Cohesion and Adhesion in Liquids: Surface Tension and Capillary Action, Fluid Statics | College Physics 2e

Learning objectives

Learning Objectives

By the end of this section, you will be able to:

Understand cohesive and adhesive forces.
Define surface tension.
Understand capillary action.

Cohesion and Adhesion in Liquids

Children blow soap bubbles and play in the spray of a sprinkler on a hot summer day. (See import-auto-id2058131.) An underwater spider keeps his air supply in a shiny bubble he carries wrapped around him. A technician draws blood into a small-diameter tube just by touching it to a drop on a pricked finger. A premature infant struggles to inflate her lungs. What is the common thread? All these activities are dominated by the attractive forces between atoms and molecules in liquids—both within a liquid and between the liquid and its surroundings.

Attractive forces between molecules of the same type are called cohesive forces. Liquids can, for example, be held in open containers because cohesive forces hold the molecules together. Attractive forces between molecules of different types are called adhesive forces. Such forces cause liquid drops to cling to window panes, for example. In this section we examine effects directly attributable to cohesive and adhesive forces in liquids.

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Original figure description: The soap bubbles that the child blows into the air maintain their shape because of the attractive force between the molecules of the soap bubble.

The soap bubbles in this photograph are caused by cohesive forces among molecules in liquids. (credit: Steve Ford Elliott)

Surface Tension

Cohesive forces between molecules cause the surface of a liquid to contract to the smallest possible surface area. This general effect is called surface tension. Molecules on the surface are pulled inward by cohesive forces, reducing the surface area. Molecules inside the liquid experience zero net force, since they have neighbors on all sides.

The model of a liquid surface acting like a stretched elastic sheet can effectively explain surface tension effects. For example, some insects can walk on water (as opposed to floating in it) as we would walk on a trampoline—they dent the surface as shown in import-auto-id2970796(a). import-auto-id2970796(b) shows another example, where a needle rests on a water surface. The iron needle cannot, and does not, float, because its density is greater than that of water. Rather, its weight is supported by forces in the stretched surface that try to make the surface smaller or flatter. If the needle were placed point down on the surface, its weight acting on a smaller area would break the surface, and it would sink.

View original figure (external; image not reproduced in this text edition)

Original figure description: A leg of an insect resting on the water surface is shown in the first figure. In the second figure an iron needle rests on the surface of water without sinking. Both are possible due to the tension on the surface of the liquid.

Surface tension supporting the weight of an insect and an iron needle, both of which rest on the surface without penetrating it. They are not floating; rather, they are supported by the surface of the liquid. (a) An insect leg dents the water surface.

F_{ST}

is a restoring force (surface tension) parallel to the surface. (b) An iron needle similarly dents a water surface until the restoring force (surface tension) grows to equal its weight.

Surface tension is proportional to the strength of the cohesive force, which varies with the type of liquid. Surface tension $γ$ is defined to be the force F per unit length $L$ exerted by a stretched liquid membrane:

γ = \frac{F}{L} .

import-auto-id1816998 lists values of $γ$ for some liquids. For the insect of import-auto-id2970796(a), its weight $w$ is supported by the upward components of the surface tension force: $w = γL sin θ$ , where $L$ is the circumference of the insect’s foot in contact with the water. import-auto-id3306680 shows one way to measure surface tension. The liquid film exerts a force on the movable wire in an attempt to reduce its surface area. The magnitude of this force depends on the surface tension of the liquid and can be measured accurately.

Surface tension is the reason why liquids form bubbles and droplets. The inward surface tension force causes bubbles to be approximately spherical and raises the pressure of the gas trapped inside relative to atmospheric pressure outside. It can be shown that the gauge pressure $P$ inside a spherical bubble is given by

P = \frac{4γ}{r},

where $r$ is the radius of the bubble. Thus the pressure inside a bubble is greatest when the bubble is the smallest. Another bit of evidence for this is illustrated in import-auto-id2054965. When air is allowed to flow between two balloons of unequal size, the smaller balloon tends to collapse, filling the larger balloon.

View original figure (external; image not reproduced in this text edition)

Original figure description: Sliding wire device which is used to measure surface tension shows the force exerted on the two surfaces of the liquid. This force remains a constant until the film’s breaking point.

Sliding wire device used for measuring surface tension; the device exerts a force to reduce the film’s surface area. The force needed to hold the wire in place is

F = γL = γ (2 l)

, since there are two liquid surfaces attached to the wire. This force remains nearly constant as the film is stretched, until the film approaches its breaking point.

View original figure (external; image not reproduced in this text edition)

Original figure description: When two balloons are attached to the ends of a glass tube air flows from one to the other if their sizes are different.

