DENSITYAND SPECIFIC GRAVITY DENSITY Maybe we have encountered this tricky question: “ Which is heavier, 1 kg of cotton or 1 kg of nails?”,or “Which is lighter, cork or plywood?” A cork is lighter than wood if the volumes of both the cork and the wood are the same. The preceding statement involves two important physical quantities: volume and mass. DENSITY Density is a measure of how matter is squeezed together in a given amount of space. It is defined as an objects mass per unit volume. It is one important property of a substance. The density can be expressed as ρ = m / V where: ρ = density (kg/m 3 ) m = mass (kg) V = volume (m 3 ) The higher the density, the tighter the particles are packed inside the substance. Density is a physical property constant at a given temperature and density can help to identify a substance. The density of solids is fairly constant, but for fluids, it is more variable because fluids are more compressible, especially gases. Volumes of solids are usually expressed in cubic meters (m 3 ) and those of fluids in liters (L). The following may be helpful in converting from liters to cubic meters and vice versa: 1 L = 0.001 m 3 1 m 3 = 1000 L 1 L = 1000 cm 3 1 mL = 1 cm 3 DENSITY UNITS OF DENSITY The SI units for density are kg/m 3 . The imperial (U.S.) units are lb/ft 3 (slugs/ft 3 ). While people often use pounds per cubic foot as a measure of density in the U.S., pounds are really a measure of force, not mass. Slugs are the correct measure of mass. You can multiply slugs by 32.2 for a rough value in pounds. DENSITIES OF SOME COMMON SUBSTANCES Material Density (kg/m 3 ) Material Density (kg/m 3 ) Air(1atm,20 0 C) 1.20 Aluminum 2.7 x 10 3 Ethanol 0.81 x 10 3 Iron, steel 7.8 x 10 3 Benzene 0.90 x 10 3 Brass 8.6 x 10 3 Ice 0.92 x 10 3 Copper 8.9 x 10 3 Water 1.0 x 10 3 Silver 10.5 x 10 3 Seawater 1.03x 10 3 Lead 11.3 x 10 3 Blood 1.06 x 10 3 Mercury 13.6 x 10 3 Glycerin 1.26 x 10 3 Gold 19.3 x 10 3 Concrete 2 x 10 3 Platinum 21.4 x 10 3 EXAMPLE 1 An irregular-shaped piece of copper has an unknown mass. When it is placed in a cup of water filled to the brim, 1.26 mL of water overflows. What is the mass of the piece of copper? (density of copper = 8.9 x 10 3 ) Solution: The volume of water that overflows is equal to the volume of the copper (1.26 mL). Convert mL to m 3 : 1.26 mL (1L/1000 mL)(0.001 m 3 /1L ) = 1.26 x 10 -6 m 3 ρ = m/V; m = ρV m = (8.9 x 10 3 )(1.26 x 10 -6 m 3 ) m = 0.0112 kg or 11.2 g Find the density of a 0.1 m 3 aluminum having a mass of 270 kg. Given: V = 0.1 m 3 m = 270 kg ρ = ? Solution: ρ = m = 270 kg = 2,700 kg/m 3 V 0.1 m 3 EXAMPLE 2 Find the density of a 51 grams gasoline that occupies 75 cm 3 . Given: V = 75 cm 3 m = 51 g ρ = ? Solution: ρ = m = 51 g = 0.68 g/cm 3 V 75 cm 3 EXAMPLE 3 Determine the mass of a solid iron wrecking ball with a radius of 20 cm. Given: r = 20 cm ρ iron = 7.8 g /cm 3 m = ? Solution: Solve first for the volume of the ball (sphere): V = 4/3 πr 3 = 4/3 π (20cm) 3 = 3.35 x 10 4 cm3 Then, solve for the mass: m = ρV = (7.8 g/cm 3 )(3.35 x 10 4 cm 3 ) m = 2.61 x 10 5 g EXAMPLE 4 EXAMPLE 5 What is the density of a box that is 1.2 m x 2.5 m x 3.0 m and weighs 50 N? Solution: ρ = m ; W = mg V ρ = m = 50N/(9.8 m/s 2 ) = 5.10 kg = 0.57 kg/m 3 V 1.2m x 2.5m x 3.0m 9 m 3 RELATIVE DENSITY Another variable, called specific gravity, is used to identify materials. Relative density, or specific gravity, is the ratio of the density of a substance to the density of a given reference material. Specific gravity usually means relative density with respect to water. The term "relative density" is often preferred in modern scientific usage. specific gravity = ρ / ρ w RELATIVE DENSITY If a substance's relative density is less than one then it is less dense than the reference; if greater than 1 then it is denser than the reference. If the relative density is exactly 1 then the densities are equal; that is, equal volumes of the two substances have the same mass. If the reference material is water then a substance with a relative density (or specific gravity) less than 1 will float in water. For example, an ice cube, with a relative density of about 0.91, will float. A substance with a relative density greater than 1 will sink. Specific gravity is often used by geologists to help determine the mineral content