Question

An air bubble of volume 1.0 cm³ rises from the bottom of a lake 40 m deep at a temperature of 12 °C. To what volume does it grow when it reaches the surface, which is at a temperature of 35 ° C?

Answer 1

initial volume of air bubble,

V₁ = 1.0 cm³

= 10^-6m³

Initial temperature, T₁ = 12°C = 285K

Initial Pressure, P₁ = P_s + ρgh

where P₁ = depth of air bubble = 40m

P_s = Pressure at lake surface 

= 1atm

ρ = density of water = 10²kg/cm³

g = acceleration due to gravity

= 9.8 ms^-2

∴ P₁ = 1.013 × 10⁵+10³ × 9.8 × 40

= 493300 Pa

= 4.933 × 10⁵ Pa

Let final volume of air bubble = V₂

Final Pressure = P_s = 1.013 × 10⁵ Pa

Final Temperature, T₂ = 35°C = 308 K

From ideal gas law,

PV = n RT

∴ \(\frac{PV}{T}\) = n R

∴ for given n, \(\frac{PV}{T}\) = constant

= 5.263 × 10^-6 m³

= 5.263 cm³.

∴ The volume of air bubble grows to 5. 263 cm³ when it reaches the lake surface.

Answer 2

Volume of the air bubble, V₁= 1 cm³ 

V₁​=1.0cm³= 1.0×10^-6  m³

Bubble rises to height, D = 40 m

Temperature at a depth of 40 m,

 T₁ = 12° C = 285 K

Temperature at the surface of the lake,

 T_{2 =} 35° C = 308 K

The pressure on the surface of the lake: 

P₂ =1atm =1×1.103×10⁵ Pa 

The pressure at the depth of 40 m:

P₁ = 1atm + Dρg

Where,

ρ is the density of water =10³ kg/m³
g is the acceleration due to gravity = 9.8m/s²

 P₁ = 493300 Pa

Let V₂ be the Volume when it reaches the surface

We know,

\(P1 V1/T1 = P2V2/T2\)

Therefore when the air  bubble reaches the surface, its volume becomes 5.263 cm³