I am sorry to tell you that your explanation of why the gas and cylinder cool is not correct. Let's start with the gas. When the gas leaves the cylinder and expands, it has to push the gas in the room out of the way, thus doing work to the gas in the room. The work the gas from the cylinder does to the gas in the room is at the expense of the internal energy of the gas escaping from the cylinder, so the gas from the cylinder cools down. It is not, as you state in the video, caused by the gas particles from the cylinder pushing on one another. In the FREE, adiabatic expansion of a gas, there is no temperature drop because no work is done to the outside world. The cylinder itself cools down due to evaporative cooling. Most of the CO2 in the tank is in liquid form with a small amount of vapor in the head space above the liquid. When you open the valve the gas leaves the cylinder and drops the pressure. This causes the most energetic particles in the liquid to escape the liquid leaving only less energetic particles behind. This means that the average kinetic energy of the remaining liquid is lower, and the liquid is therefore cooler. The relatively warmer gas in the head space then exits the cylinder and undergoes J-T cooling as discussed above. The liquid in the cylinder continues to cool through evaporative cooling as long as the valve remains open releasing gas. Your explanation doesn't really explain why a cooling gas ejected from the cylinder would also cool the cylinder itself. Enjoyed the video, none the less and I encourage you to re-record it with the correct physics.
Thanks Simon - that is excellent. I have pinned your comment and will refer to it in my description. You are right, I should remake this at some point! Many thanks, Anthony.
@Anthony Francis-Jones When the particles escape from the liquid in the bottle, is it because the drop in pressure allows the energetic particles to escape the surface tension of the liquid.
So does the gas that leaves the cylinder cool and condense as a result? Would that be possible if the cylinder was just in a vertical position? Would this scenario also be applicable to a Nitrogen gas cylinder?
Yeah its around 240K because that’s what happens at mach one during isentropic nozzle expansion. By T=T0/(1+ (k-1)/2* M^2). You can determine the percent dry ice of the fluid by assume adiabatic and following the enthalpy down from a pressure-enthalpy diagram from bottle pressure to atmospheric. It will be in the solid/gas phase and the chart will tell you percent gas / solid. Cheers.
Thanks again Derrick. There is a lot of interesting stuff going on here as you say and my oversimplification has not really captured it at all correctly. Thankfully many of the comments have put bits right and that has been really helpful. Thanks for watching and taking time to comment. I hope you find some of my other videos rather better!
Hi that was a wonderful explanation, May i know what happened when we applying high pressure fog to the greenhouse atmosphere which is hotter and we can find the cooling effect and its very efficient, what happens actually, does it change or affect the heat content of the Air/any affect on the air enthalpy? Looking forward to hearing, Thanks a Ton.
James, good to hear from you and a great question. So, if I understand the situation correctly, the fog is probably very fine water droplets - these are cooler than the air so will need to get to thermal equilibrium with the air (warm up!) this will take energy from the air and therefore reduce its heat content and make it cooler. Secondly water has a very high Specific Heat Capacity so large additions of energy are needed to heat it up. Thirdly if the droplets vaporise then they will take additional energy from the air (cooling it down) and this is the physics of Latent Heat of Vaporisation. Hope that helps explain your excellent question. Perhaps watch this too - it is really counterintuitive! ruclips.net/video/81cax3Tq6KM/видео.html
@@AnthonyFrancisJones it’s a lovely video really! I love the temp measurements. I always wondered how cold CO2 got like how you did it as well! Well done sir!
Thanks, it was fun to make and you probably realise it, like so many others of mine are made with 30mins to spare over a lunch hour between classes! Still, I got ahead of myself with the explanation and really need to sort that out - unlike me! Thanks again for commenting and watching. More to come!
Sure - the biggest application I can think of is in refrigeration units and air conditioning systems where cold temperatures are produced by a compressed fluid expanding through a small nozzle. Hope that helps.
You are unlikely to get paralysed but freezing of living cells is never a good thing - you can easily kill them and get frostbite! Also if you freeze your hand to the cold metal surfaces/tap you may be unable to turn it off - not a good situation to be in!
can we apply Gay Lussac's Law to predict the temperature of a sealed room when a valve quickly reduces pressure inside from an initial T1=300k, P1=23.4 PSI, P2=15.0 PSI. The result would seem to be T2=(P2/P1)*T1=210K ?
Good question. The simple answer is you can use the gas laws, but in this case the rate of expansion is required to calculate the adiabatic cooling of the gas.
I am sorry to tell you that your explanation of why the gas and cylinder cool is not correct.
Let's start with the gas. When the gas leaves the cylinder and expands, it has to push the gas in the room out of the way, thus doing work to the gas in the room. The work the gas from the cylinder does to the gas in the room is at the expense of the internal energy of the gas escaping from the cylinder, so the gas from the cylinder cools down. It is not, as you state in the video, caused by the gas particles from the cylinder pushing on one another. In the FREE, adiabatic expansion of a gas, there is no temperature drop because no work is done to the outside world.