of a rock sample. In industry, specific gravity is used to determine the concentrations of substances in aqueous solutions. RELATIVE DENSITY EXAMPLE 1 What is the specific gravity of gold? Given: ρ gold = 19,320 kg/m 3 Solution: specific gravity = ρ gold ρ w = 19,320 kg/m 3 1000 kg/m 3 = 19.32 The specific gravity of aluminum is 2.7. Find its density in kg/m 3 . Solution: Sp.gr = ρ aluminum ρ water ρ aluminum = (Sp. gr)(ρ water ) = 2.7(1000 kg/m 3 ) ρ aluminum = 2,700 kg/m 3 EXAMPLE 2 What is the mass in kilograms of 1 liter of kerosene which has specific gravity of 0.8? Given: V = 1 L Sp. gr = 0.8 m = ? Solution: Sp.gr = ρ kerosene ρ water ρ kerosene = (Sp. gr)(ρ water ) = 0.8(1000 kg/m 3 ) = 800 kg/m 3 m = ρV = (800 kg/m 3 )(0.001m 3 ) = 0.8 kg EXAMPLE 3 PRESSURE PRESSURE Which can cause greater damage to the floor: the boots or the stiletto heels? The woman wearing the stiletto heels may weigh lighter than the man wearing boots, but her stiletto heels are more damaging to the floor. This is because when the woman is standing, her weight is concentrated on the small area under the forefoot and heels of each shoe. Force applied on a smaller area produces a larger amount of pressure. How can someone lie on a bed of nails without getting hurt? A nail has a pointed tip, which concentrates a hammer’s blow onto a small area and produces pressure big enough to part wood fibers and embed the nail. However, in the case of the man lying on a bed of nails, because his weight is distributed on so many nails, the pressure created by each nail on his body is not enough to puncture him. PRESSURE PRESSURE Pressure is a scalar quantity. In standard units, pressure is measured in newtons per square meter (N/m 2 ),also known as pascal (Pa). One pascal is the pressure exerted by a force of 1 N on an area of 1m 2 . 1 N/m 2 = 1 Pa BLAISE PASCAL The unit pascal was named after Blaise Pascal (1623 – 1662), a french physicist and mathematician, for his experiment works on the pressure exerted by liquids and gases. PRESSURE Pressure is defined as the force acting per unit area (P = F/A). PRESSURE For a given magnitude of force, the pressure this force exerts can be increased or decreased by adjusting the area on which it is applied. Thus to create a smaller amount of pressure, the area should be made large. It therefore makes sense that skis and snow shoes are spread out and frogs have webbed feet. On the other hand, a small area causes a bigger amount of pressure. This is the reason why it hurts a lot when someone thrusts his or her elbow or leans over your thigh. APPLICATIONS OF PRESSURE HIGH PRESSURE Each scissors and knife has a very small surface area on its cutting edge so that a high pressure can be exerted to cut something. Each ice skate has a sharp blade that enables the skaters to glide smoothly across the surface of ice. The skater’s weight acting on a small area of contact gives a high pressure on ice. Ice will melt under high pressure. Therefore, when skating, you are actually gliding through a layer of water. LOW PRESSURE Snowshoes reduce the pressure acting on snow by increasing the area over which the weight is spread. Thus, one can walk on deep snow without sinking in. The large tires of tractors increase the area of contact with ground and this reduces the pressure exerted onto the ground. Therefore, the tractors can move without sinking into the ground. EXAMPLE 1 A 1.0 m x 1.0 m x 2.0 m block weighs 100 N. Calculate the pressure under the block when it is lying on its a. rectangular surface; and on its b. square surface. 1.0 m 2.0 m 1.0 m 1.0m 2.0 m 1.0 m SOLUTION The pressure depends on the area underneath the block. a. The area of the rectangular surface of the block is A = 2.0 m x 1.0 m = 2.0 m 2 P = _F_ = 100 N = 50 N/m 2 or 50 Pa A 2.0 m 2 b. The area of the square surface is A = 1.0 m x 1.0 m = 1.0 m 2 P = _F_ = 100 N = 100 N/m 2 or 100 Pa A 1.0 m 2 EXAMPLE 2 Consider a car with a mass of 2,000 kg. The surface area of the part of each tire that is in contact with the floor is 15 cm x 20 cm. What is the pressure under each of the car’s tire?