13.5 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?

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    Answered by

    Payal Gupta | Contributor-Level 10

    5 months ago

    13.5 Volume of the air bubble , V1 = 1.0 cm3 = 1 *10-6 m3

    Height achieved by bubble, d = 40 m

    Temperature at a depth of 40 m, T1 = 12 ? = 285 K

    Temperature at the surface of the lake, T2 = 35 ? = 308 K

    The pressure at the surface of the lake, P2 = 1 atm = 1.013 *105 Pa

    The pressure at 40m depth: P1 = 1 atm + d ?g

    Where ? is the density of water = 103 kg/ m3

    G = acceleration due to gravity = 9.8 m/ s2

    Hence P1 = 1.013 *105+(40*103*9.8) = 493300 Pa

    From the relation P1V1T1 = P2V2T2 , where V2 is the volume of bubble when it

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