The excess pressure inside a soap bubble is

Explanation

Consider a small, spherical, thin-filmed soap bubble with a radius R. Let the pressure outside the bubble be `p_(@)` and that inside be p. A soap bubble in air is like a spherical shell and has two gas-liquid interfaces. Hence, the surface area of the bubble is
`A=8piR^(2)" "`.....(1)
Hence, with an increase in radius by an infinitesimal amount dR,the differential increase in surface area and surface energy would be respectively
`dA=16piR.dR" "` and
`dW=T.dA=16piTR dR" "`.....(2)
We assume that dR is so small that the pressure inside remains the same, equal to p. All parts of the surface of the bubble experience an outward force per unit area equal to `p-p_(@)`. Therefore, the work done by this outward pressure-developed force against the surface tension force during the increase in radius dR is
dW=(excess pressure `xx` surface area).dR
`=(p-p_(@))xx4piR^(2).dR" "`....(3)
From Eqs. (2) and (3), `(p-P_(@))xx4piR^(2).dR=16piTRdR`
`:. p-p_(@)=(4T)/(R )" "`.....(4)
which is the required expression [Laplace's law].