Chapter 10.17 - Pressure Inside Drops and Bubbles

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 F_y(1), F_y(2), F_y(3), . . . be the vertical components
• Then the total vertical force will be given by:
F_y = F_y(1) + F_y(2) + F_y(3) + . . .
15. Since the hemisphere is in equilibrium, we have: F_y = 2πrS
• Thus we successfully obtain the net upward internal force F_y
16. If we divide this F_y by 'area on which Fy acts', we will get the internal pressure exerted by F_y
■ But by definition of pressure,
    ♦ the area on which the force acts
    ♦ must be perpendicular to
    ♦ the force
• In our present case, F_y 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 F_y 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 = πr^₂
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: F_y = 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 = $\mathbf\small{\rm{\frac{2S}{r}}}$

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Thus we successfully calculated the pressure inside a liquid drop. Based on that, we can write the details about the pressure inside a bubble. It can be written in 6 steps:

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 F_y arising due to the pressure inside the bubble
• So we get: F_y = 4πrS
3. F_y is the resultant of the vertical components of the red forces in fig.10.63(b)
• This F_y 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: F_y = 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 = $\mathbf\small{\rm{\frac{4S}{r}}}$

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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 S_oa
    ♦ Oil-water. This creates S_ow
    ♦ Air-water. This creates S_aw
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

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Now we will see some solved examples
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 × 10_⁵ 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 P_a
• This P_a is compressing the drop from all directions
• So the total pressure in the interior of the drop
= (1.01 × 10_⁵ Pa + 310 Pa) = 1.0131 × 10_⁵ 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 × 10_⁵ 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 × 10_⁵ 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 × 10_³ × 9.8 × 0.40)
5. So we get:
Total pressure = 1.01 × 10_⁵ + (1.2 × 10_³ × 9.8 × 0.40) + 10.0 Pa
= 1.05714 × 10_⁵ 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 × 10_⁵ 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 × 10_³ 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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In the next chapter, we will see thermal properties of matter



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Chapter 10.5 - Pascal's law

In the previous section, we saw the details about mercury barometer. In this section, we will see Pascal's law

Pascal's law can be explained in 7 steps:

1. Consider the arrangement shown in fig.10.22 below:

Fig.10.22

• A horizontal cylinder is shown in pink color

• It has a piston on it’s left end

2. Three vertical cylinders having different diameters are taken out from the horizontal cylinder

• We know that, though the vertical cylinders have different diameters, the liquid levels will be the same (in our present case, it is h₁)

• The pressure in the horizontal cylinder will be equal to 𝛒gh₁

    ♦ Where 𝛒 is the density of the liquid in the cylinder

• So initial pressure at the bottom of each vertical cylinder will be the same: 𝛒gh₁

3. Now, push the piston towards the right

• The pressure in the horizontal cylinder will increase

• The liquid will rise to a new ‘common level’ in all the vertical cylinders 

4. Note that: A new ‘common level’ (h₂) is attained

■ That means, the ‘increase in pressure’ in the horizontal cylinder is distributed equally among the vertical cylinders

    ♦ Each cylinder gets the same 'extra pressure'

    ♦ So the final pressure in each cylinder will be the same: 𝛒gh₂

■ Pascal’s law of transmission of fluid pressure states that:

Whenever external pressure is applied on any part of a fluid in a container, it is transmitted undiminished and equally in all directions

5. A simple demonstration of the law can be given using a filled balloon

• When we apply a force on any one portion of the balloon, the air inside will apply an outward force on every point of the balloon

• This is shown in fig.4.23 below:

Fig.10.23

6. The word 'undiminished' is used to indicate that, the 'applied external pressure' is completely available to act on every part of the fluid

    ♦ The distance between 'the point of application' to 'a far end of the fluid body' has no significance    

    ♦ If a pressure P is applied, both 'far ends' and 'near ends' will experience the same P

