Surface Tension Notes
Surface Tension Notes
Surface Tension Notes
Definition: Surface tension (𝛾) is the tangential force acting normally on one side of an
imaginary line of length one meter drawn in the surface of a liquid
Surface tension
Force
Length
MT 2
1. A drop of water may remain clinging on to the tap for some time before falling as if the
water particles were held in a bag.
2. Mercury forms spherical droplets when spilt on glass.
3. Rain drops are spherical due to surface tension.
4. Formation of soap bubbles and films. Consider a thread placed on a soap film which is
supported by a ring as shown below.
However if the soap film is broken in the middle of the loop in the region bounded by the
thread, there are no more forces inside the thread, and so the only forces acting on it are
outward.
5. Some insects such as pond skaters are able to walk over the surface of the water without
getting wet.
6. A steel needle can be made to float on water despite its greater density.
The floating needle creates a depression in the liquid surface so that the surface tensional forces,
F which act in the surface now have upward directed components which are capable of
supporting the weight of the needle.
Molecules in the bulk of the liquid such as D and E are surrounded by equal number of
molecules on all sides. As a result they are attracted equally in all directions by the surrounding
molecules. The average distance of these molecules is such that attractive and repulsive forces
on such molecules balance. So the net force on these molecules is equal to zero.
Molecules in the liquid surface such as A, B and C have fewer vapour molecules above than the
liquid molecules below. The resultant attractive force on such molecules acts downwards.
On the other hand, molecules in the liquid surface are more widely spaced than the molecules in
the bulk. Since the intermolecular attraction increases with separation between the molecules,
the molecules in the surface attract each other with a greater force. This puts the surface
molecules in a state of tension hence behave like an elastic skin a phenomenon called surface
tension.
In terms of energy
Molecules in the liquid surface have potential energy because if they were to be moved to
infinity, a definite amount of work is needed to overcome the net inward force on the molecules
due to those below them.
Molecules in the bulk of the liquid form bonds with more neighbours than those in the surface.
Work must be done to break these bonds to bring such a molecule into the surface.
Surfactants
These are substances that reduce the surface tension e.g impurities, detergents and temperature.
Temperature. When the temperature of a liquid increases, the mean kinetic energy of the
molecules of the liquid increases. The forces of attraction between the molecules will decrease
since the molecules spend less time in the neighborhood of a given molecule hence surface
tension decreases with increase in temperature.
N.B: At the boiling point, the surface tension of the liquid becomes zero and maximum at the
freezing point.
Impurities. The molecules of the impurity get in the spaces between the liquid molecules and
this reduces the cohesive forces between the liquid molecules. This reduces the surface tension
of the liquid.
Calmness of the liquid surface. When the liquid surface is calm, the surface tension is high due
to high intermolecular attraction whereas when the liquid surface becomes turbulent, the
intermolecular forces are reduced and thus surface tension.
Procedure
Lycopodium powder or some light dust is sprinkled on the surface of water in a flat metal dish.
One side of the dish is heated for some time and observations are made.
It is observed that the particles of the powder are swept away from the heated part, implying
that the surface tensional forces can longer hold the particles in their previous positions.
This shows that surface tension of liquids decreases with increase in temperature.
Explain why when a drop of methylated spirit (soap solution) is dropped into the centre of
a dish of water whose surface has been sprinkled with lycopodium powder, the powder
rushed out to the sides leaving a clear patch.
This is due to the surface tension of water being greater than that of methylated spirit (soap
solution) causing an imbalance between the surface tension forces at the boundary of the two
liquids. The powder is thus carried away from the centre by water.
ANGLE OF CONTACT
𝑀𝑒𝑟𝑐𝑢𝑟𝑦
𝑊𝑎𝑡𝑒𝑟
𝛉
𝛉
Angle of contact:
This is the angle between the solid surface and the tangent plane at the point of intersection of
the liquid surface and the solid surface measured through the liquid.
(OR)
When water is poured in a clean beaker or a capillary tube, the meniscus curves upwards
(concave), whereas for mercury the meniscus curves downwards (convex).
For water, is acute, i.e. less than 90 0 . This is due to the fact that the adhesion forces between
the liquid and the solid surface are greater than the cohesion forces between the liquid
molecules.
For mercury, is obtuse, i.e. 900 < 𝜃 < 1800 . This due to the fact that the cohesion forces
between the liquid molecules are greater than the adhesion forces between the liquid molecules
and the solid surface.
NOTE:
𝑊𝑎𝑡𝑒𝑟 Mercury
A clean glass plate with a provision of adjusting its angle of inclination on the liquid surface is
dipped into the liquid under investigation.
