# What is the difference between pressure in water droplet and that in an air bubble?

The question is at least supposed to be simple but yeah, I need some detailed answers. I have tried thinking about it in the lines of surface tension but it seems what is required of me is more than that so if anyone has faced a similar question, your help would be most needed.

Let us solve this problem by considering the free energy $F$ of the whole system in thermal equilibrium (temperature and particle number constant, so $d F = - p \; dV$). In fact the type of system does not matter as far as we are considering a bubble of some gaseous or liquid substance inside another substance (can be the same one). Let us for now just look at the case of an air bubble (label $a$) with a spherical surface ($s$) in water (liquid $l$).

The total free energy (differential) of the system is $$dF = dF_a + dF_l + dF_s.$$ The individual free energies of air and liquid are $dF_a = -p_a\; dV$ ($V = V_a$) and $dF_l = -p_l\; dV_l = p_l\; dV$ since an increase in volume of the liquid is equivalent to the decrease in volume of the air. The free energy of the surface can be described as $dF = \Gamma dA$ where $\Gamma$ is the surface tension and $A$ is the surface area of the bubble. Inserting this into the equation above gives

$$0 = -p_a dV + p_l dV + \Gamma dA \qquad \text{or} \qquad p_a = p_l + \frac{dA}{dV} \Gamma.$$

Assuming a spherical geometry of the bubble we can write $dA/dV = 4 \pi \; d(r^2)/(4\pi \; d(r^3)/3) = 3 \times 1/(3 r^2) \times d(r^2)/dr = 2/r$ and

$$p_a = p_l + \frac{2}{r} \Gamma.$$

The calculation is equivalent for different combinations of substance (air bubble in air (soap), water droplet in air, air bubble in water).

The differential pressure due to surface tension is the same in both cases, if the bubble and droplet have the same size, respectively. This is most easily understood by thinking of the surface tension force as a consequence of the medium in question trying to minimize its surface area, which in turn exists because it requires energy to increase it. The required energy is due to the fact that the potential of a particle on the surface is higher than the one of the same particle immersed in the fluid. Moving a particle to the surface of the air bubble requires the same energy as moving it to the surface of the droplet.

Answer is very simple.... a water droplet has only one surface i.e its outer surface where as in bubble it has two surfaces i.e both inner and outer surface... so the have pressure difference inner and outer surface in a bubble twice as that as droplet.

• I think the air bubble means a bubble of air within a continuous fluid, not a bubble containing air and floating in air. So there is only one interface. Good point though. May 16, 2017 at 5:51

It was a good thaught of you and was similar to mine,I do not have any theoretical answer but according to my understanding : An air bubble inside water will experience a lot of inward pressure than what it can exert outward as it is inside a denser medium , and a water droplet in air will exert a more pressure outwards than what it experiences from air inward as it is in rarer medium . So, pressure in water droplet should be less in air and pressure in air bubble should be more in water . It is to be noted that in both cases the height of the medium above the water drop or air bubble also matters ,like more the depth of air bubble in water more will be the pressure in that bubble and also due to pressure of water on bubble the bubble will be more smaller at the 10 metre depth than at 1 metre depth.