7. 'Equally in all directions' indicate that, the shape of the container has no significance

    ♦ The fluid may be contained inside a vessel of complicated shape

    ♦ Even then, if a pressure P is applied, all points of the fluid will experience the same P

The Hydraulic lift

• Working of the hydraulic lift is based on Pascal’s law. The details can be written in 5 steps:

1. In fig.10.24, we see two pistons separated by a liquid

    ♦ The smaller piston has an area of A₁

    ♦ The larger piston has an area of A₂

Fig.10.24

2. A downward force F₁ is applied on the smaller piston

• Then the pressure experienced by the liquid will be $\mathbf\small{\rm{\frac{F_1}{A_1}}}$

3. According to Pascal’s law, this pressure will be experienced on the whole liquid body

• The liquid body will apply this pressure on the sides and roof of the container

■ So the larger piston will experience the same upward pressure $\mathbf\small{\rm{\frac{F_1}{A_1}}}$

4. Let F₂ be the upward force experienced by the larger piston

• Then the pressure experienced by the larger piston is equal to $\mathbf\small{\rm{\frac{F_2}{A_2}}}$

• This pressure is equal to $\mathbf\small{\rm{\frac{F_1}{A_1}}}$

5. So we can write: $\mathbf\small{\rm{\frac{F_1}{A_1}=\frac{F_2}{A_2}}}$

• From this we get: $\mathbf\small{\rm{F_2=\left(\frac{A_2}{A_1} \right )F_1}}$

• We see that: F₁ is multiplied by a factor $\mathbf\small{\rm{\frac{A_2}{A_1}}}$

    ♦ 'A₂ in the numerator' is larger than 'A₁ in the denominator'

    ♦ So the factor $\mathbf\small{\rm{\frac{A_2}{A_1}}}$ will be larger than 1

    ♦ So F₂ will be greater than F₁

■ That means, we can:

    ♦ lift a large load placed on the larger piston

    ♦ by applying a small force at the smaller piston

■ The factor $\mathbf\small{\rm{\frac{A_2}{A_1}}}$ is the mechanical advantage of the hydraulic lift

---

Solved example 10.12

Two syringes of different cross sections (without needles) filled with water are connected with a tightly fitted rubber tube filled with water. Diameters of the smaller piston and larger piston are 1.0 cm and 3.0 cm respectively. (a) Find the force exerted on the larger piston when a force of 10 N is applied to the smaller piston. (b) If the smaller piston is pushed in through 6.0 cm, how much does the larger piston move out?

Solution:

Part (a):

1. The arrangement is shown in fig.10.25(a) below:

Fig.10.25

• We have: $\mathbf\small{\rm{F_2=\left(\frac{A_2}{A_1} \right )F_1}}$

2. Substituting the known values, we get:

$\mathbf\small{\rm{F_2=\left(\frac{0.25 \times \pi \times 3^2}{0.25 \times \pi \times 1^2} \right )\times 10=90\;N}}$

Part (b):

1. Water is an incompressible liquid

• So all the volume which is pushed out from the left syringe will enter the right syringe

2.Volume pushed out from the left syringe will be (A₁ × 6)

• Let x be the distance moved by the piston of the right syringe. This is shown in fig.b

• So the volume entering the right syringe will be (A₂ × x)

3. Equating the two volumes, we get:

A₁ × 6 = A₂ × x

⇒ $\mathbf\small{\rm{x=\left(\frac{A_1}{A_2} \right )6}}$

4. Substituting the known values, we get: $\mathbf\small{\rm{x=\left(\frac{0.25 \times 1^2}{0.25 \times 3^2} \right )\times 6=0.67\;cm}}$

 

Solved example 10.13

In a car lift compressed air exerts a force F₁ on a small piston having a radius of 5.0 cm. This pressure is transmitted to a second piston of radius 15 cm (Fig 10.7). If the mass of the car to be lifted is 1350 kg, calculate F₁ . What is the pressure necessary to accomplish this task? (g = 9.8 ms^-2 ).

