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

1 Answer
Feb 6, 2017

Answer:

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.