## Notes### Pressure- Pressure is force acting normally per unit area. If the amount of applied force is the same, then - Larger area --> Lower pressure
- Smaller area --> Higher pressure
### Examples of Pressure- Skis have a large area to hold up the weight of the skier on the snow
- Flat bottomed shoes are comfortable to wear due to reduced pressure acting on our feet
- A sharp knife can cut easily because the very high pressure under the cutting surface is more than the object can withstand
- Atmospheric pressure exists because of MOLECULAR BOMBARDMENT of energetic air molecules (from the air around us)
- Under normal conditions, there are large numbers of air molecules and these molecules move with high velocities. They make frequent collisions with things around us
- The pressure exerted by the air molecules is almost equivalent to putting a 1 kg mass on an area of 1 cm
^{2} - Normal atmospheric pressure= 1 atm (about 1.013 x 105 pa or 101300 pa)
- 101300 Nm
^{-2} = 10.13 Ncm^{-2} = 1.013 kgcm^{-2}
### Applications of atmospheric pressure - Drinking with a straw
- Drawing a liquid into a syringe by withdrawing the plunger
- Holding a rubber sucker on a smooth surface
- Removing dust with vacuum cleaner
### Pressure due to a liquid column- The taller the liquid column (with narrow base), the larger the amount of liquid contained, the greater the weight of the liquid to exert pressure
- The amount of pressure in the SAME liquid column is DIFFERENT at DIFFERENT DEPTHS.
- The greater the depth, the greater the weight of the liquid above it, the greater the pressure
**The pressure in a liquid depends on the ****HEIGHT of the liquid****The amount of pressure increases with ****DEPTH**
#### 2 cases of liquid pressure **1. With atmospheric pressure ****p = p0 + ρgh**- Pressure at bottom = atmospheric pressure + pressure due to liquid column In this case, when the container is open, there is atmospheric pressure acting on the liquid as well.
**2. Without atmospheric pressure **-
**p = ρgh** - Pressure at bottom = pressure due to liquid column only In this case, when the container is closed, air is removed (vacuum), so there is no atmospheric pressure.
####
Factors affecting pressure in a liquid **1.Density of liquid**
2.Depth of liquid 3. Gravitational accelerationWhen it is at equilibrium, pressure must be the same at any point along the same depth (h). Note: pressure does not depend on the shape of the liquid column. ###
Measurement of pressure **Simple Mercury Barometer **
- Used to measure atmospheric pressure **How to construct **- A thick-walled glass tube (about 1m long) is filled with mercury completely
- The open end of the tube is covered with a finger and inverted
- Place the inverted tube in a trough of mercury
**Observation**: The height of the mercury column found to be about 760mm or 76cm Atmospheric pressure = 1 atm or 760 mmHg or 76 cmHg **Reasons for using mercury in a barometer **- Mercury does not wet glass
- Mercury has a high density
#### Manometer- Used to measure gas pressure **How to construct **- The manometer consists of a U-tube containing a column of liquid
- The liquid can be mercury, water or oil
How to measure? - When both arms are open, same atmospheric pressure is exerted on the liquid surfaces (same horizontal level)
- To measure the pressure of a gas, left side is connected to a gas supply
- The gas exerts pressure on the surface at L. The gas pressure must be greater than atmospheric pressure to cause the right side to rise
**Pressure at L given by p = p0 + ρgh**
###
Hydraulic System - Pressure can be transmitted throughout a liquid in hydraulic presses
- When a small force is applied to the smaller piston, pressure is exerted on the liquid
- This pressure is transmitted in the liquid (oil) and is the same everywhere within the oil. Thus the pressure at the bigger piston must also be p.
