# I Gas Law - increasing temperature with constant volume held

#### wee VooDoo

Summary
Increasing the temperature using pressure with volume held constant.
I've set up a simple experiment to look at the ideal gas laws. My experiment is relatively simple in that I have a metal tube which is capped on one side. I am then pressurising the tube with air to 100 psi and locking it off.

My thought is that as the pressure increased, with volume held constant, the temperature of the tube should rise. However, after 30 minutes of observation, there is no rise in temperature. Am I missing something? I was expecting to see the temperature of the tube rise. The setup is not enclosed at the moment and in the workshop. I am using an infrared temperature sensor to measure the temperature of the pipe surface. I didn't think heat loss due to ambient outside air temperature would be too much of an factor, or I would at least see some temperature increase, however small.

The tube is SS 316 with a ball valve capping it off at one end and using the pump to hold the pressure once it is reached.

Thank you.

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#### jbriggs444

Homework Helper
What ideal gas law are you using? You are holding volume constant and varying pressure. Can you think of something else that you are also varying?

#### wee VooDoo

I was looking at Combined Gas Law PV/T = C

Maybe I should be looking at ideal gas law? Is it because I am adding more air to the system, rather than compressing the original volume of air? Newbie at this, sorry :) Any learnings would be appreciated

#### jbriggs444

Homework Helper
Is it because I am adding more air to the system
Right. By adding more air you are breaking the assumption of a constant amount of the gas.

PV=nRT is more general.

#### wee VooDoo

Thank you, so if I were to compress the volume using a piston instead, the temperature would increase? i.e:

P1V1/T1 = P2V2/T2

I'm having trouble thinking about how to calculate on paper what the temperature increase would be.

Inputting starting pressure (atmosphere) and temperature (ambient temperature) and keeping the volume constant on both sides. Increasing the pressure should give the resulting temperature? Or should I be changing the volume too based on the piston moving? Not sure how to calculate that as would need to calculate the pressure exerted back by the piston for the given compressed volume of air?

Basically, looking to figure out how to calculate making a pipe heat up using compressed air from a starting point of ambient.

Sorry for the confusion, I think I need to go back to school!

#### jbriggs444

Homework Helper
Thank you, so if I were to compress the volume using a piston instead, the temperature would increase? i.e:

P1V1/T1 = P2V2/T2

I'm having trouble thinking about how to calculate on paper what the temperature increase would be.
The problem is that you are controlling one parameter (volume) but you have two variables that can change as a result (pressure and temperature).

You could have an isothermal compression in which volume goes down, pressure goes up and temperature remains constant.

Or you could have an adiabatic compression in which volume goes down, pressure goes up even more and temperature also goes up.

Thermodynamics is not my strong suit -- never took a course. But as you compress the gas you will be doing work on it. The gas will have a thermal capacity -- this much work will result in that much temperature rise. That tells you how much the temperature rises for a tiny bit of volume reduction. With the new temperature in hand, The PV/T=constant equation tells you how much the pressure must rise.

The result is going to be a differential equation. But I'd better stop here so that an actual expert can step in. This is well trodden ground and there are likely lots of more efficient approaches than working from first principles. I'll send up the bat-signal for @Chestermiller

#### Chestermiller

Mentor
Can you please provide a sketch of what the experiment is? I'm having trouble visualizing it.

#### russ_watters

Mentor
I've set up a simple experiment to look at the ideal gas laws. My experiment is relatively simple in that I have a metal tube which is capped on one side. I am then pressurising the tube with air to 100 psi and locking it off.

My thought is that as the pressure increased, with volume held constant, the temperature of the tube should rise. However, after 30 minutes of observation, there is no rise in temperature.
After 30 minutes it has cooled back down again. How big and thick is the tube? How did you measure the temperature?

#### russ_watters

Mentor
Can you please provide a sketch of what the experiment is? I'm having trouble visualizing it.
I'm picturing pumping-up a bike tire. Mine's 85psi and I can definitely feel the heat of compression in the tube from the pump.

#### 256bits

Gold Member
Or you could have an adiabatic compression in which volume goes down, pressure goes up even more and temperature also goes up
At the pump that is happening.
Then through the line the pressure drops, heat flows out to ambient and temperature drops, plus an expansion from the higher pressure at the pump to the lower pressure at the metal tube.