Solution:

Part (a):

1. We have: $\mathbf\small{\rm{F_1=\left(\frac{A_1}{A_2} \right )F_2}}$

2. Substituting the known values, we get:

$\mathbf\small{\rm{F_1=\left(\frac{0.25 \times 5^2}{0.25 \times 15^2} \right )\times 1350 \times 9.8=1470\;N}}$

Part (b):

1. We are applying a force of 1470 N on the small piston

2. The small piston has an area of (𝝅 × 5²) cm²

• So we are applying a pressure of:

$\mathbf\small{\rm{\frac{1470}{\pi \times 5^2}}}$ = 18.73 N cm^-2 = 1.873 × 10⁵ N m^-2

3. Normal atmospheric pressure at sea level is 1.013 × 10⁵ N m^-2

• So 1.873 × 10⁵ N m^-2 is nearly twice the atmospheric pressure    

Solved example 10.14

A hydraulic automobile lift is designed to lift cars with a maximum mass of 3000 kg. The area of cross-section of the piston carrying the load is 425 cm². What maximum pressure would the smaller piston have to bear ?

Solution:

1. We have: $\mathbf\small{\rm{\frac{F_1}{A_1}=\frac{F_2}{A_2}}}$

2. Substituting the known values, we get: $\mathbf\small{\rm{\frac{F_1}{A_1}=\frac{3000 \times 9.8}{425 \times 10^{-4}}}}$ = 6.92 × 10⁵ Nm^-2

Hydraulic brake

• The working of the hydraulic brake system is also based on Pascal’s law. This can be explained in 4 steps:

1. When the driver apply a force on the brake pedal, he is actually pushing the piston of a ‘master cylinder’

2. The fluid inside this cylinder will move out and apply pressure on four other pistons

    ♦ These pistons are situated at the four wheels of the automobile

    ♦ These pistons have a larger area than the piston in the master cylinder

          ✰ So a small force is sufficient at the brake pedal

3. The large pistons press against the brake lining of the wheels

• Thus the wheels slow down and come to a stop

• A schematic diagram of the system can be seen here 

4. The advantage of this system can be explained in three steps:

(i) We know that:

• The pressure will be transmitted undiminished and equally in all directions

(ii) So all wheels will experience the same braking force

• Also all wheels will experience the braking force at the same time

(iii) This will enhance safety while applying brakes

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In the next section, we will see buoyancy and flotation

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Chapter 10.4 - The Mercury Barometer

In the previous section, we saw the details about hydrostatic paradox and some of it's applications. We also saw U-tube manometer and a solved example related to it. In this section, we will see a few more solved examples. Later in this section, we will see the details about mercury barometer

Solved example 10.8

A U-tube contains water and methylated spirit separated by mercury. The mercury columns in the two arms are in level with 10.0 cm of water in one arm and 12.5 cm of spirit in the other. What is the specific gravity of spirit ?

Solution:

1. The arrangement is shown in fig.10.19(a) below:

Fig.10.19

• The green dotted line indicates that mercury is at the same level in both the arms

2. At this stage, 10 cm of water and 12.5 cm of spirit are present

• It is clear that:

    ♦ Pressure due to 10 cm of water

    ♦ is balanced by

    ♦ Pressure due to 12.5 cm of spirit

3. Let us write the actual pressures:

    ♦ Pressure due to 10 cm of water = 𝛒_wgh_w = 𝛒_wg × 0.10

    ♦ Pressure due to 12.5 cm of spirit = 𝛒_sgh_s = 𝛒_sg × 0.125

4. Equating the two pressures, we get: 𝛒_wg × 0.10 = 𝛒_sg × 0.125

⇒ 𝛒_w × 0.10 = 𝛒_s × 0.125

5. 'Specific gravity' is another name for 'Relative density'

