Question

A bubble of gas released at the bottom of a lake increases to eight times its original volume when it reaches the surface. Assuming that atmospheric pressure is equivalent to the pressure exerted by a column of water $10m$ height, the depth of the lake is :

• A. $80m$
• B. $90m$
• C. $40m$
• D. $10m$
• E. $70m$

Answer 1

Answer: (E)

$70m$

hroughout 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 $P_0$ 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 $=P_0+\rho gh=\rho g\times 10+\rho gh=\rho g(h+10)$.

At the surface of water, only atmospheric pressure would be acting on the bubble i.e. $P=P_0=\rho g\times 10$

As PV = constant, $\rho g(h+10)\times V=\rho g10\times 8V$

$h+10=80$

$h=70$ meters.