Question

What volume is occupied by a `0.20*mol` quantity of gas confined in a piston at a temperature of `300*K`, and a pressure of `1*atm`?

Answer 1

About `5*L`...

Explanation:

We use the old Ideal Gas equation....

`V=(nRT)/P=(0.20*molxx0.0821*L*atm*K^-1*mol^-1xx300*K)/(1*atm)`

`=??*L`

You can repeat this with your data....I do not like using absurd units such as `kPa`, when the atmosphere is a perfectly good and intuitive unit.