When a gas is expanded by forcing it through a nozzle, the Joule-Thomson coefficient describes how much the temperature of a gas will drop as its pressure decreases.
It always helps to know the physical interpretation, as well as the math behind it. Good job, searching until you found an explanation that clicked for you.
Thank you very much! I can now understand how that actually works! You actually deserve much more subs and views than mere thousands! Btw a little fun fact: you didn't say "Joule Thomson Effect" in the whole video; anyways, it was interesting! Thanks again!
An awesome explanation ,,, you really deserve much more views than this,,,, this video helped me in understanding the effect that others just write the formula of and never tells where it comes from ,,,, I'm and JEE aspirant from India ,,,, It really helped a lot,,,, Thanks a lot
Thank you sir. I tried to study this on my text book but didn't get that clearly. Then I searched it in RU-vid and I got your video. Now it's clear to me
Thanks! But I'm writing forwards. The computer gets the credit for doing things backwards: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-YmvJVkyJbLc.html
@@PhysicalChemistry Although I have just one question, - What happens to the potential and kinetic energy after the isenthalpic expansion and how temperature changes due to the changes in KE and PE.
hello I would like to know about the difference between the Joule-Thomson effect and the hydration of gas in gas pipelines. Is there a link between them?
Thank you for the conceptual explanation. I had a question though, how can I calculate the outlet temperature for a fixed inlet and outlet pressure and an inlet temperature. I am working on a project that requires me to do isenthalpic calculations however I'm not finding any luck with the methodology behind calculating the outlet temperature. Any help or guidance would be appreciated!
The JT coefficient is μ = ∂T/∂P (at constant H). So, for an isenthalpic system, dT = μ dP. Or, approximately ΔT = μ ΔP. You can use this to calculate the temperature change. This is just one example of the sort of differential relationships that are used all the time in thermodynamics. If you want more info on this sort of relationship, see this video: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-LttWusDZLGA.html
Hi professor. Thanks for the video and the explanation. I would like to ask you about the inversion temperature of the JT coefficient and how the heating in the gas expansion can be explained in terms of kinetic and potential energy as you did with the cooling. Thank you!
Excellent question. In the video I only considered the case where the interactions between gas molecules are attractive, on average. For some gases, or some conditions, the interactions are replusive (on average). In that case, if the molecules are separated further from one another, that **lowers** their potential energy, which will **increase** their kinetic energy and their temperature. The inversion temperature is the temperature where the attractive interactions exactly balance the repulsive interactions, and the gas behaves ideally (does not heat or cool as it expands). This is equivalent to the Boyle temperature, if you're familiar with that property of real gases. Here's a video that may help: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-rE4DETh-mqA.html
No, I'm not smart enough to write right-to-left. Instead, I use the computer to flip the image: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-YmvJVkyJbLc.html
The purpose of the porous plug (or small nozzle, etc) is to make sure the gas expands slowly. Under these conditions, we can assume both the compression on one side, and the expansion on the other side, occur at constant pressure. The gas would still cool if it expanded without being throttled. But the J-T equations are specific to the situation where the gas is throttled.
Hi, well done for your video. However, the way you explain it is slightly misleading. Cooling occurs only below a threshold temperature (called inversion temperature) at which the JT coefficient is calculated. Above, you would have an increase in temperature. At room temperature only hydrogen, helium and neon are heating while expanding by this process. Maybe that's an addendum you should make. The formula is dT/dP_H = V/Cp × (T×alpha -1) (easy to derive with triple product rule and a Maxwell formula). For an ideal gas alpha = 1/T so no cooling occurs. For a real gas it depends upon the starting temperature. The only way to always have a cooling effect is to perform the process isentropically. The formula above would then be the same but without the minus one. As all terms are always positive for all gases (except quantum helium but that's very niche), a drop in pressure implies a drop in temperature. And same for ideal gases no cooling nor heating. Best regards
You're right, of course, that the JT coefficient depends on temperature, and can be either positive or negative. (Thermodynamic relationships such as the one you give are discussed a little later in the course, once we have introduced other energies and Maxwell relations.) You're also right that dT/dP_H is positive for most gases at room temperature (with a few examples given at the end of the video) and is negative only for a few like H and He. But when a real gas undergoes an *expansion*, the pressure drops. Since dT/dP_H is positive for most gases, that means dT < 0, and the gas cools. He and He and Ne are anomalous in that they *warm* when expanded at room temperature. Normal refrigerants, like Freon, stop behaving as refrigerants once they get *above* their JT inversion temperature.
