An air bubble of volume 1.0 cm3 rises fr | Class 11 Physics Chapter Kinetic Theory, Kinetic Theory NCERT Solutions

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 at a temperature of 35 °C?

Answer:

Volume of the air bubble, V1 = 1.0 cm3 = 1.0 × 10-6 m3

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, T2 = 35°C = 308 K

The pressure on the surface of the lake:

P2 = 1 atm = 1 ×1.013 × 105 Pa

The pressure at the depth of 40 m:

P1 = 1 atm + dpg

Where, p is the density of water = 103 kg/m3

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

∴P1 = 1.013 × 105 + 40 × 103 × 9.8 = 493300 Pa

We have:  P1V1 / T1  = P2V2 / T2

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

 V2  =  P1V1T/  T1P2

= 493300 x (1.0 x 10-6) 308  285 x 1.013 x 105

= 5.263 × 10-6 m3 or 5.263 cm3

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


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Welcome to the NCERT Solutions for Class 11 Physics - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 5: An air bubble of volume 1.0 cm3 rises from the bottom of a lake 40 m deep at a temperature of 12 &de....