Question

An air bubble at the bottom of a lake 32.0m deep has a volume of 1.00 cm3.

If the temperature at the bottom is 2.2 ?C and at the top 20.2 ?C, what is the radius of the bubble just before it reaches the surface?

Answer 1

The pressure at the bottom of the lake is

P₂ = 1.01 x 10⁵ N/m² + hg

Here, depth of the lake where bubble settle is h, density of the water is and acceleration due to gravity is g.

Substitute the values in the above equation,

P₂ = 1.01 x 10⁵ N/m²+ hg

    = 1.01 x 10⁵ N/m² + (32.0 m) (1000 kg/m³)(9.8 m/s²)

    = 1.01 x 10⁵ N/m² + 3.13600 x 10⁵ N/m²

     = 4.146 x 10⁵ N/m²

Using ideal gas equation,

Here, volume of the air bubble at the depth is V₂, volume of the air bubble at the surface is V₁, pressure at the surface is P₁, pressure at the depth is P₂, temperature at the surface is T₁ and temperature at the depth is T₂.

Substitute the values in the above equation,

Let r be the radius of the air bubble before reaching the surface of the lake.

Now, the volume of the bubble is

Substitute the values in the above equation,

Rounding off to three significant figures, the radius of the air bubble before reaching the surface is 2.08 cm or 0.0208 m.