Question

An air bubble of volume $1.0c{m}^3$ rises from the bottom of a lake 40 m deep at a temperature of ${12}^{\circ }C$. To what volume does it grow when it reaches the surface which is at a temperature of ${35}^{\circ }C$ ?

Answer 1

Volume of the air bubble, $V_1=1.0c{m}^3=1.0\times {10}^{-6}{m}^3$

Bubble rises to height, $d=40m$

Temperature at a depth of 40 m, $T_1={12}^{o}C=285K$

Temperature at the surface of the lake, $T_2={35}^{o}C=308K$

The pressure on the surface of the lake: $P_2=1atm=1\times 1.103\times {10}^{5}Pa$

The pressure at the depth of 40 m: $P_1=1atm+d\rho g$

Where,

$\rho$ is the density of water $={10}^{3}kg/m^3$
$g$ is the acceleration due to gravity $=9.8m/s^2$

$\therefore P_1=1.103\times {10}^5+40\times {10}^3\times 9.8=493300Pa$

We have $\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}$

Where, $V_2$ is the volume of the air bubble when it reaches the surface.

$V_2=\frac{P_1{V}_1{T}_2}{T_1{P}_2}$

$=\frac{493300\times 1\times {10}^{-6}\times 308}{285\times 1.013\times {10}^5}$

$=5.263\times {10}^{-6}{m}^3$ or $5.263c{m}^3$

Therefore, when the air bubble reaches the surface, its volume becomes $5.263c{m}^3$.