Question

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

• 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

Answer: (B)

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),
$g = 9.8 \,m\,s^{-2}$
$\therefore \quad P_{1} = 1.013 \times 10^{5} + 10^{3} \times 9.8 \times 40 = 493300 \,Pa$
Now, $\quad \frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}$
Here, $T_{1} = \left(12 + 273\right) = 285\, K$, $T_{2} = \left(35 + 273\right) = 308\, K$,
$V_{1} = 1 \times 10^{-6}\, m^{3}$
$V_{2}$ is the volume of the air bubble when it reaches the surface
$\therefore \quad V_{2} = \frac{P_{1}V_{1}T_{2}}{T_{1}P_{2}} = \frac{\left(493300 \,\times\, 1\,\times\,10^{-6}\right)}{285\,\times\,1.013\,\times\,10^{5}} \times 308$
$= 5.26 \times 10^{-6}\, m^{3} = 5.3 \times 10^{-6} \,m^{3}$