Ideal gas law, applied to non-rigid container

In summary, the conversation discusses using the ideal gas law principle to determine the internal volume of a container. The process involves pressurizing a known volume container, venting an unknown volume container, and measuring the final equilibrium pressure. However, the calculation does not work for non-rigid containers, leading to a discussion about the effects of container rigidity on the final pressure. The theoretical approach to this problem is complex and may require experimental data to fully understand.
  • #1
ARQuattr
2
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.
 
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  • #2
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.
 
  • #3
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.
 
  • #4
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!
 

Related to Ideal gas law, applied to non-rigid container

1. What is the ideal gas law and how is it applied to non-rigid containers?

The ideal gas law is a mathematical relationship that describes the behavior of ideal gases. It states that the pressure, volume, and temperature of an ideal gas are proportional to each other. This law can be applied to non-rigid containers by taking into account the changes in volume of the container as the gas expands or contracts.

2. How does the ideal gas law differ for non-rigid containers compared to rigid containers?

In rigid containers, the volume remains constant and the ideal gas law can be simplified to PV = nRT. However, in non-rigid containers, the volume can change and this must be taken into account by including a correction factor, known as the compressibility factor (Z), in the equation.

3. What are some real-life examples of non-rigid containers that follow the ideal gas law?

Some examples include balloons, inflatable tires, and the human lungs. These containers can expand or contract as the gas inside them changes in temperature and pressure, making them good examples of non-rigid containers in which the ideal gas law can be applied.

4. Can the ideal gas law be applied to all gases in non-rigid containers?

No, the ideal gas law is only applicable to ideal gases. Real gases deviate from ideal behavior at high pressures and low temperatures, and the ideal gas law cannot accurately describe their behavior. Therefore, it is important to use the ideal gas law with caution and know its limitations when applied to non-rigid containers.

5. How is the ideal gas law useful in practical applications with non-rigid containers?

The ideal gas law is used in many practical applications, such as in the design of inflatable structures and packaging materials. It can also be used in the study of respiratory physiology and the behavior of gases in the atmosphere. By understanding the relationship between pressure, volume, and temperature in non-rigid containers, we can better control and predict the behavior of gases in these systems.

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