The glass plate is tilted until the liquid surface on one side of the plate is horizontal up to the
line of contact.
The angle, 𝛼 between the glass plate and the liquid surface is measured by means of a
protractor, suitably placed against the edge of the plate.
The angle of contact, is obtained from 𝜃 = 180 − 𝛼
CAPILLARITY
When a capillary tube is immersed in a beaker with water, the water rises in the tube to a height
above the surface due to surface tension. The narrower the tube, the greater the height. This is
due to the fact that the adhesive forces between water molecules and glass molecules are greater
than the cohesive forces between the water molecules. The water therefore rises up the tube so
that more water molecules are in contact with the glass, and a concave meniscus is formed.
When on the other hand the capillary tube is placed inside mercury, the liquid is depressed
below the outside mercury level. The depression decreases as the diameter of the capillary tube
increases.
This is because the cohesive forces between mercury molecules are greater than the adhesive
forces between the mercury and the glass molecules. Mercury therefore sinks down the tube
such that more mercury molecules remain together, and a convex meniscus is formed.
F
At equilibrium, Fcos mg but m V and F 2r
2r
2r cos r 2 h g
2 cos rhg
2 cos
h
rg
Note: if 𝜃 is acute, then cos 𝜃 is positive and therefore h is positive, implying that the liquid
rises up in the tube. If 𝜃 is obtuse, 𝑐𝑜𝑠𝜃 is negative and therefore h is negative, implying that the
liquid falls in the capillary tube, below the level of the surrounding.
2𝛾
If 𝜃 = 00 like for water, ℎ = 𝜌𝑟𝑔
Examples
1. A capillary tube of diameter 0.4mm is placed vertically inside the liquid. Calculate the
height to which liquid rises in the tube if;
i) A liquid of density 800kgm 3 and surface tension 0.05Nm 1 angle of contact 300 ,
(i) 0.05Nm 1 , 30 0
2. The internal diameter of the glass tube of mercury barometer is 3.5mm . The barometer
reads 752.4mm . Find the correct reading of the barometer after allowing for the error due to
h
2
hgr
6.6 10 2 1000 9.81 0.0002
0.064746 Nm 1
gr 2 2
5. Water rises to a height of 10cm in a capillary tube dipped in water. When the same
capillary tube is dipped in mercury, it is depressed by 2.60mm. Compare the surface tension of
water and mercury. (Density of water= 1000𝑘𝑔𝑚−3 , Density of mercury = 13600𝑘𝑔𝑚 −3,
angle of contact of water= 50 , angle of contact of mercury= 1350 )
hw w gr
For water, w ............................(i)
2 cos w
hm m gr
For mercury, m ............................(ii)
2 cos m
hw w gr
w 2 cos w w h cos m
w w
m hm m gr m hm m cos w
2 cos m
A clean capillary tube which is supported by a retort stand is placed in a beaker containing a
clean liquid of known density, and angle of contact,
A pin bent at right angles at two places is attached to a capillary tube with a rubber band.
The pin is adjusted until its sharp point just touches the horizontal level of the liquid in the
beaker.
A travelling microscope is focused on the meniscus and the reading, S1 on the scale is
recorded.
The beaker is removed and the travelling microscope is focused on the tip of the pin. The scale
reading, S 2 is recorded.
grh
The surface tension , of the liquid is calculated from ; .
2cos
NOTE:
It is assumed that the weight of the small quantity of the liquid in the meniscus is negligible,
and that temperature remains constant throughout the experiment.
A bubble is a thin sphere of liquid enclosing/containing air or gas or vapour. E.g. air bubbles,
soap bubbles etc.…
N.B:
For bubbles and drops, the inside pressure is always greater than the outside pressure otherwise
the combined effect of the external pressure and surface tension would cause the bubble or drop
to collapse. (Pressure is greater on the concave side of the drop or bubble)
Pressure difference across an air bubble (Excess pressure within an air bubble)
In the figure shown F is the surface tensional force, P1 is the external pressure acting on the
P1 A 2r P2 A
P1 r 2 2r P2 r 2
2 P2 P1 r
P2 P1 2
r
2
Therefore pressure difference across an air bubble is P
r
A soap bubble has two liquid surfaces in contact with air, one inside the bubble and the other
outside the bubble.
P1 P2
P1 r 2 4r P2 r 2
P2 P1 r 4
P2 P1 4
r
4
Excess pressure, P
r
N.B:
The excess pressure of bubbles is inversely proportional to the radius of the bubble. This
explains why;
(i) The pressure needed to form a very small bubble is high.
(ii) One needs to blow hard to start a balloon growing. Once the balloon has grown, less air
pressure is needed to make it expand more.
