Question

An air bubble of volume 1.0 cm3 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 of a temperature of 35°C?

• A. 10.6 × 10–6 m3
• B. 5.3 × 10–6 m3
• C. 2.8 × 10–6 m3
• D. 15.6 × 10–6 m3

Answer 1

Answer: (B)

5.3 × 10–6 m3

Answer 2

Using , $P_{1} = P_{2} + \rho gh$

Here, $P_{2} = 1.013 \times 10^{5}atm,h = 40m$

$\rho = 10^{3}kgm^{- 3}$ (density of water), $g = 9.8ms^{- 2}$

$\therefore P_{1} = 1.013 \times 10^{5} + 10^{3} \times 9.8 \times 40$

$= 493300Pa$

Now, $\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}$

Here, $T_{1} = (12 + 273) = 285K,T_{2} = (35 + 273) = 308K$

$V_{1} = 1 \times 10^{- 6}m^{3}$

$V_{2}$is the volume of the air bubble when it reaches the surface

$\therefore V_{2} = \frac{P_{1}V_{1}T_{2}}{T_{1}P_{2}} = \frac{(493300 \times 1 \times 10^{- 6})}{285 \times 1.013 \times 10^{5}} \times 308$

$= 5.26 \times 10^{- 6}m^{3} = 5.3 \times 10^{- 6}m^{3}$