A bubble of gas released at the bottom of lake increases to 8 times , its original volume when it reaches the surface , . Assuming that atmospheric pressure is equivalent to the pressure exerted by a column of water 10 m high,find the depth of the lake ?
Answers
Throughout the entire process, the number of moles of the gas bubble remains same. So, we can apply the ideal gas equation PV = constant.
Let us assume that the bubble is h meters below the surface of the water, and P₀ is the atmospheric pressure.
Pressure only due to water at a depth of h meters would be equal to ρgh. Also, atmospheric pressure would be acting on the bubble. So, the total pressure on the gas bubble at that depth = P₀ + ρgh = ρg×10 + ρgh = ρg(h+10).
At the surface of water, only atmospheric pressure would be acting on the bubble i.e. P = P₀ = ρg×10
As PV = constant, ρg(h+10) × V= ρg10 × 8V → h+10 = 80 → h = 70 meters.
Answer:
Explanation:
Le the pressure at the bottom = P1 and volume = V1
Let the pressure at the surface = P2 and volume = V2
Temperature remains constant so
P1V1 = P2V2
Given that: V2 = 8V1
therefore, P1 = 8P2
Pressure = hρg
ρ = density of water
g = acceleration due to gravity
P2 = hρg = 10ρg
P1 = 8(10ρg)
P1 - P2 = 80ρg - 10ρg = 70ρg
From hydrostatics,
P1 - P2 = hρg
70 ρ g = hρg
h = 70 m
Depth of the lake = 70m