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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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  • $\begingroup$ Were you spraying from a can of compressed gas, or from the nozzle of a compressor, or something else? $\endgroup$
    – bigjosh
    Nov 24, 2020 at 20:48
  • $\begingroup$ In addition to all the answers here, some spray cans contain a very volatile high pressure propellant (e.g. butane) that boils when the pressure in the can drops. Boiling is a cooling process. $\endgroup$ Feb 7, 2021 at 21:50
  • $\begingroup$ The model answer is now this great Minute Physics video. I wonder how much of this post went into informing that video. $\endgroup$ Mar 30, 2021 at 0:51

19 Answers 19

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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.

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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  • $\begingroup$ The Joule-Thompson coefficient is for a constant enthalpy process which is not the case here. Please see my response. $\endgroup$
    – John Darby
    Feb 3, 2021 at 16:26
  • $\begingroup$ This answer has a lot of numbers ... JT coefficient, heat capacity, Maxwell relations, inversion temperature ... and not much physics. The physics here is in the sentence "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." If you explained what you mean by that, and how it relates to all the numbers above and below, you'd have a first-class answer. $\endgroup$
    – garyp
    Feb 5, 2021 at 12:23
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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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    $\begingroup$ Ten years later and this is still the wrong answer. Do the math. Adiabatic expansion into room air could not create enough of a temp drop to account for the OP's observation that "the gas gets very cold". There is a phase change here. See physics.stackexchange.com/questions/14140/…. $\endgroup$
    – bigjosh
    Nov 24, 2020 at 20:47
  • $\begingroup$ @bigjosh No phase change is required. Please see my response. $\endgroup$
    – John Darby
    Feb 3, 2021 at 16:34
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    $\begingroup$ Did you know that in some cases the action of spraying gas from a cylinder can lead to heating at the nozzle? It is true that elsewhere the gas cools as it expands, but in the high-pressure region (if the gas was liquified for example) things are more complicated. This is a safety issue in the use of hydrogen cylinders for example. $\endgroup$ Feb 5, 2021 at 12:03
  • $\begingroup$ @JohnDarby In your response you choose to ignore phase change, which is not necessarily warranted, as most cans can be heard 'sloshing' when they are full. Even when they are not, the gas is usually compressed enough to not be correctly described by the ideal gas law. Still, some cooling can be expected without phase change indeed. $\endgroup$
    – Nicolas
    Feb 5, 2021 at 16:13
  • $\begingroup$ I don't agree at all with the adiabatic expansion while blowing hot or cold air. When blowing slowly, mouth wide open, you blow wet air at 37°C - which feels warm. When blowing fast through tight lips, you send a little fast air (at 37°) wich gets mixed with the surrounding (dry and sometimes colder) air before reaching the target. The difference, is that in the 2nd case the evaporation of the target is increased compared to the 1st case; it "feels" colder, especially on skin. (cont'd…) $\endgroup$
    – Nicolas
    Feb 5, 2021 at 16:21
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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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  • $\begingroup$ 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? $\endgroup$ Sep 20, 2020 at 2:13
  • $\begingroup$ No, the internal energy of the canned gas is used to do all of the work here. When this process occurs very quickly (like in a duster briskly blowing a gust of air), there is little time for the environment to transfer heat to the can and for some time the can will be cold (but of course not permanently as you point out). What Ron is saying is that if instead you sprayed the aerosol into an external "vacuum" of some sort, then the can would not get cold (having significant atmospheric pressure is a requirement for the can to get cold). $\endgroup$
    – ManRow
    Feb 1, 2021 at 16:18
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Are you sure you are starting with a compressed gas and not, in fact, a compressed liquid?

DustOff says "Compressed Gas" on the label...

enter image description here

...but a look at the material safety data sheet shows that it, in fact, contains not gas but liquefied Ethane, 1,1-Difluoro...

enter image description here

...with a boiling point of -25C.

Similarly Ultra Duster "Compressed Air"...

Ultra Duster

...is also actually liquefied Difluoroethane according to the MSDS... enter image description here

...also with a boiling point of -25C.

These are actually different names for the same stuff, and that stuff is R 152a - 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).

More info on the effect on Wikipedia...

