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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 F. Deen Dec 17 '13 at 10:52

8 Answers 8

up vote 8 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
    
@AndyS: Ron is right. The adiabatic formula is exact only for quasi-static expansions. But it is a good approximation for given case. In any case +1. –  Martin Gales Aug 31 '11 at 6:31
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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

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
    
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

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

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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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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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

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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