Question

A balloon used for underwater salvage is inflated to 50.0 L at a depth of 200 feet, where the pressure is 6.89 atm and the temperature is 3 °C. The balloon rises to the surface (22°C and 0.988 atm). What is new volume of the balloon?

Answer 1

Initial volume V₁ = 50.0 L

Initial temperature, T₁ = 3 ^oC = (273 + 3) K = 276 K

Initial pressure, P₁ = 6.89 atm

Let us say that the final volume = V₂

Final temperature, T₂ = 22 ^oC = (273 + 22) K = 295 K

Final pressure, P₂ = 0.988 atm

From ideal gas law,

P₁V₁/T₁ = P₂V₂/T₂

or, (6.89 atm x 50.0 L)/276 K = (0.988 atm x V₂)/ 295 K

or, V₂ = 373 L

Hence, the new volume of the balloon = 373 L