With the valve closed, two balloons of different sizes are attached to each end of a tube. Upon opening the valve, the smaller balloon decreases in size with the air moving to fill the larger balloon. The pressure in a spherical balloon is inversely proportional to its radius, so that the smaller balloon has a greater internal pressure than the larger balloon, resulting in this flow.

Surface Tension of Some LiquidsAt 20ºC unless otherwise stated.
Liquid	Surface tension γ(N/m)
Water at $0 º C$	0.0756
Water at $20 º C$	0.0728
Water at $100 º C$	0.0589
Soapy water (typical)	0.0370
Ethyl alcohol	0.0223
Glycerin	0.0631
Mercury	0.465
Olive oil	0.032
Tissue fluids (typical)	0.050
Blood, whole at $37 º C$	0.058
Blood plasma at $37 º C$	0.073
Gold at $1070 º C$	1.000
Oxygen at $- 193 º C$	0.0157
Helium at $- 269 º C$	0.00012

Surface Tension: Pressure Inside a Bubble

Calculate the gauge pressure inside a soap bubble $2 . 00 \times 10^{- 4} m$ in radius using the surface tension for soapy water in import-auto-id1816998. Convert this pressure to mm Hg.

Strategy

The radius is given and the surface tension can be found in import-auto-id1816998, and so $P$ can be found directly from the equation $P = \frac{4γ}{r}$ .

Solution

Substituting $r$ and $γ$ into the equation $P = \frac{4γ}{r}$ , we obtain

P = \frac{4γ}{r} = \frac{4 (0.037 N/m)}{2 . 00 \times 10^{- 4} m} = 740 {N/m}^{2} = 740 Pa .

We use a conversion factor to get this into units of mm Hg:

P = (740 {N/m}^{2}) \frac{1.00 mm Hg}{133 {N/m}^{2}} = 5.56 mm Hg .

Discussion

Note that if a hole were to be made in the bubble, the air would be forced out, the bubble would decrease in radius, and the gauge pressure would reduce to zero, and the absolute pressure inside would decrease to atmospheric pressure (760 mm Hg).

Our lungs contain hundreds of millions of mucus-lined sacs called alveoli, which are very similar in size, and about 0.1 mm in diameter. (See import-auto-id774766.) You can exhale without muscle action by allowing surface tension to contract these sacs. Medical patients whose breathing is aided by a positive pressure respirator have air blown into the lungs, but are generally allowed to exhale on their own. Even if there is paralysis, surface tension in the alveoli will expel air from the lungs. Since pressure increases as the radii of the alveoli decrease, an occasional deep cleansing breath is needed to fully reinflate the alveoli. Respirators are programmed to do this and we find it natural, as do our companion dogs and cats, to take a cleansing breath before settling into a nap.

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Original figure description: This image shows the bronchioles and alveolar sacs in the lungs and depicts the exchange of oxygenated and deoxygenated blood in the pulmonary blood vessels.

Bronchial tubes in the lungs branch into ever-smaller structures, finally ending in alveoli. The alveoli act like tiny bubbles. The surface tension of their mucous lining aids in exhalation and can prevent inhalation if too great.

The tension in the walls of the alveoli results from the membrane tissue and a liquid on the walls of the alveoli containing a long lipoprotein that acts as a surfactant (a surface-tension reducing substance). The need for the surfactant results from the tendency of small alveoli to collapse and the air to fill into the larger alveoli making them even larger (as demonstrated in import-auto-id2054965). During inhalation, the lipoprotein molecules are pulled apart and the wall tension increases as the radius increases (increased surface tension). During exhalation, the molecules slide back together and the surface tension decreases, helping to prevent a collapse of the alveoli. The surfactant therefore serves to change the wall tension so that small alveoli don’t collapse and large alveoli are prevented from expanding too much. This tension change is a unique property of these surfactants, and is not shared by detergents (which simply lower surface tension). (See import-auto-id2971179.)

View original figure (external; image not reproduced in this text edition)

Original figure description: Graph of surface tension as a function of surface area for detergents and interstitial fluids.

Surface tension as a function of surface area. The surface tension for lung surfactant decreases with decreasing area. This ensures that small alveoli don’t collapse and large alveoli are not able to over expand.

If water gets into the lungs, the surface tension is too great and you cannot inhale. This is a severe problem in resuscitating drowning victims. A similar problem occurs in newborn infants who are born without this surfactant—their lungs are very difficult to inflate. This condition is known as hyaline membrane disease and is a leading cause of death for infants, particularly in premature births. Some success has been achieved in treating hyaline membrane disease by spraying a surfactant into the infant’s breathing passages. Emphysema produces the opposite problem with alveoli. Alveolar walls of emphysema victims deteriorate, and the sacs combine to form larger sacs. Because pressure produced by surface tension decreases with increasing radius, these larger sacs produce smaller pressure, reducing the ability of emphysema victims to exhale. A common test for emphysema is to measure the pressure and volume of air that can be exhaled.

