Question

An air bubble starts rising from the bottom of a lake. Its diameter is $3.6mm$ at the bottom and $4mm$ at the surface. The depth of the lake is $250cm$ and the temperature at surface is ${40}^{o}C$.
What is the temperature at the bottom of the lake? Given atmospheric pressure $=76cm$ of $Hg$ and $g=980cm/s^2$.

• A. ${11}^{o}C$
• B. ${12.36}^{o}C$
• C. ${13}^{o}C$
• D. ${10.37}^{o}C$

Answer 1

The correct option is D ${10.37}^{o}C$

At the bottom of the lake, volume of the bubble
$V_1=\frac{4}{3}\pi r^3=\frac{4}{3}\pi {\left(0.18\right)}^3$
Pressure on the bubble, $p_1=$ atmospheric pressure $+$ pressure due to a column of $250cm$ of water.
$=76\times 13.6\times 980+250\times 1\times 980$
$=(76\times 13.6+250)980dyne/{cm}^2$
At the surface of the lake, volume of the bubble
$V_2=\frac{4}{3}\pi r_2^3=\frac{4}{3}\pi {\left(0.2\right)}^3{cm}^3$
Pressure on the bubble,
$p_2=$ atmospheric pressure
$=(76\times 13.6\times 980)dyne/{cm}^2$
$T_2=273+40=313K$
Now, $\frac{p_1{V}_1}{T_1}=\frac{p_2{V}_2}{T_2}$
$\Rightarrow \frac{(76\times 13.6\times 250)980\times \left(\frac{4}{3}\right)\pi {\left(0.18\right)}^3}{T_1}$
$=\frac{(76\times 13.6)\times 980\left(\frac{4}{3}\right)\pi {\left(0.2\right)}^3}{313}$
$\Rightarrow T_1=283.37K$
$\therefore T_1=283.37-273={10.37}^{o}C$.