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

If you disagree, please try pumping up a bike tire to 120PSI, 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! :)

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    $\begingroup$ As you discuss, cooling occurs with a liquid in the container due to phase change, as in an expansion valve in a refrigeration cycle. So far no one has been able to address the phenomena that determines how much a pure gas in the container cools off; for example, we need to address your bike tire example. The degree to which the process is non-isentropic/irreversible needs to be addressed supported by real data and hopefully someone out there can do this. Surprisingly, none of my references address this. Thanks for continuing to point out that we have not addresed this:) $\endgroup$
    – John Darby
    Feb 8, 2021 at 3:38
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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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  • $\begingroup$ but is it adiabatic in fact? what about turbulence in the nozzle? $\endgroup$ Feb 6, 2021 at 0:21
  • $\begingroup$ Quite right: it's not; but the adiabatic process is the simplest model explaining why it cools down. $\endgroup$ Feb 6, 2021 at 9:03
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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.

That answers the original question.

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

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  • $\begingroup$ Temperature dropping for gas expansion has nothing to do with constant internal energy being spread over a larger volume (as demonstrated by the Joule free expansion of an ideal gas : constant energy, no temperature drop), and has all to do with the work done in the process (against walls, or against internal interactions for a real gas) $\endgroup$
    – Nicolas
    Feb 5, 2021 at 16:04
  • $\begingroup$ I have downvoted this answer for the following reasons. 1. spreading a gas out to higher volume does not, in and of itself, change the internal energy (e.g. Joule expansion). 2. temperature has no direct relation to energy density. 3. "heat content" is incorrect terminology. 4. phase change is mentioned (which is correct) but the associated impact on the temperature is not (so it does not answer the question). $\endgroup$ Feb 6, 2021 at 0:20
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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.)

At the time of writing, other answers on this site are as follows.

. 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
. leftaroundabout says "to do this properly" and briefly describes adiabatic expansion from a molecular point of view
. bigjosh says it is to do with a liquid-to-gas phase change inside the can
. ManRow says adiabatic and presents some formulae
. Prof. 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

and I have ignored a few other answers. 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 will present calculations for the case of butane (C$_4$H$_{10}$) gas. 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. $$ \begin{array}{ll} \text{adiabatic index of butane gas at $300$ K} & \gamma = 1.09 \\ \text{constant volume specific heat capacity} & c_V = 1.57 \, {\rm kJ/kg K} \\ \text{constant pressure molar heat capacity} & C_{p} = 98.5 \, {\rm J/mol K} \\ \text{ditto for liquid} & C_p = 130\,{\rm J/mol K} \\ \text{molecular weight} & 58.124 \, {\rm g/mol} \\ \text{boiling point (at 1 atm)} & -0.4^\circ\,{\rm C} \\ \text{latent heat of vapourization} & L = 22.44\,{\rm kJ/mol} \\ \end{array} $$ $$ \begin{array}{cc} \text{temperature ($^\circ$C)} & \text{vapour pressure (kPa)} \\ -0.5 & 101 \\ 10 & 170 \\ 19 & 200 \end{array} $$

Adiabatic expansion

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). Under the approximations of constant heat capacities and ideal gas, we have $pV^\gamma = $ constant, and therefore, after using $pV = RT$ per mole, we have $$ p^{1-\gamma} T^\gamma = \mbox{constant} $$ so $$ T_{\rm f} = T_{\rm i} (p_{\rm i} / p_{\rm f})^{(1-\gamma)/\gamma} $$ where subscript i and f stand for 'initial', 'final'. Here $\gamma$ (the adiabatic index) is equal to $C_p/C_V$; it has the value $1.09$ for butane. The vapour pressure of butane at room temperature is about 2 atm, so let's take the pressure ratio to be $2$. Then we find $$ T_{\rm f} / T_{\rm i} = 0.94 $$ so if the initial temperature is $293\,$K then the final temperature is $277\,$K which is about $4^\circ$C. So that's a cooling in the right ball-park, and one may think we are on the right track.

But are we?

Joule-Thomson expansion

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. (An atmosphere of pressure is $10^5$ N/m$^2$; that's a large pressure. 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. If the conditions are right then one gets cooling, quantified by the coefficient $$ \mu_{H} = \left( \frac{\partial T}{\partial p} \right)_H = \frac{1}{C_p} \left( T \left( \frac{\partial V}{\partial T} \right)_p - V \right) $$ For the ideal gas this gives $\mu_H = 0$ so the cooling is related to the degree to which the gas is not well described as an ideal gas, for example near the liquid/gas phase transition. 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. This is the model where the equation of state for one mole is $$ (p + a/V^2)(V-b) = RT $$ and you can look up the constants $a$ and $b$ for your chosen gas. It's not a perfectly accurate model but pretty good. For the van der Waals gas one finds (for one mole) $$ \mu_H = \left( \frac{\partial T}{\partial p} \right)_H = \frac{V}{C_p} \left( \frac{2 a (V-b)^2 - R T V^2 b} {R T V^3 - 2 a (V-b)^2} \right) $$ (and this is roughly $(2a/RT \; - \; b)/C_p$).

