FORMULAE  to calculate atmospheric pressure above sea level (hppa)?
912617
P=P(0)*exp(-1.1e-4*H)\n\nP(0) is the pressure at sea level. Note it's temperature dependent and it assumes a stable Boltzmann type distribution.
1002151
14.7 PSI is the atmospheric pressure at sea level. It has nothing to do with gravity.
751939
you can either do the math, look it up on a chart or go here:\n\nhttp://leebrewery.com/hydromet.htm\n\nand use the java-calculator.
360638
I'm thinking that you meant to type pressure instead of height. because millibars do not measure height. they measure pressure. and altitude and height are pretty much the same thing.\n\nthe relationship between altitude and pressure can be modeled by this equation\n\nlog(.01*P)~5-(h/50853.0185)\n\nwhen P=pressure in millibars and h=altitude in feet
111676
Actually standard pressure at sea level is taken to be 14.7 lbs/in^2. \n\nLots of things are factors in air pressure: \n\nAltitude of observations \n\nHeight of air column\n\ndensity of air, depending on temperature, and gas constitutents (constituents usually fairly slowly varying, but temperature varies rapidly)\n\nStrength of gravitational field\n\nTidal forces (small but measurable)\n\nLocal topography (orographic effects)\n\nair motion (usually driven by density differences, which are usually connected with tempearture). \n\n\n\nProbably the biggest factors are temperature, and altitude.
790378
h=-16t^2+48t+50\nfor first part put t=3\nh=-16*9+48*3+50\nsolve and subtract 50 to find height above projection\nfor second part diff the eq. to get velocity\ndh/dt=-32t+48\nat max height this velocity is zero\n0=-32t+48\nt=3/2\nput this in given eq. to find height
687925
P = Pressure in atmospheres\nV = Volume in Liters\nn = number of moles\nR = ideal gas constant, which in this equation is 0.08206 L * atm * mol^-1 * K^-1\nT = temperature in Kelvin (degrees Centigrade + 273.15)
1018146
try http://www.worldclimate.com/
969415
It's a geometry problem.  Suppose you're high up in a building on the beach overlooking the ocean.  You're at height h above sea level.\n\nNow draw a circle which represents the earth, with radius r.  Your distance from the center of the earth is r+h.  The horizon is some distance away, x.  Your line of sight to the horizon is <tangent> to the radius of the earth, forming a right triangle, where the hypoteneuse is r+h.\n\nUsing the Pythagorean Theorem,\n\nr^2 + x^2 = (r+h)^2 = r^2 + 2rh + h^2\nx^2 = 2rh + h^2\nx = sqrt(2rh + h^2)\n\nThat's the exact answer,  But for practical purposes, 2rh is <much, much> bigger than h^2.  To see this, suppose you're 500 feet in the air.  That's about a tenth of a mile, and using miles, h^2 = 0.1^2 = 0.01.  On the other hand, the earth's radius is about 4000 miles, so 2rh = 2 x 4000 x 0.1 = 800.  This is 80,000 times bigger than h^2.\n\nSo we can safely ignore the h^2, and say x = sqrt(2rh).  That can be broken down further.  Again using miles (you can also do this in kilometers),\n\nx = sqrt(2rh) = sqrt(2r) sqrt h = sqrt(8000) sqrt h = 90 sqrt h\n\n(rounding off, since sqrt 8000 = 89.44; also, h must be in miles).\n\nUsing the formula x = 90 sqrt h, suppose you're in an airplane 4 miles up in the air (21,000 feet).  Then the horizon is 90 sqrt 4 = 180 miles away.\n\nOkay, now let's work it out for h in feet instead of miles.  We'll go back to the formula x = sqrt(2rh), and use r = 3960 miles (a more accurate figure).  Then\n\nx = sqrt(2 x 3960 x h / 5280) = sqrt(3h/2) = 1.225 sqrt h\n\nwhere h is in feet and x is in miles.\n\nTwo examples:  (1) For h = 6 feet, use x = sqrt(3h/2) = 3 miles.  (2) For x = 100 feet, use x = 1.225 sqrt h = 12.25 miles.\n\nTo recap:  For h in miles, use x = 90 sqrt h.  For h in feet, either use x = sqrt(3h/2) or x = 1.225 sqrt h.\n\nIn all cases, x will be in miles.  And of course, all of this can be done using meters and kilometers.
