Question

At 22.0 m below the surface of the sea (density = 1 025 kg/m3), where the temperature is 5.00°C, a diver exhales an air bubble having a volume of 0.90 cm3. If the surface temperature of the sea is 20.0°C, what is the volume of the bubble just before it breaks the surface?

Answer 1

h = depth = 22 m

= density of sea water = 1025 kg/m³

P_o = atmospheric pressure = 101325 Pa

P = pressure at depth = P_o + gh = 101325 + (1025) (9.8) (22) = 322315

T₁ = Temperature at surface = 20 C = 20 + 273 = 293 K

T₂ = Temperature at bottom = 5 C = 5 + 273 = 278 K

V₂ = Volume at bottom = 0.90 cm³

V₁ = Volume at top = ?

Using the equation

P_o V₁/T₁ = P V₂/T₂

(101325) V₁/293 = (322315) (0.90)/278

V₁ = 3.02 cm³