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laws

Question

At a depth of 40 m, the temperature of the lake is 12 °C and an air bubble has a volume of 1.0 cm³. The air bubble is rising up to reach the surface of the lake, where the temperature is 35 °C. Find the volume of the air bubble when it reaches the surface of the lake

Moderate

A

$10.6 \times 10^{- 6} m^{3}$
B

$5.3 \times 10^{- 6} m^{3}$
C

$2.8 \times 10^{- 6} m^{3}$
D

$15.6 \times 10^{- 6} m^{3}$

Solution

$Using, P_{1} = P_{2} + ρgh$
$Here, P_{2} = 1 .013 \times 10^{5} atm, h = 40 m$
$ρ = 10^{3} {kgm}^{- 3} (density of water)$
$g = 9 .8 {ms}^{- 1}$
$∴ P_{1} = 1 .013 \times 10^{5} + 10^{3} \times 9 .8 \times 40$
= 493300 Pa
$Now, \frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$
$Here, T_{1} = (12 + 273) = 285 K, T_{2}$
$= (35 + 273) = 308 K$
$V_{1} = 1 \times 10^{- 6} m^{3}$
V₂ is the volume of the air bubble when it reaches the surface
$∴ V_{2} = \frac{P_{1} V_{1} T_{2}}{T_{1} P_{2}}$
$= \frac{(493300 \times 1 \times 10^{- 6})}{285 \times 1.013 \times 10^{5}} \times 308$
$= 5.26 \times 10^{- 6} m^{3} = 5.3 \times 10^{- 6} m^{3}$

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