For butane one finds $$ a = 1.466\;{\rm Jm}^3/{\rm mol}^2, \;\;\;\;\;\;\; b = 1.226 \times 10^{-4} \;{\rm m}^3/{\rm mol} $$ and at or just below room temperature the above formula gives $$ \mu_H = 12\,{\rm K/MPa}. $$ This is of the same order of magnitude as some values I managed to find on the web. So we find that in a pressure drop of 1 atm = $0.1\,$MPa, the temperature drop suffered by butane gas in a Joule-Thomson process near room temperature is $1.2$ K. So we conclude that the observed cooling (10 K or more) is not primarily owing to the Joule-Thomson process in the nozzle.

Evaporative cooling

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? When the pressure first falls, enough liquid evaporates to keep the system on the $p-T$ coexistence curve, which has a positive slope so the lower pressure corresponds to a lower temperature. 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. These two temperatures are $-0.5^\circ$C and $19^\circ$C. Hence if the can is left spraying out gas then the temperature will fall steadily until it reaches about $-0.5^\circ$C (in the case of butane).

This tells us how to calculate, but the original question was why does the cooling happen?

Consider the specific latent heat $L$. 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.g. by moving a piston, or by allowing gas to escape). If mass $m$ of liquid now turns to gas, then the internal energy of that part of the whole system has gone up by about $L m$ (well, to be precise, it keeps some of this energy and also transfers energy to the surroundings by doing work as it expands). Where has this energy $Lm$ come from? It came from the rest of the system as internal energy moved from the rest of the system to this part. So the temperature of the rest of the system must fall by $$ \Delta T \simeq \frac{L m}{C_p} $$ where $C_p$ is the heat capacity of the rest of the system, which can be written $C_p \simeq M c_p$ where $M$ is the mass of the whole and $c_p$ is the specific heat capacity. So we have $$ \Delta T \simeq \frac{L}{c_p} \frac{m}{M} \simeq 220 \frac{m}{M} {\rm kelvin} $$ which is valid for $m \ll M$. I am here just giving some rough feel for the sorts of numbers involved. It says we can expect a temperature drop of order $20^\circ$C if we spray out one tenth of the contents. 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 return of adiabatic expansion

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.

But what if there is no liquid in the can? What if it is simply a high pressure gas?

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.

So why does the can get cold?

In terms of overall energy movements, the can gets cold because the escaping gas does work in pushing back the surrounding atmosphere, and the energy required for this comes from internal energy of the can and its contents. In terms of physical process, the can gets cold for two different reasons depending on its contents. If there is liquid (and therefore also vapour) in the can then the cooling is evaporative cooling. If there is just vapour then the cooling is owing to adiabatic expansion inside the can of the vapour that remains. There is also a slight Joule-Thomson cooling in the nozzle.

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  • $\begingroup$ For only gas in the can, your (reversible) adiabatic expansion - hence isentropic- calculation for the final temperature gives same result as I calculated. 100 psia dry air ($\gamma$ = 1.4) initially at 388 K (115 C) going to 14.7 psia: final temp is 225 K (-48 C). 100 psia initial temp 293 K (20 C) final temp is 169 K (-104 C). @bigjosh points out, air inside a 100 psia bike tire relieving air out its valve does not cool this much; why: tire volume changes (decreases), not isentropic, ...? I can find little info on this type of problem. Appreciate your response. – John Darby 12 mins ago $\endgroup$
    – John Darby
    Feb 7, 2021 at 0:56
  • $\begingroup$ @JohnDarby For the tire it seems to me that there are two further issues: heat conduction and the changing volume. As the tire volume changes the outside air is doing work on the tire rubber. The tire in turn does work on the gas inside, thus maintaining its energy to some degree at least. And of course heat conduction can happen both for tire and for gas can, but the tire has a shape which makes its surface area comparatively larger. This is all just my first guess. $\endgroup$ Feb 7, 2021 at 10:30
  • $\begingroup$ For the bike tire I also think the change in volume is important. Also, the valve probably has a high discharge coefficient (ratio actual to maximum flow rate) reducing the rate of gas loss allowing more time for heat transfer into the gas to occur. Inside the can the expansion is to some extent irreversible due to turbulence/viscous heating from the pressure differential from the inner gas to gas close to the exit. I have updated my response accordingly. Maybe someone has some empirical data/correlations, but I cannot find any. I will keep looking:) $\endgroup$
    – John Darby
    Feb 7, 2021 at 21:09
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    $\begingroup$ Terminology reference for the physics disease you mention at the beginning of this answer: xkcd.com/356. $\endgroup$
    – tparker
    Jan 15, 2022 at 19:15
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    $\begingroup$ So, why does the can get cold? $\endgroup$
    – Innovine
    Apr 22, 2022 at 13:58
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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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    $\begingroup$ How do you know that the volume doesn't go up enough unless you already know the answer for $T$? Why doesn't the volume go up enough? $\endgroup$ Feb 6, 2021 at 0:24
  • $\begingroup$ Ideal gases that expand into a vacuum do not experience a temperature drop. $\endgroup$
    – lamplamp
    Feb 12 at 2:34
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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.

