At a depth of 40 m, the temperature of the lake is 12 °C and an air bubble has a volume of 1.0 cm³. The air bubble is rising up to reach the surface of the lake, where the temperature is 35 °C. Find the volume of the air bubble when it reaches the surface of the lake

$\text{ Using, }{\mathrm{P}}_1={\mathrm{P}}_2+\mathrm{\rho gh}$

$\text{ Here, }{\mathrm{P}}_2=1.013\times {10}^5\mathrm{atm}, \mathrm{h}=40\mathrm{m}$

$\mathrm{\rho }={10}^3 {\mathrm{kgm}}^{-3}\text{ (density of water) }$

$\mathrm{g}=9.8{\mathrm{ms}}^{-1}$

$\therefore {\mathrm{P}}_1=1.013\times {10}^5+{10}^3\times 9.8\times 40$

= 493300 Pa

$\text{ Now, }\frac{{\mathrm{P}}_1{\mathrm{V}}_1}{{\mathrm{T}}_1}=\frac{{\mathrm{P}}_2{\mathrm{V}}_2}{{\mathrm{T}}_2}$

$\text{ Here, }{\mathrm{T}}_1=(12+273)=285\mathrm{K},{\mathrm{T}}_2$

$=(35+273)=308\mathrm{K}$

${\mathrm{V}}_1=1\times {10}^{-6}{\mathrm{m}}^3$

V₂ is the volume of the air bubble when it reaches the surface

$\therefore {\mathrm{V}}_2=\frac{{\mathrm{P}}_1{\mathrm{V}}_1{\mathrm{T}}_2}{{\mathrm{T}}_1{\mathrm{P}}_2}$

$=\frac{(493300\times 1\times {10}^{-6})}{285\times 1.013\times {10}^5} \times 308$

$=5.26 \times {10}^{-6} {\mathrm{m}}^3 = 5.3 \times {10}^{-6} {\mathrm{m}}^3$