Surface Tension and Capillarity

Surface Tension

The surface of a liquid behaves somewhat like a stretched membrane under tension. This property is known as surface tension. It is a phenomenon arising from the cohesive forces between liquid molecules.

Inside the bulk of the liquid, a molecule is surrounded by other molecules from all sides, and the intermolecular forces are balanced. However, a molecule on the surface of the liquid is only surrounded by other liquid molecules from below and the sides, and by molecules of the surrounding medium (like air) from above. The cohesive forces pulling the surface molecule inwards are stronger than the forces pulling it outwards (assuming the surrounding medium's molecules have weaker attraction to the liquid molecules). This results in a net inward cohesive force on the surface molecules.

This net inward force causes the surface layer of the liquid to be under tension and to try to minimise its surface area. A liquid surface, if left to itself and not influenced by external forces (like gravity, if the volume is small), will contract to the smallest possible area for a given volume, which is a sphere (e.g., raindrops, mercury droplets on a surface).

Definition of Surface Tension ($\sigma$ or $T$ or $\gamma$)

Surface tension is defined in two ways:

As a force per unit length: Surface tension is the tangential force per unit length acting perpendicular to a line drawn in the liquid surface. This force tends to pull the surface layer tight.
$ \text{Surface Tension} = \frac{\text{Force}}{\text{Length}} = \frac{F}{L} $
The SI unit is Newton per metre (N/m).
As surface energy per unit area: This definition is related to the work done in creating or expanding a liquid surface.

The magnitude of surface tension is usually denoted by the Greek letter sigma ($\sigma$), sometimes $T$ or $\gamma$.

Surface tension depends on the type of liquid and the temperature. It generally decreases with increasing temperature and becomes zero at the critical temperature of the liquid. Adding impurities or surfactants (like soap) can also significantly affect surface tension.

Surface Energy

To create a new surface area of a liquid, work must be done against the net inward cohesive forces acting on the surface molecules. This work done is stored as potential energy in the molecules forming the new surface. This stored energy is called surface energy.

Molecules in the surface layer have a higher potential energy compared to molecules in the interior of the liquid because work was done against attractive forces to bring them to the surface from the bulk.

The surface energy of a liquid is the excess potential energy of the molecules in the surface layer compared to the molecules in the bulk of the liquid.

Surface energy is proportional to the surface area. To increase the surface area of a liquid, energy must be supplied.

Surface Energy And Surface Tension

Surface tension and surface energy are two different aspects of the same phenomenon and are quantitatively related. The magnitude of surface tension (force per unit length) is numerically equal to the surface energy per unit area.

Let's demonstrate this relationship by considering the work done in stretching a liquid film. Imagine a rectangular wire frame with a movable bar sliding along two parallel sides. If the frame is dipped in a soap solution and taken out, a soap film forms on it. The film has two surfaces (top and bottom).

Diagram illustrating the relationship between surface tension and surface energy by stretching a soap film.

(Image Placeholder: A U-shaped wire frame with a movable bar across it. A liquid film is shown stretched across the frame. The movable bar is shown displaced by a distance x, increasing the film area. A force F is needed to hold the bar against the surface tension force.)

Due to surface tension $\sigma$, the film exerts a force on the movable bar, trying to contract the film. Since the film has two surfaces, the total force pulling the bar inwards is $F_{tension} = \sigma \times (2L)$, where $L$ is the length of the bar.

To increase the area of the film by moving the bar outwards slowly by a distance $\Delta x$, an external force equal and opposite to $F_{tension}$ must be applied, $F_{ext} = 2L\sigma$. The work done by this external force is:

$ W = F_{ext} \Delta x = (2L\sigma) \Delta x $

The increase in the total surface area of the film is $\Delta A = 2L \Delta x$ (since there are two surfaces, each increasing by $L\Delta x$).

$ W = \sigma (2L \Delta x) = \sigma \Delta A $

This work done $W$ is stored as the increase in surface energy $\Delta U_{surface}$.

$ \Delta U_{surface} = \sigma \Delta A $

This means the surface energy added per unit area created is equal to $\sigma$. Thus, surface tension (force per unit length) is numerically equal to surface energy per unit area (J/m$^2$). Note that N/m = (N·m)/m$^2$ = J/m$^2$, confirming the units are consistent.

The units of surface tension can be expressed as both N/m and J/m$^2$. While N/m is often used when thinking of the surface as a stretched film, J/m$^2$ is perhaps a better reflection of the underlying molecular energy phenomenon.

Angle Of Contact

When a liquid surface meets a solid surface, the liquid surface is usually curved near the point of contact. The angle between the tangent to the liquid surface at the point of contact and the solid surface inside the liquid is called the angle of contact ($\theta_c$).

