Thread: Air pressure
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Old 05-28-2009, 11:01 AM   #4 (permalink)
telekinetic
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There are two ways to model a balloon. The simplest case is that the membrane is infinitely deformable, in which case the volume of the balloon adjusts until the pressure inside and outside the balloon are equal.

Let's will deal with that first:

PV=nRT is exactly correct, as red lemon implied (P is absolute pressure, V is volume, T is absolute temperature). We're keeping n and R constant, so we can reorder to P1V1/T1=P2V2/T2.

Now we need to make some assumptions about initial and final temperature and pressure. Due to the vast multitude of starting conditions, we'll use the ISO standard temperature and pressure, or STP: 59 degrees F and 14.7PSI. Let's assume you've filled the balloon to 1 liter. (I know I'm making a mess of the units, but it realy doesn't matter, since they all cancel--this might as well be 'one lungful')

The ever-handy Wolfram|Alpha gives us the conditions at 40,000 feet as 5.5inHg pressure (which google handily converts to 2.7PSI) and -70 degrees F. We obviously don't know the volume yet, so let's use the ideal gas law to solve for it

Before we can plug our numbers in, we need to get our temperature in absolute units. Since we're already in english units, let's just convert to Rankine by adding 459. This gives us:

(14.7*1)/(59+459)=(2.7*V2)/(-70+459)

(14.7 * 1 * 389)/(518*2.7) = V2

4.09 = V2

Therefore, your balloon, if brought from sea level in 60 degree weather up to 40,000 feet, would have swelled to 4 times its size. Do note that we canceled the gas constants out of this, so this works equally well for helium.


The second model brings the physical properties of the balloon itself into play. Assuming we are talking about a latex balloon, and not some non-deformable material like, say, a giant mylar balloon filled to less than 1/4th of its capacity, thus leaving plenty of room for the air to 'grow', then the latex exerts some nonzero force inwards on the air in the balloon, increasing the pressure slightly. If you have read this far into the wall of text that is way more than I intended to write, please make a bad science pun (possibly about hot air?) in your reply. Anyways, to determine the effect on your balloon, you would need to know what that pressure increase was at sea level compared to what it would be at your increased volume. However, as long as your balloon is big enough to not pop at altitude, it is pretty safe to assume that this effect will be negligible....if it's not, just start with a larger balloon.

Quote:
Originally Posted by Zweiblumen View Post
Maybe I'm just too lazy to Google or something, how ever I'm hoping that someone can point me in the right direction. What I want to calculate is how the pressure inside a balloon changes with altitude. Lets say that we have a balloon that is inflated to 1 psi at sealevel, what would the pressure be at 40.000 feet ?

Yours
Zweiblumen
To strictly answer your question, as I've shown above, in a deformable vessel, it is safe to assume the pressure doesn't change, and saying you are filling a balloon with 1psi has no meaning unless you know the physical properties of the balloon walls (Young's modulus of elasticity, thickness, and internal surface area would probably be enough to figure it out)

If you instead had, say, a steel canister filled with some volume of air, just raising the altitude isn't going to change the absolute pressure in the vessel--only a temperature change could do that. If you wanted to find the relative pressure change (that is, the forces on the walls of the vessel--0psi when you screw the lid on at sea level), you'd adjust the internal pressure for the temperature change, and then just do simple subtraction of the absolute pressure outside (we already found that to be 2.7PSI) from the absolute pressure inside. Both are going down, since it is getting colder, but I suspect the pressure drops faster than the temperature, so you should see an increase relative pressure.
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Last edited by telekinetic; 05-28-2009 at 01:06 PM..
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