Question

An air bubble of volume $1.0$ $c{m}^3$ rises from the bottom of a lake $40$m deep at a temperature of $12$ ${}^{o}C$. To what volume does it grow when it reaches the surface, which is of a temperature of $35$ ${}^{o}C$?

• A. $10.6\times {10}^{-6}{m}^3$
• B. $5.3\times {10}^{-6}{m}^3$
• C. $2.8\times {10}^{-6}{m}^3$
• D. $15.6\times {10}^{-6}{m}^3$

Answer 1

The correct option is B $5.3\times {10}^{-6}{m}^3$

Using, $P_1=P_2+\rho gh$
Here, $P_2=1.013\times {10}^5$atm, $h=40$m
$\rho ={10}^3$kg $m^{-3}$(density of water),
$g=9.8m$ $s^{-2}$
$\therefore P_1=1.013\times {10}^5+{10}^3\times 9.8\times 40=493300$Pa
Now, $\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}$
Here, $T_1=(12+273)=285$K, $T_2=(35+273)=308$K. $V_1=1\times {10}^{-6}{m}^3$
$V_2$ is the volume of the air bubble when it reaches the surface
$\therefore V_2=\frac{P_1{V}_1{T}_2}{T_1{P}_2}=\frac{(493300\times 1\times {10}^{-6})}{285\times 1.013\times {10}^5}\times 308$
$=5.26\times {10}^{-6}{m}^3=5.3\times {10}^{-6}{m}^3$.