• We have: $\mathbf\small{\rm{Relative \;Density\;=\;\frac{Density\;of\;substance}{Density\;of\;water}}}$

• Thus from (4), we get:

Relative density of spirit = $\mathbf\small{\rm{\frac{\rho_s}{\rho_w}=\frac{0.10}{0.125}}}$ = 0.8

Solved example 10.9

In the previous problem, if 15.0 cm of water and spirit each are further poured into the respective arms of the tube, what is the difference in the levels of mercury in the two arms ? (Specific gravity of mercury = 13.6)

Solution:

1. In this situation, there will be:

    ♦ 25 cm of water in the left arm

    ♦ 27.5 cm of spirit in the right arm

2. Let us calculate the pressures:

(i) Pressure due to 25 cm of water = 𝛒_wgh_w = 1 × 10³ × g × 0.25 = 250g

(ii) Pressure due to 27.5 cm of spirit = 𝛒_sgh_s = 0.8 × 10³ × g × 0.275 = 220g

3. So the pressure due to 27.5 cm of spirit is low

• That means, the spirit side will require some 'contribution from mercury' in order to balance the pressure from the water side

• This situation is shown in fig.10.19(b) above

• We see that: an additional x cm of mercury has risen up into the right arm

4. In this situation, we can write:

    ♦ Pressure due to 25 cm of water

    ♦ is equal to

    ♦ The sum of the pressures due to '27.5 cm of spirit' and 'x cm of mercury'

5. The pressure due to x cm of mercury = 𝛒_mgh_m = 13.6 × 10³ × g × x = 13600xg

• Equating the pressures according to (4), we get:

250g = 220g + 13600xg

⇒ x = 0.002205 m = 0.221 cm

---

• We have seen the basics about atmospheric pressure. We discussed it based on fig. 10.2 at the beginning of this chapter

• Now we will learn about an apparatus which is used to measure atmospheric pressure. It is the mercury barometer. It can be explained in 11 steps:

1. A long glass tube closed at one end is filled with mercury

2. It is then inverted into a trough containing mercury

• This is shown in fig.10.20 below

[Note that, mercury vapours, if inhaled, is dangerous. This experiment should be done only in an advanced lab under the supervision of the lab authorities]

Fig.10.20

3. We see that a column of mercury will remain inside the tube

• The reason for the formation of this column can be written in 2 steps:

(i) The atmospheric pressure P_a presses down on the mercury surface in the trough

• The P_a tries to push more and more mercury into the tube

(ii) But P_a has a certain maximum value

• It cannot push more mercury

    ♦ If the P_a is at the 'maximum possible value' (which occurs at sea level), the height of mercury column will be the maximum

    ♦ If the P_a is at a lower value, the height of mercury column will be less

4. The horizontal green dotted line in fig.b is the datum line

    ♦ Two points A and B are marked in this line

    ♦ The pressure at A must be equal to the pressure at B

5. Let us calculate those pressures

    ♦ The pressure at A = 𝛒_mgh 

    ♦ Pressure at B is the atmospheric pressure Pa

• We can equate the two pressures:

    ♦ Pa = 𝛒_mgh  

6. This equation gives us an easy method to obtain atmospheric pressure

• All we need is to calculate the quantity on the left side

    ♦ For that, we simply measure the height h

    ♦ Then multiply it with 𝛒_m and g

7. Possible values of h are:

• If the apparatus is placed at sea level on a normal day, h will be equal to 76 cm

• If h is 75 cm or less at sea level, it is a sign of approaching storm

• At top of mountains, h will be far less than 76 cm

8. So on a normal day, at sea level, we get '76 cm' by actual reading on the apparatus

• This 76 cm can be obtained theoretically also. It can be explained in steps:

(i) We have: P_a = 𝛒_mgh

(ii) We know that P_a at sea level = 1.013 × 10⁵ N m^-2

(iii) Substituting the known values in (i), we get:

