Why do expanding gases cool

Heat can be seen as the total amount of energy of all the molecules in a certain gas. Spacecraft designers sometimes need to cool equipment to very cold temperatures. They need to do this because they want to look at infared light. The reason why the equipment needs to be cooled to very low temperatures is that infrared radiation is generated by heat.

If the camera which you are watching is warm, then it will be generating "light" and will not be able to see into space. All the camera will see is its own warmth and therefore cannot look at the warmth infared radiation of other objects. In other words, to look at the warmth infared radiation of other objects, you need to be cooler than those other objects.

By being heated from 32 to , cubic inches of iron become " water " " air " Gases are, therefore, more expansable by heat than matter in the other two conditions ot liquid and solid. The reason is, that the particles of air or gas, far from being under the influence of cohesive attraction, like solids or liquids, are actuated by a powerful repulsion for each other.

The addition of heat mightily enhances this repulsive tendency, and causes great dilatation. The rate of the expansion of air and gases from increase of temperature, was long involved in considerable uncertainty. This arose from the neglect of the early experimenters to dry the air or gas upon which they operated.

The presence of a little water by rising in the state of steam into the gas, on the application of heat, occasioned great and irregular expansions.

But in , the law of the dilitation of gases was discovered by M. Gay-Lussac, of Paris, and by Dr. Dalton, of England, independently of each other. Simple: 10 m s Now imagine that the elephant is running towards you at 2 m s Certainly it will rebound more quickly, but what will its speed be?

The answer is 14 m s -1 see box. So, with the elephant coming towards you, the ball rebounds more quickly. In the same way, gas particles will speed up when they rebound off the approaching piston. And therefore the temperature of the gas will go up. If the ball is coming towards the elephant at 10 m s -1 and the elephant is moving at 2 m s -1 , then their relative velocity is 12 m s So it bounces off at a relative velocity of 12 m s What is the new average speed as a proportion of the old average speed?

What will be the new average KE as a proportion of the original KE? 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. From a molecular point of view it is straightforward if hard to calculate : the faster molecules preferentially move out of the liquid, and on their journey they slow down because they are escaping from attractive forces in the liquid.

But how to calculate this using thermodynamic properties? Eventually the system reaches a dynamic equilibrium with pressure approximately 1 atm outside the nozzle, pressure somewhere between 1 and 2 atm inside the can, and temperature somewhere between the temperature where the vapour pressure is 1 atm and the temperature where the vapour pressue is 2 atm.

This is the heat you would have to provide to cause unit mass of substance to change from liquid to gas in conditions of constant pressure and temperature. This is the case we are used to when we boil water in a kettle. But evaporative cooling is a different process. No heat is provided, but we do something which lowers the pressure e. It came from the rest of the system as internal energy moved from the rest of the system to this part.

I am here just giving some rough feel for the sorts of numbers involved. As I already said, the temperature will not fall indefinitely; it reaches a new equilibrium; the purpose of this rough calculation was merely to show that the energy movements are consistent. The summary of the above discussion is that if there is liquid in the can then the temperature drop is primarily owing to evaporative cooling of that liquid, and the vapour, because the latent heat of vapourization has to be provided by the contents of the can in the absence of heat flow from outside.

We already established that there is only a modest cooling after the gas has left the nozzle and makes its way into the room. Let's look inside the can again. To understand the effect of a leak in a can of gas, imagine a thin membrane dividing the gas which is about to escape from the gas which will remain. As the gas escapes this membrane moves and the gas within it expands. That expansion is, to good approximation, adiabatic.

To prove this we need to claim that there is no heat transfer across this membrane. There will be no heat transfer if the gas on either side of the membrane is at the same temperature. If you think it is not, then allow me to add another membrane further down, dividing the gas which will remain into two halves.

This gas is all simply expanding so there is no reason for temperature gradients within it. But this argument will apply no matter where we put the membrane. We conclude that the part of the gas which remains in the can simply expands adiabatically to fill the can. So now the initial calculation which I did, describing adiabatic expansion, is the right calculation, but one must understand that the process is happening right in the can!

So no wonder the can gets cold! Here is another intuition for this. As it moves through the nozzle, the gas that gets expelled is being worked on by the gas that gets left behind, giving it energy, and the gas left behind loses energy. If there were a big hole the gas would rush out very quickly. In the case of a narrow nozzle it is prevented from getting up to very high speed. It that case it makes its way out into the surrounding atmosphere at a similar pressure to the ambient pressure and it uses up its extra energy pushing that atmosphere back to make room for itself.

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.

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.

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. 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.