Question

An air bubble released by a remotely operated underwater vehicle, 120 m below the surface of a lake, has a volume of 1.40 cm3. The surface of the lake is at sea level, and the density of the lake water can be approximated as that of pure water. As the bubble rises to the surface, the temperature of the water and the number of air molecules in the bubble can each be approximated as constant. Find the volume (in cm3) of the bubble just before it pops at the surface of the lake.

Answer 1

Answer:

17.7 cm^3

Explanation:

depth, h = 120 m

density of water, d = 1000 kg/m^3

V1 = 1.4 cm^3

P1 = P0 + h x d x g

P2 = P0

where, P0 be the atmospheric pressure

Let V2 be the volume of the bubble at the surface of water.

P0 = 1.01 x 10^5 Pa

P1 = 1.01 x 10^5 + 120 x 1000 x 9.8 = 12.77 x 10^5 Pa

Use

P1 x V1 = P2 x V2

12.77 x 10^5 x 1.4 = 1.01 x 10^5 x V2

V2 = 17.7 cm^3

Thus, the volume of bubble at the surface of water is 17.7 cm^3.

Answer 2

The volume (in cm3) of the bubble just before it pops at the surface of the lake is mathematically given as

V2 = 17.7 cm^3

What is the volume (in cm3) of the bubble just before it pops at the surface of the lake.?

Question Parameter(s):

An air bubble released by a remotely operated underwater vehicle, 120 m

below the surface of a lake, has a volume of 1.40 cm3.

Generally, the equation for the initial Pressure  is mathematically given as

P1 = P0 + h x d x g

Where,atmospheric pressure

P0 = 1.01 x 10^5 Pa

Therefore

P1 = 1.01 * 10^5 + 120 * 1000 &* 9.8

P1= 12.77 * 10^5 Pa

In conclusion

P1 x V1 = P2 x V2

12.77 x 10^5 x 1.4 = 1.01 x 10^5 x V2

V2 = 17.7 cm^3

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