Fluids Physics Book PDF

Physics Book (Part 2) Mechanics Lesson 8: Fluids at Rest 8.1 Pressure in Fluids..........................................................................113 8.2 Archimedes Principle......................................................

Physics Book (Part 2) Mechanics Lesson 8: Fluids at Rest 8.1 Pressure in Fluids..........................................................................113 8.2 Archimedes Principle....................................................................114 8.3 Gas laws........................................................................................115 Numerical Problems........................................................................117 Experiment M6: Boyle’s Law..........................................................118 Sample LC Questions.......................................................................119 Lesson 8 Solutions..........................................................................267 © Grinds360 Page 112 Kieran Mills & Tony Kelly  Physics Book (Part 2) 8.1 Pressure in fluids Fluids are substances which can flow. Examples of fluids are liquids and gases. Definition: The density of a substance is its mass per unit volume. m r: Density (kg m–3) Formulae and Tables Book: Page 57  m: Mass (kg) V Density (Mechanics) V: Volume (m3) Definition: The pressure is the force per unit area. F P: Pressure (N m–2 or Pascal Pa) Formulae and Tables Book: Page 57 P= F: Force (N) Pressure (Mechanics) A A: Area (m2) Example 1: Find the pressure exerted by a 75 kg person (a) on a snow shoe of area 400 cm2, (b) on a stilleto heel of area 0.25 cm2. (g = 9·8 m s–2) Solution The pressure at a point in a Pressure in a liquid at any point is exerted liquid depends only on the A equally in all directions. depth and the density of the liquid. h The water in the container below is exerted F mg Vg perpendicular to its sides. This can be P     gh P A A A demonstrated by punching holes in the side and viewing the streams of water that emerge. P = rgh P: Pressure (Pa) –3 Water pressure is pushed perpendicular to the r: Density (kg m ) sides of its container and the pressure increases g: Acceleration due to gravity (m s–2) with increasing depth. h: Depth (m) Formulae and Tables Book: Page 57 Pressure in a fluid (Mechanics) This means that if a liquid is poured into a number of differently shaped interconnected vessels, the depth of liquid is the same. © Grinds360 Page 113 Kieran Mills & Tony Kelly  Physics Book (Part 2) Example 2: Find the atmospheric pressure in Pa corresponding to 760 mm of Hg. Find the total pressure at a depth of 20 m below the surface of a lake. (g = 9·8 m s–2; Density of water = 1000 kg m–3; Density of Hg = 13 600 kg m–3) Solution Example 3: What force is exerted on the bottom of a tank of uniform cross-sectional area 3 m2 by water which fills it to a depth of 0.5 m? (g = 9·8 m s–2; Density of water = 1000 kg m–3) Solution 8.2 Archimedes Principle When a body is placed in a fluid it will do one of three things: (i) stay where it is or, (ii) float to the surface or, (iii) sink to the bottom. These are all consequences of the force on the body due to the fluid itself called the buoyant force or upthrust. Definition: Archimedes Principle states that when a body is partly or wholly immersed in a fluid, the upthrust (buoyant force) is equal to the weight of the fluid displaced and acts up through the centre of gravity of the displaced fluid. (i) If B = W it stays where it is (the density of the body equals the density of the fluid) (ii) If B > W it floats up to the surface (the density of the body is less than the density of the fluid) (iii) If B < W it sinks to the bottom (the density of the body is greater than the density of the fluid) Law of Flotation This law deals with floating objects. It is a special case of Archimedes Principle. An object which floats has no apparent weight. So Archime- des Principle states: Weight = Weight of fluid displaced = Buoyant force Definition: The Law of Flotation states that a floating object displaces a weight of fluid equal to its own weight. © Grinds360 Page 114 Kieran Mills & Tony Kelly  Physics Book (Part 2) Demonstration: Archimedes Principle Apparatus: Shown in diagram. Method: 1. Weigh the glass block. 