- Since area at the bigger piston is bigger, force must also be greater
A small force applied to the smaller piston can lift a greater load on the bigger piston
## Additional Notes
- Pressure is the force acting normal or perpendicularly per unit area
- SI unit: Pascal (Pa) or N/m
^{2}
**Pressure in: **
| ** Solid** | **Liquid**
| **Gas **
| Equation | Pressure = Force/Area
| Pressure = hpg
h = depth of liquid (m) p = density of liquid (kg/m^{3}) g = gravitational field strength
| The air surrounding us exerts a pressure in all directions which is about 105 Pa. | Remarks | This formula can only be used for **solids**. | - A liquid exerts pressure because of its weight. - Liquid pressure acts equally in all directions. This is because particles of the water can flow and wrap around the object. | - A barometer is used to measure pressure. It consists of an inverted tube in a dish of mercury. The space above the mercury in the tube is vacuum. - Liquid mercury is used as its density is very high and a shorter barometer can be used to show atmospheric pressure. - An object can be bent/sucked in due to the production of vacuum and due to the difference in pressure; the atmospheric pressure will press on the object. |
## MCQ Questions**1. Stiletto heels can exert great pressure mainly due to**** ** a. the large force acting on it b. the small force acting on it c. its large surface area d. its small surface area
**2. Which of the following places has the highest atmospheric pressure?** a. on the top of a hill b. in a cable car c. on the roof top of a tall building d. at the bottom of the sea
**3. Wind blows ** a. from areas of high atmospheric pressure to low pressure areas b. from areas of low atmospheric pressure to high pressure areas c. only at areas above normal atmospheric pressure d. only at areas below normal atmospheric pressure
**4. The pressure in a liquid decreases with** a. increase in surface area b. decrease in surface area c. increase in depth d. decrease in depth
**5. A simple barometer filled with water has to have a minimum length of** a. 1cm b. 10cm c. 10m d. 100m
**6. A block of wood measuring 6m by 3m by 0.5m is placed on a table. If the mass of the block of wood is 4500kg, what is the pressure on the table due to the block? take gravitational force acting on a mass of 1kg to be 10N** a. 2500Pa b. 5000Pa c. 9000Pa d.22500Pa
7. A man stands on snow wearing a pair of skis. The total mass of the man is 60kg and each of the skis has an area of 0.2m^{2} in contact with the snow. A 1kg mass has a gravitational force of 10N acting on it. What pressure does the man exert on the snow? a. 15N/m^{2}b. 30N/m^{2}c. 1500N/m^{2}d. 3000N/m^{2}**8. Which of the following does not cause the height of the mercury column of a simple mercury barometer to vary?** a. changes in atmospheric pressure b. changes in temperature of the mercury c. changes in the value of g d. evaporation of mercury from the barometer reservoir e. leakage of air into the tube
**9. In which of the following examples is the greatest pressure exerted?** a. a barefooted person standing on the beach b. a brick resting on the ground c. a book resting on a table d. an elephant standing on the ground e. a knife cutting a piece of meat
**10. A tank 3 m long, 1 m wide, and 0.5 m deep is filled with oil which weighs 12 000 N. What is the pressure on the base of the tank due to the oil?** a. 4000 Pa b. 6000 Pa c. 8000 Pa d. 18 000 Pa e. 24 000 Pa
**11. Water of depth 10m exerts a pressure equal to atmospheric pressure. An air bubble rises to the surface of a lake which is 20m deep. When the bubble reaches the surface, its volume is 6cm**^{3}. What was the volume of the air bubble at the bottom of the lake?
a. 2cm^{3} b. 3cm^{3} c. 12cm^{3} d. 18cm^{3} ^{}
**12. A rigid tank containing air at atmospheric pressure has a capacity of 5000 cm**^{3}. A bicycle hand pump with a capacity of 500cm^{3} is used to pump more air into the rigid container. Given that the hand pump is pumped twice to push two tubes of air into the tank, what is the final air pressure in the tank? (take atmospheric pressure as 100 000 Pa)a. 100 000 Pa b. 120 000 Pa c. 500 000 Pa d. 1 000 000 Pa
**13. Oxygen can be supplied to a fish tank by bubbling air into the water. What happens to the pressure and the volume of air bubbles while they are rising?**
__pressure__ __volume__
a. decreases decreases b. decreases increases c. increases decreases d. increases increases
**14. The rudder of a large ship is operated hydraulically. Oil at high pressure exerts a force F on a piston that in turn moves a lever arm.**
**The oil is at a pressure of 500kPa and the surface area of the piston is 0.20m**^{2}. What is the size of F?
a. 100N b. 2500N c. 100 000N d. 250 000N
**15. A manometer contains water and a liquid X. The two liquids do not mix. The pressure at level P in the water is equal to the pressure at level P in liquid X. The density of water is 1000 kg/m**^{3}. What is the density of liquid X?
