An air bubble of volume 1.0 cm3 rises from the bottom of a lake 40 m deep at a temperature of 12 °C. To what volume does it grow when it reaches the surface, which is at a temperature of 35 °C?
Volume of the air bubble, V1=1.0cm3=1.0×10−6m3 Bubble rises to height, d=40m Temperature at a depth of 40 m, T1=12oC=285K Temperature at the surface of the lake, T2=35oC=308K The pressure on the surface of the lake: P2=1atm=1×1.103×105Pa The pressure at the depth of 40 m: P1=1atm+dρg Where, ρ is the density of water =103kg/m3 g is the acceleration due to gravity =9.8m/s2 ∴P1=1.103×105+40×103×9.8=493300Pa We have T1P1V1=T2P2V2 Where, V2 is the volume of the air bubble when it reaches the surface. V2=T1P2P1V1T2 =285×1.013×105493300×1×10−6×308 =5.263×10−6m3 or 5.263cm3 Therefore, when the air bubble reaches the surface, its volume becomes 5.263cm3.
