Question

The volume of an air bubble becomes three times as it rises from the bottom of a lake to its surface. 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.

Answer 1

We can assume that the process is isothermal because the temperature of the surrounding remains constant

So, by applying Boyle’s law

$P_s{V}_s=P_d{V}_d.....\left(I\right)$

Where, $s$ = surface of water

$d$ = depth

Given that,

Depth of lake $=h$

Volume $V_s=3V_d$

Now,

$P_s=75cmHg$

$P_d=75+\frac{h}{10}$

Now, put the value of equation(I)

$75\times 3=75+\frac{h}{10}$

$225-75=\frac{h}{10}$

$h=1500cm$

$h=15m$

Hence, the depth of lake is $15m$