#### wee VooDoo

Thank you all for your kind replies.

The pump is an air compressor with max pressure of 120 psi. I've threaded in a 100mm length of 3/4" stainless steel (SS 316) which has a wall thickness of around 2mm. So the pipe is just simply sticking out the air compressor outlet. The backpressure is held by a closed ball valve.

The temperature was measured with an infrared thermometer, taking readings every 30 seconds. There was no change in temperature at any stage.

I agree with the bike pump, my road bike does the same thing and I can feel a bit of heat, so I thought this would do the same.

So the effect is happening within the pump, but as the pump moves the air through the outlet and into the pipe there is a pressure drop and the temperature goes down again?

What kind of apparatus would be required to heat the pipe using compressed air? Would a piston (which is air tight) be able to do this?

Just doing this for a bit of an experiment and interest. Enjoying learning through doing! Thank you all for your comments and help, I've already learned a lot!

#### wee VooDoo

The problem is that you are controlling one parameter (volume) but you have two variables that can change as a result (pressure and temperature).

You could have an isothermal compression in which volume goes down, pressure goes up and temperature remains constant.

Or you could have an adiabatic compression in which volume goes down, pressure goes up even more and temperature also goes up.

Thermodynamics is not my strong suit -- never took a course. But as you compress the gas you will be doing work on it. The gas will have a thermal capacity -- this much work will result in that much temperature rise. That tells you how much the temperature rises for a tiny bit of volume reduction. With the new temperature in hand, The PV/T=constant equation tells you how much the pressure must rise.

The result is going to be a differential equation. But I'd better stop here so that an actual expert can step in. This is well trodden ground and there are likely lots of more efficient approaches than working from first principles. I'll send up the bat-signal for @Chestermiller
Thank you very much for your detailed reply, really appreciate the help

#### russ_watters

Mentor
I've threaded in a 100mm length of 3/4" stainless steel (SS 316) which has a wall thickness of around 2mm....

The temperature was measured with an infrared thermometer, taking readings every 30 seconds. There was no change in temperature at any stage.
Gut-check this by touching the pipe. One problem with SS is that it is reflective in IR. I've tried reading the temperature of a metal pot on a stove and it does not work.

Another possible issue is steel has both a lot of heat capacity and ability to lose heat. So it could be that it's not warming the pipe up enough to notice.

#### Chestermiller

Mentor
Do you have a good idea what the temperature and pressure of the air is at the valve entrance? Also, what is the time scale for the closed-end pipe to fill? Also, what is the final pressure within the pipe when you consider the pipe essentially filled?

#### russ_watters

Mentor
This is something I've wanted to brush-up on for a while, so I have a scenario and a calc to share;

To me the scenario seems straightforward: It's a piston-cylinder which is compressed until the pressure reaches 85psig (starting 0psig, 70F)....for my bike. We can re-do for 120psig.... The entire system is an extension of the cylinder and the fact that you are pumping multiple times to reach that pressure doesn't matter.

Assume isentropic (adiabatic and reversible) compression. This is valid for rapid compression with limited heat transfer. This is going to be a problem, but we'll start with it to provide an upper bound for the temperature.

My thermo book tabulates the ideal gas properties of air using relative values for things like pressure and specific volume vs temperature. So there's actually not much to do here: Just ratio the pressures and find the new relative pressure (Pr2) based on the Pr1 (itself based on the temperature). Then read the new temperature straight off the table. The answer is 367F.

Now, my bike pump and hose get quite warm, but clearly not 367F. Since we have a small mass of air, a large mass of pump and system, and a relatively slow process, there is a lot of heat transfer out of the system. I suppose the next step would be to time the pumping, find the energy of compression (same table) and convert to power. That would be the steady state heat transfer to the air in what is eventually actually going to be isothermal compression at a certain temperature between 70 and 367F.

#### Chestermiller

Mentor
This is something I've wanted to brush-up on for a while, so I have a scenario and a calc to share;

To me the scenario seems straightforward: It's a piston-cylinder which is compressed until the pressure reaches 85psig (starting 0psig, 70F)....for my bike. We can re-do for 120psig.... The entire system is an extension of the cylinder and the fact that you are pumping multiple times to reach that pressure doesn't matter.

Assume isentropic (adiabatic and reversible) compression. This is valid for rapid compression with limited heat transfer. This is going to be a problem, but we'll start with it to provide an upper bound for the temperature.

