Question
Derive the expression for excess pressure inside:
- A liquid drop.
- A liquid bubble.
- An air bubble.
✓
Answer
- Inside a liquid drop: Let r = radius of a spherical liquid drop of centre O.
Let $P_i$ and $P_0$ be the values of pressure inside and outside the drop.
$\therefore$ Excess pressure inside the liquid drop = $P_i - P_0$.
Let $\Delta\text{r}$ be the increase in its radius due to excess pressure. It has one free surface outside.
$\therefore$ Increase in surface area of the liquid drop
$=4\pi(\text{r}+\Delta\text{r})^2-4\pi\text{r}^2$
$=4\pi[\text{r}^2+(\Delta\text{r})^2+2\text{r}\Delta\text{r}-\text{r}^2]$
$=8\pi\text{r}\ \Delta\text{r}\dots\text{(i)}$ $(\because\Delta\text{r}$ is small $\therefore\Delta\text{r}^2$ is neglected$)$
$\therefore$ Increase in surface energy of the drop is
W = Surface tension × increase in area
$=\text{T}\times8\pi\text{r }\Delta\text{r}\dots\text{(ii)}$
Also W = Force due to excess of pressure × dispplacement
= Excess pressure × Area of drop × increase in raduis
$=(\text{P}_\text{i}-\text{P}_0)4\pi\text{r}^2\Delta\text{r}\dots\text{(iii)}$
$\therefore$ From equations (ii) and (iii), we get
$(\text{P}_\text{i}-\text{P}_0)\times4\pi\text{r}^2\Delta\text{r}$
$=\text{T}\times\text{8}\pi\text{r }\Delta\text{r}$
$\Rightarrow\text{P}_\text{i}-\text{P}_0=\frac{\text{T}}{\text{r}}$
- Inside a liquid bubble: A liquid bubble has air both inside and outside it and therefore it has two free surfaces.
$=2[4\pi(\text{r}+\Delta\text{r})^2-4\pi\text{r}]$
$=2\times8\pi\text{r}\Delta\text{r}=16\pi\text{r}\Delta\text{r}$
$\therefore\text{W}=\text{T}\times16\pi\text{r }\Delta\text{r}\dots\text{(i)}$
Also $\text{W}=(\text{P}_\text{i}-\text{P}_0)4\pi\text{r}^2\times\Delta\text{r}\dots\text{(ii)}$
From equations (i) and (ii), we get
$(\text{P}_\text{i}-\text{P}_0)\times4\pi\text{r}^2\times\Delta\text{r}=\text{T}\cdot16\pi\text{r}\Delta\text{r}$ or $\text{P}_\text{i}-\text{P}_0=\frac{4\text{T}}{\text{r}}.$
- Inside an air bubble: Air bubble is formed inside liquid, thus air bubble has one free surface inside it and liquid is outside.
$\Delta\text{r}=$ increase in its radius due to excess of pressure
$(\text{P}_\text{i}-\text{P}_0)$ inside it.
T = surface tension of the liquid in which bubble is formed.
$\therefore$ Increase in surface area $=8\pi\text{r}\Delta\text{r}.$
$\therefore\text{W}=\text{T}\times8\pi\text{r}\Delta\text{r}$
Also $\text{W}=(\text{P}_\text{i}-\text{P}_0)\times4\pi\text{r}^2\Delta\text{r}$
$\therefore(\text{P}_\text{i}-\text{P}_0)\times4\pi\text{r}^2\Delta\text{r}=\text{T}\times8\pi\text{r }\Delta\text{r}$ or $\text{P}_\text{i}-\text{P}_0=\frac{2\text{T}}{\text{r}}.$
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