# A balloon that contains 1.50 L of air at 1.00 atm is taken underwater to a depth at which the pressure is 3.00 atm. What the new volume of the balloon, assuming constant temperature?

At constant $T$, we use Boyle's Law. . Volume should decrease proportionally at treble the pressure.
${P}_{1} {V}_{1} = {P}_{2} {V}_{2}$
${V}_{2} = \frac{{P}_{1} {V}_{1}}{P} _ 2$ $=$ $\frac{1.00 \cdot a t m \times 1.50 \cdot L}{3.00 \cdot a t m}$ $=$ $0.500 \cdot L$