Question

An air bubble of $2c{m}^3$ rises from the bottom of a lake of $32m$ at a temperature of $9^{o}C$ . When the bubble reaches the surface of the lake from the bottom of the lake, what volume does it grows for which temperature is $30^{\circ }C$ (assume $g=10m/s^2$ and density $\rho =10^{3}kg/m^3$ )

• A. $5.937c{m}^3$
• B. $8.937c{m}^3$
• C. $12.937c{m}^3$
• D. $16.937c{m}^3$

Answer 1

Answer: (B)

$V_1=2c{m}^3,V_2=?,h=32m$
$p_1=\rho gh+p_0,p_2=p_0$
$T_1=9+273=282K$
$T_2=30+273=303K$
$g=10m/s^2$
$\rho =10^{3}kg/m^3$
$\Rightarrow$ From $\frac{p_1{V}_1}{T_1}=\frac{p_2{V}_2}{T_2}$
[Here, we assume air as an ideal gas]
$\Rightarrow \frac{\left(\rho gh+p_0\right)2\times 10^{-6}}{282}=\frac{p_0\times V_2}{303}$
$\Rightarrow \frac{\left(10^3\times 10\times 32+10^5\right)2\times 10^{-6}\times 303}{282\times 10^5}=V_2$
$\Rightarrow \frac{(3.2+1)\times 2\times 10^{-6}\times 303}{282}=V_2$
$\Rightarrow \frac{4.2\times 2\times 303}{282}\times 10^{-6}=V_2$
$\Rightarrow 9\times 10^{-6}=V_2$
$\Rightarrow V_2=9c{m}^3$
Here, there is nothing said about the surface tension, so we have assumed the water pressure equal to air pressure at a particular position.