The angle of contact depends on the specific liquid-solid pair and the medium above the liquid (usually air). It is determined by the balance of three surface tensions acting along the line of contact:

$\sigma_{sl}$: Surface tension between solid and liquid.
$\sigma_{la}$: Surface tension between liquid and air (or vapour).
$\sigma_{sa}$: Surface tension between solid and air (or vapour).

At equilibrium, the forces must balance along the horizontal direction (tangential to the solid surface):

$ \sigma_{sa} = \sigma_{sl} + \sigma_{la} \cos\theta_c $

$ \cos\theta_c = \frac{\sigma_{sa} - \sigma_{sl}}{\sigma_{la}} $

The angle of contact determines whether a liquid "wets" a solid surface or not.

If $\sigma_{sa} > \sigma_{sl}$, $\cos\theta_c$ is positive, and $\theta_c < 90^\circ$. The liquid wets the solid. The liquid surface is concave upwards near the contact. This happens when the adhesive forces between liquid molecules and solid molecules are stronger than the cohesive forces between liquid molecules (e.g., water on clean glass).
If $\sigma_{sa} < \sigma_{sl}$, $\cos\theta_c$ is negative, and $\theta_c > 90^\circ$. The liquid does not wet the solid. The liquid surface is convex upwards near the contact. This happens when the cohesive forces between liquid molecules are stronger than the adhesive forces (e.g., mercury on glass, water on a lotus leaf or wax).
If $\theta_c = 90^\circ$, $\cos\theta_c = 0$, $\sigma_{sa} = \sigma_{sl}$.

The angle of contact is important in phenomena like capillarity and the formation of drops.

Diagram showing angle of contact for wetting and non-wetting liquids.

(Image Placeholder: Two diagrams. One shows a liquid column in a tube, curved downwards (concave meniscus), angle of contact less than 90 degrees, liquid rising up the wall. Another shows a liquid column (like mercury) in a tube, curved upwards (convex meniscus), angle of contact greater than 90 degrees, liquid pushed down near the wall.)

Drops And Bubbles

Surface tension plays a crucial role in the formation and properties of liquid drops and bubbles. A small liquid drop tends to be spherical because, for a given volume, a sphere has the minimum surface area, and the liquid tries to minimise its surface energy by minimising area.

Excess Pressure Inside a Liquid Drop

A spherical liquid drop experiences a net inward force due to surface tension, trying to shrink it. This inward force is balanced by a slightly higher pressure inside the drop compared to the outside pressure. This difference is called the excess pressure.

Consider a spherical drop of radius $R$ with surface tension $\sigma$. Let the pressure inside be $P_i$ and the pressure outside be $P_o$. The excess pressure is $\Delta P = P_i - P_o$.

Imagine increasing the radius of the drop from $R$ to $R+dR$. The work done by the excess pressure during this expansion is $dW = (P_i - P_o) \times (\text{increase in volume})$. The increase in volume is $dV = \frac{4}{3}\pi (R+dR)^3 - \frac{4}{3}\pi R^3 \approx \frac{4}{3}\pi (R^3 + 3R^2 dR) - \frac{4}{3}\pi R^3 = 4\pi R^2 dR$ (neglecting higher order terms in $dR$).

$ dW = (P_i - P_o) (4\pi R^2 dR) = \Delta P (4\pi R^2 dR) $

This work done by the excess pressure goes into increasing the surface energy. The original surface area is $A = 4\pi R^2$. The new surface area is $A' = 4\pi (R+dR)^2 \approx 4\pi (R^2 + 2RdR) = 4\pi R^2 + 8\pi RdR$. The increase in surface area is $\Delta A = A' - A \approx 8\pi RdR$.

The increase in surface energy is $dU_{surface} = \sigma \Delta A = \sigma (8\pi R dR)$.

By energy conservation, $dW = dU_{surface}$ (assuming a reversible, isothermal process):

$ \Delta P (4\pi R^2 dR) = \sigma (8\pi R dR) $

Divide both sides by $4\pi R^2 dR$:

$ \Delta P = \frac{8\pi R \sigma}{4\pi R^2} = \frac{2\sigma}{R} $

The excess pressure inside a spherical liquid drop of radius $R$ is:

$ \Delta P = P_i - P_o = \frac{2\sigma}{R} $

Excess Pressure Inside a Soap Bubble

A soap bubble has two free surfaces (an inner surface and an outer surface) in contact with air. When a bubble is expanded, both surfaces increase in area.

Repeating the work-energy calculation for a bubble, the increase in total surface area when the radius increases from $R$ to $R+dR$ is $\Delta A = 2 \times (8\pi R dR) = 16\pi R dR$ (because there are two surfaces). The increase in surface energy is $dU_{surface} = \sigma \Delta A = \sigma (16\pi R dR)$.

The work done by the excess pressure is the same, $dW = \Delta P (4\pi R^2 dR)$.

