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When you spray gas from a compressed spray, the gas gets very cold, even though, the compressed spray is in the room temperature.

I think, when it goes from high pressure to lower one, it gets cold, right? but what is the reason behind that literally?

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In one perspective, it is simply because the heat transferring in through the container is not fast enough. Let the spray can sit there for long enough after you do some spraying. I tip and tilt the cans, and roll them around, to speed up the reheating. –  Skaperen Mar 2 '13 at 16:32
    
Are you asking if the sprayed gas gets colder or the canister itself gets colder? –  user80551 Sep 11 '13 at 11:18
    
The question says "Why does the gas get cold..." –  Kenan Deen Dec 17 '13 at 10:52

12 Answers 12

up vote 10 down vote accepted

The temperature of the gas that is sprayed goes down because it adiabatically expands. This is simply because there is no heat transferred to or from the gas as it is sprayed, for the process is too fast. (See this Wikipedia article for more details on adiabatic processes.)

The mathematical explanation goes as follows: let the volume of the gas in the container be $V_i$, and its temperature $T_i$. After the gas is sprayed it occupies volume $V_f$ and has temperature $T_f$. In an adiabatic process $TV^{\,\gamma-1}=\text{constant}$ ($\gamma$ is a number bigger than one), and so $$ T_iV_i^{\,\gamma-1}=T_fV_f^{\,\gamma-1}, $$ or $$ T_f=T_i\left(\frac{V_i}{V_f}\right)^{\gamma-1}. $$ Since $\gamma>1$ and, clearly, $V_f>V_i$ (the volume available to the gas after it's sprayed is much bigger than the one in the container), we get that $T_f<T_i$, i.e. the gas cools down when it's sprayed.

By the way, adiabatic expansion is the reason why you are able to blow both hot and cold air from your mouth. When you want to blow hot air you open your mouth wide, but when you want to blow cold air you tighten your lips and force the air through a small hole. That way the air goes from a small volume to the big volume around you, and cools down according to the equations above.

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How does the volume change when it is the same bottle? what happened to the number of particles that is not constant as we spray? –  Revo Aug 31 '11 at 5:12
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A fast expansion is not adiabatic, a slow one is. A fast expansion into vacuum keeps the temperature constant. –  Ron Maimon Aug 31 '11 at 5:25
    
@Ron Maimon Just read the second sentence of the Wikipedia article I link before you post nonsense. –  AndyS Aug 31 '11 at 5:52
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@AndyS: the Wikipedia article is using "Adiabatic" and "Isentropic" as two different words. I usually don't. I still don't believe you that STP processes are isentropic, which is what you need to get the cooling formula. my initial feeling was that half or more of the gas expansion cooling would be lost just because air is such a lousy pressure wall, and you have internal expansion in turbulent wakes that waste the work into internal heat. I am still have a feeling that this is less than 50% of perfect adiabatic cooling, despite your two sure opinions. I will think about it. –  Ron Maimon Aug 31 '11 at 14:33
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@leftaroundabout: What you two are saying is that if you fill up a stiff balloon with compressed air (which is closer to perfectly adiabatic), you get the same cooling as when you allow the same amount of compressed air to escape freely into air. I don't think this is true, I think you get less than half the cooling in free escape as compared to a perfect adiabatic wall expansion, but someone who fills up balloons would know. To calculate it precisely requires knowing the exact nonequilbirium thermodynamic profile of a highly turbulent flow –  Ron Maimon Aug 31 '11 at 16:39

your question is about large drops in pressure, and why they cool gasses. The answer is that the gas is doing work in the process of expanding, and this work releases energy to the environment.

If you prevent the gas from doing work, if there is nothing for it to push against, it doesn't get cold. If you have a dilute gas in the corner of a room and you open a barrier to a vacuum, the gas expands into the vacuum with no change in temperature. This is not what you are doing when you spray the can into air. There, the gas is encountering air, and produces a pressure wall which it then pushes against doing work. Once the equilibrium spray-profile is established, there is a pressure gradient from the can outward that accelerates the spray to its final velocity. Travelling along this pressure gradient, the gas expands and does work, and this removes energy from the gas. The cold temperature profile sneaks back towards the can, because the air is such a lousy conductor of heat, so the heat is all coming from the can. Eventually, your hand gets cold.

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+1 for no math ;-) –  Martin Gales Aug 31 '11 at 6:38
    
I didn't put math, because I didn't believe that any approximation would be accurate. –  Ron Maimon Aug 31 '11 at 14:26

This is a very confused discussion. Gas being forced through a nozzle, after which it has a lower pressure, is an irreversible process in which the entropy increases. This has nothing to do with adiabatic expansion. It has everything to do with the Joule-Thomson effect, which is discussed in this Wikipedia article.

