas and increases its internal energy. This is the reason for the temperature increase. However wouldn't that logic also lead to a temperature increase in the joule free expansion?

Any insight would be much appreciated.

$\endgroup$
3
$\begingroup$

There is a huge difference with the usual Joules experiment. Here the system is not isolated during the gas expansion into the evacuated chamber. Two things enter in competition in principle when a setup as the one considered in the article you mention is considered:

  • Whether the gas has to work to expand itself in the larger chamber
  • Is there anything from outside that helps the gas expanding?

Having an evacuated chamber deals with point 1 and tells us that the gas does not have to work (and hence loose energy) to expand in the chamber to gain extra room. The fact that the whole system is connected to a piston in contact with atmospheric pressure tells us that the system is a) not isolated and b) that the atmospheric pressure will "help" the gas expanding in the chamber.

Now, assuming an expansion that is quasi-static, the pressure on the piston has to be balanced at every time during the expansion process. Basically, the piston will work a little bit each time there is a pressure drop on the gas side (due to leaking into the chamber) so as to balance pressures again. If the process is adiabatic, the work provided by the atmosphere has nowhere to go and stays in the gas as an increase in temperature/kinetic energy.

$\endgroup$
  • $\begingroup$ Thanks very much for your explanation. I think that I understand the situation a lot better now. However I am still a bit confused. In the demonstration there isn't actually a piston. That is an analogy that is used to explain why the gas heats up. Really all that is going on is a small valve is opened and air can come back into the chamber. I guess now I am wondering why the piston logic can't be used in the joule free expansion experiment: (en.wikipedia.org/wiki/Joule_expansion). Woulden't the pressure in the left part act like a piston and "help the gas expand" into the right side? $\endgroup$ – sTr8_Struggin Mar 18 '14 at 0:30
  • $\begingroup$ In Fig.1 of the article there is a piston. It is modelled in the derivation by imposing the value of the pressure in the container. After all, when equilibrium is reached, the density has to be the same in both compartments and hence the pressure as well. Now, the pressure value is imposed on the left hand side by the atmospheric pressure and that's it. Note that the heating comes from the atmosphere that works, the gas doesn't do a thing. In the original Joules experiment, there is no imposed value for the pressure because the system is totally isolated. $\endgroup$ – gatsu Mar 18 '14 at 8:40
  • $\begingroup$ Thanks! The paper described joules free expansion as gas flowing from one constant volume container to another constant volume container. So the container that originally contains gas loses thermal energy in the expansion because it had to act like the piston and do work on the gas and the net zero change in temperature is only because the two sides were in thermal equilibrium. However this contradicts with my thermal physics textbook An Introduction to Thermal Physics By Daniel Schroeder, which describes free expansion as having no work because the gas has nothing to push on. Hmmm... $\endgroup$ – sTr8_Struggin Mar 18 '14 at 16:23
  • $\begingroup$ No in the Joules expansion problem the gas (as a whole) doesn't work at all to expand in the other container and hence its energy change is zero. In the case you point out, it is not the gas in the box that works but the atmosphere outside of the system via a piston. That is because in this case, the system {box+gas} is not completely isolated from the outside and hence there is no reason for the energy to stay the same. $\endgroup$ – gatsu Mar 18 '14 at 19:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.