<s>[INST] What is the depth in meters of the pond? 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
 [/INST] 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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