# What is the pressure in a 2 dimensional container with a known load/weight applied?July 16, 2010 11:02 AM   Subscribe

Pressure. I can find all kinds of information for a 3-dimensional flexible container, but what happens when we start talking about a container that's only flexible in 2 dimensions?

For instance, if I have a plastic, recyclable PET water bottle that's both empty and capped and I squeeze it, it's easy to observe that the pressure in the container increases. On the other hand, if I have a car tire and I measure the pressure in the tire while the tire is not mounted on the car vs. when the tire is mounted on the car and the weight of the car is on the tire, it doesn't appear as though the pressure changes.

What got me thinking about this is an air mattress I have. When I inflate it to a reasonably firm 1 psi, what happens to the pressure in the mattress when a 200lb person lies on top of it? Most such air mattresses are rated for some ridiculous internal pressure that we all know they can't withstand for prolonged periods, but at the same time I was wondering what is the actual pressure inside of the mattress under these conditions? Assume that the mattress is inflated to 1psi with no load. What will the pressure be with an average adult of 200lbs on the same mattress be?

Granted, these are special conditions because the container is not rigid in all 3 directions, but my searches online only talk about the 3 dimensional flexible container - not the much more common 2-dimensionally flexible container. I naturally thought of as many other common gas containers I could think of, but I'm still having trouble coming up with the answer.
posted by signal7 to Science & Nature (3 answers total)

For an ideal gas in an enclosed space, P*V/T is constant. If you further assume that you're not compressing it enough for the temperature to rise (or you wait for any temperature change to equalize before measuring), that means that P*V is constant, which means for pressure to go up as a result of squeezing then volume has to go down. To use the car tire example, if you look at the tire before and after it's on the car, its shape really doesn't change all that much. Yeah, the sidewalls bow out a bit but that's compensated for by the small reduction in distance between the tread and axle as the car is lowered; same volume means same pressure.

It also makes sense if you think about trying to burst a balloon -- if you take one of those long thin types that are used to make animals and try to pinch it with your fingers then the relative amount of displacement of the thickness compared to the length is small, and the ample elasticity along the longitudinal direction will let it elongate just enough to conserve volume, so that the pressure stays the same and it doesn't pop. However if you take a round balloon and try to crush it between your palms there's really no way for the balloon to reconfigure itself in order to keep the volume constant -- after all a sphere has the highest ratio of volume to surface area that a solid can have, which means pancaking to keep the volume constant would require the balloon to take an impossible shape.

If you go back to the air mattress example, consider poking it with one finger in a single spot vs taking a large plate and compressing the entire surface of the mattress all at once. In the former, the mattress can probably reconfigure its shape so that the volume remains roughly constant, but with the latter there's really no degrees of freedom left that it can expand so the pressure would increase until it split. And with the water bottle, its plastic shell can easily bend but it does resist tension, i.e. the length doesn't increase and the circumference of the walls doesn't change, so when you try to flatten the circular walls with your hand, in order to keep volume constant it would need to either increase the circumference of the walls (again: circle, highest area:circumference ratio) or increase in length, but since it can do neither you can feel the pressure increase instead.
posted by Rhomboid at 12:16 PM on July 16, 2010

Your initial question has an answer of "it's very complicated". In the regimes you're considering, the ideal gas equation will indeed obtain and you'll also be fine with a constant temperature.

All the standard thermodynamics "pV=const" examples you'll find in high-school level physics examples take "flexible container" to mean that it takes zero work to change the volume besides what is done by the expanding gas. A container like a balloon or, even more significantly, your air mattress, is different in that expanding its volume takes work that is stored as elastic energy in the container.

Now, with numbers as they are, you can probably consider the volume constant (I don't have data for the rubber walls' mechanial properties, but just check to see whether your mattress expands significantly with pressure, and you'll notice). The entire deformation aspect also plays only a minor role, so your first order model will simply be your 200lb person sitting on a rigid piston, counteracted by a pressure from below, and we can ignore the thermodynamics to simply calculate the mechanical equilibrium - if you put your puny human on a single mattress of 39in x 75in = 2 925 in^2, the gas in the piston will have to counteract this through a pressure of 0.068 psi to obtain equlibirum.

Now, in real life your human will not rest on the entire surface of a non-rigid mattress, and complications depending on the way the mattress adapts to distribute the weight will ensue, but still, you'll have to tweak the numbers significantly before you'll reach any safety limits.
posted by themel at 1:40 PM on July 16, 2010

Response by poster: It's hard to mark a best answer - you're both right.

Since I'm not looking for exact numbers, I'm just assuming the mattress does not expand with respect to pressure - at least not enough to consider the effects of such expansion. The difference in the final numbers would be negligible.

Thanks for the help.
posted by signal7 at 10:18 AM on July 21, 2010

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