In the previous section, we saw contact angle and capillarity. In this section we will see pressure inside a drop. Some basic details can be written in 21 steps:
1. Consider a drop of a liquid
♦ A molecule in the interior of the drop will not experience any net force
♦ But a molecule at the surface will experience a net inward force
• We saw this same situation when we first saw the basics about surface tension (See the first fig.10.48 of section 10.15)
2. So all the surface molecules of the drop will experience a net inward force
• As a result, the drop will roll into a spherical shape
3. Also we know that, the surface will act like stretched membrane
♦ That means, the surface of the drop tries to shrink to the least possible area
♦ That means, the drop will be trying to attain the least possible surface area
4. Which shape has the least possible surface area?
• We can find the answer in 2 steps:
(i) Let there be two samples of any substance
• Let both have the same volume V
(ii) Let us mold them into two different shapes:
♦ The first sample is made into a cubical shape
♦ The second sample is made into a spherical shape
• Then, the second sample will be having lesser surface area
• In fact, we can compare a sphere with any other shape with the same volume
■ The surface area of the sphere will always be lesser
■ So naturally, the liquid drop will be spherical
5. The force which tries to shrink the surface is the surface tension S
• This force will be tangential to the surface of the drop
6. We can draw infinite number of tangents on the surface of a sphere
• Three of them are shown in fig.10.61(a) below:
 |
| Fig.10.61 |
• Surface tension will be acting along all those tangents
7. For our present case, we are going to draw tangents at a few specific points
This can be explained in steps:
(i) Consider a sphere
(ii) Cut the sphere using a horizontal plane
♦ This horizontal plane should be passing through the centre of the sphere
♦ This is shown in fig.10.61(b) above
(iii) The horizontal plane will intersect the ‘surface of the sphere’ along a horizontal circle
♦ Draw vertical tangents of the sphere at the points on that circle
♦ These are the tangents that we need for our present problem
8. Surface tension will be acting along those tangents
♦ To indicate those surface tension forces, we give arrow heads to the tangents
♦ This is shown in fig.c
9. Using the horizontal plane, cut the sphere into two halves. So we have two hemispheres
♦ Separate the top hemisphere
♦ This is shown in fig.d
10. We see that the surface tension forces are acting downwards
• We can easily calculate the total downward force in just 2 steps:
(i) S is the surface tension force per unit length
(ii) So total surface tension force = (S × circumference of the circle) = 2πrS
♦ Where r is the radius of the drop
11. So a force of 2πrS is acting in the downward direction
• But the hemisphere is in equilibrium
• That means, there are some vertical upward forces also acting on the hemisphere
• The next four steps from (12) to (15) will explain those vertical upward forces
12. Fig.10.62(a) below shows the 2D view of the hemisphere
 |
| Fig.10.62 |
• The red arrows indicate the pressure P inside the drop
♦ Some of those arrows will be vertical. Some will be horizontal
13. Consider those arrows which are neither vertical nor horizontal
• Such red arrows will have a vertical component and a horizontal component
♦ This is shown in fig.b
✰ Red arrows which are closer to vertical will have lesser 𝜃 values
✰ Red arrows which are closer to horizontal will have greater 𝜃 values
• The horizontal components will cancel each other
♦ So the net horizontal force will be zero
14. But there will be a net vertical upward force
• Let Fy(1), Fy(2), Fy(3), . . . be the vertical components
• Then the total vertical force will be given by:
Fy = Fy(1) + Fy(2) + Fy(3) + . . .
15. Since the hemisphere is in equilibrium, we have: Fy = 2πrS
• Thus we successfully obtain the net upward internal force Fy
16. If we divide this Fy by 'area on which Fy acts', we will get the internal pressure exerted by Fy
■ But by definition of pressure,
♦ the area on which the force acts
♦ must be perpendicular to
♦ the force
• In our present case, Fy and it's components are vertical
♦ So the area on which they act must be horizontal
17. But in our present case, the components of Fy acts on the surface of the hemisphere
• Surface of the hemisphere is not horizontal. It is curved
18. So we consider the 'projected area' of the curved surface
• That is., we project the curved surface of the hemisphere onto a horizontal plane
• Obviously, the 'projected area' thus obtained, will be the base of the hemisphere
• So the projected area = Base area of the hemisphere = πr2
19. Now we can find the pressure:
• Pressure in the vertical direction = $\mathbf\small{\rm{\frac{Force}{Area}=\frac{F_y}{\pi r^2}}}$
20. But from (15), we have: Fy = 2πrS
• Substituting this in (19), we get:
Pressure in the vertical direction = $\mathbf\small{\rm{\frac{F_y}{\pi r^2}=\frac{2 \pi rS}{\pi r^2}=\frac{2S}{r}}}$
21. But inside a fluid, pressure at a point will be the same in all directions
(See fig.10.5 in section 10.1)
■ So we can write Eq.10.22: Pressure inside a liquid drop = $\mathbf\small{\rm{\frac{2S}{r}}}$
1. We know that a bubble is hollow inside. Also, the outer skin has two surfaces
• So the upper hemisphere will be as shown in fig.10.63(a) below:
 |
| Fig.10.63 |
• There are two rings of arrows
♦ The arrows in the outer ring are shown in magenta color
✰ They indicate the surface tension in the outer surface
♦ The arrows in the inner ring are shown in yellow color
✰ They indicate the surface tension in the inner surface
2. So the total downward force due to surface tension = 2×2πrS = 4πrS
