Quote:
Originally posted by daan:
round square kilometers .. you english peeps are weird.
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Hey man, I'm dutch!
I was just saying 100 is a nice round number. I did not mean a circular square kilometer, although that is also possible.
Quote:
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.
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Yep, works too. I'm glad to see the difference is quite small.
But still, a balloon that's half a mile wide....
Faceman, you care to reply?