1.013 × 10⁵ = 13.6 × 10³ × 9.81 × h

⇒ h = 0.759 m = 76 cm     

9. This apparatus was invented by the Italian scientist Evangelista Torricelli

• In his honour, a unit called torr is also used to express pressure

• The details about this unit can be written in 4 steps:

(i) We have seen that 76 cm (760 mm) of mercury corresponds to 1.013 Pa

(ii) In a similar way, we can calculate the pressure corresponding to 1 mm of mercury

    ♦ It will be equal to 𝛒_mgh = 13.6 × 10³ × 9.81 × 0.001 = 133.416 Pa 

(iii) A pressure of 133 Pa is called one torr

(iv) But while using torr, we need not mention about the corresponding value in Pa

• We can directly write the ‘torr value’ after measuring the height

• For example, on a normal day, the pressure at sea level can be written as: 760 torr

10. Another unit for measuring pressure is bar

• One bar is equal to 10⁵ Pa

• So on a normal day, the pressure at sea level can be written as: 1.013 bar

11. An important point has to be noted about the mercury column of the barometer

• It can be written in steps:

(i) In fig.10.20, we see that:

A vacant portion is present between the top end of the tube and the top surface of the mercury column

(ii) This vacant portion must be perfect vacuum

• If air is present, the weight of that air will affect the reading

(iii) Normally, since the filled tube is inverted, we do get a vacuum

• Even then some mercury vapours will be present

• This is due to a the mercury molecules separating away from the main liquid mercury

(iv) But the properties of mercury is such that, only a very few molecules separate away from the liquid body

• So the weight due to gaseous mercury will be negligible

Solved example 10.10

Toricelli’s barometer used mercury. Pascal duplicated it using French wine of density 984 kg m^-3. Determine the height of the wine column for normal atmospheric pressure.

Solution:

1. Imagine that, in fig.10.20 above, the liquid is wine instead of mercury

• We can write: P_a = 𝛒_winegh

2. We know that P_a at sea level = 1.013 × 10⁵ N m^-2

• Substituting the known values in (1), we get:

1.013 × 10⁵ = 948 × 9.81 × h

⇒ h = 10.89 m

Solved example 10.11

A manometer reads the pressure of a gas in an enclosure as shown in Fig. 10.21 (a). When a pump removes some of the gas, the manometer reads as in Fig. 10.21 (b). The liquid used in the manometers is mercury and the atmospheric pressure is 76 cm of mercury.

(a) Give the absolute and gauge pressure of the gas in the enclosure for cases (a) and (b), in units of cm of mercury.

(b) How would the levels change in case (b) if 13.6 cm of water (immiscible with mercury) are poured into the right limb of the manometer ? (Ignore the small change in the volume of the gas).

Fig.10.21

Solution:

Part (a), Fig.a:

• The gas is shown in yellow color. Initially, it is completely confined inside the enclosure (red circle)

• When the manometer is attached, some gas comes out of the red circle and pushes the mercury down

    ♦ That means, the gas expands and fill a small portion of the left arm of the manometer

    ♦ That means, there will be some gas present in the the left arm

• If it was a liquid, we must include the 'height of the liquid column in the left arm' in the calculations

• But since it is a gas, we are told to ignore the volume change due to expansion

• That means,

    ♦ the pressure at the bottom end of the gas column

    ♦ can be considered to be the same as

    ♦ the pressure at A

• Now we can write the steps:

1. The two pressures at datum (green dotted line):

• On the left side, we have the pressure at A

• On the right side we have the sum of two pressures:

    ♦ Pressure due to 20 cm of mercury

    ♦ Pressure due to atmosphere (76 cm of mercury)

2. Balancing the two sides, we get:

• Pressure at A =  (20 cm of mercury + 76 cm of mercury) = 96 cm of mercury

• This is the absolute pressure at A

3. When we subtract the atmospheric pressure from absolute pressure, we get gauge pressure

• So the gauge pressure at A = (96 -76) = 20 cm of mercury

Part (a), Fig.b:

1. Consider the situation in which the pump gradually removes gas from the enclosure

• As the gas is being removed, the pressure at A in fig.a gradually falls

• This is because, lesser gas can create only lesser pressure

• So mercury level in fig.a gradually falls

2. When the mercury levels are same in both arms, we can say:

The pressure at A is equal to the atmospheric pressure

3. But in fig.b, we see that:

The mercury level in right arm is below the level in left arm by 18 cm

• That means, the pressure at A is less than the atmospheric pressure

• The action of the pump can be written as two stages:

Stage 1:

The pump removed some gas in such a way that, the pressure at A became equal to atmospheric pressure (76 cm of mercury)

Stage 2:

The pump removed some more gas in such a way that, the pressure at A became less than atmospheric pressure

    ♦ The 'fall in pressure' in stage 2 is equal to '18 cm of mercury'

■ So the absolute pressure at A is (76-18) = 58 cm of mercury

4. As usual, gauge pressure is obtained by subtracting atmospheric pressure from absolute pressure

• So we get:

Gauge pressure = (58-76) = -18 cm of mercury

• The negative sign indicates that, the pressure is below atmospheric pressure

• Note:

We had learnt that, gauge pressure corresponds to the reading in the manometer. It has become true in the case of negative pressure also

Part (b):

1. Imagine that, 13.6 cm of water is poured into the right limb

Now the pressures at the datum in fig.b are:

• On the left side we have the sum of two pressures:

    ♦ Pressure at A

    ♦ Pressure due to 18 cm of mercury

• On the right side also, we have the sum of two pressures:

    ♦ Atmospheric pressure (76 cm of mercury)

    ♦ Pressure due to 13.6 cm of water

2. But the pressure due to 13.6 cm of water is an 'excess pressure'

• That 'excess pressure' will disturb the equilibrium shown in fig.b

• The 'excess pressure is on the right side

• In order to balance that excess pressure on the right side, some mercury will rise up in the left side

• When such a rise of mercury occurs, a new datum will be attained

• This is shown in fig.c

3. Let the mercury rise by x cm in the left arm

The pressures at the datum in fig.c are:

• On the left side we have the sum of three pressures:

    ♦ Pressure at A

    ♦ Pressure due to 18 cm of mercury

    ♦ Pressure due to x cm of mercury

• On the right side, we have the sum of two pressures:

    ♦ Atmospheric pressure (76 cm of mercury)

    ♦ Pressure due to 13.6 cm of water

4. Equating the two sides, we get:

58 cm + 18 cm + x cm = 76 cm + pressure (in cm of mercury) due to 13.6 cm of water

5. Let us calculate the pressure (in cm of mercury) due to 13.6 cm of water. it can be done in 4 steps:

(i) In a vessel, water is taken upto a height 13.6 cm

• The pressure at the bottom of the vessel will be: 𝛒_watergh_water = 1 × 10³ × 9.81 × 0.136 Pa

(ii) In another vessel, mercury is taken upto a height of h cm

• The pressure at the bottom of the vessel will be: 𝛒_mercurygh_mercury = 13.6 × 10³ × 9.81 × h × 0.01 Pa

(iii) We want the pressures in (i) and (ii) to be equal. So we equate them:

1 × 10³ × 9.81 × 0.136 = 13.6 × 10³ × 9.81 × h × 0.01

⇒ h = 1 cm

(iv) So we can write:

13.6 cm of water will produce the same pressure as: 1 cm of mercury

6. So the equation in (4) becomes:

58 cm + 18 cm + x cm = 76 cm + 1 cm

⇒ x = 1 cm

7. So we can write:

The mercury will rise by 1 cm in the left arm and thus the reading will become 19 cm

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In the next section, we will see Pascal's law

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