(Assuming that the car’s weight is distributed evenly by the four wheels.) Solution: P = _F_ = ¼ W = W = mg = (2,000 kg)(9.8 m/s 2 ) = A A 4A 4A 4(.15 m x .20 m) P = 1.63 x 10 5 Pa EXERCISE A box has a mass of 18 kg. It has a height of 0.55 m. and bottom dimensions of 0.70 m x 0.60 m. a. What force does the box exert on the floor? b. What pressure does the box exert on the floor? Solution: a. F = mg = 18 kg(9.8 m/s 2 ) = 176.4 N b. P = F = 176.4 N = 420 Pa A 0.70 m x 0.60 m The atmosphere is a deep ocean of gases piled several kilometers above us. The weight of all those gases causes a downward pressure on us. The pressure of the atmosphere helps you sip your drink using a straw. By emptying your mouth of air, you create a partial vacuum, allowing the atmosphere to push your drink up the straw and making the liquid move against the flow of gravity. ATMOSPHERIC PRESSURE ATMOSPHERIC PRESSURE Atmospheric pressure is the force per unit area exerted into a surface by the weight of air above that surface in the atmosphere of Earth (or that of another planet). In most circumstances atmospheric pressure is closely approximated by the hydrostatic pressure caused by the weight of air above the measurement point. Low-pressure areas have less atmospheric mass above their location, whereas high-pressure areas have more atmospheric mass above their location. Likewise, as elevation increases, there is less overlying atmospheric mass, so that pressure decreases with increasing elevation. The pressure of the atmosphere at sea level is about 1.013 x 10 5 Pa. This means that every square meter on the Earth’s surface is weighed down by a force of 101.3 kilonewtons (kN). This is like being squashed under the weight of two male elephants. ATMOSPHERIC PRESSURE UNITS OF PRESSURE The pressure at any point in the Earth’s atmosphere may be expressed in a non- standard unit called atmosphere (atm). A pressure of 1.013 x 10 5 Pa is equal to 1 atm. 1.013 x 10 5 Pa = 1 atm Other non-standard units of pressure are bar; millimeter of mercury (mm Hg), which is also known as torr; and pound per square inch (lb/in 2 or psi). The bar and millibar (mbar) are widely used in meteorology. EVANGELISTA TORRICELLI The unit mm Hg is used in measuring blood pressure and is derived from the use of the mercury barometer, a pressure gauge invented by the Italian inventor, Evangelista (1608-1647). Evangelista Torricelli discovered Torricelli's Law, regarding the speed of a fluid flowing out of an opening, which was later shown to be a particular case of Bernoulli's principle. The following equivalences are useful in the conversion of pressure measurements: 1 bar = 10 5 Pa 1 millibar = 10 2 Pa 1 atm = 1.013 x 10 5 Pa = 1.013 bar = 760 mm Hg = 14.70 psi (lb/in 2 ) 1 kPa = 1000 Pa 1 mm Hg = 1 torr = 133 N/m 2 EXAMPLE The pressure of the atmosphere is 1.013 x 10 5 Pa. How much force does the atmosphere exert on the roof of a house, assuming that it is a flat roof with dimensions of 9.0 m x 12.0 m? Solution: F = PA F = (1.013 x 10 5 N/m 2 )(9.0 m x 12.0 m) F = (1.013 x 10 5 N/m 2 )(108m 2 ) F = 1.09 x 10 7 N F F F F A 1 A 2 A 3 A 4 A 6 A 5 Force acts perpendicularly on each face. Pressure is the force acting on a unit area. Air pressure on the top face of the cube is due to the weight of the air above it. Thus, the force on the top face is equal to the weight W of the air column above it. Hence, F = W = mg Where: m = mass of air column and g = acceleration due to gravity But: mass m = volume V x density ρ and volume V = area A x height h Therefore: m = Vρ = Ahρ. Since F = mg = Ahρg. From P = F = Ahρg A A We now have P = hρg. PRESSURE DEPENDS ON HEIGHT AND DENSITY From the equation P = ρgh, since g is constant for a specific location, air pressure P depends on height h and density ρ of air. As height increases, or ad one goes higher into the atmosphere, atmospheric pressure decreases. Hence, atmospheric pressure is higher in lowlands than on mountains. The human body is used to atmospheric pressure on the earth’s surface. Airplane passengers eventually feel pressure changes in their ears when the airplane ascends or descends. The ears are built-in body pressure measurers. They feel air pressure changes as painful sensations. Air pressure inside aircrafts should approximate atmospheric pressure on the earth’s surface to ensure the passengers’ comfort and well-being. PRESSURE DEPENDS ON HEIGHT AND DENSITY LIQUID PRESSURE Do you feel the pressure of water acting against your eardrums when you swim under water? Observe