Given a 500 m sample of H_2 at 2.00 atm pressure. What will be the volume when the pressure is changed to 720. torr?

Well, ${P}_{1} {V}_{1} = {P}_{2} {V}_{2}$........so ${V}_{2} = 1055.6 \cdot m L$
${V}_{2} = \frac{{P}_{1} {V}_{1}}{P} _ 2$
$= \frac{2 \cdot a t m \times 500 \cdot m L}{\left(720 \cdot {\text{Torr")/(760*"Torr"*"atm}}^{-} 1\right)} = 1055.6 \cdot m L$.