Effusion rates

Comparing this with equation (A.2.2), it is seen to predict exactly the same dependence of che effusion rate on pressure and temperature. Furthermore, Che ratio of specific heats y depends relatively weakly on che nature of the gas, through its molecularity, so che prediction chat dV/dt 1/M, which follows from equation (A-2.2) and agrees with Graham s results, is not markedly inconsistent with equation (A.2.3) either. 

Vapor pressures were determined by using the Knudsen effusion technique. Effusion rates through and orifice contained in each sample cell were measured as a function of temperature by use of a mass spectrometer/target collection... 

The enrichment procedure uses the small mass difference between the hexafluorides of uranium-235 and uranium-238 to separate them. The first procedure to be developed converts the uranium into uranium hexafluoride, UFfl, which can be vaporized readily. The different effusion rates of the two isotopic fluorides are then used to separate them. From Graham s law of effusion (rare of effusion l/(molar mass)1/2 Section 4.9), the rates of effusion of 235UFfe (molar mass, 349.0 g-mol ) and 238UF6 (molar mass, 352.1 g-mol ) should be in the ratio... 

The effusion rate Am/At at identical orifice diameter and temperatures was slowest for the cylindrical hole with long effusion channel and fastest for a cylindrical orifice with very short effusion channel. This means that the back scattering of the vapor molecules is strongly reduced if the ratio L/D between length L of the effusion channel and diameter D of the orifice is small, which is in agreement with theoretical calculations by Clausing66,67. There are three contributions to the Knudsen cell current which originate from the cell interior, from the channel wall, and the cell lid. 

For example, suppose you wanted to calculate the ratio of effusion rates for hydrogen and nitrogen gases. Remember that both are diatomic, so the molecular mass of H2 is 2.016 g/mol and the molecular mass of N2 would be 28.02 g/mol. Substituting into the Graham s law equation ... 

Hydrogen gas would effuse through a pinhole 3.728 times as fast as nitrogen gas. The answer is reasonable, since the lower the molecular mass, the faster the gas is moving. Sometimes we measure the effusion rates of a known gas and an unknown gas, and use Graham s law to calculate the molecular mass of the unknown gas. 

In comparing two gases at the same temperature and pressure, we can set up an equation showing that the ratio of the effusion rates of the two gases is inversely proportional to the ratio of the square roots of their masses ... 

For the evaporation of CdS powder, dp/dt is constant at about 0.17 x 103 g/cm3-s (5). Neglecting the density change in this example would introduce an error of only about 10% in the calculated effusion rates. 

Figure 4. CdS effusion rate versus charge temperature. (Reproduced with permission from reference 5. Copyright 1982 American Institute of Chemical...

High-Rate Sources. Sources such as those shown in Figures 1 and 2 are used in many semiconductor applications and operate at sufficiently high effusion rates to ensure transition or laminar flow through the nozzle or orifice. 

Probability distributions have been measured for Cd and Zn (1) and CdTe (15) for several nozzle dimensions and temperatures yielding effusion rates of 0.01-0.33 g/min. A representative distribution is shown in Figure 11. The curve drawn through the data points in Figure 11 illustrates agreement between the probability distribution F(< )), which was predicted by equations 25 and 26, and the experimental data. The root-mean-square difference between the experimental and predicted probability distribution values is 1-3% of the center line value. This agreement is as good as empirical fits described in the literature that use two or more adjustable parameters (16). 

The rate at which a gas effuses (flows out) through a small hole in the container into a vacuum is exactly the rate at which the molecules would collide with a wall area equal to the area of the hole. From Eq. (5-10), the effusion rates of two gases, both at the same pressure and temperature, are in the ratio... 

Calculate the ratio of effusion rates of oxygen (02) to hydrogen (H2). 

Methane (CH4) effuses at a rate of 2.45 mol/s. What will be the effusion rate of argon (Ar) under the same conditions ... 

The effusion rate of hydrogen sulfide (H2S) is 1.50 mol/s. Another gas under similar conditions effuses at a rate of 1.25 mol/s. What is the molar mass of the second gas ... 

The kinetic theory of gases was briefly discussed. It enables the mean or thermal velocity (c) of gas molecules at a given temperature to be obtained and gas flux to be calculated. From the latter, effusion rates, area-related condensation rates and conductances under molecular flow can be determined (see Examples 1.5 and 1.7-1.10). Calculation of collision frequency (obtained from c, n and the collision cross-section of molecules), enables the mean free path (f) of particles to be determined. The easily obtained expression for Ip is a convenient way of stating the variation of / withp (Examples 1.11-1.15). 

Orifice Average Effusion Rate, Flux IAgram/hr. No. of... 

To determine the effect of orifice size on vapor pressure, five different orifices were used to measure the effusion rates at 25 °C. of one compound, 2-chlorcMt-aminopyrimidine. Carson, Cooper, and Stranks (6) have shown that the net rate of effusion is directly proportional to the flux (rate of effusion per unit area) if other factors are constant. That is, a graph of... 

From a conceptual standpoint, the equation tells you that larger molecules will effuse more slowly than smaller particles. Mathematically, the equation provides a number that tells you the ratio and the effusion rates of gases 1 and 2. If the ratio is greater than 1.00, gas 1 is effusing more quickly than gas 2. If the ratio is less than 1.00, gas 2 is effusing more quickly than gas 1. 

Answer The first thing to notice here is that the equation for calculating the ratio of effusion rates requires a distinction between gas 1 and gas 2. Since no such distinction is made in the problem, the decision about what to label each gas is arbitrary. However, because the lighter gas is typically labeled as gas 1, that s how we shall proceed. 

PROBLEM Calculate the effusion rate of nitrogen gas to oxygen gas. How does this compare to the density of the gases ... 

Only (c) The average kinetic energies of their molecules must be the same, must be true. The average speeds and their effusion rates depend on the molar masses (Graham s law). Their pressures and nnmbers of moles (and thus molecules) are governed by the ideal gas law. 

---