The details are in the picture How would I find the mass of helium required?

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1 Answer
May 23, 2018

See Below

Explanation:

You can use some form of the Combined Gas Law, but I'm just going to pull it apart.
I'm going to assume this goes on at 298K.

First we'll determine the volume for the O_2O2, then use this volume to determine how much He is needed to get 8.25 atm. We'll assume ideal behavior of gases throughout.

Volume of O_2O2 container:
moles of O_2O2 = "38,000g" / "32g/mole"38,000g32g/mole = 1187.5 moles
PV=nRTPV=nRT
V="nRT"/PV=nRTP
V = ((1187.5"mole")(0.0821)(298))/(7.8"atm")V=(1187.5mole)(0.0821)(298)7.8atm = 3724.8L

The tank is 3724.8L

Now, in this tank with He, we want the pressure to be 8.25 atm.
So, we'll solve for the number of moles required to get that pressure.

n="PV"/"RT"n=PVRT
n = ((8.25"atm")(3724.8))/((0.0821)(298))n=(8.25atm)(3724.8)(0.0821)(298) = 1256 moles He

1256 "moles He" xx 4"g/mole"1256moles He×4g/mole = 5024 g, or 5.024 kg He

You can also do:
(P_1)/(n_1) = (P_2)/(n_2)P1n1=P2n2 and solve for n_2n2. This is the number of moles of He, then multiply by molar mass. This is an easier approach, but assumes you are cool with playing around with Ideal Gas Law.