My thermo book tabulates the ideal gas properties of air using relative values for things like pressure and specific volume vs temperature. So there's actually not much to do here: Just ratio the pressures and find the new relative pressure (Pr2) based on the Pr1 (itself based on the temperature). Then read the new temperature straight off the table. The answer is 367F.

Now, my bike pump and hose get quite warm, but clearly not 367F. Since we have a small mass of air, a large mass of pump and system, and a relatively slow process, there is a lot of heat transfer out of the system. I suppose the next step would be to time the pumping, find the energy of compression (same table) and convert to power. That would be the steady state heat transfer to the air in what is eventually actually going to be isothermal compression at a certain temperature between 70 and 367F.
Did this calculation take into account that the number of moles of air in the tube increases substantially, of did it just do an adiabatic reversible compression of the original air by a piston.

#### Chestermiller

Mentor
If we assume that the air enters the globe value at 70 F and 120 psig, and that the gas exhibits close to ideal gas behavior, we can use the exact same equations derived in the following thread to determine the final temperature: https://www.physicsforums.com/threads/unsteady-filling-of-a-vacuum-tank.971992/. The key equation is the relationship derived in post #8 for the final temperature. For the present problem, the pressure ratio is 1/6. Of course, this all depends on the gas exiting the compressor and entering the globe valve at 70 F. Based on this, the predicted final temperature is only 236 F, still pretty high.

#### russ_watters

Mentor
The OP hasn't been back, but...
Did this calculation take into account that the number of moles of air in the tube increases substantially, of did it just do an adiabatic reversible compression of the original air by a piston.
The first one, more or less. I simplified the problem to be a single piston/cylinder, compressing once. I believe that assumption is valid because all of the air the ends up in the cylinder started at room temperature and pressure and I've included the pump in the system to make it easier. Otherwise you have to model the pump's output temperature as the pressure in the cylinder rises.
If we assume that the air enters the globe value at 70 F and 120 psig...
I don't believe this will work because my read of the scenario is that the pipe is hard-connected to the compressor, with no intermediate receiver vessel. In that case, neither the pressure nor temperature of the air coming out of the compressor are constant.

But it would be a valid assumption if the compressor includes a receiver vessel and the air is allowed to cool back down to room temperature before starting the experiment. In that case, only the air that enters the pipe is being compressed; the air leaving the receiver tank and entering the pipe is actually expanding and cooling.

#### russ_watters

Mentor
Gut-check this by touching the pipe. One problem with SS is that it is reflective in IR. I've tried reading the temperature of a metal pot on a stove and it does not work.
@wee VooDoo, if you're still around, I'd like to reiterate the above point. Here's a picture of three containers of boiling water that just came off a stove. It should be obvious which is stainless steel and why using an IR gun on a stainless steel pipe won't work.

#### Chestermiller

Mentor
The OP hasn't been back, but...

The first one, more or less. I simplified the problem to be a single piston/cylinder, compressing once. I believe that assumption is valid because all of the air the ends up in the cylinder started at room temperature and pressure and I've included the pump in the system to make it easier. Otherwise you have to model the pump's output temperature as the pressure in the cylinder rises.
It can't involve a single compression. The air pressure drops to the receiving tube pressure after exiting the compressor. Initially the receiving tube is at 1 atm, but, eventually all the air in the receiving tube will be at the compressor exit pressure. So the part of the air that exits the compressor early on is recompressed from 1 atm all the way up to the 100 psi. But the air that exits the compressor later is recompressed from a higher pressure up to the 100 psi. So not all the air in the tube has been recompressed the same amount. This is what is captured in the model we developed in the other thread.

#### russ_watters

Mentor
It can't involve a single compression. The air pressure drops to the receiving tube pressure after exiting the compressor.
I don't think so. My understanding is that a reciprocating compressor is nothing more than a piston-cylinder with a check valve to atmosphere and another to the system. It generates the exact pressure the system needs, plus just a touch to open the check valve to the system and make the air flow. When the system is at 25psig it pumps in a touch over 25psig air. When the system is 50psig it pumps in about half as much air at just over 60psig.

Also note that if the target vessel is small enough and compressor cylinder big enough it can indeed pump-up the target vessel in one stroke.