Equating work and energy change, $dW = dU_{surface}$:

$ \Delta P (4\pi R^2 dR) = \sigma (16\pi R dR) $

Divide by $4\pi R^2 dR$:

$ \Delta P = \frac{16\pi R \sigma}{4\pi R^2} = \frac{4\sigma}{R} $

The excess pressure inside a spherical soap bubble of radius $R$ is:

$ \Delta P = P_i - P_o = \frac{4\sigma}{R} $

The excess pressure in a bubble is twice that in a drop of the same radius because it has two surfaces. For a given liquid and surrounding medium, smaller drops and bubbles have larger excess pressures.

Example 1. Calculate the excess pressure inside a water droplet of radius 1 mm. The surface tension of water is $72 \times 10^{-3}$ N/m.

Answer:

Radius of the water droplet, $R = 1$ mm $= 1 \times 10^{-3}$ m.

Surface tension of water, $\sigma = 72 \times 10^{-3}$ N/m.

For a liquid drop, the excess pressure is $\Delta P = \frac{2\sigma}{R}$.

$ \Delta P = \frac{2 \times (72 \times 10^{-3} \text{ N/m})}{1 \times 10^{-3} \text{ m}} $

$ \Delta P = \frac{144 \times 10^{-3}}{1 \times 10^{-3}} $ Pa

$ \Delta P = 144 $ Pa.

The excess pressure inside the water droplet is 144 Pascals.

Capillary Rise

When a narrow tube (called a capillary tube) is dipped into a liquid, the liquid surface inside the tube is usually observed to be higher or lower than the liquid surface outside the tube. This phenomenon is called capillarity or capillary action.

If the liquid wets the material of the tube (angle of contact $\theta_c < 90^\circ$, like water in a glass tube), the liquid rises in the capillary tube. If the liquid does not wet the tube (angle of contact $\theta_c > 90^\circ$, like mercury in a glass tube), the liquid level falls in the capillary tube.

Explanation of Capillary Rise (or Fall)

Capillarity is due to the combined effects of surface tension and the angle of contact. When a wetting liquid is in a capillary tube, the liquid surface forms a concave meniscus. Surface tension acts along the circumference of this meniscus, tangential to the liquid surface at the wall. The vertical component of this surface tension force pulls the liquid column upwards.

Consider a capillary tube of radius $r$ dipped vertically into a liquid of density $\rho$ and surface tension $\sigma$. Let the liquid rise to a height $h$ in the tube, forming a concave meniscus with an angle of contact $\theta_c$. The circumference of the meniscus where it touches the tube wall is $2\pi r$. The surface tension force acting tangentially upwards along this circumference is $\sigma$. The total upward force due to surface tension is the vertical component:

$ F_{up} = (\sigma \cos\theta_c) \times (2\pi r) $

This upward force is balanced by the weight of the liquid column in the tube above the outside level. The volume of the liquid column is approximately the volume of the cylinder of height $h$ and radius $r$, $V = \pi r^2 h$. (We neglect the small volume of the meniscus itself for simplicity). The mass of this column is $m = \rho V = \rho \pi r^2 h$. The weight of the liquid column is $W = mg = \rho \pi r^2 h g$.

In equilibrium, the upward force balances the downward force:

$ F_{up} = W $

$ 2\pi r \sigma \cos\theta_c = \rho \pi r^2 h g $

Solving for the height of the capillary rise $h$:

$ h = \frac{2\pi r \sigma \cos\theta_c}{\rho \pi r^2 g} = \frac{2 \sigma \cos\theta_c}{\rho r g} $

This is the formula for capillary rise. Key points:

Capillary rise $h$ is inversely proportional to the radius $r$ of the tube. Narrower tubes result in higher capillary rise.
$h$ is directly proportional to the surface tension $\sigma$.
$h$ is inversely proportional to the density $\rho$ of the liquid.
$h$ depends on the angle of contact $\theta_c$. If $\theta_c < 90^\circ$ (wetting liquid, $\cos\theta_c > 0$), $h$ is positive, and the liquid rises. If $\theta_c > 90^\circ$ (non-wetting liquid, $\cos\theta_c < 0$), $h$ is negative, meaning the liquid level falls below the outside level.

Capillary action is important in many natural phenomena and applications, such as the rise of water in plants through narrow vessels, the absorption of ink by blotting paper, and the flow of wax in a candlewick.

Example 2. Water rises to a height of 2.0 cm in a capillary tube of radius 0.4 mm. If the surface tension of water is $7.2 \times 10^{-2}$ N/m and the angle of contact for water with glass is $0^\circ$, find the density of water. (Take $g = 9.8 \, \text{m/s}^2$).

Answer:

Capillary rise, $h = 2.0$ cm $= 2.0 \times 10^{-2}$ m.

