Concept of surface tension: excess pressure inside an air bubble in a tank of water

Question

I know the expression for excess pressure inside a LIQUID DROP is : $$ P_2 - P_1 = 2S/R $$ where $P_2$ is the pressure just inside the liquid drop and $P_1$ is the pressure outside the liquid drop and $S$ is surface tension of water.

I considered equilibrium of one hemisphere of the liquid drop and balanced the forces which gave me the above result.

However, upon reading some illustrations I observed that the pressure inside an AIR BUBBLE in a tank of water is also $$ P_2 - P_1 = 2S/R \tag{1} $$ where $P_2$ is the pressure just inside the air bubble and $P_1$ is the pressure outside the air bubble and $S$ is surface tension of water.

Using the same method of considering equilibrium of one hemisphere of AIR BUBBLE, I have the following :

(i) $F_2$ due to water outside the surface of hemisphere of air bubble

(ii) $F_3$ due to air inside the surface of hemisphere of air bubble

Now the third force in the case of LIQUID DROP would have been:

(iii) $F_1$ due to surface tension of water of the other hemispherical surface in contact with the hemisphere chosen and it will act along the points of periphery of the hemisphere in contact with the other hemisphere directed towards the other hemisphere

However, for an AIR BUBBLE, how will this $F_1$ will be present as there is no water inside the air bubble. Water is present only outside the air bubble.

Is $F_2$ responsible for this force due to surface tension (since water is present outside and air inside) and how will it act? Then, what is corresponding $F_1$ for air bubble and how will it act? How will the balance of $F_1$, $F_2$ and $F_3$ yield me equation $(1)$ for air bubble?

The above was my try at deriving $(1)$ which I couldn’t complete as I encountered many problems to which I didn’t have answers as shown above.

So, to summarise the question :

How is the excess pressure inside an AIR BUBBLE in a tank of water $2S/R$ where $S$ is surface tension of water?

Answer 1

This can be done by work and energy. Consider a spherical air bubble in water. The surface interface is air and water just like a water drop in air.

The net outward pressure on the bubble:

$$P_{net}=P_{in}-P_{out}$$

where $P_{in}$ is the pressure of the air bubble and $P_{out}$ is the water pressure just outside the bubble.

The work done by the net pressure to increase the radius of the bubble by $\mathrm dR$ is

$$dW=P_{net}\ A\ \mathrm dR=(P_{in}-P_{out})\ 4\ \pi\ R^2\ \mathrm dR$$

The change in surface area with the radius going from $R$ to $R+\mathrm dR$ is

$$dA=4\ \pi(R+\mathrm dR)^2-4\ \pi\ R^2$$

which when taking $$\mathrm dR^2=0$$ simplifies to

$$dA=8\ \pi\ R\ \mathrm dR$$

The change in surface energy of the bubble is

$$dE=dA\ T=8\ \pi\ R\ \mathrm dR\ T$$

where T is the surface tension. Now equate work and energy.

$$4\ \pi\ R^2\mathrm dR (P_{in}-P_{out})=8\ \pi\ R\ \mathrm dR\ T$$

or $$P_{in}-P_{out}=\frac{2T}{R}$$

Answer 2

There is actually surface tension between the air particles themselves, roughly 0.076 N/m, which is the reason for $F1$.