Welcome to the NCERT Solutions for Class 11 Physics - Chapter Kinetic Theory. This page offers a step-by-step solution to the specific question from Exercise 1, Question 5: . With detailed answers and explanations for each chapter, students can strengthen their understanding and prepare confidently for exams. Ideal for CBSE and other board students, this resource will simplify your study experience.
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?
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 = P1V1T2 / 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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Read MoreWelcome 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....
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