Question

How many kilograms of `O_2` gas are needed to give an oxygen pressure of 3.48 atm at 22°C?

Answer 1

Just to retire this question, let us specify a volume of `1*m^3`; and I get a mass of dioxygen gas of under `5*kg`.

Explanation:

We use the Ideal Gas Equation, with `R=0.0821*L*atm*K^-1*mol^-1`.

And thus `n=(PV)/(RT)=(3.48*cancel(atm)xx1*cancel(m^3)xx10^3*cancelL*cancel(m^-3))/(0.0821*cancel(L*atm)*cancel(K^-1)*mol^-1xx295*cancelK)`

`=143.7*mol` (note that the answer has units `1/(mol^-1)=1/(1/(mol))=mol` `"as required"`

And we convert this molar quantity into a mass:

`143.7*cancel(mol)xx32.00*cancel(g*mol^-1)xx10^-3*kg*cancel(g^-1)`

`=4.60*kg.`

Please review my arithmetic. There are no money-back guarantees.

Note that `1*m^3` is a very large volume, and is equal to `1000*L`. Chemists tend to deal with `"litres"` and `mL` and `cm^3` (`1*mL-=1*cm^3-=10^-3*L-=10^-6*m^3`). Of course all the measurements should be equivalent dimensionally, but it is so easy to divide instead of multiply and vice versa.