round square kilometers .. you english peeps are weird.
And though I suppose you mean that if you get the pressure inside the balloon 10 times the pressure outside, it would "explode" covering a wider area, we are talking about a piece of cloth and not a metal container which is unexpandable. Therefor I doubt that raising the pressure inside the piece of cloth to 10 times the pressure outside the piece of cloth would acutally make that balloon shrink
But the point is still valid if the balloon can hold the pressure though [img]smile.gif[/img]
I'm going to mimick your calculations, but in Mol.
2 H2 + O2 = 2 H2O
H2 + 0.5 O2 = H20
We want 10^8 kg H20 --> 1 Mol H20 = 18.016 gram
So we need 1*10^11 gram / 18.016 gram = 5550621670 Mol H20 and thus also that much Mol H2.
Im gonna assume I'm letting my Balloon up to 5 km height. Air Pressure there is 5*10^4 Pascal, assuming that the balloon is flexible and thus adjusts, we can use the Gay-Lusac : ((p*V)/ T ) = n*R
We know everything but V:
((5*10^4*V)/ +-250 K ) = 5550621670 * 8,3145
V = 230753219,4 m^3 = 2,3 * 10^8 m^3
This difference can probably be contributed to the fact that I used the lower pressure and temperature of 5 km height.
EDIT:
Assuming for ease we're putting it in a perfecty sphere-shaped balloon and it stays that way:
Volume = 4/3 Pi*r^3 = 230753219,4
r = 380,4985913 m ( A span of about 761 meters )
[ 08-20-2003, 06:14 AM: Message edited by: daan ]