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At 20 °C and 101.325 kPa, dry air has a density of 1.2041 kg/m 3. At 70 °F and 14.696 psi, dry air has a density of 0.074887 lb/ft 3. The following table illustrates the air density–temperature relationship at 1 atm or 101.325 kPa: [citation needed]
The density of air at sea level is about 1.2 kg/m 3 (1.2 g/L, 0.0012 g/cm 3). Density is not measured directly but is calculated from measurements of temperature, pressure and humidity using the equation of state for air (a form of the ideal gas law). Atmospheric density decreases as the altitude increases.
Densities using the following metric units all have exactly the same numerical value, one thousandth of the value in (kg/m 3). Liquid water has a density of about 1 kg/dm 3, making any of these SI units numerically convenient to use as most solids and liquids have densities between 0.1 and 20 kg/dm 3. kilogram per cubic decimetre (kg/dm 3)
at each geopotential altitude, where g is the standard acceleration of gravity, and R specific is the specific gas constant for dry air (287.0528J⋅kg −1 ⋅K −1). The solution is given by the barometric formula. Air density must be calculated in order to solve for the pressure, and is used in calculating dynamic pressure for moving vehicles.
= molar mass of Earth's air: 0.0289644 kg/mol; The value of subscript b ranges from 0 to 6 in accordance with each of seven successive layers of the atmosphere shown in the table below. The reference value for ρ b for b = 0 is the defined sea level value, ρ 0 = 1.2250 kg/m 3 or 0.0023768908 slug/ft 3.
The relative density of gases is often measured with respect to dry air at a temperature of 20 °C and a pressure of 101.325 kPa absolute, which has a density of 1.205 kg/m 3. Relative density with respect to air can be obtained by =, where is the molar mass and the approximately equal sign is used because equality pertains only if 1 mol of the ...
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In case of air, using the perfect gas law and the standard sea-level conditions (SSL) (air density ρ 0 = 1.225 kg/m 3, temperature T 0 = 288.15 K and pressure p 0 = 101 325 Pa), we have that R air = P 0 /(ρ 0 T 0) = 287.052 874 247 J·kg −1 ·K −1. Then the molar mass of air is computed by M 0 = R/R air = 28.964 917 g/mol. [11]