Question -
Answer -
Volume of the air bubble, V1 = 1.0 cm3
= 1.0 x 10-6 m3
Air bubble rises to height, d = 40 m
Temperature at a depth of 40 m, T1 = 120 C= 285 K
Temperature at the surface of the lake, T2 = 350 C= 308 K
The pressure on the surface of the lake:
P2 =1 atm = 1 x 1.013 x 105 Pa
The pressure at the depth of 40 m:
P1= 1atm + dρg
Where,
ρ is the density of water = 103 kg/ m3
g is the acceleration due to gravity = 9.8 m/s2
Hence,
P1 =1.013 x 105 + 40 x 103 x 9.8
We get,
= 493300 Pa
We have
P1V1 / T1 = P2V2 / T2
Where, V2 isthe volume of the air bubble when it reaches the surface
V2 =P1V1T2 / T1P2
= 493300 x 1 x 10-6 x308 / (285 x 1.013 x 105)
We get,
= 5.263 x 10-6 m3 or 5.263 cm3
Hence, when the air bubble reaches the surface, its volumebecomes 5.263 cm3