An air bubble of volume 1.0 cm^{3} 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?

#### Solution 1

Volume of the air bubble, *V*_{1} = 1.0 cm^{3} = 1.0 × 10^{–6} m^{3}

Bubble rises to height, *d* = 40 m

Temperature at a depth of 40 m, *T*_{1} = 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:

*P*_{2} = 1 atm = 1 ×1.013 × 10^{5} Pa

The pressure at the depth of 40 m:

*P*_{1} = 1 atm + *dρ*g

Where,

*ρ* is the density of water = 10^{3} kg/m^{3}

g is the acceleration due to gravity = 9.8 m/s^{2}

∴*P*_{1} = 1.013 × 10^{5} + 40 × 10^{3} × 9.8 = 493300 Pa

We have : `(P_1V_1)/T_1 = (P_2V_2)/T_2`

Where, *V*_{2} is the volume of the air bubble when it reaches the surface

`V_2 = (P_1V_1T_2)/(T_1P_2)`

`= ((493300)(1.0 xx 10^(-6))308)/(285 xx 1.013 xx 10^5)`

= 5.263 × 10^{–6} m^{3} or 5.263 cm^{3}

Therefore, when the air bubble reaches the surface, its volume becomes 5.263 cm^{3}

#### Solution 2

Volume of the bubble inside, `V_1 = 1.0 cm^3 = 1xx10^(-6) m^3`

Pressure on the bubble, `P_1` = Pressure of water + Atmospheric pressure

= `pgh + 1.01 xx 10^5 = 1000 xx 9.8 xx 40 + 1.01 xx 10^5`

`=3.92 xx 10^5 + 1.01 xx 10^5 = 4.93 xx 10^5` Pa

Temperature, `T_1 = 12 ^@C = 273 + 12 = 285 K`

Also, pressure outside the lake, `P_2 = 1.01 xx 10^5 n m^(-2)`

Temperature, `T_2 = 35 ^@C = 273 + 35 = 308` K, Volume `V_2` = ?

Now `(P_1V_1)/T_1 = (P_2V_2)/T_2`

`:. V_2 = (P_1V_1)/T_1. T_2/P_2 = (4.93xx10^5xx1xx10^(-6)xx 308)/(285 xx 1.01 xx 10^5) = 5.3 xx 10^(-6) m^(-3)`