Ideal gas law, applied to non-rigid container

AI Thread Summary
The discussion focuses on determining the internal volume of a container using the ideal gas law, specifically addressing challenges with non-rigid containers. The calculations for rigid containers yield accurate results, but non-rigid containers, like plastic jugs, lead to discrepancies due to their ability to expand under pressure, which affects final pressure measurements. Participants note that the pressure dynamics and the container's elasticity complicate the theoretical framework, suggesting that the problem may involve complex fluid dynamics. Experimental data collection is recommended to better understand the effects of container rigidity on pressure equilibrium. Theoretical exploration is acknowledged as challenging, potentially involving coupled partial differential equations.
ARQuattr
Messages
2
Reaction score
0
I'm trying to determine the internal volume of a container using the ideal gas law principle.

1. A container of known volume is pressurized with air to some known value relative to ambient.
2. The container of unknown volume is vented to ambient.
3. Temperature is constant for all air masses and throughout the experiment.
4. The vent on the container under test is sealed and the containers are allowed to communicate air mass.
5. The final equilibrium pressure of the combined air mass is measured.
6. Determine the unknown volume, given the control volume, initial pressure, final pressure.

Using PV = nRT, I have

PcVc = ncRTc, PuVu = nuRTu, and PfVf = nfRTf, or

nc = PcVc/RTc, nu = PuVu/RTu, and nf = PfVf/RTf

for subscripts c (control container), u (unknown container), and f (final), where

Vf = Vc + Vu, and nf = nc + nu

Then,

PfVf/RTf = PcVc/RTc + PuVu/RTu

and assuming T constant,

PfVf = PcVc + PuVu

Since the unknown container starts at ambient pressure, Pu = 0, so

PfVf = PcVc, or Pf(Vc+Vu) = PcVc

(I've tried this using absolute pressures and it works out the same.)

So the unknown quantity is

Vu = Vc(Pc - Pf)/Pf

Experimentally, this works fine for rigid unknown containers, but when I use a non-rigid container it doesn't. Why not?

By non-rigid, I'm referring to something like a plastic jug with a fairly well-defined volume, but if pressurized, even to only a fraction of a psig, the sides bulge out slightly. This drastically affects the final pressure, but I don't see how. I understand that the volume will increase slightly because of the measurement process, (and I'm able to accept the error in volume measurement due to the expansion of the container,) but the calculation yields a result that is way off.

I can also imagine how the container walls act like springs to increase resistance as it grows, but I don't see how that changes things. The pressure introduced by the control vessel causes the container to expand which allows the pressure to drop until it all reaches equilibrium and my end result should be the final container volume, right?

What am I missing?

Thanks in advance.
 
Physics news on Phys.org
At first glance it sounds like you have a nasty semi-free-boundary problem that I don't even begin to know how to pose mathematically. Certainly the final pressure depends on the rigidity of the containing vessel.
 
Thanks tankFan86 for your reply. Experimentally and intuitively, you're right, but I can't see how the theory supports it. Do you know what I might look for to research this further?

Thanks again.
 
I think you have posed an insanely hard question. If immediate results are required, I think collecting some experimental data might tell you lot about this problem. If you really want to explore the problem theoretically, you are dealing with the dynamics of a compressible fluid in a crazy geometry, something that I do not think has many good theoretical results. The theory must involve many coupled PDEs!
 
Thread 'Question about pressure of a liquid'
I am looking at pressure in liquids and I am testing my idea. The vertical tube is 100m, the contraption is filled with water. The vertical tube is very thin(maybe 1mm^2 cross section). The area of the base is ~100m^2. Will he top half be launched in the air if suddenly it cracked?- assuming its light enough. I want to test my idea that if I had a thin long ruber tube that I lifted up, then the pressure at "red lines" will be high and that the $force = pressure * area$ would be massive...

Similar threads

Back
Top