Question

A bubble having a volume of 1.00 cm^3 is released from the bottom of a swimming pool where the depth is 5.00 m. What will the volume of the bubble be when it reaches the surface? (The density of water is 1.00 times 10^3 kg/m^3, air pressure is 1.00 times 10^5 Pa) 1.00 0.49 2 1.49 infinity

Answer 1

Find the absolute pressure at depth (P1) and at the surface (P2)
According to the ideal gas law,
P1V1/T1 = P2V2/T2 (But T1 =T2)

=> P1V1 = P2V2

P1( water pressure below 5 m)

Fresh water has a density of 1 g per cm^3
or 1000 Kg /m^3

So at a depth of 5 m then a 1 m^2 square has a weight of water equivalent to a column 5 m * 1m^2
= 5 m^3 acting on it .
This has a mass of 5,000 Kg

Now multiply by g ( F = mg) 5,000 * 9.8 = 4.9 * 10 ^4 Pa

Standard atmospheric pressure is 1.00 * 10 ^5 Pa
So the pressure at this depth in atmospheres is (4.9 * 10 ^4/(1.00 * 10^5)) +1 = 1.49 ATM

V1 = 1.00 cm3 (given)
And P2( surface pressure ) = 1 ATM ( pressure on surface by air pressure)

Thus, 1.49 ATM x 1.00 cm3 = 1ATM x X cm3
So
X = 1.49 cm3

The bubble volume will be 1.49 cm3 at the water surface.