# A gas occupies 12.3 L at a pressure of 40.0 mm Hg. What is the volume when the pressure is increased to 60.0 mm Hg?

At constant temperature ${P}_{1} {V}_{1} = {P}_{2} {V}_{2}$.
So, ${V}_{2} = \frac{{P}_{1} {V}_{1}}{P} _ 2$ $=$ $\left(40 \cdot \text{mm Hg"xx12.3*L)/(60*"mm Hg}\right)$ $\cong$ $8$ $L$.