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Oxygen molecules, velocity

Calculate the temperature at which oxygen molecules have the same average velocity as hydrogen molecules have 273 K. [Pg.207]

Calculate the detonation velocity in a gaseous mixture of 75% ozone (03) and 25% oxygen (02) initially at 298 K and 1 atm pressure. The only products after detonation are oxygen molecules and atoms. Take the AffjfO-,) I40kj/mol and all other thermochemical data from the JANAF tables in the appendixes. [Pg.307]

The velocity toward the left is determined by the number of collisions of three different molecules or one oxygen molecule with two SO 2 molecules... [Pg.129]

For low space velocities of synthesis gas, the products from nitrided catalysts in fixed-bed reactors are somewhat similar to those of the Synol process employing high space velocities and low temperatures. Possibly the yields of alcohols and other oxygenated molecules obtained from nitrided catalysts under the conditions employed in the Synol process would be even higher than those obtained from nitrides at low space velocities. [Pg.381]

Example Calculate the value of p(c) for oxygen molecules at 27 C when c = cp, where c is the most probable velocity. [Pg.99]

Calculate the ratio of the velocity of neon atoms to the velocity of oxygen molecules at the same temperature. [Pg.279]

Since the axial component of velocity near the channel/backing layer interface is zero, each consumed oxygen molecule removes momentum from the flow. Each incoming water molecule also removes momentum, since it must be accelerated to the flow velocity. The momentum balance (the Euler equation), therefore, has a form [192]... [Pg.515]

The velocity in the cathode channel, therefore, increases. Physically, oxygen molecules are replaced with water molecules in the flow the total mass of... [Pg.515]

Alveoli are the tiny sacs of air in the lungs (see Problem 5.122) whose average diameter is 5.0 X 10 m. Consider an oxygen molecule (5.3 X 10 kg) trapped within a sac. Calculate the uncertainty in the velocity of the oxygen molecule. (Hint The maximum uncertainty in the position of the molecule is given by the diameter of the sac.)... [Pg.284]

Analysis of hydrodynamic equations for the flow in the fuel cell channel shows that this flow is incompressible [13]. In other words, the variation of pressure (total molar concentration) along the channel is small. Consider first the case of zero water flux through the membrane. Each oxygen molecule in the cathode channel is replaced with two water molecules. Pressure is proportional to the number of molecules per unit volume. To support constant pressure, the flow velocity in the channel must increase. The growth of velocity provides expansion of elementary fluid volume the expansion keeps pressure in this volume constant. [Pg.214]

Maxwell derived an equation which linked the viscosity of a gas to the density, the mean free path and the average velocity of the molecules. From measurements of viscosity made with the help of his wife he estimated the mean free path of oxygen molecules at 0 C to be 5.6 x 10" cm (modern value 6.6 x 10" cm). Maxwell was also able to estimate the mean free path from diffusion experiments, and the two values were in satisfactory agreement. [Pg.207]

For example, oxygen stoichiometry A = 2 means that in a fuel cell, 50% of oxygen flux is converted to current. If flow velocity in the channel is constant, A = 2 also means that 50% of oxygen molecules are consumed. [Pg.18]

Figure 1.7 In the channel, each oxygen molecule consumed is replaced by two water molecules. To keep pressure in the elementary volume constant, the front face velocity of this volume must exceed the rear face velocity. Figure 1.7 In the channel, each oxygen molecule consumed is replaced by two water molecules. To keep pressure in the elementary volume constant, the front face velocity of this volume must exceed the rear face velocity.
On the right side of Eq. (4.1), we should add a term describing the momentum flux through the channel/GDL interface. Each oxygen molecule consumed removes momentum from the cathode flow and each water molecule injected into the flow must be accelerated to the average flow velocity v. [Pg.119]

This effect can be explained as follows. Even at = 0, one oxygen molecule in the cathode flow is replaced by two water molecules, that is, the total mass and the number of molecules in the flow increase. If > 0, the water flux from the anode side further increases the flow mass. Flow density, however, remains almost constant along z. This occurs due to the growth of velocity with 2 an elementary volume of gas expands along 2 while it moves. [Pg.122]

Calculate the root mean square velocity of oxygen molecules at 25 °C. [Pg.228]


See other pages where Oxygen molecules, velocity is mentioned: [Pg.103]    [Pg.403]    [Pg.142]    [Pg.159]    [Pg.91]    [Pg.91]    [Pg.122]    [Pg.139]    [Pg.148]    [Pg.160]    [Pg.174]    [Pg.142]    [Pg.159]    [Pg.21]    [Pg.281]    [Pg.279]    [Pg.280]    [Pg.610]    [Pg.54]    [Pg.82]    [Pg.216]    [Pg.128]    [Pg.216]    [Pg.253]    [Pg.124]    [Pg.529]    [Pg.10]    [Pg.447]    [Pg.231]    [Pg.236]   
See also in sourсe #XX -- [ Pg.83 ]




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