The cylinder itself cools down due to evaporative cooling. Most of the CO2 in the tank is in liquid form with a small amount of vapor in the head space above the liquid. When you open the valve the gas leaves the cylinder and drops the pressure. This causes the most energetic particles in the liquid to escape the liquid leaving only less energetic particles behind. This means that the average kinetic energy of the remaining liquid is lower, and the liquid is therefore cooler. The relatively warmer gas in the head space then exits the cylinder and undergoes J-T cooling as discussed above. The liquid in the cylinder continues to cool through evaporative cooling as long as the valve remains open releasing gas. Your explanation doesn't really explain why a cooling gas ejected from the cylinder would also cool the cylinder itself. Enjoyed the video, none the less and I encourage you to re-record it with the correct physics.
Thanks Simon - that is excellent. I have pinned your comment and will refer to it in my description. You are right, I should remake this at some point! Many thanks, Anthony.
@Anthony Francis-Jones When the particles escape from the liquid in the bottle, is it because the drop in pressure allows the energetic particles to escape the surface tension of the liquid.
Thanks, it cleared a lot of doubts of mine
Nice Simon....
So does the gas that leaves the cylinder cool and condense as a result?
Would that be possible if the cylinder was just in a vertical position?
Would this scenario also be applicable to a Nitrogen gas cylinder?
It's just about wondering and asking......thanks again for feeding our curiosity
Pleasure - I am planning another few mercury experiments for the future too. Thanks for taking the time to comment. Enjoy your physics!
@@AnthonyFrancisJones actually i should thank your for your content sir
@@starstuff5813 It's a pleasure!
Superb explanation
Love from India... .
Thank you and glad to hear all the way from India!
Yeah its around 240K because that’s what happens at mach one during isentropic nozzle expansion. By T=T0/(1+ (k-1)/2* M^2). You can determine the percent dry ice of the fluid by assume adiabatic and following the enthalpy down from a pressure-enthalpy diagram from bottle pressure to atmospheric. It will be in the solid/gas phase and the chart will tell you percent gas / solid. Cheers.
Thanks again Derrick. There is a lot of interesting stuff going on here as you say and my oversimplification has not really captured it at all correctly. Thankfully many of the comments have put bits right and that has been really helpful. Thanks for watching and taking time to comment. I hope you find some of my other videos rather better!
I loved your lectures Sir, I learned a lot during this lecture.thank you soo much🙏❤ Sir. ☺
Excellent, glad it was useful. Do read the top comment as well which makes some corrections!
Hi that was a wonderful explanation, May i know what happened when we applying high pressure fog to the greenhouse atmosphere which is hotter and we can find the cooling effect and its very efficient, what happens actually, does it change or affect the heat content of the Air/any affect on the air enthalpy? Looking forward to hearing, Thanks a Ton.
James, good to hear from you and a great question. So, if I understand the situation correctly, the fog is probably very fine water droplets - these are cooler than the air so will need to get to thermal equilibrium with the air (warm up!) this will take energy from the air and therefore reduce its heat content and make it cooler. Secondly water has a very high Specific Heat Capacity so large additions of energy are needed to heat it up. Thirdly if the droplets vaporise then they will take additional energy from the air (cooling it down) and this is the physics of Latent Heat of Vaporisation. Hope that helps explain your excellent question. Perhaps watch this too - it is really counterintuitive! ruclips.net/video/81cax3Tq6KM/видео.html
Superman is the best example
Super practical experience sir
Pleasure Nerella. Glad you enjoyed it and found it interesting.
Really nice...
Great - thanks. Glad you liked it.
Very useful thanks alot ❤
Thanks Ahamad. Glad you liked it. Do read the pinned comment too which is extremely helpful. Good luck with your studies.
@@AnthonyFrancisJones ok sir. I wish happy long life for you..💛
It’s not going to a lower pressure bruv. It’s evaporating the liquid which is at the same pressure. The heat loss is just q=mdot L
Thanks Derrick. Yes, I made a bit of a mess of this video! I need to redo it really!
@@AnthonyFrancisJones it’s a lovely video really! I love the temp measurements. I always wondered how cold CO2 got like how you did it as well! Well done sir!
Thanks, it was fun to make and you probably realise it, like so many others of mine are made with 30mins to spare over a lunch hour between classes! Still, I got ahead of myself with the explanation and really need to sort that out - unlike me! Thanks again for commenting and watching. More to come!
Can I ask about some industrial applications that use the concept of adiabatic expansion of gases?
Sure - the biggest application I can think of is in refrigeration units and air conditioning systems where cold temperatures are produced by a compressed fluid expanding through a small nozzle. Hope that helps.
can one get paralyzed if they don't wear gloves in this experiment?
You are unlikely to get paralysed but freezing of living cells is never a good thing - you can easily kill them and get frostbite! Also if you freeze your hand to the cold metal surfaces/tap you may be unable to turn it off - not a good situation to be in!
Great thanks sir
Thanks and glad you liked it!
can we apply Gay Lussac's Law to predict the temperature of a sealed room when a valve quickly reduces pressure inside from an initial T1=300k, P1=23.4 PSI, P2=15.0 PSI. The result would seem to be T2=(P2/P1)*T1=210K ?
Good question. The simple answer is you can use the gas laws, but in this case the rate of expansion is required to calculate the adiabatic cooling of the gas.
@@AnthonyFrancisJones The room is about 200 cubic feet, it vents in 20-30 seconds.
Subscribed the chanell ❤️❤️
Thanks. Hope you enjoy it and find it useful.
6:26 that’s not ice on that sock
Xd
sorry but your explanation is non-intuitive
No problem - it's not an easy topic - is there a bit you would like me to go over?
You explained very well 👍