Making Connections: Take-Home Investigation

(1) Try floating a sewing needle on water. In order for this activity to work, the needle needs to be very clean as even the oil from your fingers can be sufficient to affect the surface properties of the needle. (2) Place the bristles of a paint brush into water. Pull the brush out and notice that for a short while, the bristles will stick together. The surface tension of the water surrounding the bristles is sufficient to hold the bristles together. As the bristles dry out, the surface tension effect dissipates. (3) Place a loop of thread on the surface of still water in such a way that all of the thread is in contact with the water. Note the shape of the loop. Now place a drop of detergent into the middle of the loop. What happens to the shape of the loop? Why? (4) Sprinkle pepper onto the surface of water. Add a drop of detergent. What happens? Why? (5) Float two matches parallel to each other and add a drop of detergent between them. What happens? Note: For each new experiment, the water needs to be replaced and the bowl washed to free it of any residual detergent.

Adhesion and Capillary Action

Why is it that water beads up on a waxed car but does not on bare paint? The answer is that the adhesive forces between water and wax are much smaller than those between water and paint. Competition between the forces of adhesion and cohesion are important in the macroscopic behavior of liquids. An important factor in studying the roles of these two forces is the angle $θ$ between the tangent to the liquid surface and the surface. (See import-auto-id1439052.) The contact angle $θ$ is directly related to the relative strength of the cohesive and adhesive forces. The larger the strength of the cohesive force relative to the adhesive force, the larger $θ$ is, and the more the liquid tends to form a droplet. The smaller $θ$ is, the smaller the relative strength, so that the adhesive force is able to flatten the drop. import-auto-id3245851 lists contact angles for several combinations of liquids and solids.

View original figure (external; image not reproduced in this text edition)

Original figure description: Water is seen to make beads on the waxed surface of car paint and it remains flat on the surface without wax. The beads are due to the greater force of attraction between the water molecules than between the water molecules and the surface. On the surface without wax the force of attraction between the water molecules and paint is greater.

In the photograph, water beads on the waxed car paint and flattens on the unwaxed paint. (a) Water forms beads on the waxed surface because the cohesive forces responsible for surface tension are larger than the adhesive forces, which tend to flatten the drop. (b) Water beads on bare paint are flattened considerably because the adhesive forces between water and paint are strong, overcoming surface tension. The contact angle

θ

is directly related to the relative strengths of the cohesive and adhesive forces. The larger

θ

is, the larger the ratio of cohesive to adhesive forces. (credit: P. P. Urone)

One important phenomenon related to the relative strength of cohesive and adhesive forces is capillary action—the tendency of a fluid to be raised or suppressed in a narrow tube, or capillary tube. This action causes blood to be drawn into a small-diameter tube when the tube touches a drop.

If a capillary tube is placed vertically into a liquid, as shown in fs-id1333088, capillary action will raise or suppress the liquid inside the tube depending on the combination of substances. The actual effect depends on the relative strength of the cohesive and adhesive forces and, thus, the contact angle $θ$ given in the table. If $θ$ is less than $90º$ , then the fluid will be raised; if $θ$ is greater than $90º$ , it will be suppressed. Mercury, for example, has a very large surface tension and a large contact angle with glass. When placed in a tube, the surface of a column of mercury curves downward, somewhat like a drop. The curved surface of a fluid in a tube is called a meniscus. The tendency of surface tension is always to reduce the surface area. Surface tension thus flattens the curved liquid surface in a capillary tube. This results in a downward force in mercury and an upward force in water, as seen in fs-id1333088.

View original figure (external; image not reproduced in this text edition)

Original figure description: Mercury kept in a container into which a narrow tube is inserted lowers its level inside the tube relative to the level in the rest of the container. In a similar situation, water rises in the tube so that the water level in the tube is above the water level in the rest of the container. This phenomenon is due to the large contact angle of mercury with glass and the smaller contact angle of water with glass.

(a) Mercury is suppressed in a glass tube because its contact angle is greater than

90º

. Surface tension exerts a downward force as it flattens the mercury, suppressing it in the tube. The dashed line shows the shape the mercury surface would have without the flattening effect of surface tension. (b) Water is raised in a glass tube because its contact angle is nearly

0º

. Surface tension therefore exerts an upward force when it flattens the surface to reduce its area.

