| MadSci Network: Physics |
Dear Jen: Thank you very much for your question and your kind compliments. In order to answer properly your question, I need to address some issues beforehand. In the very nice site http://kids.earth.n asa.gov/archive/air_pressure/ we found the following information: Air pressure is the force exerted on you by the weight of tiny particles of air (air molecules). Although air molecules are invisible, they still have weight and take up space. Since there's a lot of "empty" space between air molecules, air can be compressed to fit in a smaller volume. When it's compressed, air is said to be "under high pressure". Air at sea level is what we're used to, in fact, we're so used to it that we forget we're actually feeling air pressure all the time! Weather forecasters measure air pressure with a barometer. Barometers are used to measure the current air pressure at a particular location in "inches of mercury" or in "millibars" (mb). A measurement of 29.92 inches of mercury is equivalent to 1013.25 millibars. How much pressure are you under? Earth's atmosphere is pressing against each square inch of you with a force of 1 kilogram per square centimeter (14.7 pounds per square inch). The force on 1,000 square centimeters (a little larger than a square foot) is about a ton! Why doesn't all that pressure squash me? Remember that you have air inside your body too, that air balances out the pressure outside so you stay nice and firm and not squishy. Air pressure can tell us about what kind of weather to expect as well. If a high pressure system is on its way, often you can expect cooler temperatures and clear skies. If a low pressure system is coming, then look for warmer weather, storms and rain. The Mercurial Barometer: Measuring Pressure Based on a principle developed by Evangelista Torricelli in 1643, the Mercurial Barometer is an instrument used for measuring the change in atmospheric pressure. It uses a long glass tube, open at one end and closed at the other. Air pressure is measured by observing the height of the column of mercury in the tube. At sea level, air pressure will push on the mercury at the open end and support a column of mercury about 30 inches high. If you used water instead of mercury, you would need a glass tube over 30 feet in length. As atmospheric pressure increases, the mercury is forced from the reservoir by the increasing air pressure and the column of mercury rises; when the atmospheric pressure decreases, the mercury flows back into the reservoir and the column of mercury is lowered. The site has a virtual barometer, and the following instructions for making your own barometer: Make Your Own Barometer Materials Needed: Drinking straw (clear plastic). Narrow-neck glass bottle. A rubber or cork stopper which fits in the neck of the bottle Instructions Insert a drinking straw into the bottle. Fill the bottle about half-way full of water. Seal the neck of the bottle around the straw either with the rubber stopper or a cork. Make sure the end of the straw is immersed in the water and that the water level in the straw is above the top of the bottle. As the air pressure outside the bottle decreases, the trapped air inside the bottle will push the water up the straw. As the air pressure outside the bottle increases, it will push the water farther down the straw. Please Note: You'll need to keep your barometer's temperature constant, since temperature will also affect the water level. What Happens if Air Pressure Changes? Why do my ears pop? If you've ever been to the top of a tall mountain, you may have noticed that your ears pop and you need to breathe more often than when you're at sea level. As the number of molecules of air around you decreases, the air pressure decreases. This causes your ears to pop in order to balance the pressure between the outside and inside of your ear. Since you are breathing fewer molecules of oxygen, you need to breathe faster to bring the few molecules there are into your lungs to make up for the deficit. As you climb higher, air temperature decreases. Typically, air temperatures decrease about 3.6° F per 1,000 feet of elevation. Now, in another interesting place http://www.usatoday.com /weather/wstdatmo.htm we learn that: The standard atmosphere The standard atmosphere can be thought of as the average pressure, temperature and air density for various altitudes. It is useful for engineering calculations for aircraft. It also shows in a general way the pressures and temperatures to be expected at various altitudes. The standard atmosphere is based on mathematical formulas that reduce temperature and pressure by certain amounts as altitude is gained. But, the results are close to averages of balloon and airplane measurements at various altitudes. The table below uses metric units, which scientists use Height Temperature Pressure Density (m) (C) (hPa) (kg/m3) 0000 15.0 1013 1.2 1000 8.5 900 1.1 2000 2.0 800 1.0 3000 -4.5 700 0.91 4000 -11.0 620 0.82 5000 -17.5 540 0.74 6000 -24.0 470 0.66 7000 -30.5 410 0.59 8000 -37.0 360 0.53 9000 -43.5 310 0.47 10000 -50.0 260 0.41 11000 -56.5 230 0.36 12000 -56.5 190 0.31 13000 -56.5 170 0.27 14000 -56.5 140 0.23 15000 -56.5 120 0.19 16000 -56.5 100 0.17 17000 -56.5 90 0.14 18000 -56.5 75 0.12 19000 -56.5 65 0.10 20000 -56.5 55 0.088 21000 -55.5 47 0.075 22000 -54.5 40 0.064 23000 -53.5 34 0.054 24000 -52.5 29 0.046 25000 -51.5 25 0.039 26000 -50.5 22 0.034 27000 -49.5 18 0.029 Source: Aerodynamics for Naval Aviators Note: This is a summary table. For a more detailed table or a table in metric units, consult an aerodynamics textbook. Many weather textbooks also have tables of the standard atmosphere. Finally in the very didactic site http:// www.met.tamu.edu/class/Metr151/tut/pres/hydroequ.html we will find the answer to your question. But first we need to consider an equation: We'll start with a mathematical representation of the vertical balance of forces and then construct the equation from the following two statements: Force equals mass times acceleration The gravitational force equals the vertical pressure gradient force. Recall that