Pressure Question

Document Title

1. Water of depth 10 m exerts a pressure equal to atmospheric pressure. An air bubble rises to the surface of the lake which is 20 m deep. When the bubble reaches the surface, its volume is 6 cm3.

                                    

Calculate
(a) the pressure at the bottom of the lake
(b) the volume of the air bubble at the bottom of the lake.

(a) Analysis:
Remember to include atmospheric pressure into calculation unless it is in vacuum! A very common mistake is not to include this!

pressure = pressure due to water depth + atmospheric pressure
                 = Patm * 2 + Patm
                 = 3Patm

(b) Note that you might not be able to do this question now, but this is a typical integrated question in pressure. Once you know PV is a constant under Kinetic Model of Matter, you can always come back to try this.

Ptop * Vtop = Pbot * Vbot
   Patm * 6 = 3Patm * Vbot
      Vbot = 2cm3

2. Two barometers are set up below using identical tubes of uniform cross-sectional area at sea level.
                                    

(a) Given that the empty space above the mercury column in barometer P is a vacuum, state the atmospheric pressure as measured by this barometer in mmHg.
(b) What is the pressure at point E? Give your answer in mm Hg and show essential working on how you arrive at your answer.
(c) If barometer P is brought up to the mountains, what do you expect to the column of vacuum in P? Give a reason for that.
(d) Suggest a possible reason why barometer Q gives a shorter mercury column compared to P.
(e) Calculate the pressure in the empty space above the mercury column in barometer Q.

(a) Since the empty space above the mercury column is a vacuum, the pressure there is 0Pa.

So, Patm = pressure due to mercury column
            = 1000 - 250 = 750 mm Hg
(b) Recall that When finding pressure at point E, the height is from the top surface to point E, not from the surface of trough of mercury to point E.

pressure = 1000 - 250 - 400 = 350 mmHg
(c) At the mountain, the atomspheric pressure is lower thus the mercury column is shorter, and thus column of vacuum is longer.
(d) Analysis:
To have a shorter liquid column, we can replace the mercury with a liquid with higher density as P = pgh. However, in this case, the liquid is still mercury. Thus we have to think about other alternative.
In order to have a shorter mercury column, the empty space above the mercury column cannot be vacuum. It has to be air which contributes some pressure, and hence creates a lower pressure difference and thus shorter mercury column.

Answer:
The empty space above the mercury column is air.
(e) Patm = pressure due to mercury column + Pair
         750 = (1000 - 280) + Pair
         Pair = 30 mmHg

3. A gas pipe is connected to two mercury manometers as shown. The two manometers contain the same amount of mercury, except that the diameter of the right-hand tube in manometer B is double that of the right-hand tube in manometer A.

                                    

(a) Determine the liquid level difference in manometer B.
(b) Is the gas pressure smaller or greater than the atmospheric pressure?
(c) Calculate the excess pressure of the gas supply in Pascal.
(Density of mercury = 13 600 kgm-3, atmospheric pressure = 76 cmHg)

(a) Analysis:
Remember that liquid level difference in manomater depends on the pressure difference. Since the pressure difference between two arms in B is same as the pressure difference between two arms in A, thus the liquid level difference is the same as manometer A.

Answer:
Liquid level difference = 30 cm
(b) Analysis:
To determine which arm has a higher pressure, we have to look at the liquid level on each arm. The higher the liquid level, the lower the pressure is.

Answer:
Since the liquid level on left arm is lower than that on the right arm, the gas supply has a higher pressure than atomspheric pressure.
(c) Analysis:
Remember that the liquid level difference contributes to the pressure difference (excess pressure).

excess pressure = 13600 * 10 * 0.3
                              = 40800 Pa

4. A uniform capillary tube, closed at one end, contained air which is trapped by a thread of mercury 85 mm long. When the tube was held vertically with the closed end downwards, the length was 45 mm.

                                    

(a) Calculate the pressure of the air in Pa given that the atmospheric pressure is 760 mmHg. Take the density of mercury to be 13600 kgm3
(b) The tube is now slowly rotated 90o as shown in second Figure. Assuming that there is no change in temperature, state whether the length L will remain as 45 mm, longer than 45 mm or shorter than 45 mm. Explain your answer.

(a) Pair = Patm + pressure due to 85mm of mercury
            = 760 + 85
            = 845 mmHg
            = 0.845 * 13600 * 10
            =1.15 * 105 Pa

Comment: We cannot assume atomspheric pressure is 1.01 * 105 Pa. We have to use the value given in the question.
(b) Initially, the pressure of air is greater than the atmospheric pressure, thus the mercury column will move out until the pressure due to air is same as the atmospheric pressure. Thus, the length L will be longer than 45 mm.

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