Examples
7. (a) Define the term surface tension and derive its dimensions
(b) Explain using the molecular theory the occurrence of surface tension.
4
(c) Show that the excess pressure in a soap bubble of radius, r is given by P
r
8. Calculate the pressure inside a spherical air bubble of diameter 0.1cm blown at a depth
of 20cm below the surface of a liquid of density 1.26 10 3 kgm 3 and surface tension
0.064 Nm1 .(Given: Height of mercury barometer is 76cmHg and density of mercury is
1.36 10 4 kgm3 )
Pressure due to the liquid column 0.2 1.26 10 3 9.81 2472 .12Pa
2
Excess pressure, P2 P1
r
2 2 0.064
P2 P1 103868 .28
r 0.05 10 2
9. Calculate the total pressure within an air bubble of radius 0.1mm in water if the bubble is
formed 10cm below the water surface given that the surface tension of water is 7.27 10 2 Nm1 ,
density of water is 1000kgm 3 and atmospheric pressure is 1.01 10 5 Pa . 103435 Nm2
10. A soap bubble has a radius of 0.005m. Calculate the excess pressure in the bubble if the
surface tension of soap solution is 2.5 10 2 Nm1 20 Nm2
11. A soap bubble has a radius of 0.001m. Calculate the excess pressure in the bubble if the
surface tension of soap solution is 2.5 10 2 Nm2 . 10 Nm 2
12. A soap bubble has a diameter of 4mm. calculate the pressure inside it, if the atmospheric
pressure is 10 5 Nm2 and that the surface tension of soap solution is 2.8 10 2 Nm1 .
4 4
P2 P1 P2 P1
r r
P2 10 5
4 2.8 10 2
100056 Pa
2 10 3
Pressure difference across a spherical liquid surface
Consider the two situations as illustrated bellow.
The pressure on the concave side of each liquid surface exceeds the pressure on the convex side
2
by where, r is the radius of curvature of the surface provided the angle of contact is zero.
r
H hg H 2cos h
2cos
r gr
Pressure difference across a liquid meniscus in terms of angle of contact, and radius, r of
the capillary tube.
𝜽
𝑃2 𝐫
𝑃1
𝛉
𝛉
R
cos R rcos ............................(i)
r
2
P2 P1 ........................................................(ii)
R
2 cos
P2 P1
r
H H
H
𝐏𝟏 𝐏𝟐 𝐏𝟑
liquid
P1 P3 H
H H
𝐏𝟒
h
𝐏𝟏 𝐏𝟐 𝐏𝟑
liquid
At equilibrium,
P1 P2 P3 H..................................(i)
P2 P4 hg H P4 hg
P4 H hg .........................................(ii)
2 cos
Also, H P4 ..........................(iii )
r
2 cos 2 cos
H H hg hg
r r
2 cos
h
gr
1
h . The larger the capillary tube the smaller the rise and vice versa.
r
Consider a capillary tube of radius, r dipped in a liquid which makes an obtuse angle of contact,
with glass.
𝐏𝟐
𝐏𝟏 𝐏𝟑
Mercury
P1 P3 H
H H H
𝐏𝟏 𝐏𝟑
𝐡
𝐏𝟒 𝐏𝟐
At equilibrium,
P1 P3 H........................................(i)
P2 P4 ...................................................(ii)
P4 hg P1 But P4 P2
2cos
Also, P2 H ..........................(iv)
r
hg H H 2cos
r
2 cos 2 cos
hg h
r gr
Examples
1. A U-tube with limbs of diameter 5.00mm and 2.00mm contains water of surface tension
0.07Nm 1 , angle of contact zero and density 1000kgm 3 , find the difference in water levels.
Use g 10ms 2
r D R
H
C
A
2 cos 2 cos
H H hg
R r
2 cos 1 1
h
g r R
2 0.07 cos 0 1 1
h 3
3
8.4 10 3 m 8.4mm
1000 10 1 10 2.5 10
2. A U-tube with limbs of diameter 7.0mm and 4.0mm contains water of surface tension
7.0 10 2 Nm1 angle of contact zero and density 1000 kgm 3 . Find the difference in water
levels.