the vertical pressure gradient force is a result of air pressure, which pushes in all directions. A vertical pressure gradient represents a situation in which the pressure decreases with height, so that the air pressure pushing upward on the bottom of a parcel of air is larger than the air pressure pushing downward on the top of the parcel of air. The force per unit volume is just the rate of change of pressure with height. Mathematically, that is written as dp/dz, where p is pressure, z is height, and the entire expression represents the change in pressure over a tiny vertical distance divided by the length of that distance. The gravitational force is usually expressed mathematically in a very simple way, by writing it as the mass of the object times the acceleration gravity causes. This acceleration is a constant, equal to 9.8 m/s2. (Sooner or later, you'll need to memorize this number). To compare to pressure, we need the force per unit volume: m g / V or g, where m is the mass of a volume of air, g is the gravitational acceleration, V is the size of the volume, and , which is m / V, is the density of air. So the mathematical statement that gravity is balanced in the atmosphere by the vertical pressure gradient force is: dp/dz = - g The minus sign is there because the force of gravity is oriented downwards, in the negative z direction. This mathematical statement is called the "hydrostatic equation", and the balance of forces it represents is called "hydrostatic balance". We'll start with the simplest interpretation: since the right-hand side of the equation is always negative (g is a constant, and there's no such thing as negative density), the left-hand side must always be negative, too. Recall the definition of dp/dz: the change in pressure over a tiny vertical distance divided by the length of that distance. This change (or any derivative, for that matter) must always be measured in the direction in which z (or whatever variable is in the denominator) increases. So if dp/dz is to be negative, p must decrease as z increases. All of this is just a fancy way of saying that the higher you go, the smaller the pressure. But you probably knew that by now. Second, you may remember from high school physics that weight is mass times the acceleration of gravity. So g turns out to be the weight of air per unit volume. Thus, the hydrostatic equation states that the vertical change in air pressure is equal to the weight of the air. Imagine a layer of air 100 m thick. According to the hydrostatic equation, if we know the average weight or mass of the air per unit volume, we can calculate the change in air pressure between the top and the bottom. A typical value for the density of air near sea level is 1.3 kilograms per cubic meter. (To me, that's surprisingly heavy!) With that number, you have everything you need to know to compute the change in air pressure over 100m using the hydrostatic equation. From the preceding section, you may realize that the density of the air has a big effect on pressure. In a sense, the air pressure at a given level of the atmosphere is equal to the average density of all the air above that level. The denser the air, the more rapidly pressure increases as you go downward in the atmosphere. Careful, though: this does not mean that dense air implies high pressure. It means that dense air corresponds to a rapid change of pressure with height. I like to draw an analogy to stacking boxes. Suppose each box is equally filled with (brown) popcorn, so that each box weighs the same. Suppose you stack three boxes. The weight at the bottom of the stack is equal to the total weight of the three boxes, while the weight two-thirds of the way up is equal to the weight of just one box. Just like weight increases downward, so pressure increases downward. And the amount by which pressure or weight change over a given vertical distance depends directly on the density of the intervening layers The same principle applies to the atmosphere. Imagine two regions of the atmosphere, one with dense air on top of light air, and one with light air on top of dense air. The pressure at the ground at these two places might very well be equal, if there's the same total amount of air above them. Well, the atmosphere is like that: Near the ground, the air is warmer and less dense near the equator than near the pole, while at a height of 15 to 20 km, the air is actually warmer and less dense near the pole than near the equator. The surface pressures at the equator and pole are almost identical. If you think about it, it should be obvious that the lower you are in the atmosphere, the more air you have above you, and therefore the higher the air pressure. Indeed, the change in pressure with height is so regular that meteorologists use pressure as a vertical coordinate instead of height. So instead of seeing weather maps at, say, 1.5 km, you would see a map at 850 mb, which is approximately 1.5 km above sea level. Since pressure and density are so closely related, you might expect that temperature is closely related to pressure too. It is. The three quantities are tied together through the ideal gas law. You are probably familiar with the ideal gas law in the following form: pV = nRT (pressure times volume equals the number of molecules times the gas constant times temperature). Through a little bit of mathematical manipulation which I won't go into here, you can convert the terms involving volume and number of molecules into an expression involving density: p = R* T Now R* is a constant. Imagine two air parcels at the same pressure (side by side). Suppose that parcel 1 is warmer than parcel 2. Since the ideal gas law must be satisfied for both air parcels, and the pressure is the same for each of them, the right-hand side of the equation must be the same for each of them, too. Since R* is constant, if one has a higher density, it must also have a lower temperature, and vice versa. In order to determine the "normal" temperature, you need to know the "normal" pressure. I hope this help Regards Jaime Valencia
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