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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 ...
R ∗ = 8.314 32 × 10 3 N⋅m⋅kmol −1 ⋅K −1 = 8.314 32 J⋅K −1 ⋅mol −1. Note the use of the kilomole, with the resulting factor of 1000 in the constant. The USSA1976 acknowledges that this value is not consistent with the cited values for the Avogadro constant and the Boltzmann constant. [ 13 ]
Where: R is the Ideal gas constant (8.314 Pa·m 3 /mol·K); T is the absolute temperature (K); H is the Henry's law constant for the target chemical (Pa/m 3 mol); K ow is the octanol-water partition coefficient for the target chemical (dimensionless ratio); P s is the vapor pressure of the target chemical (Pa); and v is the molar volume of the ...
The value used for γ is typically 1.4 for diatomic gases like nitrogen (N 2) and oxygen (O 2), (and air, which is 99% diatomic). Also γ is typically 1.6 for mono atomic gases like the noble gases helium (He), and argon (Ar). In internal combustion engines γ varies between 1.35 and 1.15, depending on constitution gases and temperature. ^ b.
Macroscopically, the ideal gas law states that, for an ideal gas, the product of pressure p and volume V is proportional to the product of amount of substance n and absolute temperature T: =, where R is the molar gas constant (8.314 462 618 153 24 J⋅K −1 ⋅mol −1). [4]
1 Nm 3 of any gas (measured at 0 °C and 1 atmosphere of absolute pressure) equals 37.326 scf of that gas (measured at 60 °F and 1 atmosphere of absolute pressure). 1 kmol of any ideal gas equals 22.414 Nm 3 of that gas at 0 °C and 1 atmosphere of absolute pressure ... and 1 lbmol of any ideal gas equals 379.482 scf of that gas at 60 °F and ...
p w is the partial pressure of gaseous water during condition 1 and 2, respectively; For example, calculating how much 1 liter of air (a) at 0 °C, 100 kPa, p w = 0 kPa (known as STPD, see below) would fill when breathed into the lungs where it is mixed with water vapor (l), where it quickly becomes 37 °C (99 °F), 100 kPa, p w = 6.2 kPa (BTPS):
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]