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. 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. 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 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? 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.
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. Are you sure you are starting with a compressed gas and not, in fact, a compressed liquid? Similarly Ultra Duster "Compressed Air" These are actually different names for the same stuff, and that stuff is R a - literally a refrigerant used to cool things! It is the same for every kind of "compressed gas" or "compressed air" that I could find on Amazon.
So it would seem that we are not dealing with an expanding gas here, we are dealing with a liquid evaporating into a gas.
This liquid is normally maintained at high pressure inside the can, but when you release it through the nozzle you have effectively created an air conditioner where the nozzle is the metering device and the liquefied gas is the refrigerant. This is a straightforward reason why the gas that makes it out of the nozzle is so cold. This is a simple and very well understood effect - and you likely have many other places in your house where this happens regularly like inside your refrigerator or air conditioner.
If you disagree, please try pumping up a bike tire to PSI, letting the gas inside reach equilibrium temperature, and then releasing the high pressure gas though the value and measuring the temperature using a thermometer or Flir fingers do not work because even a warm breeze can feel cool due to other effects and report back if it is very cold a couple degree drop is not "very cold" as per OP.
Only actual empirical results not theories about why the gas should be cold accepted! 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.
The joule Thompson coefficient is negative as the gas is coming out through the small hole, for negative coefficient the tempr. However, as Ron Maimon pointed out , the above formulae are only valid for reversible adiabatic processes, so they will only give the "maximum possible" cooling of the spray can. This is a long answer.
Last night I suffered a bout of physics disease. This is the nervous compulsion to solve a physics problem because it bugs you. AndyS says it is because the gas expands adiabatically. Markus Deserno says it is a Joule-Thomson expansion constant enthalpy. Ron Maimon describes the gas pushing back the surrounding atmosphere; this is adiabatic expansion again, apparently in the room containing the can. ManRow says adiabatic and presents some formulae.
Gopal Lohar says Joule-Thomson. John Darby acknowledges that there is irreversibility at the nozzle and then offers another treatment of adiabatic expansion. Anthony X says the cooling is wholly due to evaporation inside the can, and explicitly not any subsequent expansion. So most people are arguing for adiabatic expansion. However, the process inside the nozzle of any spray can is certainly not adiabatic in the sense of not isentropic , and that is where most of the pressure drop occurs.
Therefore the answer cannot be that simple. Nevertheless in the following I will first present the adiabatic expansion of a gas, to see what it offers, and then move on. It will turn out that this calculation will be valuable at the end. I have often used butane gas cylinders while camping and they do indeed get quite cold after a few minutes of use.
Here are some properties. First let's consider adiabatic expansion. The physical basis of the cooling in adiabatic expansion is energy and momentum conservation. The gas pushes against the rest of the atmosphere, doing work on it. Therefore its internal energy falls since no heat is coming in. In a system as simple as a gas, such a drop in internal energy results in a drop in temperature the molecules slow down.
So that's a cooling in the right ball-park, and one may think we are on the right track. The trouble with the previous calculation is that it seems to be assuming that while we spray the gas, the pressure just outside the nozzle is 2 atmospheres. But that is an impossible statement.
You can't create a whole atmosphere of pressure difference in air just using a spray can! It would cause the gas just outside the nozzle to rush explosively outwards, reaching the speed of sound in a microsecond or so.
It is quite impossible. Recall that weather charts record pressure variations in millibars; pressure changes in weather are at the percent level.
So how does the pressure go from 2 atm the vapour pressure in the can to 1 atm the ambient pressure in the surroundings? The answer is simple: the pressure drop is almost entirely in the nozzle. The pressure outside the nozzle is somewhat above one atmosphere, so that gas does spray away from the nozzle, but compared to one atmosphere this does not present much of a ratio, so the adiabatic expansion as the gas moves away from the nozzle is not able to account for much of the observed cooling at the can.
So let's consider what happens inside the nozzle. Here the process is, to good approximation, a pressure drop in a constriction without heat exchange. This process is called the Joule-Thomas process. It is a well-studied process much used for cooling and liquifying gases. I tried to find values for this coefficient for butane online. I got some numbers but was unsure, so for added confidence I used the van der Waals model.
It's not a perfectly accurate model but pretty good. So we conclude that the observed cooling 10 K or more is not primarily owing to the Joule-Thomson process in the nozzle. So far we have found that the answer is not adiabatic expansion after the nozzle, nor is it Joule-Thomson isenthalpic expansion in the nozzle.
So we have to look inside the can for our answer. I will first treat the case where there is liquid and vapour in the can, and then the case where only gas is involved. With liquid in the can, let's suppose in the first instance that there is not enough time for significant heat flow into the walls of the can.
In this case we have evaporative cooling. When some vapour escapes through the nozzle, the pressure in the can falls, and consequently some liquid evaporates.
Both liquid and vapour then cool. For a long time I found this cooling puzzling from a thermodynamic point of view. Shake the can and fresh, cooler content comes into contact with the metal next to your hand. This increases the temperature gradient, and heat is quickly transferred from your hand and the aerosol again feels cold.
The cooling effect is also felt with cans of fizzy drinks but is seldom noticed as we are trained as children not to shake them, for obvious reasons. Aerosol instructions, on the other hand, often request vigorous shaking to mix the propellant and active ingredients, so chilling is commonly felt.
The cooling from shaking is not to be confused with the coldness you get when spraying. Aerosol propellant is often liquefied gas, which readily boils at room temperature and is only liquid inside the can due to the high pressure. When the nozzle is pressed, that pressure is released and the propellant boils, taking heat from its surroundings. With continual spraying, condensation may form on the outside of the can and might even freeze. The evaporation of the liquid inside the can results in a drop in its internal heat energy, and it absorbs a large amount of heat from the surrounding air and environment -- in this case, the metal can.
When the liquid inside absorbs the heat from the can's metal body, the can cools down rapidly. As the expanding gas leaves the can, it also absorbs heat energy from the nozzle and straw, and anything else the gas comes into contact with. For example, if you spray a keyboard, you'll see a thin white layer of frost form on the keys briefly.
With prolonged use of the canned air, you might notice that the force of the air stream weakens over time, and the can becomes too cold to comfortably hold in your hand. The heat energy from the can has all gone into evaporating the liquid inside; when the can itself becomes cold, not enough heat remains to vaporize more liquid.
To remedy the "out of breath" condition, set the can down and let it warm for a few minutes. This restores the strength of the air bursts.
The can carries a warning label telling you to avoid spraying onto your skin; the rapid absorption of heat can easily cause frostbite. The frost that forms on the can and nozzle comes the condensation of water vapor in the surrounding air. Ryan Maxwell began his professional freelance writing career in He is a former U.
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