# If pressure decreases, does temperature then decrease?

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So, quick disclaimer: i am very young. only 12. i might also misspell some thing or not translate it correctly. i just like physics, so dont expect much.

So, i was reading in my science textbook (as a hobby) and it asked to describe some examples, in wich i should find out wether the work on the object was positive, negative or 0, and if the thermal energy recieved was negative, positive or 0. and i one of the questions it asked to describe and explain some things about the gas in a spray paint when released, and i was wondering, since the pressure decreases in the paint can, does the tempurature decrease with it? im not sure about it, im sure that the gas laws should tell me, but im not sure how to use them.

Thanks alot.

russ_watters and berkeman

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anorlunda
Staff Emeritus
Welcome to PF. We are very glad to have a young man interested in things like gas laws at the age of 12.

What gas laws do you know about?

PV=nRT is called the perfect gas law. P is pressure, V is volume, nR you can think of as the quantity of gas, T is temperature. So, can you try to use that for the gas inside the spray can to answer your own question?

russ_watters and berkeman
berkeman
Mentor
Welcome to the PF.
it asked to describe and explain some things about the gas in a spray paint when released, and i was wondering, since the pressure decreases in the paint can, does the tempurature decrease with it?
As another example to help show how the pressure and temperature of an ideal gas are related, consider that when you go to the scuba diving store to get your air tank refilled, they often put it in a tub of water while they pump up the pressure of the air in the tank. Can you comment on why they do that?

pinball1970 and russ_watters
Welcome to the PF.

As another example to help show how the pressure and temperature of an ideal gas are related, consider that when you go to the scuba diving store to get your air tank refilled, they often put it in a tub of water while they pump up the pressure of the air in the tank. Can you comment on why they do that?

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im guessing it would be to keep it at a constant tempurature, since a pressure increase means the molocules are moving faster?

berkeman
sophiecentaur
Gold Member
What goes on in a paint can is more complicated than can be described by Ideal Gas Laws because the propellant actually undergoes a change of state (from liquid to gas). That involves Energy (in the same way that boiling a kettle dry requires a lot of energy from the heating element for perhaps many minutes). The energy involved with a change of state is a lot more than in just heating or cooling but a 'reasonable' explanation why reducing pressure can reduce the temperature of a gas is as follows.

The temperature of a gas corresponds to the average Kinetic Energy of the molecules (basic definition of temperature). If you have a cylinder of gas with a piston and you gradually pull the piston out. The molecules bouncing against the piston surface will end up going a bit slower on impact because the piston is moving away from them. That corresponds to a reduction in temperature of the gas because the KE of each molecule gets reduced at each collision. It doesn't matter how slowly you move the piston; KE will still be lost - it will just take longer.

There is also the Energy vs Work argument. As you compress a gas, you do work on it and, as energy is conserved, the Work (mechanical Energy is Force times Distance) will turn up as Heat (Thermal Energy).

Welcome to PF. We are very glad to have a young man interested in things like gas laws at the age of 12.

What gas laws do you know about?

PV=nRT is called the perfect gas law. P is pressure, V is volume, nR you can think of as the quantity of gas, T is temperature. So, can you try to use that for the gas inside the spray can to answer your own question?
so, if i fired the spray paint till the pressure was 20% lower, the volume inside would stay the same. also, the amount of gas inside would have been reduced by 20%, and if P was lower, and nR was lower, but V had stayed the same, then im guessing that T would stay the same? but at the same time the movement of the molocules inside had been lowered cause of the lower pressures and therefore, weaker collisions it would be corrosponding to a loss of tempurature,so at the same time i guess it would have been... or maybe im just interpreting it wrong?

sophiecentaur
Gold Member
the volume inside would stay the same.
You need to be careful here. The volume of the original mass of gas is very much increased; some of it has left the cylinder and is now at atmospheric pressure. The 'simple' law doesn't apply any more. The remaining 80% of gas, inside will have expanded to fill the can so its pressure will be lower - but the pressure drop will be a lot less than the pressure drop for the escaped gas.

But, as I have already pointed out, the 'boiling off' of the liquid propellant will have had a vastly greater effect on the temperature inside. Try to understand a much simpler model, involving just a gas before you venture into the paint can.

Try to understand a much simpler model, involving just a gas before you venture into the paint can.
so, im gonna think of a bit simpler, as you said to make me understand some. so if i have a box with gas, and i increase the volume, then the pressure would decrease, and the amount of gas would stay the same, and if PV=nRT and nR is the same, P is lower than before, and V is higher, then if im correct then T would be the same, because the volume change is equal to the pressure change, right? what would be the next step in order to understand the paint can, because im not sure.

also, about that boiling thing, so, you mean the tempurature inside the spray can is going to increase because of the boiling of the liquid, right? im not quite sure, please tell me, but: where is the energy coming from? is it coming from friction, or does it happen in the manufactering process, or somewhere else? im not quite sure in all of this, so please tell me.

