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Derive an expression for the excess pressure inside a soap bubble. OR Derive Laplace's law for spherical membrane of a bubble due to surface tension.

Vidyadip · Accepted Answer

Consider a small, spherical, thin-filmed soap bubble with a radius $R$. Let the pressure outside the drop be Po 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
$\mathrm{A}=8 \pi R^2$
Hence, with a hypothetical increase in radius by an infinitesimal amount $\mathrm{dR}$, the differential increase in surface area and surface energy would be
$d A=16 \pi R \cdot d R$ and
$d W=T \cdot d A=16 \pi T R d R$
We assume that $d R$ is so small that the pressure inside remains the same, equal to $p$. All parts of the surface of the bubble experiences an outward force per unit area equal to $p-p_0$. Therefore, the work done by this outward pressure-developed force against the surface tension force during the increase in radius $d R$ is
$\mathrm{dW}=$ (excess pressure $\times$ surface area) $\cdot \mathrm{dR}$
$=\left(p-p_0\right) \times 4 \pi R^2 \cdot d R$
From Eqs. (2) and (3),
$
\begin{aligned}
& \left(\mathrm{p}-\mathrm{p}_0\right) \times 4 \pi \mathrm{R}^2 \cdot \mathrm{dR}=16 \pi \mathrm{TRdR} \\
& \therefore \mathrm{p}-\mathrm{p}_0=\frac{4 T}{R} \ldots \ldots \ldots \ldots . .(4)
\end{aligned}
$
which is the required expression.
[Note : The excess pressure inside a drop or bubble is inversely proportional to its radius : the smaller the bubble radius, the greater the pressure difference across its wall.]

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