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

## 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 ${O}_{2}$, 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 ${O}_{2}$ container:
moles of ${O}_{2}$ = $\text{38,000g" / "32g/mole}$ = 1187.5 moles
$P V = n R T$
$V = \frac{\text{nRT}}{P}$
V = ((1187.5"mole")(0.0821)(298))/(7.8"atm") = 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.

$n = \text{PV"/"RT}$
$n = \frac{\left(8.25 \text{atm}\right) \left(3724.8\right)}{\left(0.0821\right) \left(298\right)}$ = 1256 moles He

$1256 \text{moles He" xx 4"g/mole}$ = 5024 g, or 5.024 kg He

You can also do:
$\frac{{P}_{1}}{{n}_{1}} = \frac{{P}_{2}}{{n}_{2}}$ and solve for ${n}_{2}$. 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.