820723
Atmospheric pressure is the pressure at any point in the Earth's atmosphere.\n\nIn 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. Similarly, as elevation increases there is less overlying atmospheric mass, so that pressure decreases with increasing elevation. A column of air 1 square inch in cross section, measured from sea level to the top of the atmosphere, would weigh approximately 14.7 lb. A 1 m² column of air would weigh about 100 kilonewtons.\n\nStandard atmosphere (symbol: atm) is a unit of pressure and is defined as being precisely equal to 101,325 Pa. This value is intended to represent the mean sea level pressure at the latitude of Paris, France, and as a practical matter, approximates the mean sea level pressure for many of the industrialized nations (those with latitudes similar to Paris). One standard atmosphere is standard pressure used for pneumatic fluid power (ISO R554), and in the aerospace (ISO 2533) and petroleum (ISO 5024) industries.\n\nIn 1982, the International Union of Pure and Applied Chemistry (IUPAC) recommended that for the purposes of specifying the physical properties of substances, “the standard pressure” should be defined as precisely 100 kPa (≈750.062 torr) rather than the 101.325 kPa value of “one standard atmosphere”.[1] This value is used as the standard pressure for the compressor and the pneumatic tool industries (ISO 2787).[2] (see also Standard temperature and pressure)\n\nIn the United States, compressed air flow is often measured in "standard cubic feet" per unit of time, where the "standard" means the equivalent quantity of air at standard temperature and pressure. However, this standard atmosphere is defined slightly differently: temperature = 68 °F (20 °C), air density = 0.075 lb/ft³ (1.20 kg/m³), altitude = sea level, and relative humidity = 0%. In the air conditioning industry, the standard is often temperature = 32 °F (0 °C) instead. For natural gas, the petroleum industry uses a standard temperature of 60 °F (15.6 °C).\n\nAtmospheric pressure is often measured with a mercury barometer, and a height of approximately 760 mm (30 inches) of mercury is often used to teach, make visible, and illustrate (and measure) atmospheric pressure. However, since mercury is not a substance that humans commonly come in contact with, water often provides a more intuitive way to conceptualize the amount of pressure in one atmosphere.\n\nOne atmosphere (101.325 kPa or 14.7 lbf/in²) is the amount of pressure that can lift water approximately 10.3 m (33.9 feet). Thus, a diver at a depth 10.3 meters under water in a fresh-water lake experiences a pressure of about 2 atmospheres (1 atm for the air and 1 atm for the water).\n\nIn terms of city water pressure, one atmosphere is approximately one-half to one-fifth the pressure of typical city water mains (i.e., water pressure is around 2 to 5 atmospheres).\n\nThe most common type of barometer used in homes is the aneroid barometer (Figure 3). Inside this instrument is a small, flexible metal capsule called an aneroid cell. In the construction of the device, a vacuum is created inside the capsule so that small changes in outside air pressure cause the capsule to expand or contract. The size of the aneroid cell is then calibrated and any change in its volume is transmitted by springs and levers to an indicating arm that points to the corresponding atmospheric pressure.
