# What pressure (in atmospheres) is required to compress 1.00 L of gas at 760. mm Hg pressure to a volume of 50.0 mL?

$\text{20 atmospheres........}$
${P}_{1} {V}_{1} = {P}_{2} {V}_{2}$, now ${P}_{1} = 1 \cdot a t m$, and so we solve for ${P}_{2}$:
P_2=(P_1V_1)/V_2=(1.00*cancelLxx1*atm)/(0.050*cancelL)=??atm