4 ms·

One of my favorite things about the JT effect is how gases not being “perfect” is actually “better” than if they were. What I mean is, if gases all behaved as

One of my favorite things about the JT effect is how gases not being “perfect” is actually “better” than if they were. What I mean is, if gases all behaved as some kind of perfect, platonic* ideal of a gas and followed the ideal game law exactly, there would be no temperature change. But because they don’t, the Joules-Thomson effect is what allows for refrigeration. *Helium is probably the closest to some platonic ideal of a gas.

pas 5y ago
Could you explain this a bit more? If every gas blob in every situation, system, vessel, pipe were to behave according to the ideal gas law a heat pump that cools down the inside of a box and radiates the heat away would be impossible? :o
- Pulcinella 5y ago
  Yes. One of the assumptions of the Ideal Gas Law is that particles don’t interact with each other or even collide with each other. They only collide with the walls of the container. So if the ideal gas law was true then a heat pump wouldn’t be able to refrigerate anything. When the gas would expand, the volume would go up, pressure would go down, and temperature would remain the same because there wouldn’t be any reason for the gas particles to slow down (because under the ideal gas law the particles don’t interact with eachother).
selimthegrim 5y ago
If you use fugacity can’t you rescale and collapse all gas properties into basically universal curves? https://en.wikipedia.org/wiki/Compressibility_factor https://en.wikipedia.org/wiki/Compressibility_factor
- chermi 5y ago
  Yeah, one of the kind of early hints of universality https://en.m.wikipedia.org/wiki/Theorem_of_corresponding_states https://en.m.wikipedia.org/wiki/Theorem_of_corresponding_sta.... Weird, I've never seen it called the "theorem" of corresponding states, always the "law". You get it straight from VDW EoS Edit-- I guess there's multiple laws of corresponding state, duh. My last sentence above should say you can a law of corresponding states from VDW.
bernulli 5y ago
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 https://en.wikipedia.org/wiki/Isentropic_process [edit: typo]
- hashimotonomora 5y ago
  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.
  - chermi 5y ago
    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_Introduction_to_Thermostatistics https://en.m.wikipedia.org/wiki/Thermodynamics_and_an_Introd...
    - bernulli 5y ago
      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!
mattb314 5y ago
> Joules-Thomson effect is what allows for refrigeration I don't think this is true. The "simple" model of refrigeration taught in highschool is just a carnot cycle running backwards, and this can be modeled with an ideal gas. The author of the post covers this the section on "the Thermodynamics 101 Answer"[1], where all you need to drop the temperature of a gas is to let it do work on the piston. That's not to say that JT is not useful, just that we can explain a theoretical refrigerator without it. [1] https://mattferraro.dev/posts/joule-thomson#the-thermodynamics-101-answer https://mattferraro.dev/posts/joule-thomson#the-thermodynami...
- chermi 5y ago
  Yes, you can explain refrigeration with ideal gas. But... You need non-ideal gas (attraction) to get temperature inversion. Then you just need compressor and voila, look at PT charts to find what temperature range you need. With ideal gas reverse carnot refrigerator your refrigeration effect is bounded on low temperature side by the available low temperature source. So yes, you can refrigerate with ideal gas, but it's not very helpful in warm areas or if you need to get something super cold.

pas 5y ago

Could you explain this a bit more? If every gas blob in every situation, system, vessel, pipe were to behave according to the ideal gas law a heat pump that cools down the inside of a box and radiates the heat away would be impossible? :o

Pulcinella 5y ago
Yes. One of the assumptions of the Ideal Gas Law is that particles don’t interact with each other or even collide with each other. They only collide with the walls of the container. So if the ideal gas law was true then a heat pump wouldn’t be able to refrigerate anything. When the gas would expand, the volume would go up, pressure would go down, and temperature would remain the same because there wouldn’t be any reason for the gas particles to slow down (because under the ideal gas law the particles don’t interact with eachother).

selimthegrim 5y ago

If you use fugacity can’t you rescale and collapse all gas properties into basically universal curves? https://en.wikipedia.org/wiki/Compressibility_factor https://en.wikipedia.org/wiki/Compressibility_factor

chermi 5y ago
Yeah, one of the kind of early hints of universality https://en.m.wikipedia.org/wiki/Theorem_of_corresponding_states https://en.m.wikipedia.org/wiki/Theorem_of_corresponding_sta.... Weird, I've never seen it called the "theorem" of corresponding states, always the "law". You get it straight from VDW EoS Edit-- I guess there's multiple laws of corresponding state, duh. My last sentence above should say you can a law of corresponding states from VDW.

bernulli 5y ago

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 https://en.wikipedia.org/wiki/Isentropic_process [edit: typo]

hashimotonomora 5y ago
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.
- chermi 5y ago
  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_Introduction_to_Thermostatistics https://en.m.wikipedia.org/wiki/Thermodynamics_and_an_Introd...
  - bernulli 5y ago
    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!

mattb314 5y ago

> Joules-Thomson effect is what allows for refrigeration I don't think this is true. The "simple" model of refrigeration taught in highschool is just a carnot cycle running backwards, and this can be modeled with an ideal gas. The author of the post covers this the section on "the Thermodynamics 101 Answer"[1], where all you need to drop the temperature of a gas is to let it do work on the piston. That's not to say that JT is not useful, just that we can explain a theoretical refrigerator without it. [1] https://mattferraro.dev/posts/joule-thomson#the-thermodynamics-101-answer https://mattferraro.dev/posts/joule-thomson#the-thermodynami...

chermi 5y ago
Yes, you can explain refrigeration with ideal gas. But... You need non-ideal gas (attraction) to get temperature inversion. Then you just need compressor and voila, look at PT charts to find what temperature range you need. With ideal gas reverse carnot refrigerator your refrigeration effect is bounded on low temperature side by the available low temperature source. So yes, you can refrigerate with ideal gas, but it's not very helpful in warm areas or if you need to get something super cold.