Physics > Fluid Mechanics > 8.0 Intermolecular forces
Fluid Mechanics
1.0 Introduction
1.1 Ideal liquid
1.2 Density of a liquid $\left( \rho \right)$
1.3 Relative density of a liquid $(RD)$
1.4 Density of a mixture of two or more liquid
1.5 Density variation with temperature
1.6 Density variation with pressure
2.0 Fluid pressure
2.1 Atmospheric pressure
2.2 Pressure variation with depth
2.3 Measurement of pressure
2.4 Pressure difference in accelerating fluids
3.0 Pascal's law
4.0 Buoyant force
5.0 Flow of fluids
6.0 Viscosity
7.0 Stoke's law
8.0 Intermolecular forces
9.0 Angle of contact
8.4 Excess pressure
1.2 Density of a liquid $\left( \rho \right)$
1.3 Relative density of a liquid $(RD)$
1.4 Density of a mixture of two or more liquid
1.5 Density variation with temperature
1.6 Density variation with pressure
2.2 Pressure variation with depth
2.3 Measurement of pressure
2.4 Pressure difference in accelerating fluids
Surface tension causes a pressure difference between the inside and outside of liquid drop or a soap bubble.
8.4.1 Excess pressure inside liquid drop
Consider a liquid drop of radius $R$ and surface tension $T$.
Each half of the soap bubble is in equilibrium because the upward force of surface tension balances the downward force due to pressure difference as shown in the figure.
As we know that the liquid drop has only one surface.
So at equilibrium we can write, $$\left( {{P_2} - {P_1}} \right)\pi {R^2} = 2\pi RT$$ or $$\left( {{P_2} - {P_1}} \right) = \frac{{2T}}{R}$$
Also,pressure on the concave size of the liquid surface is greater than the pressure on the convex side. This difference of pressure is known as excess pressure.
8.4.2 Excess pressure inside an air bubble in a liquid
Consider an air bubble of radius $R$ inside a liquid and surface tension $T$.
As we know that an air bubble has only one surface.
So at equilibrium we can write, $$\left( {{P_2} - {P_1}} \right)\pi {R^2} = 2\pi RT$$ or $$\left( {{P_2} - {P_1}} \right) = \frac{{2T}}{R}$$
Also, we can write, $${P_1} = {P_O} + \rho gh$$
From the above equations we get, $$\begin{equation} \begin{aligned} {P_2} - ({P_O} + \rho gh) = \frac{{2T}}{R} \\ {P_2} = {P_O} + \rho gh + \frac{{2T}}{R} \\\end{aligned} \end{equation} $$
If ${P_O} = 0$ then, $${P_2} = \rho gh + \frac{{2T}}{R}$$ where,
$\rho $: Density of the liquid
$h$: Depth from the free surface
The above equation gives the excess pressure inside a soap bubble at a depth $h$ from the free surface.
8.4.3 Excess pressure inside a soap bubble
Consider a soap bubble of radius $R$ and surface tension $T$.
As we know that the soap bubble has two surfaces.
So at equilibrium we can write, $$\left( {{P_2} - {P_1}} \right)\pi {R^2} = 2 \times 2\pi RT$$ or $$\left( {{P_2} - {P_1}} \right) = \frac{{4T}}{R}$$
8.4.4 Additional explanations
- Two soap bubbles of radii $R_1$ and $R_2$ are in contact with each other and $R$ is the radius of interface
From the knowledge of excess pressure we can write the following equations from the figure, $$\begin{equation} \begin{aligned} \left( {{P_1} - {P_3}} \right) = \frac{{2T}}{R}\quad ...(i) \\ \left( {{P_1} - {P_O}} \right) = \frac{{2T}}{{{R_1}}}\quad ...(ii) \\ \left( {{P_2} - {P_O}} \right) = \frac{{2T}}{{{R_2}}}\quad ...(iii) \\\end{aligned} \end{equation} $$
Subtracting equation $(ii)$ and $(iii)$ we get, $$\left( {{P_1} - {P_2}} \right) = \left( {\frac{{2T}}{{{R_1}}} - \frac{{2T}}{{{R_2}}}} \right)\quad ...(iv)$$
From equation $(i)$ and $(iv)$ we get, $$\begin{equation} \begin{aligned} \frac{{2T}}{R} = \frac{{2T}}{{{R_1}}} - \frac{{2T}}{{{R_2}}} \\ \frac{1}{R} = \left( {\frac{1}{{{R_1}}} - \frac{1}{{{R_2}}}} \right) \\\end{aligned} \end{equation} $$ or $$R = \left( {\frac{{{R_1}{R_2}}}{{{R_1} - {R_2}}}} \right)$$
For excess pressure inside the soap bubble, $$\begin{equation} \begin{aligned} {P_1} - {P_O} = \frac{{4T}}{{{R_1}}}\quad {\text{or}}\quad V = \frac{4}{3}\pi R_1^3 \\ {P_2} - {P_O} = \frac{{4T}}{{{R_2}}}\quad {\text{or}}\quad V = \frac{4}{3}\pi R_2^3 \\\end{aligned} \end{equation} $$
For new soap bubble of radius $R$ we can write, $$P - {P_O} = \frac{{4T}}{R}\quad {\text{or}}\quad V = \frac{4}{3}\pi {R^3}$$
Under isothermal condition we can write, $$PV = {P_1}{V_1} + {P_2}{V_2}$$
Let ${P_O} = 0$ So, $$\begin{equation} \begin{aligned} \left( {\frac{{2T}}{R}} \right)\left( {\frac{4}{3}\pi {R^3}} \right) = \left( {\frac{{2T}}{{{R_1}}} \times \frac{4}{3}\pi R_1^3 + \frac{{2T}}{{{R_2}}} \times \frac{4}{3}\pi R_2^3} \right) \\ {R^2} = R_1^2 + R_2^2 \\ R = \sqrt {R_1^2 + R_2^2} \\\end{aligned} \end{equation} $$
Two soap bubbles of radii $R_1$ and $R_2$ are connected $\left( {{R_1} < {R_3}} \right)$ by means of a tube
As we know that the excess pressure inside a soap bubble is given by, $$P = \frac{{4T}}{R}$$ or $$P \propto \frac{1}{R}$$
So, as pressure is inversely proportional to the radius of the tube.
Therefore, air passes from smaller bubble to larger bubble because pressure inside the smaller bubble is higher than the larger bubble.
- Two soap bubbles of radii $R_1$ and $R_2$ coalesce to form a new soap bubble of radius $R$ under the isothermal condition