2. Weigh the beaker. 3. Immerse the block in water. Read off the weight of block in water on the spring balance. 4. Find the weight of displaced water. Observation: Loss in the weight of the block = Weight of water Demonstration: Law of Flotation Apparatus: Shown in diagram. Method: 1. Weigh a wooden block which floats. 2. It displaces a certain volume of water. Weigh the displaced water. Observation: Weight of block = Weight of displaced water STS: Hydrometer This is a device based on the Law of Flotation for measuring the densities of liquids. The hydrometer floats in the liquid whose density it is measuring. The more dense the liquid the smaller the volume displaced and the more the stem is above the surface of the liquid. The density can be read off the scale at the level of the liquid surface. Uses: 1. To check the density of the acid in a car battery. 2. To measure the density of milk in dairies - the density is lower as the percentage of fat increases. 8.3 The Gas Laws The behaviour of gases can be described in terms of three quantities: Pressure, P (Pa), Volume, V (m3) and Temperature, T (K). The formula connecting them was discovered by an Irish scientist, Robert Boyle. Definition: Boyle’s Law states that for a given mass of gas at constant temperature the pressure is inversely proportional to the volume. Experiment M6: To verify Boyle’s law. 1 P: Pressure (Pa) Formulae and Tables Book: Page 57 P∝ V: Volume (m3) Boyle’s law (Mechanics) V © Grinds360 Page 115 Kieran Mills & Tony Kelly  Physics Book (Part 2) 1 P∝ at constant temperature V or PV = k (constant) at constant temperature or P1V1 = P2V2 P1: Pressure of the gas in situation 1 V1: Volume of the gas in situation 1 P1V1 = P2V2 P2: Pressure of the gas in situation 2 V2: Volume of the gas in situation 2 Example 4: A gas at constant temperature is in a container with a piston. If the volume of the gas is 100 cm3 and the pressure is 1 × 105 Pa, find the volume if the pressure is increased to 2 × 105 Pa. Solution Atmospheric Pressure The earth is covered with a blanket of air to a height of about 150 km. It is called the atmosphere. The atmospheric pressure decreases with altitude because gravity causes the density and hence the pressure to be greater in the lower regions (aeroplane cabins at high altitudes have to be pressurised). Demonstration: Atmospheric Pressure Apparatus: Metal can with a screw-on cap, water, heat source Method: Put some water in a can and heat to boiling. When steam is gushing out for a few minutes turn off the heat and screw the cap firmly back on. Observation: The can collapses due to the pressure of the atmosphere. STS: Atmospheric pressure and Weather Falling pressure indicates that the weather is likely to become cloudier and wetter while rising pressure indicates the arrival of finer and drier weather. The “bends” in diving The bends is a painful, crippling and sometimes fatal affliction of divers. Because of increased pressure underwater the nitogen in their air tanks goes into solution in the blood. As the diver returns to the surface the nitrogen comes out of the blood as bubbles which can clog veins Answers to Examples Example 1: (a) 18 375 Pa, (b) 2∙94 × 107 Pa Example 2: 2∙973 × 105 Pa Example 3: 1∙47 × 104 N Example 4: 50 cm3 © Grinds360 Page 116 Kieran Mills & Tony Kelly  Physics Book (Part 2) Numerical Problems 1. The pressure of the hydraulic fluid in a car 7. A gas at constant temperature is expanded braking system rises to 1020 kPa when a from a volume of 20 cm3 to 80 cm3 by man puts his foot on the brake pedal. If the releasing the pressure on the piston. If the area of the piston in the brake cylinder is pressure was 3 × 105 N m-2. at 20 cm3, 9 cm2, find the force the brake pad exerts what is it at 80 cm3? on the disc. 8. A gas at constant temperature has its 2. The density of mercury is volume increased by 40%. What is the 13∙6 × 103 kg m-3. Calculate the liquid percentage decrease in the pressure? pressure 0.76 m below the surface of the liquid. (g = 9∙8 m s–2) Answers 3. A block of metal of density 4000 kg m-3 1. 816 N is 3 m high and stands on a square base of 2. 