0
$\begingroup$

My lecture notes state that if gas particles sit in an insulating container, with a partition separating the gas from vacuum is removed, no change in internal energy is done, and it is stated that it is because the thing separating the gas is vacuum. I'm confused as to why this implies that the gas particles do no work on expansion, and I think this is touching on a general confusion of mine.

Without thinking about maths, why is it that a gas expanding into a vacuum does no work, but expanding into something that isn't a vacuum does do work?

Improve this question
$\endgroup$

2 Answers 2

Reset to default
1
$\begingroup$

Think in this way. The gas particles in the container, before the separator is removed, just doing their thing, that is, moving all over the place, hitting each other, hitting to the walls of the container, etc. When the separator is removed then suddenly for gas particles there are new places to explore. Since there is nothing in the new place, gas particles would feel no resistance while moving into the vacuum. Of course because of expansion the volume will increase and the pressure will decrease but considering that the container is insulating, that is, ho heat exchange can occur, the temperature of the gas would not change.

Improve this answer
$\endgroup$
5
  • $\begingroup$ Would this be the explanation if there was an atmosphere and not just vacuum? And why then, does increasing the volume of a container cause a pressure change in general? How is possible? How is not always the case you've described? $\endgroup$
    – sangstar
    Dec 2, 2017 at 20:06
  • $\begingroup$ No, the case in which the gas expands to vacuum is called Free Expansion. The case you are talking about now would be mixing of two gases. Google for "free expansion" and "thermodynamics of mixing" and you would get a lot of stuff to read or watch. Why increase of V cause decrease of P, or vice versa? Well, this thing is called Boyle's law. Think about you have a syringe and you close the whole with your finger. As you push the plunger, 1st it gets harder to push as the volume gets lower and 2nd your finger that you used for closing the whole hurts more, both because of pressure increase. $\endgroup$
    – physicopath
    Dec 2, 2017 at 21:17
  • $\begingroup$ Right okay, because the particles are hitting the walls more times in a given time interval? And this explains the reverse with volume expansions? $\endgroup$
    – sangstar
    Dec 2, 2017 at 21:27
  • $\begingroup$ But if this is true, how is work being done on the gas when volume decreases? What's giving the particles kinetic energy? $\endgroup$
    – sangstar
    Dec 2, 2017 at 21:29
  • $\begingroup$ have a look at this video youtu.be/Qsa4aAdpHfY $\endgroup$
    – physicopath
    Dec 2, 2017 at 22:55
0
$\begingroup$

This is the case of free expansion, in which now work is performed by gas.

Think it this way: a gas is enclosed in a piston tube. Let's take gas as a system. The pressure is applied on piston surface. This pressure we can measure, and use in calculation of work.

If there were no, piston surface (or any other surface) to apply pressure on, the concept of pressure becomes fuzzy\meaningless. So we can confidently say that:

$$dW=P_{\text{external}}\cdot dV$$

where our $P_{ext}$ is pressure applied by some surface on the gas.

So in your gas expanding in vaccum, there is obviously no $p_{ext}$ so there is no work done by gas.

For more information you can refer the Joule Expansion experiment.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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