Laws effusivity

Graham s law effusion rate A effusion rate B 1 molecular mass of B V molecular mass of A Effusion and diffusion... 

Ba.rrier Flow. An ideal separation barrier is one that permits flow only by effusion, as is the case when the diameter of the pores in the barrier is sufficiently small compared to the mean free path of the gas molecules. If the pores in the barrier are treated as a collection of straight circular capillaries, the rate of effusion through the barrier is governed by Knudsen s law (eq. 46) ... 

Graham s law tells us qualitatively that light molecules effuse more rapidly than heavier ones (Figure 5.8, p. 119). In quantitative form, it allows us to determine molar masses of gases (Example 5.11). 

Use Graham s law to relate rate of effusion to molar mass. 

Graham s Law The relation stating that the rate of effusion of a gas is inversely proportional to the square root of its molar mass, 119-121 Grain alcohol, 592 Gram, 55-56 Graphite, 241-242,250 Grey tin, 250... 

This observation is now known as Graham s law of effusion. The rate of effusion is proportional to the average speed of the molecules in the gas because the average speed of the molecules determines the rate at which they approach the hole. Therefore, we can conclude that... 

Use Graham s law to account for relative rates of effusion (Self-Test 4.15). 

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... 

Graham s law of effusion The rate of effusion of a gas is inversely proportional to the square root of its molar mass. 

Since the masses of the molecules are proportional to their molecular weights and the average velocity of the molecules is a measure of the rate of effusion or diffusion, all we have to do to this equation to get Graham s law is to take its square root. (The square root of v2 is not quite equal to the average velocity, but is a quantity called the root mean square velocity. See Problem 12.18.)... 

We can use Graham s law to determine the rate of effusion of an unknown gas knowing the rate of a known one or we can use it to determine the molecular mass of an unknown gas. For example, suppose you wanted to find the molar mass of an unknown gas. You measure its rate of effusion versus a known gas, H2. The rate of hydrogen effusion was 3.728 mL/s, while the rate of the unknown gas was 1.000 mL/s. The molar mass of H2 is 2.016 g/mol. Substituting into the Graham s law equation gives ... 

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 experiments on Graham s law, time is measured. The amount of time required for a sample to effuse is the measurement. The amount of material effusing divided by the time elapsed is the rate of effusion. 

Graham s law—The lower the molecular mass of a gas, the faster it will effuse/diffuse. 

Graham s law of effusion / ley de la efusion de Graham establece que la tasa de efusion de un gas es inversamente proporcional a la rarz cuadrada de su masa en moles, (pig. 387)... 

H2 because hydrogen is a diatomic element. Consult your periodic table (or your memory, if you re that good) to obtain the molar masses of hydrogen gas (2.02 g/mol) and neon gas (20.18 g/mol). Finally, plug those values into the appropriate places within Grahcim s law, and you can see the ratio of effusion speed. 

Hydrogen, H2. The question states that the ratio of the rates is 4.0. Oxygen gas is a diatomic element, so it s written as O2 and has a molar mass of 32.00 g/mol. Substitute these known values into Graham s law to determine the molar mass of the unknown gas. The problem states that the unknown gas effuses at a rate 4.0 times faster than oxygen, so put the unknown gas over oxygen for the ratio. (In short, the unknown gas is A, and the oxygen is B). 

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