Q.Diameter of a water drop is 0.6 mm. Calculate the pressure inside a liquid drop. (T = 72 dyne/cm, atmospheric pressure = 1.013×10⁵ N/m²)

The excess pressure inside a spherical liquid drop (one free surface) is $2T/r$, added to the outside atmospheric pressure.

For a spherical liquid drop of radius $r$ (single free surface), the excess pressure inside due to surface tension $T$ is

$\Delta P = \dfrac{2T}{r}$

Given diameter $= 0.6\text{ mm} \Rightarrow r = 0.3\text{ mm} = 3\times10^{-4}\text{ m}$, and $T = 72\text{ dyne/cm} = 72\times10^{-3}\text{ N/m} = 0.072\text{ N/m}$:

$\Delta P = \dfrac{2\times 0.072}{3\times10^{-4}} = \dfrac{0.144}{3\times10^{-4}} = 480\text{ Pa}$

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Laplace's law is derived by equating the work done against surface tension in a small expansion to the work done by the excess pressure over the same expansion.

Consider a spherical bubble (or drop) of radius $r$, surface tension $T$, with the excess pressure inside over outside being $\Delta P$. Let its radius increase by a small amount $dr$.

Increase in surface area: The surface area of a sphere is $A = 4\pi r^2$, so the increase is

$dA = \frac{d}{dr}(4\pi r^2)\, dr = 8\pi r\, dr.$

For a soap bubble, which has two free surfaces (inner and outer film boundaries), the total increase in area is $2\times 8\pi r\,dr = 16\pi r\,dr$.

Work done against surface tension in creating this extra area:

$dW_{\text{surface}} = T\, dA_{\text{total}} = T\times 16\pi r\, dr \quad \text{(bubble)}.$

Work done by the excess pressure in expanding the bubble's volume by $dV = 4\pi r^2\,dr$:

$dW_{\text{pressure}} = \Delta P \times dV = \Delta P \times 4\pi r^2\, dr.$

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Balance the work done by the excess pressure in a small expansion of the bubble against the increase in its surface energy.

Consider a spherical soap bubble of radius $r$ and surface tension $T$. A soap bubble has two free surfaces (inner and outer, since the soap film has negligible thickness but two air–liquid interfaces).

Let the bubble's radius increase by a small amount $dr$ due to the excess pressure $P$ inside it (relative to outside).

Work done by the excess pressure in this small expansion equals (excess pressure) $\times$ (increase in volume):

$dW=P\times dV=P\times 4\pi r^2\,dr$

(since $V=\frac43\pi r^3\Rightarrow dV=4\pi r^2\,dr$).

Increase in surface energy equals surface tension $\times$ increase in total surface area. Since the bubble has two surfaces, total surface area $=2\times 4\pi r^2=8\pi r^2$, so

$dA=8\pi\times 2r\,dr=16\pi r\,dr$

$dE=T\,dA=16\pi Tr\,dr$

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