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?

Asked by Pragya Singh | 1 year ago |  114

##### Solution :-

Volume of the air bubble, V1 = 1.0 cm3

= 1.0 x 10-6 m3

Air 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= 35°C

= 308 K

The pressure on the surface of the lake:

P2 = 1 atm = 1 x 1.013 x 105 Pa

The pressure at the depth of 40 m:

P1= 1 atm + dρg

Where,

ρ is the density of water = 103 kg / m3

g is the acceleration due to gravity = 9.8 m/s2

Hence,

P1 = 1.013 x 105 + 40 x 103 x 9.8

We get,

= 493300 Pa

We have

$$\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}$$

Where, V2 is the volume of the air bubble when it reaches the surface

$$V2 = \dfrac{P_1V_1T_2}{T_1P_2}$$

$$\dfrac{493300 \times 1 \times 10^{-6} \times 308 }{(285 \times 1.013 \times 10^5)}$$

We get,

= 5.263 x 10-6 m3 or 5.263 cm3

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

Answered by Abhisek | 1 year ago

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