Question

An air bubble starts rising from the bottom of a lake. Its diameter is $3.6\text{mm}$ at the bottom and $4\text{mm}$ at the surface. The depth of the lake is $250\text{cm}$ and the temperature at the surface is $40{}^{\circ }\text{C}$. What is the approximate temperature at the bottom of the lake in ${}^{\circ }\text{C}$.?
Given atmospheric pressure$=76\text{cm of Hg}$

Answer 1

Given,
Radius of the bubble at the bottom of lake,
$r_1=\frac{3.6}{2}\text{mm}=1.8\text{mm}$
Radius of the bubble at the surface of lake,
$r_2=\frac{4}{2}\text{mm}=2\text{mm}$
Depth of the lake, $H=250\text{cm}$,

At the bottom of the lake, volume of the bubble
$V_1=\frac{4}{3}\pi r_1^3=\frac{4}{3}\pi (0.18{)}^3{\text{cm}}^3$

Pressure on the bubble $p_1$= Atmospheric pressure + Pressure due to column of $250\text{cm}$ of water
$p_1=76\times 13.6\times 980+250\times 1\times 980$
$\Rightarrow p_1=(76\times 13.6+250)980{\text{dyne/cm}}^2$

At the surface of the lake, volume of the bubble
$V_2=\frac{4}{3}\pi r_2^3=\frac{4}{3}\pi (0.2{)}^3{\text{cm}}^3$

Pressure on the bubble,
$p_2$ = atomspheric pressure
$\Rightarrow p_2=(76\times 13.6\times 980){\text{dyne/cm}}^2$

Temperature, $T_2=273+40=313\text{K}$

Now from ideal gas equation, we have
$\frac{p_1{V}_1}{T_1}=\frac{p_2{V}_2}{T_2}$

Substituting the values, we get
$\frac{(76\times 13.6+250)980\times \frac{4}{3}\pi (0.18{)}^3}{T_1}=\frac{(76\times 13.6)\times 980\left(\frac{4}{3}\right)\pi (0.2{)}^3}{313}$

$\Rightarrow T_1=283.37K$

$\therefore T_1=283.37-273=10.37{}^{\circ }\text{C}\approx 10{}^{\circ }\text{C}$

Hence, temperature at the bottom of the lake is approximately $10{}^{\circ }\text{C}$.