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  • $\begingroup$ The Joule-Thompson coefficient is for a constant enthalpy process which is not the case here. Please see my response. $\endgroup$
    – John Darby
    Feb 3, 2021 at 16:36
  • $\begingroup$ More to the point, the Joule Thomson effect may be cooling or not, depending on the material, and is not related to the hole itself. $\endgroup$
    – Nicolas
    Feb 5, 2021 at 15:57
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As Andy S pointed out, spraying an aerosol may be modeled as an adiabatic expansion for which the "final temperature" of the system can be obtained as $$T_f = T_i\left(\frac{V_i}{V_f}\right)^{\gamma-1}$$

It may be more "convenient" to express this as a function of spray can pressure (instead of the total aerosol "volume" of the system), so by substituting $V = nRT/P$ per the ideal gas law we can also express the final temperature as $$ T_f = T_i\left(\frac{P_f}{P_i}\right)^{1-1/\gamma} = T_i\left(\frac{P_f}{P_i}\right)\left(\frac{P_i}{P_f}\right)^{1/\gamma} $$ 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
  • $R$ is the universal gas constant, and $\gamma$ is the heat capacity ratio

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. So, perhaps a "more realistic" value for the actual amount of cooling could come from treating this as an irreversible adiabatic process instead, wherein which the aerosol expands against some "constant external/atmospheric pressure" $P_{atm} < P_f$.

Since the internal energy of an ideal gas depends only on its temperature and adiabatic processes exchange no heat with surroundings, then according to First Law of Thermodynamics we have that $$\Delta U = Q - W \Rightarrow nC_v\Delta T = -P_{atm}\Delta V $$ And by substituting $V = nRT/P$ per the ideal gas equation together with $C_v = R/\left(\gamma-1\right)$ per Mayer's Relation, we can obtain the final temperature as $$T_f = T_i \left(\frac{P_f}{P_i}\right)\frac{P_i + P_{atm}\left(\gamma-1\right)}{P_f + P_{atm}\left(\gamma-1\right)}$$ which will always be "warmer" than the reversible case since $$\frac{P_i + P_{atm}\left(\gamma-1\right)}{P_f + P_{atm}\left(\gamma-1\right)} > \left(\frac{P_i}{P_f}\right)^{1/\gamma}$$

And by substituting $C_p = R\gamma/\left(\gamma-1\right)$ per Mayer's Relation, we can also obtain the entropy generated by our irreversible process as $$\Delta S = nC_p\ln\left(\frac{T_{f,irr}}{T_{f,rev}}\right) = \frac{nR}{\gamma-1}\left(\gamma\ln\left(\frac{P_i + P_{atm}\left(\gamma-1\right)}{P_f + P_{atm}\left(\gamma-1\right)}\right) - \ln\left(\frac{P_i}{P_f}\right)\right)$$ which will likewise always be less than the entropy of free expansion (where $P_{atm} = 0$) $$\Delta S_{free} = nR\ln\left(\frac{P_i}{P_f}\right)$$

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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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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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  • $\begingroup$ ideal gases that expand into a vacuum do not experience a drop in temperature $\endgroup$
    – lamplamp
    Feb 12 at 2:41
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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.

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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.

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  • $\begingroup$ ideal gases expanding into a vacuum do not experience a drop in temperature $\endgroup$
    – lamplamp
    Feb 12 at 2:42
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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.

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Second updated answer based on numerous comments

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 is equivalent to calculations by others using the pressure/temperature relationship for reversible adiabatic expansion of a gas $$ T_{\rm f} = T_{\rm i} (p_{\rm i} / p_{\rm f})^{(1-\gamma)/\gamma} $$ (For example, see answers by @Andrew Steane and @ManRow)

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. Inside the can, for a rapid release of gas the expansion is to some extent irreversible due to turbulence/viscous heating due to the pressure differential from the inner gas to gas close to the exit. 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 0.97 or higher, while for a sharp orifice it is about 0.61. The lower the rate of gas loss, the greater the heat transferred from the surroundings to the container/gas. Maybe someone has some empirical data/correlations to clarify the important irreversibilities, but I cannot find any in my references. (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.)

