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Question: A spherical air bubble of radius 2cm is released 30 m below the surface of a pond at 280 K. Its volu...

A spherical air bubble of radius 2cm is released 30 m below the surface of a pond at 280 K. Its volume when it reaches the surface, which is at 300 K assuming it is in thermal equilibrium the whole time, is V. Ignore the size of the bubble compared to other dimensions like 30 m. Then V is equal to–

A

1.4 × 10–4 m3

B

2.8 × 10–4 m3

C

0.7 ×10–4 m3

D

4.2 × 10–4 m3

Answer

1.4 × 10–4 m3

Explanation

Solution

PiViTi\frac{P_{i}V_{i}}{T_{i}} = PfVfTf\frac{P_{f}V_{f}}{T_{f}}

Vf = (PiViPfTfTi)\left( \frac{P_{i}V_{i}}{P_{f}}\frac{T_{f}}{T_{i}} \right)

Pf = Patm

Pi = Patm + rgh