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

Using, P1=P2+ρgh Here, P2=1.013×105atm, h=40mρ=103 kgm−3 (density of water) g=9.8ms−1∴   P1=1.013×105+103×9.8×40= 493300 Pa Now, P1V1T1=P2V2T2 Here, T1=(12+273)=285K,T2=(35+273)=308KV1=1×10−6m3V2 is the volume of the air bubble when it reaches the surface∴ V2=P1V1T2T1P2=(493300×1×10-6)285×1.013×105 × 308=5.26 × 10−6 m3 = 5.3 × 10−6 m3