Timeline for Why does Joule-Thomson expansion of a gas do no (external) work?
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
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| Aug 2, 2022 at 10:20 | comment | added | Antonios Sarikas | @N.Virgo If it is like free expansion ("expansion into an evacuated space") why is it said that Joule-Thomson expansion is not the same as adiabatic free expansion? | |
| Dec 5, 2016 at 12:59 | answer | added | Dave E | timeline score: 0 | |
| Jul 3, 2012 at 14:59 | vote | accept | Bernd Jendrissek | ||
| Jul 2, 2012 at 10:35 | history | tweeted | twitter.com/#!/StackPhysics/status/219741011872714752 | ||
| Jul 2, 2012 at 9:47 | answer | added | N. Virgo | timeline score: 6 | |
| Jul 2, 2012 at 9:36 | answer | added | John Rennie | timeline score: 5 | |
| Jul 2, 2012 at 9:27 | comment | added | Bernd Jendrissek | @Nathaniel I get that it can't possibly be reversible, otherwise one could make compressed air for free. Okay, so an isentropic process is not necessarily adiabatic which probably resolves part of my confusion - I guess the "missing" work goes into disordering the expanded gas. | |
| Jul 2, 2012 at 9:19 | comment | added | Bernd Jendrissek | I actually started writing this question mindful of the Joule-Thomson effect, but then realized my question is more general than that - being relevant also to ideal gases. I hope the question as edited doesn't seem too confused. | |
| Jul 2, 2012 at 9:15 | comment | added | N. Virgo | No, wait - a gas expanding into a vacuum should stay the same temperature, not cool down. I guess I don't fully understand the Joule-Thompson effect after all. I'll look into it and get back to you. | |
| Jul 2, 2012 at 9:11 | comment | added | N. Virgo | I'm not sure if the above answers your question - if you feel it does then say so and I'll post it as an answer. | |
| Jul 2, 2012 at 9:11 | comment | added | N. Virgo | The Joule-Thompson effect kind of is an expansion into an evacuated space though - imagine the pressure on one side of the plug is zero. Then the gas coming out of the plug is expanding into a vacuum, and it's intuitively reasonable that it should cool down. Having a positive pressure on both sides is just a less extreme version of this. But note that Joule-Thompson expansion is very much not isentropic - it's adiabatic and it's a distinctly irreversible process ($dS/dt>0$). This is why the Hampson-Linde cycle is inefficient. | |
| Jul 2, 2012 at 8:58 | history | edited | Bernd Jendrissek | CC BY-SA 3.0 |
edited title
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| Jul 2, 2012 at 8:46 | history | asked | Bernd Jendrissek | CC BY-SA 3.0 |