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The relationship between the two constants is R s = R / m, where m is the molecular mass of the gas. The US Standard Atmosphere (USSA) uses 8.31432 m 3 ·Pa/(mol·K) as the value of R. However, the USSA in 1976 does recognize that this value is not consistent with the values of the Avogadro constant and the Boltzmann constant. [49]
The standard unit is the meter cubed per kilogram (m 3 /kg or m 3 ·kg −1). Sometimes specific volume is expressed in terms of the number of cubic centimeters occupied by one gram of a substance. In this case, the unit is the centimeter cubed per gram (cm 3 /g or cm 3 ·g −1). To convert m 3 /kg to cm 3 /g, multiply by 1000; conversely ...
Until 1982, STP was defined as a temperature of 273.15 K (0 °C, 32 °F) and an absolute pressure of 101.325 kPa (1 atm). Since 1982, STP is defined as a temperature of 273.15 K (0 °C, 32 °F) and an absolute pressure of 100 kPa (1 bar). Conversions between each volume flow metric are calculated using the following formulas: Prior to 1982,
where P and V are pressure and volume respectively, and n, R and T are the number of moles of gas, the gas constant (8.314 J/(mol·K)), and T is temperature in kelvins (K). The number density of air is the number of molecules or atoms per unit volume: =, and when plugged into the real gas law, the number density of air is found by using ...
The ideal gas equation can be rearranged to give an expression for the molar volume of an ideal gas: = = Hence, for a given temperature and pressure, the molar volume is the same for all ideal gases and is based on the gas constant: R = 8.314 462 618 153 24 m 3 ⋅Pa⋅K −1 ⋅mol −1, or about 8.205 736 608 095 96 × 10 −5 m 3 ⋅atm⋅K ...
p is the pressure; V is the volume; n is the amount of substance of gas (moles) R is the gas constant, 8.314 J·K −1 mol −1; T is the absolute temperature; To simplify, a volume of gas may be expressed as the volume it would have in standard conditions for temperature and pressure, which are 0 °C (32 °F) and 100 kPa. [2]
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At standard temperature and pressure (100 kPa and 273.15 K), we can use Avogadro's law to find the molar volume of an ideal gas: V m = V n = R T P ≈ 8.3145 J m o l ⋅ K × 273.15 K 100 k P a ≈ 22.711 L / m o l {\displaystyle V_{\text{m}}={\frac {V}{n}}={\frac {RT}{P}}\approx {\frac {\mathrm {8.3145\ {\frac {J}{mol\cdot K}}\times 273.15\ K ...