Contact Angles of Some Substances
Interface	Contact angle Θ
Mercury–glass	$140 º$
Water–glass	$0 º$
Water–paraffin	$107 º$
Water–silver	$90 º$
Organic liquids (most)–glass	$0 º$
Ethyl alcohol–glass	$0 º$
Kerosene–glass	$26 º$

Capillary action can move liquids horizontally over very large distances, but the height to which it can raise or suppress a liquid in a tube is limited by its weight. It can be shown that this height $h$ is given by

h = \frac{2γ cos θ}{ρ gr} .

If we look at the different factors in this expression, we might see how it makes good sense. The height is directly proportional to the surface tension $γ$ , which is its direct cause. Furthermore, the height is inversely proportional to tube radius—the smaller the radius $r$ , the higher the fluid can be raised, since a smaller tube holds less mass. The height is also inversely proportional to fluid density $ρ$ , since a larger density means a greater mass in the same volume. (See import-auto-id2953655.)

View original figure (external; image not reproduced in this text edition)

Original figure description: The left image shows liquid in a container with four tubes of progressively smaller diameter inserted into the liquid. The liquid rises higher in the smaller-diameter tubes. The right image shows two containers, one holding a dense liquid and the other holding a less-dense liquid. Identical tubes are inserted into each liquid. The less-dense liquid rises higher in its tube than the more-dense liquid does in its tube.

(a) Capillary action depends on the radius of a tube. The smaller the tube, the greater the height reached. The height is negligible for large-radius tubes. (b) A denser fluid in the same tube rises to a smaller height, all other factors being the same.

Calculating Radius of a Capillary Tube: Capillary Action: Tree Sap

Can capillary action be solely responsible for sap rising in trees? To answer this question, calculate the radius of a capillary tube that would raise sap 100 m to the top of a giant redwood, assuming that sap’s density is $1050 kg {/m}^{3}$ , its contact angle is zero, and its surface tension is the same as that of water at $20.0º C$ .

Strategy

The height to which a liquid will rise as a result of capillary action is given by $h = \frac{2γ cos θ}{ρ gr}$ , and every quantity is known except for $r$ .

Solution

Solving for $r$ and substituting known values produces

\begin{matrix} r & = & \frac{2γ cos θ}{ρ gh} = \frac{2 (0.0728 N/m) cos (0º)}{(1050 {kg/m}^{3}) (9.80 {m/s}^{2}) (100 m)} \\ = & 1.41 \times 10^{- 7} m. \end{matrix}

Discussion

This result is unreasonable. Sap in trees moves through the xylem, which forms tubes with radii as small as $2 . 5 \times 10^{- 5} m$ . This value is about 180 times as large as the radius found necessary here to raise sap $100 m$ . This means that capillary action alone cannot be solely responsible for sap getting to the tops of trees.

How does sap get to the tops of tall trees? (Recall that a column of water can only rise to a height of 10 m when there is a vacuum at the top—see fs-id2937185.) The question has not been completely resolved, but it appears that it is pulled up like a chain held together by cohesive forces. As each molecule of sap enters a leaf and evaporates (a process called transpiration), the entire chain is pulled up a notch. So a negative pressure created by water evaporation must be present to pull the sap up through the xylem vessels. In most situations, fluids can push but can exert only negligible pull, because the cohesive forces seem to be too small to hold the molecules tightly together. But in this case, the cohesive force of water molecules provides a very strong pull. import-auto-id2051381 shows one device for studying negative pressure. Some experiments have demonstrated that negative pressures sufficient to pull sap to the tops of the tallest trees can be achieved.

View original figure (external; image not reproduced in this text edition)

Original figure description: When the piston is raised the liquid stretches somewhat, which results in negative pressure.

(a) When the piston is raised, it stretches the liquid slightly, putting it under tension and creating a negative absolute pressure

P = - F / A

. (b) The liquid eventually separates, giving an experimental limit to negative pressure in this liquid.

Section Summary

Attractive forces between molecules of the same type are called cohesive forces.
Attractive forces between molecules of different types are called adhesive forces.
Cohesive forces between molecules cause the surface of a liquid to contract to the smallest possible surface area. This general effect is called surface tension.
Capillary action is the tendency of a fluid to be raised or suppressed in a narrow tube, or capillary tube which is due to the relative strength of cohesive and adhesive forces.

Conceptual Questions

The density of oil is less than that of water, yet a loaded oil tanker sits lower in the water than an empty one. Why?

Is surface tension due to cohesive or adhesive forces, or both?

Is capillary action due to cohesive or adhesive forces, or both?