The change in temperature following the drop in pressure behind the nozzle is proportional to the Joule-Thomson coefficient, which can be related to the (isobaric) heat capacity of the gas, its thermal expansion coefficient, and its temperature. This is a famous standard example in thermodynamics for deriving a nontrivial thermodynamic relation by using Maxwell relations, Jacobians, and whatnot. Interestingly, it is not certain that the temperature drops. For an ideal gas – which seems to be the only example discussed so far in this thread – it wouldn't, because the Joule-Thomson coefficient exactly vanishes. This is because the cooling results from the work which the gas does against its internal van der Waals cohesive forces, and there are no such forces in an ideal gas.

For a real gas cooling can happen, but only below the inversion temperature. For instance, the inversion temperature of oxygen is about $1040$ $K$, much higher than room temperature, so the JT expansion of oxygen will cool it. $\text{CO}_2$ has an even higher inversion temperature (about $2050$ $K$), so $\text{CO}_2$ fire extinguishers, which really just spray $\text{CO}_2$, end up spraying something that is very cold. Hydrogen, on the other hand, has an inversion temperature of about $220$ $K$, much smaller than room temperature, so the JT expansion of hydrogen actually increase its temperature.

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Joule-Thomson expansion is usually described as happening through "a porous plug", which causes it to occur smoothly over some non-negligible distance or time (which doesn't mean you can't have high flow rates). It is not clear to me that the spray from a pressurized can is a good approximation to those conditions. –  dmckee Mar 2 '13 at 20:11
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The nozzle of a spray can is indeed not what one would picture as a porous plug, but then, what is a porous plug? Presumably a device with many holes, such that gas meets sufficient resistance upon being pushed through. A nozzle really only has one hole, but it does serve that purpose. –  Markus Deserno Mar 17 '13 at 13:12
    
@Markus I also think that the major factor behind the phenomenon is the JT effect, but it would be nice to have a simple calculation on the degree of temperature change due to JT or due to the adiabatic expansion. I assume that there is a few orders of discrepancy. Could you (or anyone) do the maths? –  István Zachar Aug 23 '13 at 14:17

The ideal gas equation states that 'PV = nRT'. P is Pressure, V is volume, n is the number of moles (the amount of 'stuff') R is the universal gas constant, and T is temperature. Of course the gas in your can is not an ideal gas, but it's close. So, when you release it from its compressed state in the container, you relieve a huge amount of pressure. P drops. The volume doesn't go up enough to compensate for the pressure drop, so something else has to change. That something can't be the amount of stuff, nor the gas constant, so the temperature drops as well.

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Correct, but this doesn't really answer the question at all: why does the volume not go up enough to compesate for the pressure drop? Ideal gas law is not enogh here, you actually need the internal energy. –  leftaroundabout Aug 30 '11 at 19:23
    
But the number of moles in that equation is not constant when u start spraying –  Revo Aug 31 '11 at 5:10
    
The ideal gas law does not describe this process, which depends crucially on the work done by the gas as it expands. This answer is incorrect. –  Ron Maimon May 7 '12 at 6:29

To do this properly, you need to describe the gas by a statistical ensemble. But this answers the question, too:

We can describe the gas as a... well, gas of classical particles, flying around in the can quite fast and in a randomish fashion. The temperature is essentially the average kinetic energy per particle (more precisely: per degree of freedom) $$ E = \sum_{\text{particles}}\!\!\frac{mv^2}2 = \tfrac32n_\text{particles}\bar{\cdot}\!\!\!\!kT $$ Where $\bar{\cdot}\!\!\!\!k$ is the Boltzmann constant.

They bounce off the walls elastically, so the average squared velocity stays constant, and therefore also the temperature.

But what happens now if we expand the volume, that is, move one of the walls outwards? While we do this, particles bouncing off the moving wall will not be reflected back in the volume with constant velocity anymore – the $v^2$ will stay constant as seen from the wall, but as it is moving itself, the velocity towards the wall was actually higher than it is back after the collision. So the average $v^2$ will also decrease, and so will the temperature.

Note that this process is called adiabatic expansion.

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What wall gets moved when u start spraying? –  Revo Aug 31 '11 at 5:09
    
@Revo: in a sense, the wall that was the valve. Rather than gas actually leaving our system, we can also describe the system "gas that't still in the container + gas that has just left the container", and this system does expand with "the valve wall moving through the nozzle outside". It's obviously not an ideal description, but nevertheless qualitatively correct. –  leftaroundabout Aug 31 '11 at 9:17
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Is this expansion adiabatic? I don't think so still. The air is not a good pressure wall for the outgassing spray. –  Ron Maimon May 7 '12 at 6:27

The joule Thompson coefficient is negative as the gas is coming out through the small hole, for negative coefficient the tempr. decreases and for positive coefficient the tempr. increases.

