Answer to Question #235280 in Molecular Physics | Thermodynamics for Jimmy

Question

An air bubble from the bottom of a lake becomes triple itself as it reaches on the water surface. How deep is the lake?

Depth = _____ meters

Answer 1

We know that

$P_sV_s=P_dV_d$

Assuming atmospheric pressure to be 75 cm of Hg and the density of water to be 1/10 of the density of mercury, the depth of the lake.

$P_s=1atm\\P_d=1+\frac{h}{10}$

$V_s=3V\\V_d=V$

Put value

$1\times 3V=(1+\frac{h}{10}) \times V$

$3=1+\frac{h}{10}$

$2=\frac{h}{10}\\h=20m$

Depth(h)=20m