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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.
For simplicity, the gas is assumed to be an ideal gas. The gas flow is isentropic. The gas flow is constant. The gas flow is along a straight line from gas inlet to exhaust gas exit. The gas flow behavior is compressible. There are numerous applications where a steady, uniform, isentropic flow is a good approximation to the flow in conduits.
= 10 parts per million by volume = 10 ppmv = 10 volumes/10 6 volumes NO x molar mass = 46 kg/kmol = 46 g/mol Flow rate of flue gas = 20 cubic metres per minute = 20 m 3 /min The flue gas exits the furnace at 0 °C temperature and 101.325 kPa absolute pressure. The molar volume of a gas at 0 °C temperature and 101.325 kPa is 22.414 m 3 /kmol.
Compressible flow (or gas dynamics) is the branch of fluid mechanics that deals with flows having significant changes in fluid density.While all flows are compressible, flows are usually treated as being incompressible when the Mach number (the ratio of the speed of the flow to the speed of sound) is smaller than 0.3 (since the density change due to velocity is about 5% in that case). [1]
By the equipartition theorem, internal energy per mole of gas equals c v T, where T is absolute temperature and the specific heat at constant volume is c v = (f)(R/2). R = 8.314 J/(K mol) is the universal gas constant, and "f" is the number of thermodynamic (quadratic) degrees of freedom, counting the number of ways in which energy can occur.
For his work with gases a century prior, the physical constant that bears his name (the Avogadro constant) is the number of atoms per mole of elemental carbon-12 (6.022 × 10 23 mol −1). This specific number of gas particles, at standard temperature and pressure (ideal gas law) occupies 22.40 liters, which is referred to as the molar volume .
The reactions may differ in their stoichiometry. For example, the methylation of benzene (C 6 H 6), through a Friedel–Crafts reaction using AlCl 3 as a catalyst, may produce singly methylated (C 6 H 5 CH 3), doubly methylated (C 6 H 4 (CH 3) 2), or still more highly methylated (C 6 H 6−n (CH 3) n) products, as shown in the following example ...
The graph on the right is the square (0,0),(1.1,1.1) of the left graph expanded to display the overlap between the inversion and saturation curves. In terms of b / v {\displaystyle b/v} the equation has a simple positive solution b / v = 1 − R T b / ( 2 a ) {\displaystyle b/v=1-{\sqrt {RTb/(2a)}}} which, for b / v = 0 {\displaystyle b/v=0 ...