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Comment by bernulli | Hacker News Reader

Why wouldn't there be any temperature change in ideal gases? When I compress an ideal gas in a bike pump, it heats up; when I expand it through a nozzle, it cools down. Often I can approximate these processes as isentropic, and then the temperatures are uniquely determined by the expansion as either a function of pressure ratio T2 = T1 (p2/p1)^(g-1/g) or volume ratio; T2 = T1 (V2/V1)^g, where g is gamma, the ratio of specific heats.

https://en.wikipedia.org/wiki/Isentropic_process

[edit: typo]

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JT is isoenthalpic not isoentropic. For an ideal gas, h = u + pv = CT + RT = f(T). So an isoenthalpic process is necessarily isothermic for an ideal gas.

Expansion through a nozzle is extremely chaotic and generates entropy. You can’t model JT that way.

When you compress air in an air pump you are doing work against the system and increasing its internal energy u = q - w, which can be explained using ideal gases by knowing that u = u(T). But this is not because the pressure increases but because of your work.

The emphasis on the entropy generation kind of confuses the point from my POV. What's important about the nozzle setup is that, by construction, it generates the JT throttling process. That is, the procedure is isenthalpic. Focus on the thermodynamic consequences of that.

Sorry for butting in. It took me a long time to get comfortable with throttling. Non-equilibrium stat mech stuff can really throw you (well, at least me) off if you come at it too microscopically at first.

Edit -- H. Callen's thermo book has a great little section on it. Best book on thermo out there if you're into a real postulate-and-construct approach. One of my favorite books of all time. https://en.m.wikipedia.org/wiki/Thermodynamics_and_an_Introd...

You can very much model expansion in a nozzle as isentropic, but I did miss the part where the original comment was restricted to JT. Thanks!

I didn’t mean that there would be no temperature changes. But that in the example where you allow the gas to expand into a new volume there would be no temperature change. Remember, temperature relates to the speed the gas particles are moving. If the barrier between the two sides of the container was removed and the gas allowed to expand into that new volume, why would the ideal gas particles slow down and decrease the temperature?

I did miss the part where the original comment was restricted to JT. Thanks!

But the isentropic part is what is being violated here... (I was surprised that the article didn't mention that even though it managed to demonstrate the real behaviour in another way.)

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