# Why does the gas get cold when I spray it?

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?

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.

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.

• The statement that "this work releases energy to the environment" seems counter-intuitive. Doesn't the work draw energy from the environment and thus cool it? – Tom Russell Sep 20 at 2:13

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.

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.

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. Now the can of compressed gas IS cooled by the expansion and by the evaporation of the gas/liquid in the can.

Ethelred Hardrede

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.

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.

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.

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.

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

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.

The best and simple answer- All the K.E of the gas molecules is lost in getting out of tight nozzle and expanding. At last u are left with lowK.E gas molecules.

To explain why the cooling occurs you have to look at what’s happening to the particles themselves. In the can, the molecules’ velocities are randomly distributed vectors. Their energies form a Gaussian distribution about the temperature of the gas there. The nozzle is a sort of Maxwell’s daemon. It is a sorting mechanism, selecting only certain molecules. In order to exit, a molecule must be at the nozzle orifice inside the can but in addition it’s velocity must have a component in the direction of the inner plane of the nozzle orifice and collimated so, without hitting the sides of the nozzle tube too many times, it will exit into the atmosphere. The collection of these molecules will have, more or less, a collective exit velocity. As they exit, these molecules do work against the molecules in the atmosphere. By the Furst Law, they will be cooler than the temperature inside the can.

Are you sure you are starting with a compressed gas and not, in fact, a liquid?

DustOff says "Compressed Gas" on the label... ...but a look at the material safety data sheet shows that it, in fact, contains not gas but liquefied Ethane, 1,1-Difluoro... ...with a boiling point of -25C. This liquid is maintained at high pressure inside the can, but when you release it though 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).

https://en.wikipedia.org/wiki/Vapor-compression_refrigeration

The contents of a spray can are in mostly liquid form (both the product to be sprayed and the propellant). There is some head space above the liquid which is essentially the propellant in gas phase. As the mixture is released from the container, the volume of liquid decreases, with a corresponding increase in head space. To maintain equilibrium, some of the liquid propellant evaporates, which requires heat, so the temperature drops slightly.

Note that the mixture (product and propellant) leaving the can remains in liquid form until it passes through an orifice at which point the pressure drops, and the propellant evaporates, drawing heat from its surroundings. This has no effect on the temperature of the can because it has effectively already left it.

The can gets colder because of evaporation inside the can to maintain equilibrium between liquid and gas phases of the propellant, not because of what may be going on at the nozzle.

AndyS gave a great answer as to why the gas gets cools down upon entering the room (or outdoors, wherever it was sprayed!) If anyone is also interested in why the spray can itself also gets cold, it is for much the same reason as well. From the ideal gas equation ($$PV=nRT$$), we can express the new spray can temperature as $$T_f = T_i\left(\frac{P_f}{P_i}\right)^{1-1/\gamma} = T_i\left(1-n_\text{lost}\frac{RT_i}{P_iV_\text{can}}\right)^{\gamma-1}$$ where

• $$T_i$$ and $$T_f$$ are the initial and final spray can temperatures, respectively
• $$P_i$$ and $$P_f$$ are the initial and final spray can pressures, respectively
• $$n_\text{lost}$$ is the moles of gas lost to the atmosphere
• $$V_\text{can}$$ is the volume of the spray can
• $$R$$ is the universal gas constant, and $$\gamma$$ is the heat capacity ratio

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.