The diameter of bubble at the surface is 4mm and at the bottom of the lake is 1mm. If atmospheric pressure is 1 atmosphere and the temperature of the lake water and the atmosphere are equal. What is the depth of the lake. The density of the lake water and mercury are 1g\ml and 3.6g\ml respectively. Also neglect the contribution of pressure due to surface tension

We know from the Charles Law that the pressure is directly proportional to the temperature at a given volume.

So, we get that P1V1/T1 = P2V2/T2

Case1- When the bubble is a t surface.

The pressure is P1 = 1 atm.

And the volume is V1 = 4π*(4)^2 .

Case2- When the bubble is at the bottom then we will get that.

The pressure will be P2 = 1 atm + pgh which will be 1 + (1000*9.8*h)/10^5 atm.

The volume is V2 = 4π*1^2.

Since, the temperature is same given in the question, so on substituting the values in the equation we will get.

16 = 1+ 0.098h

On solving we will get h = 153.03 meter.