Question

An air bubble is formed at a depth $h$below the surface of water. What is the pressure inside the bubble? (${P_0} =$atmospheric pressure, $r =$radius of the bubble)

Answer 1

In order to solve this question, we are going to first analyze the information as given in the question and find the pressure at the depth as given inside the water. Then, the excess pressure of the bubble is added to the pressure inside water which gives the total pressure inside the bubble.

Formula used: The pressure under the water at this depth inside the water is given by:
$P' = {P_0} + \rho gh$
Where, $\rho$is equal to the density of water, $g$is the acceleration due to gravity and$h$is the depth inside the water.
The excess pressure which is given by
$\Delta P = \dfrac{{2T}}{r}$
Where, $T$is the tension inside the bubble, and $r$is the radius of the bubble.

Complete step-by-step solution:
It is given in the question that the bubble is present at a depth $h$below the surface of the water.
Here, the pressure at the surface of the water is equal to the atmospheric pressure as is given equal to${P_0}$
Now, the pressure under the water at this depth inside the water is given by:
$P' = {P_0} + \rho gh$
Where, $\rho$is equal to the density of water, $g$is the acceleration due to gravity and$h$is the depth inside the water.
The pressure inside the bubble is equal to the sum of the pressure at the depth $h$inside the water and the pressure difference inside the bubble.
Therefore, it is given by:
$P = P' + \Delta P$
Here, the excess pressure which is given by
$\Delta P = \dfrac{{2T}}{r}$
Where, $T$is the tension inside the bubble, and $r$is the radius of the bubble.
Putting the values in the equation for pressure inside the bubble:
$P = P' + \Delta P = {P_0} + \rho gh + \dfrac{{2T}}{r}$
Hence, the pressure inside the bubble is ${P_0} + \rho gh + \dfrac{{2T}}{r}$

Note: It is to be noted that the pressure inside a spherical bubble is greater than the pressure outside. The way in which the excess pressure $\Delta P$ depends on the radius $r$of the drop, the surface tension and the density remains constant with respect to the details of the type of the bubble inside the water.