__Question 13.5__:
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?

__Solution__:*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 X 1 X 10

_{}

^{-6}X 308 / (285 X 1.013 X 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}.

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