145449
The density of air, ρ (Greek: rho) (air density), is the mass per unit volume of Earth's atmosphere, and is a useful value in aeronautics. In the SI system it is measured as the number of kilograms of air in a cubic meter (kg/m3). At sea level and at 20 °C dry air has a density of approximately 1.2 kg/m3. varying with pressure and temperature. Air density and air pressure decrease with increasing altitude.\n\nThe formula for the density of air is given by:\n\n P=p/R*T\nwhere ρ is the air density, p is pressure, R is the gas constant, and T is temperature.\n\nThe individual gas constant R for dry air is:\n\n R( dry air )=287.05J/kg*K\nTherefore:\n\nAt standard temperature and pressure (0 °C and 101.325 kPa), dry air has a density of ρSTP = 1.293 kg/m3. \nAt standard ambient temperature and pressure (25 °C and 100 kPa) dry air has a density of ρSATP = 1.168 kg/m3. \n[edit]\nEffect of water vapor\nFor moist air, the partial pressure of the water vapor must be considered as well. In this case, the density of the air is the sum of the density of the dry air and the density of the water vapor:\nP=pdry air /Rdry air*T+p water vapour/R water vapour*T\n \nThe gas constant for water vapor is:\nR water vapour=461.495 J/kg*K\n \nEffects of altitude\nTo calculate the density of air as a function of altitude, one requires additional parameters. They are listed below, along with their values according to the International Standard Atmosphere, using the universal gas constant instead of the specific one:\n\nsea level atmospheric pressure p0 = 101325 Pa = 1013.25 mbar or hPa = 101.325 kPa (= 101325 kg/(m*s2)) \nsea level standard temperature T0 = 288.15 K \nEarth-surface gravitational acceleration g = 9.80665 m/s2. \ndry adiabatic lapse rate L = −0.0065 K/m \nuniversal gas constant R = 8.31447 J/(mol·K) \nmolecular weight of dry air M = 0.0289644 kg/mol \nTemperature at altitude h metres above sea level is given by the following formula (only valid below the tropopause):\n\n T=To+L*H\nImportance of temperature\nThe below table demonstrates that the properties of air change significantly with temperature.\n\nTable — speed of sound in air c, density of air ρ, acoustic impedance Z vs. temperature °C\n\nImpact of temperature \n°C c in m/s ρ in kg/m³ Z in Pa·s/m \n- 10 325.4 1.341 436.5 \n- 5 328.5 1.316 432.4 \n0 331.5 1.293 428.3 \n+ 5 334.5 1.269 424.5 \n+ 10 337.5 1.247 420.7 \n+ 15 340.5 1.225 417.0 \n+ 20 343.4 1.204 413.5 \n+ 25 346.3 1.184 410.0 \n+ 30 349.2 1.164 406.6
449240
Atmospheric pressure will support a column of air about 10 metres high.   The exact height varies from day to day depending on atmospheric pressure; high and low pressure systems.   To start with, partly fill your container with coloured water, with the hole at the bottom submerged in more coloured water.   If there was no air in your transparent ball, it would need to be 10 metres in diameterfor you to see a boundary between the empty space at the top and the water column supported by the atmosphere pushing down on the liquid outside, but because you've trapped some air inside, this air pushes down on the water so the water isn't as high.
94789
Use Boyle's Law for that. Assuming the temperature is always constant, the product of pressure and volume is always constant.\n\nP * V = k\nor\nP1 * V1 = P2 * V2\n\nIf temperature is included in the equation, you could either use the combined gas law or the ideal gas law.\n\nNote: Elevation plays a significant role in Boyle's Law so it will have a difference in the gauge readings if you relocate it from sea level to the mountaintop.
641078
I think this is how you would solve this problem. Use the equation PV=nRT, re-arrange to solve for V. \n\n(STP = standard temperature and pressure = 1 atm of pressure and 273 Kelvins)\n\nV = nRT/P = (1 mol)(0.0821 L*atm/k*mol)(273)/1 atm = 22.4 L
543227
55.280 kPa - 101.325 kPa = -45.045 kPa\n\nAbsolute pressure is gauge pressure plus atmospheric pressure which is 101.325kPa.
347099
I have used the Bernoulli equation to calculate pressure drop through piping and to size pumps (determine required head pressure).  You can't get much more real than that.
944167
PV=nRT\nV = nRT/P\nV = (114g/44g/mol)(0.08206L-atm/mol-K)(28+273K)/0.987atm\nV = 64.84L or 0.0648 m^3