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What volume is needed to store 10kg of o2 at 7.5 ATM pressure and 294k?
Wiki User
∙ 11y ago
(PV / nT )a = (PV / nT )b where the a and represent different conditions
At STP, 1 mole of a gas occupies 22.4 L --- these are conditions b
conditions a are the 7.5 atm, 294 K, but the 10 Kg must be converted to moles (n).
1 mole O2 = 32 g, so 10 kg x (1000 g / 1 kg) x (1 mole / 32 g) = 312.5 moles
(7.5 atm)x(V) / [(312.5 moles)(294 K)] = (1 atm)(22.4L)/[(1 mole)(273 K)]
Rearrange and solve for V
V = (1 atm)(22.4L)/[(1 mole)(273 K)] x (312.5 moles)(294 K)/(7.5 atm)
V = 1005.1 Liters
Wiki User
∙ 11y ago
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