@@PhysicalChemistry working as a design engineer at Linde I felt obliged to add this comment. I hope my English is correct though and I think clearly mentioned all you have highlighted in your answer, I was just giving more insights to other readers. I was just adding that the inversion temperature has not effect whatsoever if the process is performed at dS=0 (which in reality is hard to achieve with better yield than 90%). Apart from that point well done to explain especially for newbies I would say. That's a very challenging problematic in the industry and it's well summarised and I think I may use part of your material to share to our non technical staff who is always eager to understand more of what we do. Cheers from France 👍
Salut et merci. As an introductory video on the subject, the main goal is to explain the idea, as well as a little insight into the physical mechanism behind the effect -- the latter is often a confusing point for physical chemistry students. It's always nice to have comments from professionals using PChem in the real world, so that students understand these are not merely textbook problems. (And also so that they know that the real world usually requires additional complications beyond the first-order models we study in class.)
No, that's a different phenomenon, called evaporative cooling. In that case, the temperature drop comes from the kinetic energy of the liquid being used to pay for the enthalpy of vaporization.
Thanks for the explanation! Something that doesn’t make sense to me is that, if the cooling for a real gas happens via a direct conversion from kinetic energy to potential, then where is internal energy lost? Part of the workings shows that there is a change in internal energy, but if potential energy is just converted into kinetic, then where is energy lost? And where is it lost to?
The gas molecules are (weakly) attracted to each other. When they are (relatively) close to one another, their potential energy is low. To move them farther apart requires that some energy be supplied, in order to raise their potential energy. As an analogy, consider moving two magnets apart from one another. This is increasing their potential energy. (To be clear: molecules are not typically attracted to one another by magnetism. But this might help give you a physical, intuitive sense for how energy is required to change the potential energy of objects that are attracted to one another.)
So is Joule-Thomson expansion like adiabatic free expansion with the difference being that P2 is not zero (as in free expansion)? Also what is the role of the porous plug? Why all explanations skip that part if it is part of the experiment?
The JT process is adiabatic, but the porous plug (or throttling restriction, as illustrated in this video) is an important difference. Its role is to allow the initial pressure P₁ to be constant during the entire expansion. In the single-container free epxansion, the pressure drops as the volume of the conainer increases. Here, the pressures P₁ and P₂ remain constant while the volumes of the two sides of the system change, as gas passes from side 1 to side 2.
Very informative video. Thx for the explanation. One question about the reason gas cools down. What if JT coefficient is negative (or we can say the expansion process starts outside of the inversion curve)? In this scenario, gas temperature actually goes up until expansion reach inversion curve. But it is still expansion so I guess potential energy still goes up which cause kinetic energy goes down. Could u explain a bit about where I get it wrong?
If the JT coefficient is negative, then the gas will heat up during the expansion. The physical reason for this is that the interaction between the gas molecules is repulsive, rather than attractive. So as the gas expands, and the molecules move farther apart, their potential energy *decreases*. This decrease in PE causes an increase in KE, so the gas heats up.
I did some juggling and got that (∂T/∂p)@constant H = ((αT−1)V)/Cp (or (∂T/∂p)@constant H = ((αT−1)V)/(Cv+R)?). Where would this expression be useful? Thanks, Marcus
Yes, that's right. That would be a way to calculate the value of the JT coefficient, using other properties of the gas. For example, it could be used to determine the numerical values of μ that I listed in this video, or values for these gases at other conditions, or for other gases.
