Excess Pressure Inside A Curved Membrane:
Let us consider, there is a curved liquid surface. This surface is at rest. There are two pressures acting on it. One is inward pressure due to the surface tension and the other is outward pressure. This inward pressure is balanced by that equal outward pressure.
To find the value of excess pressure, let us consider a small curvilinear rectangular element EFGH of the liquid. The length of the side EF is \( x \) and the radius of curvature is \( r_1 \) with centre at \( O_1 \). The length of the side FG is \( y \) and the radius of curvature is \( r_2 \) with centre \( O_2 \), as shown in the adjoining Fig.1.
So the area of the element EFGH is \( xy \).
Let \( p \) be the excess pressure which acts on the concave side of the element in a vertically upward direction. So the outward force on it due to excess pressure is \( p\times{xy} \).
Now the liquid surface is moved outwards parallel to itself through a small distance \( \delta{z} \) and the new position of the liquid surface is E’F’G’H’. It is moved in such a way that the curvature remains unchanged, then the work done during the displacement is \( dW=p\cdot{xy}\cdot\delta{z} \).
The lengths of the sides E’F’ and F’G’ are \( x+\delta{x} \) and \( y+\delta{y} \) respectively. So the new area of the element is
\( {(x+\delta{x})}\cdot{(y+\delta{y})}\\=xy+y\delta{x}+x\delta{y}+\delta{x}\delta{y} \)Since, \( \delta{x} \) and \( \delta{y} \) are very small, so their product \( \delta{x}\delta{y} \) is also very small and is negligible.
So the area of E’F’G’H’ is \( xy+y\delta{x}+x\delta{y} \).
Now the increase in surface area of the element is
\( xy+y\delta{x}+x\delta{y}-xy\\=y\delta{x}+x\delta{y} \)If \( T \) is the surface tension of the liquid, then increase in surface energy of the element is
\( T(y\delta{x}+x\delta{y}) \)The work done in moving the surface outward is equal to the increase in surface energy.
\( p\cdot{xy}\cdot\delta{z}=T(x\delta{y}+y\delta{x})\\or,\ p=T(\frac{1}{y}\frac{\delta{y}}{\delta{z}}+\frac{1}{x}\frac{\delta{x}}{\delta{z}})\tag{1} \)From, \( \Delta{E'F'O_1} \) and \( \Delta{EFO_1} \), we get
\( \frac{E'F'}{EF}=\frac{E'O_1}{EO_1}\\or,\ \frac{x+\delta{x}}{x}=\frac{r_1+\delta{z}}{r_1}\\or,\ 1+\frac{\delta{x}}{x}=1+\frac{\delta{z}}{r_1} \\or,\ \frac{\delta{x}}{x}=\frac{\delta{z}}{r_1}\\or,\ \frac{\delta{x}}{\delta{z}}=\frac{x}{r_1}\tag{2}\)
Similarly, from \( \Delta{F'G'O_2} \) and \( \Delta{FGO_2} \) we get,
\( \frac{F'G'}{FG}=\frac{G'O_2}{GO_2}\\or,\ \frac{y+\delta{y}}{y}=\frac{r_2+\delta{z}}{r_2}\\or,\ 1+\frac{\delta{y}}{y}=1+\frac{\delta{z}}{r_2}\\or,\ \frac{\delta{y}}{\delta{z}}=\frac{y}{r_2}\tag{3} \)
From equations (1), (2), and (3) we get,
\( p=T(\frac{1}{y}\cdot\frac{y}{r_2}+\frac{1}{x}\cdot\frac{x}{r_1}) \\or,\ \displaystyle{p=T(\frac{1}{r_1}+\frac{1}{r_2})} \)
In case of air or soap membrane, when there are two surfaces, an inner and an outer, the increase in surface area is twice of that of a membrane having one surface.
\( \displaystyle{p=2T(\frac{1}{r_1}+\frac{1}{r_2})} \)
Special Cases:
(I)
In case of a spherical soap bubble floating in air, there are two surfaces and \( r_1=r_2=r\ (say) \), therefore
\( p=2T(\frac{1}{r}+\frac{1}{r})\\or,\ \displaystyle{p=\frac{4T}{r}} \)
(II)
In case of a liquid drop in a gas or an air bubble inside a liquid, there is one surface and \( r_1=r_2=r\ (say) \), therefore
\( p=T(\frac{1}{r}+\frac{1}{r})\\or,\ \displaystyle{p=\frac{2T}{r}} \)
(III)
When the centres of curvature lie on opposite sides of the surface, one of the surfaces is convex and the other is concave. The radii of curvature \( r_1 \) and \( r_2 \) of the two surfaces will have opposite signs. The shorter radius is taken as positive and the larger one negative. So in this case, \( \displaystyle{p=2T(\frac{1}{r_1}-\frac{1}{r_2})} \), if \( r_2>r_1 \)