Birds such as ducks, geese, and swans have greater densities than water, yet they are able to sit on its surface. Explain this ability, noting that water does not wet their feathers and that they cannot sit on soapy water.

Water beads up on an oily sunbather, but not on her neighbor, whose skin is not oiled. Explain in terms of cohesive and adhesive forces.

Could capillary action be used to move fluids in a “weightless” environment, such as in an orbiting space probe?

What effect does capillary action have on the reading of a manometer with uniform diameter? Explain your answer.

Pressure between the inside chest wall and the outside of the lungs normally remains negative. Explain how pressure inside the lungs can become positive (to cause exhalation) without muscle action.

Problems & Exercises

What is the pressure inside an alveolus having a radius of $2 . 50 \times 10^{- 4} m$ if the surface tension of the fluid-lined wall is the same as for soapy water? You may assume the pressure is the same as that created by a spherical bubble.

$592 {N/m}^{2}$

(a) The pressure inside an alveolus with a $2 . 00 \times 10^{- 4}$ -m radius is $1 . 40 \times 10^{3} Pa$ , due to its fluid-lined walls. Assuming the alveolus acts like a spherical bubble, what is the surface tension of the fluid? (b) Identify the likely fluid. (You may need to extrapolate between values in import-auto-id1816998.)

What is the gauge pressure in millimeters of mercury inside a soap bubble 0.100 m in diameter?

$2 . 23 \times 10^{- 2} mm Hg$

Calculate the force on the slide wire in import-auto-id3306680 if it is 3.50 cm long and the fluid is ethyl alcohol.

import-auto-id2953655(a) shows the effect of tube radius on the height to which capillary action can raise a fluid. (a) Calculate the height $h$ for water in a glass tube with a radius of 0.900 cm—a rather large tube like the one on the left. (b) What is the radius of the glass tube on the right if it raises water to 4.00 cm?

(a) $1 . 65 \times 10^{- 3} m$

(b) $3 . 71 \times 10^{–4} m$

We stated in fs-id2066941 that a xylem tube is of radius $2 . 50 \times 10^{- 5} m$ . Verify that such a tube raises sap less than a meter by finding $h$ for it, making the same assumptions that sap’s density is $1050 {kg/m}^{3}$ , its contact angle is zero, and its surface tension is the same as that of water at $20.0º C$ .

What fluid is in the device shown in import-auto-id3306680 if the force is $3 . 16 \times 10^{- 3} N$ and the length of the wire is 2.50 cm? Calculate the surface tension $γ$ and find a likely match from import-auto-id1816998.

$6 . 32 \times 10^{- 2} N/m$

Based on the values in table, the fluid is probably glycerin.

If the gauge pressure inside a rubber balloon with a 10.0-cm radius is 1.50 cm of water, what is the effective surface tension of the balloon?

Calculate the gauge pressures inside 2.00-cm-radius bubbles of water, alcohol, and soapy water. Which liquid forms the most stable bubbles, neglecting any effects of evaporation?

$\begin{matrix} P_{w} & = & 14.6 {N/m}^{2}, \\ P_{a} & = & 4.46 {N/m}^{2}, \\ P_{sw} & = & 7.40 {N/m}^{2} . \end{matrix}$

Alcohol forms the most stable bubble, since the absolute pressure inside is closest to atmospheric pressure.

Suppose water is raised by capillary action to a height of 5.00 cm in a glass tube. (a) To what height will it be raised in a paraffin tube of the same radius? (b) In a silver tube of the same radius?

Calculate the contact angle $θ$ for olive oil if capillary action raises it to a height of 7.07 cm in a glass tube with a radius of 0.100 mm. Is this value consistent with that for most organic liquids?

$5.1º$

This is near the value of $θ = 0º$ for most organic liquids.

When two soap bubbles touch, the larger is inflated by the smaller until they form a single bubble. (a) What is the gauge pressure inside a soap bubble with a 1.50-cm radius? (b) Inside a 4.00-cm-radius soap bubble? (c) Inside the single bubble they form if no air is lost when they touch?

Calculate the ratio of the heights to which water and mercury are raised by capillary action in the same glass tube.

$- 2 . 78$

The ratio is negative because water is raised whereas mercury is lowered.

What is the ratio of heights to which ethyl alcohol and water are raised by capillary action in the same glass tube?

Glossary

adhesive forces: the attractive forces between molecules of different types
capillary action: the tendency of a fluid to be raised or lowered in a narrow tube
cohesive forces: the attractive forces between molecules of the same type
contact angle: the angle $θ$ between the tangent to the liquid surface and the surface
surface tension: the cohesive forces between molecules which cause the surface of a liquid to contract to the smallest possible surface area

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