a. 500 kg/m^{3} b. 800 kg/m^{3} c. 1250 kg/m^{3} d. 2000 kg/m^{3} ### MCQ Answers1. d 2. d 3. a 4. d 5. c 6. a
7. c 8. d ^{}9. d 10. a 11. a 12. b total vol of air at atm = 5000 + (2 x 500) = 6000cm^{3} P_{1}V_{1} = 100 000 x 6000 Vol of air in rigid container after pumping 2 tubes of air into it = 5000cm^{3} P_{2}V_{2} = P_{2} x 5000 P_{1}V_{1} = P_{2}V_{2} 100 000 x 6000 = P_{2} x 5000 P_{2} = 120 000 Pa 13. b 14. c 15. b ## Structured Question Worked Solutions
**1. The figure shows a mercury barometer on a day when the atmospheric pressure is 750mmHg. What is the pressure at point B, at the bottom of the mercury reservoir?**
**Solution**
Pressure = 750 + 80 = 830 mmHg
**2. The figure shows a manometer with limbs of cross-sectional area of 0.0015m**^{2}. It contains a liquid which exerts a pressure of 5000Nm^{-3}. Calculate i. the volume of liquid between the levels PQ and RS in the left-hand tube ii. the weight of the volume of liquid in i. iii. the excess pressure, in Nm^{-2}, of the gas supply above the surrounding atmospheric pressure
**Solution**
i. volume = 0.0015 x 0.5 = 0.00075m^{3}ii. weight = 0.00075 x 5000 = 3.75N iii. pressure = 3.75/0.00015 = 2500Nm^{-2}
**3. When a block of metal of mass 1.2kg stands on a horizontal surface, the area of contact between the block and the surface is 8.0cm2. Assuming that the force of gravity acting on a mass of 1kg is 10N, calculate the pressure exerted by the block on the surface.**
**Solution**
pressure = force/area = (1.2 x 10)/8 = 1.5Nm^{-24. The figure shows a U-tube manometer connected to a gas cylinder of large volume. The atmospheric pressure is 76cm of mercury.ai. What is the pressure at A in the right-hand tube?aii. What is the pressure at B in the left-hand tube?b. The tap is opened and mercury is run out until the level in the left-hand tube drops to the 60cm mark. i. assuming that the pressure in the gas cylinder remains constant, what is the new position of the level in the right-hand tube?ii. Explain how you arrived at your answer.}**Solution**^{ai. 76cmHg. (because A is exposed to the atmosphere so it experiences atmospheric pressure)aii. 106cmHg (because the pressure at B equals the point in line with it in the other tube. The pressure there is atmospheric pressure + pressure due to the column of mercury) = 76 + 30bi. The new position in the right-hand tube is at the 30cm markbii. The pressure of the gas cylinder remains the same, hence the difference between the levels of mercury in A and B remains constant. Since B drops by 20cm, A likewise will drop by 20cm from the 10cm mark to the 30cm mark.}**5. The figure shows two vertical tubes P and Q, each closed at the upper end. The pressure in the space above the mercury meniscus in tube P is negligibly small. There is a small amount of air in this space in the tube Q. ** The density of mercury is 13.6 x 10 kg/m^{3}. The gravitational force on a mass of 1.00kg is 10.0N. Determine i. the atmospheric pressure, in Pa, at that time. ii. the pressure, in Pa, exerted by the air in the space at the top of tube Q^{Solution}i. Atmosphere pressure = hpg = (75.0/100)(13.6 x 10^{3})(10.0) = 1.02 x 10^{5 }Pa^{} ii. air pressure at top of tube Q = atmospheric pressure - liquid pressure due to 60.0cm of mercury = 1.02 x 105 - (60.0/100)(13.6 x 10^{3})(10.0) = 2.04 x 10^{4} Pa^{}
**6. The tyres of a car are in contact with the ground over a total area 3.0 x 10**^{-2} m^{2}. The total weight of the car is 6300N. Calculate the pressure exerted by the tyres on the ground.
Why would you expect the temperature of the tyres to have risen after the car has been in motion for some time?
**Solution**Pressure exerted by the tyres = weight/area = 6300/ 3.0 x 10^{-2} = 210 000 N/m ^{2}The temperature of the tyres rises because the work done in overcoming friction with the road is transformed to heat energy |