Calculate the final pressure using Boyle's Law:

$\overline{){\mathbf{P}}_{\mathbf{1}}{\mathbf{V}}_{\mathbf{1}}\mathbf{=}{\mathbf{P}}_{\mathbf{2}}{\mathbf{V}}_{\mathbf{2}}}\phantom{\rule{0ex}{0ex}}\frac{{\mathbf{P}}_{\mathbf{1}}{\mathbf{V}}_{\mathbf{1}}}{{\mathbf{V}}_{\mathbf{2}}}\mathbf{=}\frac{{\mathbf{P}}_{\mathbf{2}}\overline{){\mathbf{V}}_{\mathbf{2}}}}{\overline{){\mathbf{V}}_{\mathbf{2}}}}\phantom{\rule{0ex}{0ex}}{\mathbf{P}}_{\mathbf{2}}\mathbf{=}\frac{{\mathbf{P}}_{\mathbf{1}}{\mathbf{V}}_{\mathbf{1}}}{{\mathbf{V}}_{\mathbf{2}}}\phantom{\rule{0ex}{0ex}}{\mathbf{P}}_{\mathbf{2}}\mathbf{=}\frac{(1.0\mathrm{atm})(20\overline{)\mathrm{mL}})}{(7.6\overline{)\mathrm{mL}})}$

**P _{2} = 2.63 atm**

A snorkeler takes a syringe filled with 20 mL of air from the surface, where the pressure is 1.0 atm, to an unknown depth. The volume of the air in the syringe at this depth is 7.6 mL .

If the pressure increases by an additional 1 atm for every 10 m of depth, how deep is the snorkeler?

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