1∙013 × 105 Pa side 0.4 m. (g = 9∙8 m s–2) 3. (a) 0.16 m2 (b) 0.48 m3 (c) 1920 kg Find the (a) base area of the block, (d) 18 816 N (e) 1 176 000 Pa (b) volume of the block, 4. 1 × 105 Pa (c) mass of the block, 5. (a) 323 000 Pa (b) 425 400 Pa (d) weight of the block, (c) 8.8 m (d) 10∙9 m (e) pressure exerted by the weight of the block on the square base. 6. 31 360 N 7. 75 000 Pa 4. The atmospheric pressure on a particular 8. 28.6% day was 750 mm of Hg. Find the pressure in Pascals. (Density of Mercury = 13∙6 × 103 kg m-3; g = 9∙8 m s–2) 5. If the density of sea water is 1100 kg m-3. Find (a) the pressure below 30 m of sea water due to the water above, (b) the pressure at the same depth if the atmospheric pressure is 102 kPa, (c) the depth at which the pressure due to the water above is 95 kPa, (d) the depth at which the total pressure is 220 kPa. 6. What force is exerted on the bottom of a tank of uniform cross-sectional area 4 m2 by water which fills to a depth of 0∙8 m. (Density of water = 1000 kg m-3, g = 9∙8 m s–2) © Grinds360 Page 117 Kieran Mills & Tony Kelly  Physics Book (Part 2) M6: Experiment to verify Boyle’s law Diagram Volume scale 3 Volume V graduated in cm Pressure gauge (Bourdon gauge) Valve A To pump Reservoir of oil Method 1. Set up apparatus. 2. Connect a pump to apparatus and increase pressure until the pressure gauge reaches its maximum value. Close valve A. 3. Open valve to set pressure P to a certain value. Read off the corresponding value of the volume V of the air. Repeat for other values of P. 4. Record the results for P and V. 5. Draw a graph of pressure against one over volume. Results Pressure P (Pa) Volume V (cm3 ) 1 cm −3 V P (Pa) 1 (cm −3 ) V Sources of error and precautions 1. Wait a short time before taking a reading in order for the pressure and temperature to become uniform in the gas volume and the temperature to come back to room temperature. 2. Do not increase the pressure so that the density of the gas increases to too high a value. This is because at high values of pressure the gas departs from ideal gas behaviour. Boyle’s law only holds for ideal gases or real gases at low pressure. 3. Ensure that the temperature of the surroundings does not change appreciably. 4. Be careful when reading the volume of the enclosed gas. Make sure your eye is level with the horizontal which passes through the top of the oil meniscus. 5. Avoid the error of parallax in reading from the pressure scale and the volume scale. © Grinds360 Page 118 Kieran Mills & Tony Kelly  Physics Book (Part 2) LC 205: ([SHULPHQW0 %R\OH VODZ 1. In an experiment to verify Boyle’s law, a student measured the volume V of a fixed mass of gas at different values of the pressure p. The temperature of the gas was the same for each measurement. The following data were recorded. V (cm3) 80 120 160 200 240 280 p (kPa) 324 214 165 135 112 100 Describe, with the aid of a labelled diagram, how the student obtained the data. (12) Draw a suitable graph to show the relationship between the pressure of the gas and its volume. Explain how the graph verifies Boyle’s law. [Next page] (15) Use your graph to estimate the pressure of the gas at a volume of 250 cm3. Why might the temperature of the gas have changed significantly during the experiment? How did the student ensure that the temperature of the gas was the same for each measurement? (13) © Grinds360 Page 119 Kieran Mills & Tony Kelly  Physics Book (Part 2) Graph V (cm3) 80 120 160 200 240 280 p (kPa) 324 214 165 135 112 100 1 / cm -3 V © Grinds360 Page 120 Kieran Mills & Tony Kelly  Physics Book (Part 2) LC 2006: Pressure, Boyle's law 12. (a) Define pressure. (6) Is pressure a vector quantity or a scalar quantity? Justify your answer. (6) State Boyle’s law. (6) surface of lake A small bubble of gas rises from the bottom of a lake. The volume of the bubble increases threefold when it reaches the surface of the lake where the atmospheric pressure is 1.01 × 105 Pa. The temperature of the lake is 4 oC. Calculate (i) the pressure at the bottom of the lake; bottom of lake (ii) the depth of the lake. (10) (acceleration due to gravity = 9.8 m s–2; density of water = 1.0 × 103 kg m–3) © Grinds360 Page 121 Kieran Mills & Tony Kelly