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A small spherical gas bubble of diameter $d= 4$ μm forms at the bottom of a pond. When the bubble rises to the surface its diameter is $n=1.1$ times bigger. What is the depth in meters of the pond?

Note: water's surface tension and density are $σ= 73 \times 10^{-3} \mbox{ N}$ and $ρ= 10^3 \mbox{ kg/m}^3$, respectively. The gas expansion is assumed to be isothermal.

My attempts:

I used the equation of pressure:

$P_1V_1=P_2V_2$

where $P_1$ is the pressure at the top and $V_1$ is the volume of bubble at the top and $P_2$ is the pressure at the bottom, and $V_2$ is the volume of the bubble at the bottom

Because the bubble at the bottom, it received the hydrostatic pressure, so the equation became:

$P_1V_1$= $(P_0+\rho g d) V_2$

Since $V_1$ and $V_2$ is sphere, we can use the sphere volume. And $P_1$ is same with atmospheric pressure= $10^5 \mbox{ Pa}$

$10^5 \cdot (\frac{4}{3} \pi r_1^3) = (10^5 + 10^3 \cdot 9.8 \cdot d) \cdot (\frac{4}{3} \pi r_2^3)$

Cancel out the $\frac{4}{3}\pi$ and we get:

$10^5 \cdot (r_1)^3 = (r_2)^3 \cdot (10^5 + 9800d)$

Substitute $r_1 = 4 \times 1.1= 4.4 \mbox{ μm}= 4.4\times10^{-6} \mbox{ m}$ and $r_2 = 4 \times 10^{-6} \mbox{ m}$

$10^5 \cdot (4.4 \times 10^{-6})^3 = (4 \times 10^{-6})^3 \cdot (10^5 + 9800d)$

$10^{-13} \cdot (4.4)^3 = 64 \times 10^{-18} \times 10^5 + 64 \times 10^{-18} \times 9800d$

$85.184 \times 10^{-13} = 64 \times 10^{-13} + 627200 \times 10^{-18} d$

$21.184 \times 10^{-13} = 627200 \times 10^{-18} d$

$d= 3.37 \mbox{ m}$

So, the depth of the pond is $3.37 \mbox{ m}$. My question: what is the useful of $σ = 73 \times 10^{-3} \mbox{ N}$ ?

I really confused about it. Thanks

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closed as off-topic by David Z Nov 1 '13 at 17:09

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  • $\begingroup$ We have the MathJax rendering engine running on the site so that you can write math in a LaTeX-alike language. $\endgroup$ – dmckee Oct 31 '13 at 9:10
  • $\begingroup$ @dmckee Where can I find MathJax to write my equation? $\endgroup$ – akusaja Oct 31 '13 at 9:33
  • $\begingroup$ It is already running on the site so you just start using it. By hitting edit on this post you can see what has been done. You'll probably get it right away. $\endgroup$ – dmckee Oct 31 '13 at 9:35
  • $\begingroup$ @akusaja A basic overview is here. A more thorough guide covering many more features is here. $\endgroup$ – user10851 Oct 31 '13 at 10:14
  • $\begingroup$ The pressure inside the bubble is larger than the corresponding pressure of the water surrounding it because of the surface tension. In fact, the additional pressure is $P=\frac{2\sigma}{r}$. You need to take that into account to properly estimate the depth of the pond. Since the "extra pressure" becomes less as the bubble rises, the actual depth will be less than it would be if there was no surface tension, for the same change in bubble size. $\endgroup$ – Floris Jan 1 '15 at 23:01
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The surface tension of the bubble would help you to find out how much work the bubble is doing while it increases its volume.

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  • $\begingroup$ where can I insert the surface tension of bubble in my equation? $\endgroup$ – akusaja Oct 31 '13 at 9:15
  • $\begingroup$ $P_1$ is not atmospheric because the surface tension changes the pressure inside the bubble. See hyperphysics.phy-astr.gsu.edu/hbase/surten2.html $\endgroup$ – John Rennie Oct 31 '13 at 9:35
  • $\begingroup$ @JohnRennie, so the P1 of the bubbles is not same with P0, but same with Pi-Po? In your link, it says that (Pi-Po) is equal with 4T/r. In this case, what is the meaning of T? $\endgroup$ – akusaja Oct 31 '13 at 9:43
  • $\begingroup$ That's the surface tension. $\endgroup$ – Ignacio Vergara Kausel Oct 31 '13 at 9:45

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