Ncert Exercises & Solutions

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?

Given that:

Intial volume of the air bubble,

$V_{1}$ = 1.0 c

$m^{3}$ = 1.0 ×

$10^{- 6} m^{3}$

Temperature at a depth of 40 m,

$T_{1}$ = 12°C = 285 K

The 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 + ρdg

Where ρ is the density of water =

$10^{3}$ kg/

$m^{3}$

g is the acceleration due to gravity = 9.8 m/

$s^{2}$

$P_{1} = 1.013 \times 10^{5} + 40 \times 10^{3} \times 9.8 = 493300 Pa$

According to the ideal gas equation:

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

Then the final volume of the air bubble:

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

$= \frac{(493300) \times (1.0 \times 10^{- 6}) \times 308}{285 \times 1.013 \times 10^{5}}$

= 5.263 ×

$10^{- 6}$

$m^{3}$ or 5.263

${cm}^{3}$

Therefore, when the air bubble reaches the surface, its volume becomes 5.263

$c m^{3}$

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