Graham’s law

Graham s law of diffusion This law states that the rates at which two gases diffuse are inversely proportional to their densities, i.e. 

Graham s Law of Diffusion. The rates at which gases diffuse under the same conditions of temperature and pressure are inverseiy proportionai to the square roots of their densities ... 

This relation in a somewhat different form was discovered experimentally by the Scottish chemist Thomas Graham (1748-1843) in 1829. Graham was interested in a wide variety of chemical and physical problems, among them the separation of the components of air. Graham s law can be stated as... 

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. 

Average translational energy Average speed Graham s law... 

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... 

This relation is an important clue about the motion of molecules in a gas, and we shall use it shortly, but first we look at some practical applications ot Graham s law. 

If we were to write Graham s law for two gases A and B with molar masses MA and Mb, and divide one equation by the other, we would obtain,... 

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. 

Graham s law may be explained in terms of the kinetic molecular theory as follows Since the two gases are at the same temperature, their average kinetic energies are the same ... 

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.)... 

Would it be possible to separate isotopes using the principle of Graham s law Explain what factors would be important. 

Arts. Since the temperatures are the same, so are the average kinetic energies of their molecules. From Graham s law,... 

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. 

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