that the deeper you swim, the greater the pressure you feel. What causes this pressure? It is caused by the weight of the water above you that pushes against you. If you swim three times as deep, the weight of the water above will also be three times. Therefore the water pressure will be three times as great. Thus, liquid pressure varies with depth. PRESSURE VARIES WITH DEPTH Consider a beaker which contains water at a height h. If you double the height of water (thus doubling the weight), the pressure against the bottom of the beaker becomes twice as great. Furthermore, if the liquid is two or three times as dense, liquid pressure is correspondingly two or three times as great at any depth. For instance, if you swim in the sea, which is denser than ordinary water, the pressure will be greater proportionally. Thus, liquid pressure varies with density. LIQUID PRESSURE PRESSURE VARIES WITH DENSITY These vessels differ in shape but contain water of the same height. The pressure at the bottom is independent of the shape of the container provided that the heights of the liquids are exactly the same. This means that the higher the column of liquid, the greater is its pressure: and the more dense the liquid is, the greater too is the pressure. This is the formula for hydrostatic pressure. The height h is always the distance from the surface of the fluid down to the point that is being studied. Thus, it is better to use the word depth to refer to the variable h in the equation. Since both ρ and g are constants, it is clear that hydrostatic pressure varies only with the depth of the fluid column. A scuba diver, for example, experiences greater pressure as he/she dives deeper. This increase of pressure with depth puts limits on scuba diving. P = ρgh Another implication of the equation, P = ρgh, is that for the same fluid, pressure at the same level is the same everywhere. This is what we mean when we say that “water seeks its own level”, as shown in the figure. The water pressure at the faucets in our home depends on the height of the water level at the reservoir. However, pumps are often used to help increase the water pressure. The water level at the reservoir during rainy season is higher than it is during dry season. In some areas where there is no water system, people erect their own water tanks which they fill by pumping water from deep wells. The higher the tank, the greater the pressure available at the water taps. It is clear that the water pressure does not depend on the diameter of the tank. Suppose a rectangular tank is filled with a liquid of density ρ to a depth h. The liquid pressure is uniform at the bottom because all points are at the same depth from the surface the liquid. The pressure is not constant at the sides of the tank, however. The pressure varies from zero at the top, where h is zero, to maximum at the bottom where the pressure is equal to ρgh. To get the force acting on the side, the area is multiplied by the average pressure. For a rectangular side, the average pressure is the average of the pressure at the top and the pressure at the bottom. Or the average pressure is: P ave = ½ ρgh EXAMPLE A rectangular tank 4 meters long, 2 meters wide, and 1 meter high is filled with water. Find the: a. liquid pressure at the bottom; b. force on the bottom due to the liquid F 1 ; c. total force on one side of the tank due to the liquid F 2 . F 1 F 2 1 m 2 m 4 m SOLUTION a. P = ρgh = (1000 kg/m 3 )(9.8 m/s 2 )(1m) = 9,800 N/m 2 b. F 1 = PA = (9,800 N/m 2 )(4m x 2m) = 78,400 N c. The total force on one side of the tank is the average pressure multiplied by the area of the side. The average pressure is: P ave = ½ ρgh = ½ (9,800 N/m 2 ) = 4,900 N/m 2 The area of the side is the area of a rectangular of length 4m and of width 1 m. Hence the force is: F 2 = P ave A = (4,900 N/m 2 )(4m x 1m) = 19,600 N EXAMPLE Calculate the pressure on a scuba diver at a depth of 45.0 m. Given: ρ = 1030 kg/m 3 h = 45 m P = ? Solution: P = ρgh P = (1030 kg/m 3 )(9.8m/s 2 )(45 m) P = 454,230 Pa EXAMPLE A man dove 35 m under the sea. If the seawater has a density of 1.03 g/cm 3 , how much pressure in N/m 2 acts on the body of a man? Given: ρ = 1.03 g/cm 3 h = 35 m P = ? Solution: P = ρgh P = (1030 kg/m 3 )(9.8 m/s 2 )(35 m) P = 353,290 N/m 2 EXAMPLE Calculate the pressure