If the expansion is too fast, it will be irreversible and non-equilibrium, so these equations wouldn't necessarily apply. Also, if the expansion is through a narrow throttle, or sent through a porous plug, then the expansion is slow enough that the low-pressure gas can be removed, to keep it at low pressure.
Because the molecules have a (weak) attractive interaction. For any two objects that feel an attraction towards one another, it requires energy to move them farther apart. When they move farther apart, their potential energy increases.
Great question! The explanation in this video uses intermolecular attraction to explain the cooling. But real gases have two important differences from ideal gases: intermolecular interactions and finite volume. When the attraction between molecules is more important, it costs energy to separate the molecules, and the gas cools as it expands. But if the finite volume effects are more important, then the energy of the gas decreases as the molecules move farther apart, so the gas heats up as it expands. This will be the case for molecules with particularly weak interactions (like He) or at high temperatures (where interactions are negligible compared to kT).
The JT effect is well explained, thank you. However, refrigeration using Freon or other refrigerants isn't normally based on the JT effect. The refrigerants are compressed to the pressure at which they are liquid at ambient temperature. They are then cooled to ambient temperature to remove latent heat and heat of compression. The liquid refrigerant is then allowed to evaporate at lower pressure to provide cooling through evaporative cooling.
Point taken, thank you. Of course you're right that the condensation and evaporation parts of the refrigeration cycle are more important than the JT effect. I've added this to my list of planned edits / updates. Thanks again for the correction
There is a growing interest in CO2 used as a refrigerant and in the process of condensing a characteristic of CO2 called transcritical (pretty sure) which is neither liquid nor vapor. While very fascinating it's difficult to understand, but I might be over thinking it. Is this something you might consider as a viable topic for one of your video presentations? If so perhaps it would be interesting enough to attract a reasonable number of views, and I would enjoy getting your take on it. 👍 Thanks Oh wait, I see you haven't made a video in a years time. Oh well...,,,,,🤓
It's supercritical fluids (not transcritical). I do have a video that discusses supercritical fluids a little: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-V82Ve9S7gyg.html And they are definitely fascinating.
@@PhysicalChemistry oh thanks so much! I will immediately watch it and more than once, as I need to understand, or wrap my head around the concept, as it doesn't have the familiar pressure/temperature reactions of the many refrigerants I have worked with. Even at my age I intend to learn something new every day.,,,🤓👍
How about it, indeed! Entropy is one of my favorite topics, and the driving force behind a lot of physical chemistry processes, as you might know if you have watched other videos of mine. The same is true here, of course. The gas will spontaneously expand through the nozzle from high pressure to low pressure, because of entropic effects. But the Joule-Thomson coefficient is primarily concerned with the temperature change during the expansion, rather than the spontaneity.
@@PhysicalChemistry Thanks, I mean how entropy will change in Joule-Thomson effect ? This process is an adiabatic and irreversible process, according to the definition of entropy changes: heat to temperature changes are removed from the equation and we only produce entropy due to irreversibility, so entropy increases in this process. what you think ?
You talked about only the cooling effect. What about the heating effect and inversion temperature? This is not a complete Joule-Thomson Effect explanation...
That's right, it's certainly not a complete explanation. But the heating effect is the inverse of the cooling effect. Cooling happens when it costs energy to separate molecules. The heating effect occurs when the density of the (real) gas is such that the interactions are repulsive on average, so that the expansion lowers their potential energy. The inversion temperature is where neither is true: the PE of the gas does not change with expansion. This is the same as the Boyle temperature, and is the temperature where the (real) gas behaves somewhat ideally due to a cancellation of repulsive and attractive intermolecular forces.
You must be an engineer, I'm guessing. Engineering "work" is defined with the opposite sign convention from chemistry work. We (chemists, physicists) define work from the point of view of the system. Work done on the system is positive. The first law is ΔU = q + w. Engineers like to define work with the opposite sign. More detail on the chemistry sign convention is here: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-EkiNa0IoFqQ.html
