Bubbles

[mummymonkey]

Help me out here. If a bubble of gas in a fish tank is cut in half to produce two bubbles, what is the surface area of the two bubbles compared with the original single bubble?
If anybody wants to explain this to me I'd be greatful. Please be aware I am numerically challenged though.

[Benguin]

Well if it is cut into two the volume is halfed between them.

The initial volume is calculated by 4/3 pi r cubed and it's surface area is 4 pi r squared.
So for your problem we need the new value of r relative to the old one ... that is 2 * (4/3 pi s cubed) = 4/3 pi r cubed
( where r is the original radius and s the new one for the split pair)

or s cubed = 1/2 r cubed

therefore the new radius (s) is = 3 cube root ((r cubed)/2)

so the new surface area for one is 4 pi (cube root ((r cubed)/2) /3

for both would be 8 pi (cube root ((r cubed)/2) /3
if r = 1 metre ( damn big bubble)

initial surface area for one bubble; 12.6m3
for the two split new surface area; 4.193m3 each

But it is late, and I am on the whisky.

edited for the first maths cockup I spotted 23:43

[phildonnia]

Original Volume is
V = 4/3 pi r^3

Orifinal Surface Area is
S = 4 pi r^2

Solving each of these for r:
r = cuberoot(3V / 4pi)
r = squareroot(S / 4pi)

Since these are both equal to r,
cuberoot(3V / 4pi) = squareroot(S / 4pi)

Now solve this for S:
S = 4 pi * (cuberoot(3V / 4pi))^2

---

Now lets say we cut the volume in half, v = V/2 The new surface area s is

s = 4 pi * (cuberoot(3v / 4pi))^2

Substituting v = V/2

s = 4 pi * (cuberoot(3V/2 / 4pi))^2
s = 4 pi * (cuberoot(3V/ 4pi) * cuberoot(1/2))^2
s = 4 pi * (cuberoot(3V/ 4pi))^2 * (cuberoot(1/2))^2

And finally, since S = 4 pi * (cuberoot(3V / 4pi))^2, we have

s = S * (cuberoot(1/2))^2
s ~~ S * 0.63

[pupdog]

Math was never my strong field, but I hope you find the new total surface area is greater than the one original area!

[mummymonkey]

If I understand phildonnia, the new bubbles are each a bit more than half the surface area of the original bubble.
So splitting a bubble in half will increase the surface area by approx 25%?

That'll do for me. Many thanks all.

[Benguin]

Originally posted by pupdog
Math was never my strong field, but I hope you find the new total surface area is greater than the one original area!

Actually, I'm inclined to think you are right.

I'll check through my dodgy maths.

[Benguin]

OK, here is the correction, although phildonnia laid out a clearer answer (and got it right!)


so the new surface area for one is 4 pi (cube root ((r cubed)/2)squared /3

for both would be 8 pi (cube root ((r cubed)/2)squared /3
if r = 1 metre ( damn big bubble)

initial surface area for one bubble; 12.6m3
for the two split new surface area; 7.921m3 each

But it is late, and I am on the whisky.


It is now early, with added clarity of a hangover to help me focus.

[wollery]

Originally posted by Benguin

initial surface area for one bubble; 12.6m3
for the two split new surface area; 7.921m3 each
That's what I got, but I think you mean m2 !!!

[Benguin]

Absolutely.

3D Algebra and Glenfiddich don't mix well.

[American]

Have you considered pressure differentials? It's more fun with calculus. But I guess as long as they're at the same depth, it doesn't matter.

[BPSCG]

Originally posted by mummymonkey
If I understand phildonnia, the new bubbles are each a bit more than half the surface area of the original bubble.
So splitting a bubble in half will increase the surface area by approx 25%?

That'll do for me. Many thanks all. I'm confused. Doesn't it make a difference whether or not you specify the amount of gas being contained in the two new bubbles?

Assume a bubble floating in air, or to make it more visual, a basketball. Slice the basketball in half, then sew up the two hemispheres into two new mini-basketballs.

The surface area of the basketballs remains the same.

[wollery]

Originally posted by BPSCG
I'm confused. Doesn't it make a difference whether or not you specify the amount of gas being contained in the two new bubbles?

Assume a bubble floating in air, or to make it more visual, a basketball. Slice the basketball in half, then sew up the two hemispheres into two new mini-basketballs.

The surface area of the basketballs remains the same.But if you do that they wont be spherical!

[Occasional Chemist]

Originally posted by BPSCG
I'm confused. Doesn't it make a difference whether or not you specify the amount of gas being contained in the two new bubbles?


The way the question was formulated, the amount of gas was specified, wasn't it? Take a bubble and break it half. The amount of gas in each new bubble is half the amount of the gas in the original bubble.


Assume a bubble floating in air, or to make it more visual, a basketball. Slice the basketball in half, then sew up the two hemispheres into two new mini-basketballs.

The surface area of the basketballs remains the same.

... but the volume inside the basketballs will change. In the bubble case, the volume remains constant (barring pressure changes), and the surface area varies.