of water at the bottom of a swimming pool which is 1.8 m deep. Given: ρ = 1000 kg/m 3 h = 1.8 m P =? Solution: P = ρgh P = (1000 kg/m 3 )(9.8 m/s 2 )(1.8 m) P = 17,640 N/m 2 or 17,640 Pa EXERCISE A swimming pool is 50 m long and 25 m wide. The deep end has a depth of 2.0 m and the shallow end, 1.22 m. What are the pressures at the bottom of a. the deep end; and b. the shallow end. Solution: a. P = ρgh = (1000kg/m 3 )(9.8 m/s 2 )(2 m) = 19,600 Pa b. P = ρgh = (1000kg/m 3 )(9.8 m/s 2 )(1.22 m) = 11,956 Pa EXERCISE At a depth of 5.0 m, what is the pressure experienced by a fish? (Assuming that a fish is swimming at freshwater) Given: h = 5.0 m ρ w = 1000 k/m 3 Solution: Neglecting the effect of atmosphere, P = ρ w gh P = (1000 k/m 3 )(9.8 m/s 2 )(5 m) P = 49,000 Pa or 49 kPa Eugene is an expert diver and is exploring an underwater wreck. If he is at the surface, at what depth will he experience double the pressure that is on him now? Given: P i = 1.01 x 10 5 Pa Solution: For Eugene to experience double the pressure: P f = 2 P i Therefore, ∆P = P f – P i = 1.01 x 10 5 Pa And the change in depth should be: ∆h = ∆P = 1.01 x 10 5 Pa ______ ρ w g (1000kg.m 3 )(9.8m/s 2 ) ∆h = 10.31 m EXERCISE MEASURING FLUID PRESSURE The pressure of fluids is usually measured in terms of gauge pressure and not absolute pressure. The gauge pressure is the pressure determined by a measuring- device and is equal to the difference between the pressure of the fluid and the pressure of the atmosphere. For example, when you use a tire gauge, the reading on the gauge tells the difference between the pressure of the air inside the tire (absolute pressure of the air) and the pressure of the atmosphere. This pressure difference is the pressure gauge. P gauge = P a – P atm MANOMETER Another instrument which is used to measure gauge pressure is the manometer. A manometer may be made of a U-shaped tube containing a liquid, either mercury or water, which we will call the gauge fluid. One end of the tube, the open end, is exposed to the atmosphere; the other end, the closed end, is connected to a vessel containing the fluid whose pressure is being measured, or the test fluid. Both the atmosphere and the test fluid exert pressure on the gauge fluid. The difference between the levels of the gauge fluid in the two arms of the U-tube is the pressure of the test fluid. MANOMETER At the bottom of the closed arm, the pressure is the sum of the absolute P a of the test fluid and the pressure due to the column of gauge fluid in the arm. Since the column has a height of y 1 , the pressure exerted by this column is ρgy 1 . Thus, P closed arm = P a + ρgy 1 At the bottom of the open arm, the pressure is the sum of the pressure of the atmosphere P atm and the pressure exerted by the column of gauge fluid with height y 2 . P open arm = P atm + ρgy 2 The pressures at the bottom of the closed arm and that at the open arm are equal because they are the pressures at the same point on the U-tube. We can thus equate the expressions for these two pressures: P closed arm = P open arm P a + ρgy 1 = P atm + ρgy 2 Rearranging the equation to isolate Pa, we get: P a = P atm + ρgy 2 - ρgy 1 P a = P atm + ρg(y 2 - y 1 ) Thus the difference between the gauge fluid levels in the two arms of the tube, y 2 – y 1 , determines the absolute pressure of the test fluid. If we let y 2 – y 1 = h, we get another expression for absolute pressure. P a = P atm + ρgh The difference P a – P atm is the gauge pressure of the manometer. Equation can be rewritten as: P a – P atm = ρgh P gauge = ρgh MERCURY BAROMETER Invented by Evangelista Torricelli in 1643, this instrument is essentially a glass tube of mercury that is inverted into an open dish. The mercury spills into the dish but not completely because the atmosphere pushes on the mercury, so there is always some mercury left inside the tube. The height of the remaining mercury inside the tube depends on the pressure of the atmosphere. The top of the mercury column contains only mercury vapor, which has negligible(essentially zero) pressure. At the level of the mercury in the dish, the pressure due to the column of mercury is ρgh, where h is the height of the mercury column above the mercury dish. This same pressure is equal to atmospheric pressure since the dish is open to the