what mass of helium in grams is required to fill a 6.0: balloon to a pressure of 1.3 atm at 25 degrees C?

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what mass of helium in grams is required to fill a 6.0: balloon to a pressure of 1.3 atm at 25 degrees C?

2007-12-16 02:11:29 · 2 answers · asked by Steve C 1 in Science & Mathematics ➔ Chemistry

2 answers

Use the ideal gas law:

PV = nRT
PV = gRT/mm

g = PVmm / RT = (1.3 atm)(6.0 l)(4.0 g/mol) / (.0821 l atm/ mol K)(298 K)

g = 1.28 grams Helium

2007-12-16 02:19:07 · answer #1 · answered by Dennis M 6 · 0⤊ 0⤋

Once again, use the ideal gas equation. PV = nRT

When you solve for moles (n), multiply by the molecular mass of He (which is 4) to get grams.

Don't forget T is in Kelvin

2007-12-16 10:15:26 · answer #2 · answered by reb1240 7 · 0⤊ 0⤋