r D R
H
h C
A
3. A glass U-tube is such that the diameter of one limb is 4.0mm while that of the other is
8.0mm. The tube is inverted vertically with the open ends below the surface of water in a
beaker. Given that surface tension of water is 0.072Nm 1 , angle of contact between water and
glass is zero, and that density of water is 1000kgm 3 , what is the difference between the
heights to which water rises in the two limbs
𝑅 = 4 × 10−3 𝑚 r = 2 × 10−3 m
A
B
h
h1
H D H H
C h2
F E
PB H h1 g.....................................................(ii)
Put equation (ii) in equation (i) gives
2
PA H h1 g .............................................(iii)
r
For the wider limb
2 2
PD PC PD PC ........................(iv)
R R
PF h2 g PC But PF H H h2 g PC
PC H h2 g ..................................................(v)
h1 h2 g 2
2 1 1
hg 2
r R r R
2 1 1 2 0.072 1 1
h 3
3.6697 10 3 3.67 mm
g r R 1000 9.81 2 10 3
4 10
OR
2 cos 1 1
h
g r R
2 0.072 cos 0 1 1 3
h 3
3.6697 10 3.6697 mm
1000 9.81 2 10 4 10 3
2 cos 1 1
N.B: For mercury the expression for the fall, h is modified as follows h
g R r
2 cos 1 1 2 0.52 cos140 1 1 3
h 4.98 10 m 4.98mm
g R r 13600 9.81 6.0 10 3
1 10 3
6. A clean glass capillary tube of diameter 0.04cm is held with its lower end dipped in water
in the beaker with 12cm of the tube above the liquid surface. Given that the surface tension of
water is 7 10 2 Nm1 and its density is 1000kgm 3 ,
(i) To what height will water rise in the tube
(ii) What will happen if the tube is now depressed until only 4cm of its length is above the liquid
surface.
SURFACE ENERGY
Surface energy is the work done to increase the area of the liquid surface by 1m 2 under
isothermal conditions.
(OR)
Surface tension is the amount of work done (energy required) to produce a fresh surface of
liquid film of area 1m 2 under isothermal conditions.
Work done
Surface energy
Change in Area
The S.I unit is joules per square metre Jm 2
Consider a liquid film of surface tension, stretched across a rectangular wire frame.
Suppose the film is stretched isothermally from BC to B1C1 through a distance x against the
surface tensional force, F so that the surface area of the film increases
Surface tensional force, F l . However since there are two liquid surfaces in contact with air,
F 2 l 2l
Work done to stretch the film from BC to B1C1 Force distance 2l x
Iinitial area, A1 0m 2
Final area, A 2 2 4r22 2 4 7.5 10 3
2
1.414 10 3 m 2
Work done 0.03 1.414 10 3 4.24 10 5 J
2. (a)(i) Define surface tension
(ii) Explain the origin of surface tension using molecular theory.
(b) Aspherical drop of mercury of radius 2mm falls to the ground and breaks into 10
drops of equal size.
(i) Calculate the amount of work that has to be done to achieve this.
(ii) What is the minimum speed with which the original drop would have to hit the
ground? Density of mercury 13600kgm -3 , Surface tension of mercury 0.472Nm 1
(b)(i) R 2.0 10 3 m , Let, r be radius of the small drops
Volume of the big drop Total volume of the small drops
4 3 4 3 R3
R 10 r r 3
3 3 10
r
2.0 10 3 3
9.283 10 4 m
10
Work done 0.472 1.083 10 4 5.0265 10 5 2.739 10 5 J
(ii) Kinetic energy of the bg drop before impact Work done to split the drop into 10 small drops
S.5 PCB/PCM/PEM TR. SSEMPIIRA GEORGE
You are required to copy the notes in your mechanics book after writing the notes for elasticity.
1 2
mv 2.739 10 5 v
2.739 10 2 5
2 m
4
But m R 3 v
2.739 10 2 3
5
3 4 R 3
v
2.739 10 2 3 0.3467 ms
5
1
4 2.0 10 13600
3 3
3. Calculate the amount of work done in breaking up a drop of water of radius 0.5cm into
tiny droplets of water, each of radius 1mm, assuming isothermal conditions. Also determine the
number of droplets formed given that the surface tension of water is 7 10 2 Nm1
R 0.5cm 5 10 3 m and r 1mm 110 3 m
4 3 4
R n r 3
3 3
5 10 3 3
n 110 3 3
n 125
Total surface area of 125 small droplets, A 2 125 4r 2 125 4 110 3
2
1.571 10 3 m 2
Change in area, A A 2 A1 1.571 10 3 3.14 10 4 1.257 10 3 m 2
Work done Surface tension Change in area 7 10 2 1.257 10 3 8.799 10 5 J
4. A liquid drop of diameter 0.5cm breaks up into 27 tiny droplets all of the same size. If
the surface tension of the liquid is 0.07Nm 1 , calculate the resulting change in energy.
