Question
Derive an expression for excess pressure inside a drop of liquid.
✓
Answer
Consider a spherical drop as shown in the figure. Let $p_i$ be the pressure inside the drop and $p_0$ be the pressure out side it. As the drop is spherical in shape, the pressure, $p _{ i }$, inside the drop is greater than $p _0$, the pressure outside. Therefore, the excess pressure inside the drop is $p_i-p_0$.
Let the radius of the drop increase from $r$ to $r+\Delta r$, where $\Delta r$ is very small, so that the pressure inside the drop remains almost constant.
Let the initial surface area of the drop be $A_1=4 \pi r^2$, and the final surface area of the drop be
$A_2=4 \pi(r+\Delta r)^2$
$\therefore A_2=4 \pi\left(r^2+2 r \Delta r+\Delta r^2\right)$
$\therefore A_2=4 \pi r^2+8 \pi r \Delta r+4 \pi \Delta r^2\left(A s \Delta r \text { is very small, } \Delta r^2 \text { can be neglected }\right)$
$\therefore A_2=4 \pi r^2+8 \pi r \Delta r$
Thus, increase in the surface area of the drop is
$d A=A_2-A_1=8 \pi r \Delta r$
Work done in increasing the surface area by $d A$ is stored as excess surface energy.
$\therefore d W=T d A=T(8 \pi r \Delta r) \ldots(1)$
This work done is also equal to the product of the force F which causes increase in the area of the drop and the displacement $\Delta r$ which is the increase in the radius of the bubble.
$\therefore d W=F \Delta r \ldots \text { (2) }$
The excess force is given by,
(Excess pressure) $\times$ (Surface area)
$\therefore F=\left(p_i-p_0\right) 4 \pi r^2 \ldots(3)$
Equating Eq. (1), and Eq. (2), we get,
$T(8 \pi r \Delta r)=F \Delta r$
$\therefore T(8 \pi r \Delta r)=\left(p_i-p_0\right) 4 \pi r^2 \Delta r \ldots \text { (using Eq. (3)) }$
$\therefore\left(p_i-p_0\right)=\frac{2 T}{r}$
This equation gives the excess pressure inside a drop of liquid.
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