atmosphere, so that in a mercury barometer, P atm = ρgh MERCURY BAROMETER At sea level (at normal level), the height of the column in a mercury barometer is always equal to 760 mm. Thus the value of atmospheric pressure may be given as 760 mmHg. The use of the mercury barometer ha also resulted to the use of millimeter of mercury as a unit of pressure. MERCURY BAROMETER ANEROID BAROMETER An aneroid barometer, invented in 1843 by French scientist Lucien Vidie uses a small, flexible metal box called an aneroid cell (capsule), which is made from an alloy of beryllium and copper. The evacuated capsule (or usually more capsules) is prevented from collapsing by a strong spring. Small changes in external air pressure cause the cell to expand or contract. This expansion and contraction drives mechanical levers such that the tiny movements of the capsule are amplified and displayed on the face of the aneroid barometer. Many models include a manually set needle which is used to mark the current measurement so a change can be seen. In addition, the mechanism is made deliberately "stiff" so that tapping the barometer reveals whether the pressure is rising or falling as the pointer moves. ANEROID BAROMETER OLD ANEROID BAROMETER MODERN ANEROID BAROMETER ANEROID BAROGRAPH A barograph is a recording aneroid barometer. It produces a paper or foil chart called a barogram that records the barometric pressure over time. Barographs use one or more aneroid cells acting through a gear or lever train to drive a recording arm that has at its extreme end either a scribe or a pen. A scribe records on smoked foil while a pen records on paper using ink, held in a knob. The recording material is mounted on a cylindrical drum which is rotated slowly by clockwork. Commonly, the drum makes one revolution per day, per week, or per month and the rotation rate can often be selected by the user. Since the amount of movement that can be generated by a single aneroid is minuscule, up to seven aneroids (so called Vidie-cans) are often stacked "in series" to amplify their motion. It was invented in 1843 by the Frenchman Lucien Vidie (1805–1866). ANEROID BAROGRAPH BOURDON GAUGE It is a type of aneroid pressure gauge consisting of a flattened curved tube attached to a pointer that moves around a dial. As the pressure in the tube increases the tube tends to straighten and the pointer indicates the applied pressure. It is named after Eugène Bourdon (1808-84), French hydraulic engineer, who invented it. BOURDON GAUGE SPHYGMOMANOMETER A sphygmomanometer (SFIG-moh-mə-NOM-i-tər) or blood pressure meter (also referred to as a sphygmometer) is a device used to measure blood pressure, composed of an inflatable cuff to restrict blood flow, and a mercury or mechanical manometer to measure the pressure. It is always used in conjunction with a means to determine at what pressure blood flow is just starting, and at what pressure it is unimpeded. Manual sphygmomanometers are used in conjunction with a stethoscope. The word comes from the Greek sphygmós (pulse), plus the scientific term manometer (pressure meter). The device was invented by Samuel Siegfried Karl Ritter von Basch in 1881. Scipione Riva-Rocci introduced a more easily used version in 1896. In 1901, Harvey Cushing modernized the device and popularized it within the medical community. A sphygmomanometer consists of an inflatable cuff, a measuring unit (the mercury manometer, or aneroid gauge), and inflation bulb and valve, for manual instruments. EXERCISE A barometer is constructed with water inside an inverted glass tube. The pressure of the atmosphere is equal to 1.03 x 10 5 Pa. How high is the water column in meters? Solution: P atm = ρgh h = P atm = __1.03 x 10 5 N/m 2 ___ ρg 1000kg/m 3 (9.8m/s 2 ) h = 10.34 m EXERCISE A 1.22-m tall tank is filled with water. What is the gauge pressure and the absolute pressure at the bottom of the tank? Solution: The gauge pressure is equal to the pressure due to the water at any point. The water sits 1.22 m above the bottom of the tank, so the gauge pressure is: P gauge = ρgh P gauge = (1000 kg/m 3 )(9.8 m/s 2 )(1.22 m) P gauge = 11,956 Pa The absolute pressure can now be calculated as the sum of the gauge pressure and the atmospheric pressure: P a = P gauge + P atm P a = 11,956 Pa + 1.01 x 10 5 Pa P a = 112,956 Pa