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}

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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