Physics help? Air bubble?

Question

An air bubble with a diameter of 0.001 mm is released underwater at a depth of 15m. How large will the air bubble's diameter be at the surface? (There is an increase of 1 atm for every 10.3 m in depth in water; for a sphere, V= 4/3 x 3.14 x r^3. there is already 1 atm at the surface)

How do you solve this problem?

Answer 1

Take my answer to the other question, but then also adjust it for surface tension, since your bubble's only a micron.

surface tension pressure = sigma * surface area / volume = sigma * 3 / r.

Answer 2

This will assume no change in temperature.

Use Boyle's Law (pressure at the surface is less, so volume will be more)

V1 * P1 = V2 * P2

Let's figure out the pressure at 15m under water.

1 atm at surface + [(1atm/10.3m)* 15m] = 2.46 atm is P1

Let's calculate the volume of the bubble under water.

V1 = 4/3 * 3.14 * (0.0005 mm)^3 = 5.23 x10^-10 mm^3

So V2 = P1*V1/P2

V2 = 2.46 atm * 5.23 x 10^-10 mm^3 / 1 atm = 1.29 x10^-9 mm^3

Now go back and solve for the new radius:

r = [V / (4/3) / 3.14]^(1/3)
r = [1.29x10-9 / (4/3) / 3.14]^(1/3) = 6.74 x 10-4 mm

then d = 2r = 2 * 6.74 x 10-4 mm = 1.34 * 10-3 mm or 0.00134 mm

Answer 3

Use the ideal gas law, which says that the volume is inversely proportional to the pressure. So:

P1/P2 = V2/V1

where P1, V1 are the bubble's pressure and volume at 15m deep; and P2, V2 are its pressure and volume at the surface.

You are given V1 (at least you can calculate it knowing the bubble's diameter and the formula for a sphere, both of which are given).

You are given P2 (it is 1 atm).

You can figure out P1 because the problem tells you how much the pressure changes with depth, and how deep you have to be.

So given those 3 variables, you can solve for V2.

Finally, remember that you have to find the DIAMETER, not the volume. You can use the sphere formula to help you there.

Answer 4

OK use:

PV = nrT

Therefore:

PV = constant

P= constant / V

Depth of 15m means a pressure of (15/10.3)+1 (1 atm at surface)

Volume = 4 pi r^3/3 = 5.236 * 10 ^-10 mm^3

Therefore constant = 1.286 * 10^-9

By rearranging and using pressure of 1atm you get a volume of:

1.286 * 10^-9 mm^3

Therefore diameter = 0.00135 mm (3 sig fig)

(This sounds about right as it is slightly bigger than the original one)

Answer 5

there's a 1.5 times change in pressure.
(P1V1) = (P2V2)
1.5 times less pressure means 1.5 times greater volume
air bubble with a diamter of .0001 has a volume of

.0005 ^3 * pi * 1.333333

at the surface. the volume will be
.0005^3 * pi * 1.33333 * 1.5 = r^3 * pi * 1.3333333

r^3 = 1.5 * (.0005)^3
r = cube root of 1.5 * .0005

Answer 6

Is the expansion process adiabatic? Isentropic?