4 3 4 3 R3
R n r r
3
3 3 n
r 3
R 3 3 2.5 10 3
3
8.33 10 4 m
n 27
Change in area, A 2.35 10 4 7.85 10 5 1.56 10 4 m 2
Work done Surface tension Change in surface areea
Work done 7 10 2 1.56 10 -4 1.096 10 5 J
Change in surface area, A 6.28 10 2 2.51 10 3 6.03 10 2 m 2
Work done surface tension change in area
Work done 2 10 -2 6.03 10 2 1.206 10 3 J
Explain why water dripping out of the tap does so in spherical shapes.
Small mercury droplets are spherical while large ones flatten out
A small drop takes on a spherical shape to minimize the surface energy which tends to be
greater than the gravitational potential energy. Therefore the gravitational force cannot distort
the spherical shape due to the very small mass of tiny droplets.
A large drop flattens out in order to minimize the gravitational potential energy, which tends to
exceed the surface energy. Due to its large weight, gravitational force distorts the spherical
shape of large drops. For the shape of the drop to conform to the principle that the sum of
gravitational potential energy and surface energy must be a minimum, the centre of gravity of
the drop moves down as much as possible hence flattening.
COMBINED BUBBLES
1. Case 1
Consider two soap bubbles, A and B of radii r1 and r2 respectively, where r2 r1 . If the two
soap bubbles come into contact and have a common interface, then the radius of curvature, r of
the common interface can be calculated using pressure differences.
Let, 𝛾 be the surface tension of soap solution
For A
4 4
P1 H P1 H .....................(i)
r1 r1
For B
4 4 4 1 1 1
H H
r1 r2 r r1 r2 r
1 r2 r1 r1r2
r
r r1r2 r2 r1
4 4 r2 r1
Pressure difference across the common interface
r r1r2
Question.
A soap bubble of radius, 𝑟1 is attached to another bubble of radius, 𝑟2 . If 𝑟1 is less than 𝑟2 , show
r1r2
that the radius of curvature of the common interface is .
r2 r1
2. Case 2
Consider two soap bubbles, A and B of radii r1 and r2 which come together and coalesce to
form a single bubble. To find the radius, r of the resulting soap bubble, we use conservation of
surface energy.
Let, 𝛾 be the surface tension of soap solution.
Work done in forming a new bubble orf radius, r Surface energy 2 4r 2 ......(iii)
By conservation of energy,
Original work done Work done in forming a new bubble
r12 r22 r 2 r r 1
2
r22
4 4
Pressure difference Excess pressure
r r
1
2
r22
Examples
1. Two soap bubbles of radii 2cm and 4cm respectively coalesce under isothermal
conditions. If the surface tension of the soap solution is 2.5 10 2 Nm1 , calculate the excess
pressure inside the resulting soap bubble.
Let, r be the radius of the bubble formed.
work done Surface energy Surface tension Change in area
By conservation of energy;
Original work done Work done in forming a new bubble
r r r
1
2
2
2 2
r r r 1
2 2
2
r 2 10 4 10
-2 2 2 2
0.0447 m
4 4 2.5 10 2
Excess pressure 2.237 Nm 2
r 0.0447
2. Two soap, A and B of radii 67cm and 10cm respectively coalesce so as to have a portion
of their surfaces in common. Calculate the radius of curvature of this common surface and
hence the pressure difference if the surface tension of soap solution is 2.8 10 2 Nm1 .
Let r be the radius of curvature of the common interface
r
r1r2
6.7 10 1 1.0 10 1
0.118 m
r2 r1
6.7 10 1 1.0 10 1
S.5 PCB/PCM/PEM TR. SSEMPIIRA GEORGE
You are required to copy the notes in your mechanics book after writing the notes for elasticity.
4 4 2.8 10 2
Pressure difference 0.94915 Nm 2
r 0.118
Exercise
1) A soap bubble of radius 0.03m and another bubble of radius 0.04m are brought together so
that they combine. Calculate the radius of curvature of the common interface. 0.12 m
2) A soap bubble in vacuum has a radius of 3cm and another soap bubble in vacuum has a
radius of 6cm. If the two bubbles coalesce under isothermal conditions, calculate the radius of
the bubble formed. r 6.7cm
3) (a) How can you measure the angle of contact of a liquid in the laboratory?
(b) Define the term surface tension and deduce its dimensions.
(b) A clean glass capillary tube of diameter 0.04cm is held with its lower end dipped in
water in a beaker and with 12cm of the tube above the liquid surface. If the surface tension of
water is 7 10 2 Nm1 and density of water is 1000kgm 3 ,
(i) to what height will the water rise in the tube.
(ii) what will happen if the tube is now depressed until only 4cm of its length is above the liquid
surface.