What volume is occupied by a #2*mol# quantity of oxygen, nitrogen, and fluorine gases, at #300*K# under a pressure of #1.3*atm#?

1 Answer
Sep 8, 2017

Answer:

Approx. #40*L#..........

Explanation:

We use the old ideal gas law.....

#V=(nRT)/P#

We must simply KNOW that oxygen, and nitrogen, and fluorine, and chlorine, are DIATOMIC gases, i.e. #X_2#...(this is the next refinement of this problem, they could have mentioned a mass of #64*g#....)

And so.......

#V=(2*molxx0.0821*L*atm*K^-1*mol^-1xx300*K)/(1.3*atm)#

#=??L#....Do the units in the calculation cancel out to give an answer in litres? They should, and if they don't then somewhere we have made an error.