Question

An air bubble of $ 2\text{ }c{{m}^{3}} $ rises from the bottom of a lake of $ 32\text{ }m $ 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=10\,m/s^{2} $ and density $ \rho =10^{3}\, kg/m^{3} $ )

• A. $ 5.937\,cm^{3} $
• B. $ 8.937\,cm^{3} $
• C. $ 12.937\,\,cm^{3} $
• D. $ 16.937\,\,cm^{3} $

Answer 1

Answer: (B)

$V_{1}=2 \,cm ^{3}, V_{2}=?, h=32\, m$
$p_{1}=\rho g h+p_{0}, p_{2}=p_{0}$
$T_{1}=9+273=282\, K$
$T_{2}=30+273=303 \,K$
$g=10 \,m / s ^{2}$
$\rho=10^{3} \,kg / m ^{3}$
$\Rightarrow $ From $\frac{p_{1} V_{1}}{T_{1}}=\frac{p_{2} V_{2}}{T_{2}}$
[Here, we assume air as an ideal gas]
$\Rightarrow \frac{\left(\rho g h+p_{0}\right) 2 \times 10^{-6}}{282}=\frac{p_{0} \times V_{2}}{303}$
$\Rightarrow \frac{\left(10^{3} \times 10 \times 32+10^{5}\right) 2 \times 10^{-6} \times 303}{282 \times 10^{5}}=V_{2}$
$\Rightarrow \frac{(3.2+1) \times 2 \times 10^{-6} \times 303}{282}=V_{2}$
$\Rightarrow \frac{4.2 \times 2 \times 303}{282} \times 10^{-6}=V_{2}$
$\Rightarrow 9 \times 10^{-6}=V_{2}$
$\Rightarrow V_{2}=9 \,cm ^{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.