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:
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| 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 = 𝛒wghw = 𝛒wg × 0.10
♦ Pressure due to 12.5 cm of spirit = 𝛒sghs = 𝛒sg × 0.125
4. Equating the two pressures, we get: 𝛒wg × 0.10 = 𝛒sg × 0.125
⇒ 𝛒w × 0.10 = 𝛒s × 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 = 𝛒wghw = 1 × 103 × g × 0.25 = 250g
(ii) Pressure due to 27.5 cm of spirit = 𝛒sghs = 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 = 𝛒mghm = 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]
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| 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 Pa presses down on the mercury surface in the trough
• The Pa tries to push more and more mercury into the tube
(ii) But Pa has a certain maximum value
• It cannot push more mercury
♦ If the Pa is at the 'maximum possible value' (which occurs at sea level), the height of mercury column will be the maximum
♦ If the Pa 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 = 𝛒mgh
♦ Pressure at B is the atmospheric pressure Pa
• We can equate the two pressures:
♦ Pa = 𝛒mgh
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 𝛒m 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: Pa = 𝛒mgh
(ii) We know that Pa 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 𝛒mgh = 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: Pa = 𝛒winegh
2. We know that Pa 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).
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| 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: 𝛒waterghwater = 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: 𝛒mercuryghmercury = 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
In the next section, we will see Pascal's law
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