### Q.5:- 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?

**Answer:-**

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*_{1}*V*_{1} / *T*_{1} = *P*_{2}*V*_{2} / *T*_{2}

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

*V*_{2} = *P*_{1}*V*_{1}*T*_{2} / *T*_{1}*P*_{2}

= 493300 × 1 × 10^{-6} × 308 / (285 × 1.013 × 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}.