Thanks.

berkeman
Mentor
im guessing it would be to keep it at a constant tempurature, since a pressure increase means the molocules are moving faster?
Yes, basically to keep the tank cool while it is filled. Otherwise it gets a bit hotter than you would like. Also, in case of an accidental rupture of the tank (which is rare thankfully), the water would help to dampen the explosion some.

sophiecentaur
Gold Member
T would be the same, because the volume change is equal to the pressure change, right?
Boyle's Law says PV is a constant for constant Temperature. That is if you keep the temperature constant (in a water bath etc.) The water bath takes or adds Energy to the system to keep temperature constant. This is called Isothermal conditions. In your case you have Adiabatic conditions and Boyle's Law is not followed. Obvs, if you squeeze a gas, it will tend to get hotter. Boyle and others were trying to isolate the effects of the various variable.
In PV = nRT, T is not constant and the nR bit takes account of the Energy content of the gas and shows where Boyle's Law behaviour stops, due to the interaction between P and V.

you mean the tempurature inside the spray can is going to increase because of the boiling of the liquid, right?
No - quite the reverse. the temperature will drop because of the lost energy. Energy will flow in from the hotter surroundings; the can will feel colder. This is the same as wet clothes making you cold as the water evaporates. The temperature will drop until gas stops boiling off. LPG cylinders can stop supplying gas to heaters if they get too cold in an already cold building. That can be a real problem.
This is all a good example of how you need to learn to interpret equations very carefully. When you are young, you keep coming across things that don't quite make sense but asking your such good questions on PF can help. (Not only a problem for the young, of course!)

sophiecentaur
Gold Member
the water would help to dampen the explosion some.
A very small amount, perhaps. The energy stored in the compressed air in a diving cylinder can take out the back end of a car. 300bar for a compressed volume of 10l is a lotta Joules. I would guess it could empty a small swimming pool.
Pressure testing is always done by filling the cylinder with water under pressure; the amount of energy stored in an almost incompressible fluid is almost zero.

berkeman
anorlunda
Staff Emeritus
@AttoBlaze ,

Congratulations on your curiosity. I think you can see two lessons from this thread.

1. The ideal gas law is very useful. You can imagine many scenarios where P, V, n, and T vary, and use PV=nRT to figure out how the others vary with it. That is a principle in science.
2. There can be many complications in a real life experiment. The paint evaporates, heat leaks through the walls. Should nR be just the gas remaining in the can, or should it include the part sprayed? Complications go on and on. Engineers deal with the complications.
3. It is not easy to ask the perfect question on a topic like this. Are you asking about the principle? the engineering? If the question is not perfect, you get a variety of answers because different people reply to different parts of your question. Sometimes, working to perfect your question leads to more understanding. A question well asked is half answered.
So, to ask or answer a simple question like yours takes science, engineering, and English writing skills. I think you are doing well in all three.

256bits
In your case you have Adiabatic conditions and Boyle's Law is not followed
wait, so if its an adiabatic process, wich gas law does it then take to find out? does such a gas law even exist?
Obvs, if you squeeze a gas, it will tend to get hotter.
well, yes, but since the gas is also being released, wouldnt it alter in some other way?
No - quite the reverse. the temperature will drop because of the lost energy. Energy will flow in from the hotter surroundings; the can will feel colder.
oooohhh, ok. so, the tempurature would decrease because of entropy, right?
This is all a good example of how you need to learn to interpret equations very carefully.
ok, i will try to be more carefull next time.

Thanks

@AttoBlaze ,

thanks!
It is not easy to ask the perfect question on a topic like this. Are you asking about the principle? the engineering? If the question is not perfect, you get a variety of answers because different people reply to different parts of your question. Sometimes, working to perfect your question leads to more understanding. A question well asked is half answered.
i will try to improve my questions next time, and i will try to be more specific, thanks!
So, to ask or answer a simple question like yours takes science, engineering, and English writing skills. I think you are doing well in all three.
Thank you alot, kind sir!

berkeman
sophiecentaur
Gold Member
well, yes, but since the gas is also being released, wouldnt it alter in some other way?
Yes. There are other variables involved if some gas is lost. Each small volume of gas will behave according to the local conditions. (That's one way of looking at it.)
Overall, I think @anorlunda has identified the problem with your original question. It is not answerable without being more specific. If you follow the process through from the simplest scenario and try to tackle each of the 'Laws'.
You might find it interesting to take a stroll through this website. It has some good pages, dealing with a lot of different aspects of Physics.

Yes. There are other variables involved if some gas is lost. Each small volume of gas will behave according to the local conditions. (That's one way of looking at it.)
Overall, I think @anorlunda has identified the problem with your original question. It is not answerable without being more specific. If you follow the process through from the simplest scenario and try to tackle each of the 'Laws'.
You might find it interesting to take a stroll through this website. It has some good pages, dealing with a lot of different aspects of Physics.
ok, i will look at the other gas laws, try some pages here and try to interpret it agian tommorow, and specify it better and more thorougly.

its almost 12 PM, so i probably gotta go to sleep soon. ill probably reply in 12-15 hours or so, im really sleep deprived at the moment.

sophiecentaur
Gold Member
im really sleep deprived at the moment.
You need to deal with that problem now, rather than spend too much time of Physics. We've all been there and suffered the results of lack of sleep. I'm being boring boring, I know but treat your body and brain like a valuable piece of equipment and it will reward you.

https://1drv.ms/v/s!AtuYIumiulODhjf6cXEcvnR0-Zcc?e=mRGFMv

This is a great video that shows the relationship between temperature and pressure as well as pressure with velocity.

It shows a 747 taking off in foggy conditions.

As the airplane accelerates, the airflow velocity increases over the wing. The top surface of the wing is more curved than the bottom, causing the air to speed up over the top, creating lift so that the airplane can fly.

As the air accelerates over the wing, the temperature of the air decreases. Due to the low temperature/dew point spread, the humidity condensates out of the air creating increased fog over the top of the wing. This is similar to how fog forms when you open a freezer on a hot, humid day.

Watch the video to the end. As the airplane flies through the fog bank at the end of the runway, the wake of the plane and the heat from the engine thrust creates a wake tunnel through the fog.

I hope you enjoy!

Perhaps some of the physics experts can shed more light for me on what's going on! I'm just a pilot! LOL