My earlier response assuming an isentropic process follows.

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.

enter image description here

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

$m_iu_i - m_fu_f = (m_i - m_f)h + \int_{0}^{t} {v(t)^2 \over 2} dm$

where $m_f$ is the final mass of gas in the container, $m_i$ the initial mass, $u_f$ the final specific internal energy of the gas, $u_i$ the initial specific internal energy, $h$ the specific enthalpy for gas out the hole, and $v(t)$ the time-dependent velocity of gas out the hole. $h$ is constant for an isentropic process since Pout is fixed, and the two properties s and Pout uniquely specify the thermostatic state; therefore, Tout, the temperature of the gas out the hole, is constant. The velocity $v(t)$ decreases as the pressure in the container decreases. 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 0.6896 MPa (100 psia) and 388 K (240 F). For this state, s is 6.59 kJ/kg K. For the air out the hole Pout is 0.1013 MPa (14.7 psia) and the (s, P) state of the air out the hole is constant at (6.59, 0.1013) and at this state the temperature is 225 K ( - 55 F). So assuming an isentropic process the temperature of the gas out the hole is 225 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.

A couple of comments with regards to some of the earlier responses follow.

(1) Joule-Thompson expansion is for a constant enthalpy process which is not the case here since the enthalpy of the exiting fluid is not the enthalpy of the fluid in the container (while gas is ejected) due to the high velocity of the ejected fluid. The Joule-Thompson coefficient is defined as the partial derivative of temperature with respect to pressure at constant enthalpy. It can be + or - and is zero for an ideal gas. 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.

(2) It is not required that the fluid in the container be a liquid that changes phase. As discussed above, a pure gas cools off for this process.

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  • $\begingroup$ I do not follow your math, but I think empirical experiment should be good enough here. Pump up your bike tire to 100PSI, then push on the valve stem to let the air out . Measure the temp of the air coming out and report back here with your findings! $\endgroup$
    – bigjosh
    Feb 5, 2021 at 5:11
  • $\begingroup$ Early in your discussion you make the assumption that the process is isentropic. So your discussion deals with that case, which could in principle be realised in a careful experiment. However the process inside a typical nozzle is, in practice, far from isentropic. $\endgroup$ Feb 5, 2021 at 12:09
  • $\begingroup$ Indeed, you assume that the process is isentropic and not isenthalpic, while I'd have assumed the opposite. 1. It seems clearly irreversible to me, 2. for such a fluid forced through an immobile opening (with no heat transfer) because of a pressure drop I'd assume isenthalpic process. $\endgroup$
    – Nicolas
    Feb 5, 2021 at 15:57
  • $\begingroup$ I see that the process can be far from isentropic, and I updated my response. I do not know how to address the irreversibility and am looking for others to respond. I do not believe the process is isenthalpic because of the high velocity of the ejected gas. $\endgroup$
    – John Darby
    Feb 5, 2021 at 23:00
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The cooling of a high-pressure ideal gas (population H) upon expansion into a volume occupied by a lower-pressure ideal gas (population L) is explained as collision-mediated transfer of momentum from a faster-moving particle to a slower-moving particle.

Even though the bulk phase of population H experiences a temperature drop, at the particle level there are examples of two kinds of momentum-transferring collisions through the random motion of atoms of H and L:

  1. collisions between faster moving H atoms into slower moving L atoms, transferring momentum from H atoms to L atoms and thus speeding up L atoms
  2. collisions between faster moving L atoms into slower moving H atoms, transferring momentum from L atoms to H atoms and thus speeding up H atoms (note head-on collisions between H and L with same speed but opposite velocity would not transfer any momentum)

Both populations of gas start at the same temperature and thus have the same average speed (Maxwell Distribution), but because of the larger number of H atoms compared to L atoms, there will be more collisions of the first type compared to the second.

This answer only addresses particle-level interactions of ideal gases without invoking 1) bulk thermodynamic parameters and 2) mechanical work from the modelling of the lower pressure gas as a solid piston. The scenarios of real gases with van der Waals interactions, let alone the air spray can containing compressed liquid, needs the thermodynamic analyses presented in other answers.

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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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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.

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    $\begingroup$ For the record, I'm afraid this answer is plain false (as are all "maxwell's daemon" based arguments). For such a process to work, the noozle should be colder than the gas it selects (ejects) ; otherwise the thermal motion of the noozle particles defeat the purpose. $\endgroup$
    – Nicolas
    Feb 5, 2021 at 15:49

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