A spherical bubble rises from the bottom of a lake whose temperature is 6 oC at the bottom and 27 oC at the surface. If the bubble doubles its volume by the time it reaches the surface, how deep is the lake?
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From gas laws:
P1V1/T1 = P2V2/T2
P1V1/(279 K) = (1 atm)(2V1)/(300 K)
P1 = (279/300)(1 atm)(2) = (279/150 atm)
(279/150 atm) = 1 atm + (rho)gh
(129/150 atm) = (1000 kg/m^3)*(9.8 m/s/s)*(h) = 9800 kg/(m^2s^2) * h
(129/150 atm)*(101325 Pa/atm) = (9800 Pa/m)*h
h = 8.89 m
P1V1/T1 = P2V2/T2
P1V1/(279 K) = (1 atm)(2V1)/(300 K)
P1 = (279/300)(1 atm)(2) = (279/150 atm)
(279/150 atm) = 1 atm + (rho)gh
(129/150 atm) = (1000 kg/m^3)*(9.8 m/s/s)*(h) = 9800 kg/(m^2s^2) * h
(129/150 atm)*(101325 Pa/atm) = (9800 Pa/m)*h
h = 8.89 m