Question

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`?

Answer 1

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.