Question

A scuba diver releases a balloon containing 181.8 L of helium attached to a tray of artifacts at an underwater archeological site. When the balloon reaches the surface it has expanded to 478.5 L. The pressure at the surface is 1.098 atm. Pressure increases by 1.0 atm for every 10 m of depth. At what depth was the diver working? Assume the temperature of the helium is constant. Type your response in m to the nearest tenth of a m.

Answer 1

Initial volume (V₁) = 181.8 L

Let us say that the initial pressure = P₁

Final volume (V₂) = 478.5 L

Final pressure (P₂) = 1.098 atm

From Boyle's law,

P₁V₁ = P₂V₂

or, P₁ x 181.8 L = 1.098 atm x 478.5 L

or, P₁ = 2.890 atm

Thus, pressure difference = 2.890 atm - 1.098 atm = 1.792 atm

Given that the pressure increases by 1.0 atm for every 10 m of depth

Hence, the driver was working at a depth of = 1.792 atm x 10 m/1 atm= 17.9 m