An air bubble of volume 4cm\(^3\) is formed 20m under water. What will be its volume when it rises to just below the surface of the water if the atmospheric pressure is equivalent to a height of 10m of water?
The bubble is formed 20 m below the surface.
Pressure at that depth = atmospheric pressure + water pressure due to 20 m of water.
Given:
Atmospheric pressure = pressure of 10 m of water
Therefore,
P\(_1 = 20 + 10 = 30 \text{ m of water}\)
Just below the surface, the bubble is under atmospheric pressure only:
P\(_2 = 10 \text{ m of water}\)
Apply Boyle's Law
Given:
V\(_1 = 4\ \text{cm}^3\)
P\(_1V_1 = P_2V_2\)
30\(\times 4 = 10 \times V_2\)
V\(_2 = \frac{30 \times 4}{10}\)
V\(_2 = 12\ \text{cm}^3\)
The volume of the air bubble just below the water surface is 12 cm\(^3\).
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