|MadSci Network: Physics|
Hi Tim...this is a good question that has 2 answers (I think). We can consider a gas that when compressed is a liquid (CO2, I think) or one that remains a gas when compressed (like helium). For the first case, let's imagine you open up a cylinder of compressed CO2. The CO2 goes from a liquid to a gas as it exits the cylinder (due to collisions with air molecules...these collisions impart energy to the CO2 "droplets", allowing them to become gaseous...as such, the surrounding air has less energy and is "cooled"). This is the same thing that happens when you get acetone (used to be a component of nail polish remover..maybe still is?) or alcohol on your skin. The liquid "steals" heat from your skin and evaporates, thereby cooling the skin off. This is also the principle behind the operation of your refrigerator...a Freon is sprayed out of a small nozzle (it's a high vapor pressure liquid...which means that it evaporates easily, much like acetone and methanol (see a connection?) Freon collisions with the wall of a copper tube that the expansion takes place within remove energy, thereby cooling it. A compressor collects the gaseous Freon and compresses it back to a liquid..this generates heat, which must be removed (usually by a fan). So far so good? expansion from a tank of helium is something that I'm not as familiar with, but here goes... This type of expansion is called "isentropic"...entropy (randomness) is conserved. If you call helium a "perfect gas"...i.e. one with no exchange of energy between gaseous particles (clearly impossible), then you can can combine the equations dH = TdS + VdP (Maxwell eq I think) and dH=Cp dT , then work through some integrals to establish that: (T2/T1) = (P2/P1)^(R/Cp) I think that a reasonable way to think about this is that to maintain constant entropy (randomness), as the pressure decreases in the "system" (cylinder and gas), energy must leave the system as well (cooling) to balance the increase in entropy that you get as the pressure drops (with a lower pressure, there is more "room" to move)...I don't expect this cooling to be as noticeable as that for the first case because of the exponent...if you take the ln of both sides, for a 20% decrease in pressure (i.e. P2=.8*P1), T2 will go down by quite a bit less than 20%. I hope that this answers your question without making it too unreadable. Please feel free to email me if I can clarify anything or help at all. Best Regards, Mike Weibel
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