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OK, but why is the coefficient negative? You haven't explained the reason why the gas gets colder, only the way equations predict the temperature change. –  Dave Coffman Apr 3 at 12:49

Ron Maimon is basically correct when he attributes the drop in temperature to the work being done. Note that the gas would not come out of the can if the external pressure was the same or greater than the internal pressure in the can.

As to the applicability of the ideal gas law, that depends on the uniformity of the system (the can of gas). The pressure is less at the nozzle than in the bulk of the gas, but that difference disappears in roughly the time it takes a sound wave to make a couple of trips through the can. If the pressure gradient is substantial, the system is not uniform, and we are in the realm of hydrodynamics and not thermodynamics.

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When gas molecules rush out from the can, in the can they were tightly packed, but in the room they will be now loosely packed and have longer flight distances before bouncing to other gas molecules.

In the can and shortly after being released, there are more molecules of sprayed gas in the cubic centimeter than air molecules in the surrounding air per cm3 (pressure higher), but with lower molecule flying speed times higher molecule count (same temp.), then those gas molecules move away from each other (gas volume expanding) to the extent that molecule count is the same that in the surrounding air per cm3 (pressure equalised, molecule density equal) but now in the gas cloud molecule speed still low. Gas cold. Total sum average speed is less and temperature cooler than surrounding air. Its simpler to think what happens when compressed and why it heats, then deduce what happens, when it expands and why it cools

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Pretty much how an A/C system works. The rapid expansion of the refrigerant through an orifice in the A/C system is where the cooling comes from. In this case the liquid propellant (which is similar to the refrigerant in an A/C system) in the can is released when the can is held upside down instead of just gas when the can is held upright, the rapid expansion of the propellant causes the cooling.

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The nozzle is irrelevant to the physics involved. The basics of temperature vs pressure in a gas is simple. The compressed gas, not what is going on in the video many saw before coming here, has a heat content as well as a temperature. Increase the volume of the gas, however it is done, if done rapidly as in the question, and it still has the same heat content but in a larger volume, thus the temperature will drop. Basically the energy is now taking up a larger volume and the energy density is decreased, thus the temperature drops.

That answers the original question.

Many are here though came from a video with an ignorant clown that is saying words with no actual relation to what was going on the video.

He was not spraying a gas in the video, it was a liquid. It was a gas before being compressed into the can, for example the can of compressed 'air' I have next to my PC has 1,1 difloroethane in it. At STP, standard temperature and pressure, its a gas. At the high pressure its at in the can, when full, its a liquid for most of the volume of the can. Its in gas form at the top of the can.

Shake the can a bit and you will detect the sloshing. A mostly empty can will have little of the difloroethane in a liquid state. But at the start its almost all in a liquid state. Spray the can according to the instructions and the gas at the top is released. Turn the can upside down and the diflorethane that is in a liquid state is released. The liquid hits the bottle and then evaporates from the heat of the bottle, drawing off heat in the process of being converted from a liquid to a gas. Same thing that happens with water only much faster.

So the bottle is being cooled by rapid evaporation not by the expansion of a gas. Now the can of compressed gas IS cooled by the expansion and by the evaporation of the gas/liquid in the can.

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The above are not correct reasons for cooling of gas. The gas flowing from a high pressure inside to outside through a small nozzle or mouth. The cooling happens when it passes through the smaller section for accommodating the volume as per PV=nRT, and the pressure remains the same. When the gas suddenly releases to the outside, there is not any energy available; the pressure reduction adjusted with volume rather than temperature in this case. So the cooled air temperature remains the same.

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All this sounds great, but why is the temperature of the air substantially lower when the can is tilted upside-down, versus spraying it upright?

I would gather that for this to be considered an adiabatic process', the reaction of the compressed air should be virtually the same regardless of the cans position. The position of the can shouldnt alter the volume of air/speed of air coming out of the nozzle when sprayed, at least I wouldnt think.

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This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post - you can always comment on your own posts, and once you have sufficient reputation you will be able to comment on any post. –  Kyle Oman 4 hours ago
    
If you have a new question, please ask it by clicking the Ask Question button. Include a link to this question if it helps provide context. –  John Rennie 3 hours ago
    
"but why is the temperature of the air substantially lower when the can is tilted upside-down, versus spraying it upright?" You seem to have a particular situation in mind; perhaps one involving a consumer product. Given that most of the answers to this question assume a simple compressed gas you have a new question. –  dmckee 2 hours ago

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