IMO, since all of the air in the scenario as I have described it starts at one state and ends at another state, it is path independent. The fact that you compress it a little bit at a time, in steps (or even smoothly for one stroke), can be modeled of one wants, but it shouldn't be necessary if all you want to know is the final state.
Initially the receiving tube is at 1 atm, but, eventually all the air in the receiving tube will be at the compressor exit pressure. So the part of the air that exits the compressor early on is recompressed from 1 atm all the way up to the 100 psi. But the air that exits the compressor later is recompressed from a higher pressure up to the 100 psi. So not all the air in the tube has been recompressed the same amount. This is what is captured in the model we developed in the other thread.
The other thread has a much more complicated answer because the OP specifically asked to model the entire scenario's evolution over time. In this thread, the OP has only asked for the before and after states. If it's path independent, you don't have to dynamically model it. They're also just different scenarios for the reasons discussed above.

Reading it now, the scenario itself looks to me to be quite similar, though with an interesting twist. If we were to place a movable partition by the opening of the container, then what you'd have is a piston-cylinder executing a single stroke compression. It's a very dynamic situation, but with a cute simplifier in assuming a constant air pressure to move the "piston" even though it's a mismatch against the pressure in the cylinder.

....frankly I was expecting it to be about loss through the hole, but if that's ignored it is pretty easy. It also allows you to ignore the decompression and re-compression of the air entering the hole.

The main difference between the scenarios is in this thread all of the air finally in the container is being compressed and in that thread only the air originally in the container is being compressed.

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#### Chestermiller

Mentor
Reading it now, the scenario itself looks to me to be quite similar, though with an interesting twist. If we were to place a movable partition by the opening of the container, then what you'd have is a piston-cylinder executing a single stroke compression. It's a very dynamic situation, but with a cute simplifier in assuming a constant air pressure to move the "piston" even though it's a mismatch against the pressure in the cylinder.

....frankly I was expecting it to be about loss through the hole, but if that's ignored it is pretty easy. It also allows you to ignore the decompression and re-compression of the air entering the hole.

The main difference between the scenarios is in this thread all of the air finally in the container is being compressed and in that thread only the air originally in the container is being compressed.
This does not accurately characterize what was done in the other thread.

It is not only the air originally in the container that is being compressed. The air entering the container suffers a pressure drop in passing through the hole, and is recompressed by subsequent air that enters the container behind it.

Also, there are two separate parts to the analysis. The first part does not address the flow through the inlet orifice and the time dependence. It is independent of the time variation. It determines the pressure and temperature in the container as a function of the number of moles of air that have entered the container (including at the final equilibrium). So it essentially does for that system what your analysis does for the present system.

The second part of the analysis addresses the pressure drop vs flow rate relationship for the air passing through the inlet orifice, and quantifies the tank pressure and temperature vs time.

#### russ_watters

Mentor
This does not accurately characterize what was done in the other thread.

It is not only the air originally in the container that is being compressed. The air entering the container suffers a pressure drop in passing through the hole, and is recompressed by subsequent air that enters the container behind it.

Also, there are two separate parts to the analysis.
I don't see anything there that doesn't match my understanding of that problem (and yes, I know you did it two different ways and I only commented on the first). What I find kind of odd is that you seem to be harping on the pressure drop and recompression of the air entering the container even though your first pass analysis ignores that and only deals with the starting and ending states.

I think we're going to have to disagree to agree here though because I see no point in arguing over something I agree with and this really wasn't my point anyway.

I just wanted to make it clear that we're talking about two differente scenarios in *this* thread, based on different assumptions about what the system setup and boundary is. It appears to me that you challenged my approach based on a different setup, not an error in my analysis. You might be right in that I interpreted the OP wrong, but I believe my analysis works for the setup I chose.

Furthermore, the more I look at my version of this thread's setup and and the other thread's setup, the more I think they might be exactly identical. I may be curious enough to run the numbers later when I have more time.

#### Chestermiller

Mentor
What I find kind of odd is that you seem to be harping on the pressure drop and recompression of the air entering the container even though your first pass analysis ignores that and only deals with the starting and ending states.
Just to clarify, even though it is not explicitly stated in the first pass analysis, the analysis does also include all states between the starting and end states. This is used in the second part of the problem dealing with the time-dependence.

"Gas Law - increasing temperature with constant volume held"

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