Minimum radius of bubbles in a soda bottle (surface tension)

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1. Nov 16, 2016

omoplata

1. The problem statement, all variables and given/known data

The question is from chapter 9 of "Exercises in Introductory Physics" by Leighton and Vogt.

The answer given in the book is $R = 4.9 \times 10^{-5} \rm{cm}$.

2. Relevant equations

$$\sigma = \frac{\Delta P \cdot R}{4}$$
Where,
$\sigma$ is the surface tension between the surfaces
$\Delta P$ is the difference of pressure between the inside and outside of the bubble
$R$ is the radius of the bubble

3. The attempt at a solution

From the given equation, $$R = \frac{4 \sigma}{\Delta P}$$

Since the given pressure is the gauge pressure, $\Delta P = 3.00 \times 10^6\, \rm{dynes \, cm^{-1}}$.

Replacing the values,
$$R = \frac{4 ( 73\,\rm{dynes\,cm^{-1}})}{3.00 \times 10^6\,\rm{dynes\,cm^{-1}}}=9.73\times 10^{-5} \, \rm{cm}$$
which is half the value given for the answer.

What did I do wrong?

2. Nov 16, 2016

Merlin3189

You said, "ΔP is the difference of pressure between the inside and outside of the bubble"
The value you used was, "the gauge pressure inside <the> bottle"

3. Nov 16, 2016

omoplata

But there is no other pressure value given.

And 'gauge pressure' is the pressure above the atomospheric pressure, right. So I don't need to subtract the atmospheric pressure from the given value to find the pressure difference.

From what I understand, this is what's happening. Before the bottle is opened, the pressure of the air above the liquid surface is the given gauge pressure ( $3.00\times 10^6\rm{dynes\,cm^{-2}}$). After the bottle is opened, the pressure of the air above the liquid surface goes down to atmospheric pressure, or zero gauge pressure. But the pressure of the gas trapped inside the liquid remains the same. If a bubble is made near the liquid surface ( we have to consider the surface here because no depth is given. If a depth was given we would have to consider the hydrostatic pressure of the liquid at that depth too ), the pressure inside will be the previously given gauge pressure $3.00\times 10^6\rm{dynes\,cm^{-2}}$, while the pressure outside the bubble is the pressure at the liquid surface, which is atmospheric pressure, or zero gauge pressure. So the pressure difference is $\Delta P = 3.00\times 10^6\rm{dynes\,cm^{-2}}$.

4. Nov 16, 2016

haruspex

Seems right. Gauge pressure is the excess above atmospheric. When the bottle is opened, that should be the excess pressure in any bubbles.
The formula you quote for the pressure in a bubble is for a bubble out in the open air. What is different for a bubble immersed in the liquid?

5. Nov 16, 2016

Merlin3189

When you put it like that, maybe I'm wrong!

I'd thought, before the bottle is opened, the pressure inside and outside the bubble is 3 Mdyne/cm2
After it is opened, the pressure inside is 3 Mdyne/cm2 and outside is atmospheric, about 1 Mdyne/cm2, giving a difference of 2 Mdyne/cm2.
But I think you are correct in your interpretation of gauge pressure, so ΔP remains 3 Mdyne/cm2

I see Haruspex has confirmed that.

6. Nov 16, 2016

omoplata

Ah, I see now! For a bubble in air there are two interfaces. From the inside air to the liquid film and from the liquid film to the outside air. But the bubble immersed in a liquid has only one interface!

So that is why you multiply by two to get 4 instead of 2 when deriving the surface tension equation! I didn't really understand it until now!

For a bubble in liquid, $\sigma = \frac{\Delta P \, R}{2}$, which gives the correct answer!

Thanks for helping everyone!

7. Nov 16, 2016

haruspex

OK.
There is another source of confusion to watch out for in this topic. For a nonspherical surface, one considers the min and max radii. For a thick-walled bubble in air, there is an internal and external radius, or in the nonspherical case two of each. I came across a Wikipedia article which confused the two radii of a nonspherical surface with the inner and outer radii for a thick-walled spherical bubble.