Excess pressure of soap bubble in different medium I have been wondering that why excess pressure of soap bubble remains
$${4S}/{R}$$
Even if it is in water or in air?
 A: Let $p_i$ be pressure inside the bubble and $p_o$ be the pressure outside. Excess pressure is then
$$p = p_i - p_o$$ whether the bubble is in the air or in the water. That is the meaning of  excess pressure. Why the expression is given by $\frac{4S}{R}$ as above can be calculated as follows below.

The work done by this pressure is force$\times$displacement if we consider the radius of the bubble to be expanded (isothermally) by increment $dr$ : $$W=\underbrace{pA}_{force=pressure\times area}\times dr$$ so
$$W=p(4\pi r^2) dr\tag1$$ We also know that the increase in potential energy of expansion to $dr$ is $$dU=2F_T[4\pi(r+dr)^2-4\pi r^2]=16F_T\pi rdr\tag2$$ which is surface tension$\times$change in surface area$^1$.
Since work done is equal to the change in potential energy, we can equate (1) and (2) and then integrate over $r$, to give a final expression for pressure excess, $$p=\frac{4F_T}{R}$$
$^1$Note the factor of two in the front since there are two surfaces (inner surface and outer surface) of the bubble which is why in the final expression you have a factor of 4.