Radius of tube, $r = 0.4$ mm $= 0.4 \times 10^{-3}$ m.

Surface tension, $\sigma = 7.2 \times 10^{-2}$ N/m.

Angle of contact, $\theta_c = 0^\circ$. $\cos\theta_c = \cos(0^\circ) = 1$.

$g = 9.8 \, \text{m/s}^2$.

Using the formula for capillary rise, $ h = \frac{2 \sigma \cos\theta_c}{\rho r g} $. We need to find $\rho$. Rearrange the formula for $\rho$:

$ \rho = \frac{2 \sigma \cos\theta_c}{h r g} $

$ \rho = \frac{2 \times (7.2 \times 10^{-2} \text{ N/m}) \times 1}{(2.0 \times 10^{-2} \text{ m}) \times (0.4 \times 10^{-3} \text{ m}) \times (9.8 \text{ m/s}^2)} $

$ \rho = \frac{14.4 \times 10^{-2}}{(2.0 \times 0.4) \times 10^{-5} \times 9.8} $ kg/m$^3$ (Units: $\frac{\text{N/m}}{\text{m} \cdot \text{m} \cdot \text{m/s}^2} = \frac{\text{kg} \cdot \text{m/s}^2 \cdot \text{m}}{\text{m}^3 \cdot \text{m/s}^2} = \text{kg/m}^3$)

$ \rho = \frac{14.4 \times 10^{-2}}{0.8 \times 9.8 \times 10^{-5}} = \frac{14.4 \times 10^{-2}}{7.84 \times 10^{-5}} $

$ \rho = \frac{14.4}{7.84} \times 10^{(-2 - (-5))} = \frac{14.4}{7.84} \times 10^3 $

$ \rho \approx 1.8367 \times 10^3 $ kg/m$^3$.

The calculated density of water is approximately 1837 kg/m$^3$. This is significantly higher than the expected value of 1000 kg/m$^3$. Let's recheck the calculation or assumption. Ah, the force balance method assumes the meniscus is part of a sphere. A more accurate analysis shows it's slightly different. Let's re-evaluate the calculation itself.

$ \rho = \frac{14.4 \times 10^{-2}}{7.84 \times 10^{-5}} = 1.8367 \times 10^3 $ looks numerically correct based on the numbers. Maybe the input values are not standard? Let's assume $\rho_{water} = 1000 \, \text{kg/m}^3$ and check consistency with other values.

If $\rho=1000$, $h = \frac{2 \sigma \cos\theta_c}{\rho r g} = \frac{2 \times 7.2 \times 10^{-2} \times 1}{1000 \times 0.4 \times 10^{-3} \times 9.8} = \frac{14.4 \times 10^{-2}}{400 \times 10^{-3} \times 9.8} = \frac{0.144}{0.4 \times 9.8} = \frac{0.144}{3.92} \approx 0.0367$ m = 3.67 cm. The given $h$ is 2 cm. So, the input values are not perfectly consistent with the standard density of water and the formula, or there might be some experimental error assumed in the problem statement. Let's trust the formula and the given values to calculate $\rho$. The calculation $\rho \approx 1837 \, \text{kg/m}^3$ is mathematically correct for the given inputs. However, this value of water density is unusual. It's possible the problem intends to work backward from a standard water density, or the radius/height values are unrealistic for these properties.

Let's proceed with the calculated value based on the given numbers, but acknowledge it's not the standard density of water.

The density of the liquid (water as stated) is approximately 1837 kg/m$^3$ based on the provided data.

Detergents And Surface Tension

Detergents and soaps are substances that significantly affect the surface tension of water. They are examples of surfactants (surface-active agents).

When a detergent is added to water, the surface tension of the water decreases. This property is crucial for the cleaning action of detergents.

How it helps in cleaning:

Wetting Power: Pure water has relatively high surface tension, which can prevent it from easily wetting hydrophobic surfaces like oily dirt or fabric fibres. Detergents reduce the surface tension, allowing the water to spread out and penetrate the fabric and surround the dirt particles more effectively. The angle of contact between water and dirt/fabric decreases, increasing the wetting.
Emulsification: Detergents help in emulsifying grease and oil (which are non-polar) in water (which is polar). Detergent molecules have a hydrophilic (water-attracting) head and a hydrophobic (oil/grease-attracting) tail. The tails attach to the oil droplets, while the heads remain in the water, forming micelles around the oil. This reduces the interfacial tension between oil and water and keeps the oil suspended as tiny droplets that can be washed away by the water.
Lifting Dirt: By reducing surface tension, the water can flow into small crevices and under dirt particles. The detergent molecules can then surround the dirt, helping to lift it away from the surface.

In essence, detergents make water "wetter" and better able to interact with and remove dirt, particularly oily and greasy stains, by lowering its surface tension and aiding emulsification and dispersion.