The average speed of a gas is also called the kinetic energy of the gas. When you combine the average speed for each molecule with all the molecules in the gas you get heat. Heat is basically the average speed of a single gas molecule multiplied by all the molecules in the gas.
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. Summary Close gases cool when they expand the particles lose kinetic energy as they do work. The kinetic theory of matter.
Cooling by expansion. Picture 2. Releasing CO 2 from a cylinder into a beaker. You can see the clouds on the bench where the cold CO 2 has condensed water in the air. The CO 2 is so cold it solidifies in the beaker to form dry ice. Sometimes, it is desirable that sprays do feel cool. There are two ways that these sprays cool down: evaporation of a liquid expansion of a gas. A very dramatic illustration of a gas cooling as it expands is the formation of dry ice from a carbon dioxide fire extinguisher.
As the CO 2 gas leaves the nozzle of the cylinder, it expands into the air and cools enough to solidify, forming dry ice. So why does the expanding gas cool? If you pull the pump out quickly, then the gas will cool down.
The gas particles are moving with a speed that is determined by the temperature of the gas. What effect does this have on the speed of the particles?
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.
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. Many comments pointed out that the actual process can be significantly non-isentropic and irreversible. My earlier treatment using the first law assuming an isentropic process provides the maximum cooling case for gas only in the can no liquid. This estimate can be off, as pointed out in a good comment by bigjosh about release from a bicycle tire.
I think a more realistic evaluation should consider the following. The rate of gas flow out the opening depends on the discharge coefficient ratio actual to maximum gas flow ; for a machined nozzle the discharge coefficient can be 0. For the case of the bicycle tire, the coefficient of discharge through the valve may be very high. Also the tire tube volume decreases with gas loss, doing work on and raising the temperature of the gas in the tube.
This question can be answered using the first law of thermodynamics for an open system. The following example is based on an example in the text Elements of Thermodynamics and Heat Transfer, by Obert and Young. Consider the system shown in the figure below consisting of compressed gas in a container with a hole.
There is mass flow out the system to the atmosphere which is at constant pressure Pout. As mass leaves the container, the pressure and temperature of the gas inside the container, Pin t and Tin t change with time, t. There is no heat added and no work done on the gas in the container. To simplify the evaluation for discussion here, assume the process is isentropic; that is, the specific entropy s is constant. Using the first law, after time t.
Given the initial pressure and temperature of gas in the container, the specific enthalpy, s, can be calculated. The temperature of the gas out the hole is that for the state s, Pout. Also, once the pressure inside the container reduces to Pout, the gas inside the container is at state s, Pout and has the same temperature as the gas out the hole.
For example, assume the gas inside the container is dry air initially at 0. For this state, s is 6. For the air out the hole Pout is 0. So assuming an isentropic process the temperature of the gas out the hole is K - 55 F which is also the temperature of the gas inside the contained once it reaches atmospheric pressure. The gas inside the contained cools substantially as it ejects gas. To calculate the rate of depressurization, an evaluation of the area of the hole and the velocity out the hole is required; the flow out the hole may be choked flow while the container pressure is sufficiently high.
The Joule-Thompson coefficient is defined as the partial derivative of temperature with respect to pressure at constant enthalpy. See a good thermodynamics text such as one by Obert, or Sonntag and Van Wylen.
Joule-Thompson expansion is typically explained using a throttling process in which the change in fluid velocity is negligible. As discussed above, a pure gas cools off for this process. 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. 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. Their energies form a Gaussian distribution about the temperature of the gas there. It is a sorting mechanism, selecting only certain molecules.
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. Sign up to join this community. The best answers are voted up and rise to the top. Stack Overflow for Teams — Collaborate and share knowledge with a private group. Create a free Team What is Teams?
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