• This is balanced by the force Fy arising due to the pressure inside the bubble
• So we get: Fy = 4πrS
3. Fy is the resultant of the vertical components of the red forces in fig.10.63(b)
• This Fy acts on a 'projected area' which is equal to
4. So the pressure = $\mathbf\small{\rm{\frac{Force}{Area}=\frac{F_y}{\pi r^2}}}$
5. But from (2), we have: Fy = 4πrS
• Substituting this in (4), we get:
Pressure in the vertical direction = $\mathbf\small{\rm{\frac{F_y}{\pi r^2}=\frac{4 \pi rS}{\pi r^2}=\frac{4S}{r}}}$
6. But inside a fluid, pressure at a point will be the same in all directions
■ So we can write Eq.10.23: Pressure inside a bubble = $\mathbf\small{\rm{\frac{4S}{r}}}$
Detergents and Surface tension
We know that detergents, when used with water, help to remove grease and oil from fabric. We can write an explanation in 7 steps:1. First let us try to clean the oily fabric using ordinary water (with out any detergents)
• In such a situation, we have the following three interfaces:
♦ Oil-air. This creates Soa
♦ Oil-water. This creates Sow
♦ Air-water. This creates Saw
2. We have seen the details about contact angle 𝜃 (previous section) when three different materials meet
• In our present case, 𝜃 will be such that, the water rolls into spherical drops
♦ So it will not wet the oil
♦ That is., there will not be any attraction between oil and water
• So, neither running water, nor rinsing action, will remove the oil
3. Now we add some detergent to the water
• Each detergent molecule has a special shape as shown in fig.10.64(a) below:
 |
| Fig.10.64 |
• The rounded end has attraction towards water
♦ But this rounded end has repulsion towards oil
• The pointed end has attraction towards oil
♦ But this pointed end has repulsion towards water
4. Thus the pointed ends of the detergent stick to the oil particles
• They soon form spheres around the oil particles
♦ This is shown in fig.b
5. The spheres thus formed are favorable because:
• The pointed ends are protected from water as they are inside the spheres
• The rounded ends are in contact with the water
♦ The rounded ends do have attraction towards water
♦ So water need not repel away from the spheres
6. The spheres thus formed can be washed away by running water or rinsing
• Thus the detergents help to clean oily fabric
7. If we use hot water along with detergents, cleaning will become even more easy
• This is because, oil will melt into smaller particles and so more sphere will be formed
Solved example 10.35
What is the pressure inside the drop of mercury of radius 3.00 mm at room temperature ? Surface tension of mercury at that temperature (20 °C) is 4.65 × 10-1 N m-1 . The atmospheric pressure is 1.01 × 105 Pa. Also give the excess pressure inside the drop
Solution:
1. Pressure inside a liquid drop is given by Eq.10.22: $\mathbf\small{\rm{P=\frac{2S}{r}}}$
2. Substituting the known values, we get: $\mathbf\small{\rm{P=\frac{2(4.65\times 10^{-1})}{(3\times 10^{-3})}}}$ = 310 Pa
• This pressure is exerted by the surface towards the interior of the drop
♦ The surface exerts such a pressure because, it is trying to shrink
3. The drop is already subjected to atmospheric pressure Pa
• This Pa is compressing the drop from all directions
• So the total pressure in the interior of the drop
= (1.01 × 105 Pa + 310 Pa) = 1.0131 × 105 Pa
4. But in the given data, atmospheric pressure has only three significant figures
• So the result must have only three significant figures
• Thus the result in (3) becomes: 1.01 × 105 Pa
Solved example 10.36
What is the excess pressure inside a bubble of soap solution of radius 5.00 mm,
given that the surface tension of soap solution at the temperature (20 °C) is 2.50 × 10-2 N m-1 ? If an air bubble of the same dimension were formed at depth of 40.0 cm inside a container containing the soap solution (of relative density 1.20), what would be the pressure inside the bubble ? (1 atmospheric pressure is 1.01 × 105 Pa).
Solution:
Part(a):
1. Excess pressure inside a bubble is given by Eq.10.22: $\mathbf\small{\rm{P=\frac{4S}{r}}}$
2. Substituting the known values, we get: $\mathbf\small{\rm{p=\frac{4(2.5\times 10^{-2})}{(5\times 10^{-3})}}}$ = 20.0 Pa
Part (b):
1. When air bubble is formed inside, there is only one surface
• Then excess pressure inside the bubble will be given by: $\mathbf\small{\rm{P=\frac{2S}{r}}}$
2. Substituting the known values, we get: $\mathbf\small{\rm{p=\frac{2(2.5\times 10^{-2})}{(5\times 10^{-3})}}}$ = 10.0 Pa
3. Total pressure will be equal to the sum of three items:
(i) Atmospheric pressure (ii) Pressure due to the soap solution above the bubble (iii) Excess pressure inside the bubble
4. Pressure due to soap solution above the bubble = ρgh = (1.2 × 103 × 9.8 × 0.40)
5. So we get:
Total pressure = 1.01 × 105 + (1.2 × 103 × 9.8 × 0.40) + 10.0 Pa
= 1.05714 × 105 Pa
6. But in the given data, atmospheric pressure has only three significant figures
• So the result must have only three significant figures
• Thus the result in (5) becomes: 1.06 × 105 Pa
Solved example 10.37
Mercury has an angle of contact equal to 140° with soda lime glass. A narrow tube of radius 1.00 mm made of this glass is dipped in a trough containing mercury. By what amount does the mercury dip down in the tube relative to the liquid surface outside ? Surface tension of mercury at the temperature of the experiment is 0.465 N m-1 . Density of mercury = 13.6 × 103 kg m-3
Solution:
1. Capillary rise is given by: $\mathbf\small{\rm{h=\frac{2S \cos \theta}{\rho g r}}}$
2. Substituting the known values, we get: $\mathbf\small{\rm{h=\frac{(2)(0.465) \cos 140}{(13.6\times 10^{3})(9.8)(1\times 10^{-3})}}}$ = -0.00534 m
3. So the fall of mercury = 5.34 mm
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