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## Dan Price

Physics...
I was staring at the blimp flying around between periods at the RIT Men's Hockey game tonight (6-1, way to go!) and got pondering about the physics of how things float...

Most people know helium, hydrogen, and hot air can be used for lighter-than-air flight because they are ... lighter than air. But the more basic concept is that a volume of matter in a container having a total density less than the density of its surrounding will be pushed up by the pressure of that surrounding matter beneath it. So I thought... instead of the container holding a matter with less density at the same pressure as its surrounding, what if it held a vacuum?

Granted, the container would have to be quite strong to withstand the external pressure and still light enough that it doesn't increase the total density beyond that of its surrounding matter. But if those requirements were met, it should rise, right?

And then my imagination really wandered and I thought... what if we were able to build a spherical structure as big as the moon out of a super-light and super-strong material that contains a vacuum. Would it float in Earth's atmosphere rather than falling all the way to the ground? And wouldn't that be a freaky sight?

And pondering in the other direction... wouldn't it also be possible to fill a non-stretchy balloon with enough helium that the pressure and density goes up to the point that the balloon would be heavier than air? that would have to be one strong-ass balloon to become that dense :-p but of course, if you're going to that direction - why don't you just use hydrogen and create a megasuperdense sphere of hydrogen and create your own black hole ;) Hehe, ok, so the pressure required to give helium or hydrogen a higher density than the atmosphere at sea level would also make it as hot as the sun? That could be a problem. :) I want to solve that problem, but in the mean time I found a really cool question:

How does relativity affect the Ideal Gas Law? (PV=nRT) Very interesting, but they seem to be ignoring one important word: "...the length of the box will appear to change..."

To me, that means that all their calculations are only true in appearance to an observer passing at the speed of light, while someone sitting still next to it would not agree. I don't think an observeration can affect matter like that. If I was sitting next to the box and someone else passed it at the speed of light, I would not see the box get shorter as the person approached and longer after they pass it, causing gas inside to compress or expand, cool down or heat up, or change state.

I think what the moving observer would find is that the Ideal Gas Law is broken - meaning that yes the gas appears compressed, but since it's really not, it won't change temperature or state.

And if I'm wrong, who cares? The observer will be long gone. :) Let me put it another way:

If I was to construct a bomb in such a box that is triggered by the presence of liquid water, and the box contained water at 101°C, would an observer approaching the box fast enough to make it appear to shorten enough to compress the water enough to lower its temperature to 99°C see1 the box explode while I, sitting next to it, see the box remain unchanged?

1assuming the observer is capable of seeing the wavelengths of light that are also shortened2

2wouldn't an observer traveling at the speed of light be subjected to all sorts of high-frequency radiation as normal wavelengths of light coming from in front are shortened? Well he said at the end of the article:
At this point I'm mostly guessing, but I would make the hypothesis that in relativistic thermodynamics (a far from completely understood field of physics) one needs to define temperature differently depending on how one is using the parameter. For thermodynamic analysis, one might need to define T the first way (lowering it by 1/gamma); but for phase-space diagrams one might need to define it the second way (keeping it constant). When the box is at rest, the two definitions of temperatures just happen to be the same, so this distinction is not normally necessary; only from relativistic reference frames does this distinction between different types of "temperature" emerge

I don't know what the thermodynamic stuff is about. But time changes speed for the molecules in the box relative to the obvserver going near c, right? Maybe that messes with something.

I'm not sure about all of this, but it's a fun question.

About the length looking like its changing... I wish i remeber the example used where a long car can fit in a small garage. I can't remeber which way the conclusion went. Okay, here we go:

Question: A massless, very thin cubic meter shell is filled with helium that is neutrally bouyant in air. What is the temperature of the helium if the shell rests in air at 20 C, 760 torr?

Ideal gas law: PV = nRT. and relative changes: (P1*V1)/(T1*n1) = (P1*V1)/(T1*n1)

P = Pressure (torr)
V = Volume (m^3)
T = Temperature (K)
n = # of moles (mol)
R = molar gas constant = 8.3145 J/(mol*K)

The molar mass of air = 28.9 g/mol. (Air is roughly 80% N and 20% O, somehow the mm works out this way, i ended up looking it up)

For the shell to be stationary in air, the density of the shell and helium inside must be the same as the density of air. Just like floating in water, it works in the same in air. They are both fluids.

The first step is finding of the air displaced by the sphere. The sphere is 1 m^3 (V), the temperature is 20C (T), and the pressure is 760 torr. Using the ideal gas law, and solving for n:

n = PV/(RT) = 1 m^3 * 760 torr / R / 20 C.
Convert temperature to Kelvins( 20+273 = 293).
n = 1 m^3 * 760 torr / R / 293 K = 41.6 moles

Find the mass of the air:
mass = moles * molar mass
m = 41.6 mol * 28.9 g/mol = 1.20 kg

Now we know the mass and volume of the air displaced. For the density to be the same, the mass of helium must be the same. So we need to fill the shell with 1.20 kg of helium.

Next, we find the number of moles of helium used.
moles = mass / molar mass
moles = 1200g / (4 g/mol) = 300 moles

So we have 300 moles of gas inside a cubic meter. That's got to be a lot of pressure. Let's see how much.
One mole of any gas at STP take up a volume of 22.4 liters. If we keep the temperature and volume constant and increase the number of moles, what happens to the pressure?

V1*P1/(T1*n1) = V2*P2/(T2*n2)

We can cancel the volume and temperature since they'll be the same. Then solve for P2

P1*n2/n1 = P2

P2 = 760 torr * 300 mol / 1 mol = 228 000 torr

So the pressure will go from 760 torr to 228000 torr when increasing the number of moles to 300.

We now know the pressure, number of moles, and volume of helium. We can now solve for the temperature using the ideal gas law.

T = PV/(nR)
T = (228000 torr)(1 m^3)/( 300 mol * R ) = 12200 K

For a reference, the surface temperature of the sun is 5800 K

:-D Yay! I think my title as the guy with too much time on his hands is being threatened... :-P

That's cool though. I can only follow it well enough to believe it, but not well enough to check your work. I was never good with all those formulas and stuff. I had totally forgotten about the unit "moles". :) Most of it seems alright to me. The only part where I think I could have messed up was the P1*n2/n1 = P2 near the end. I dunno if I can say the temp can stay the same to find the pressure then use that pressure to find the temp. 