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Graham s law of diffusion and effusion

In diffusion, two gases gradually mix with each other. In effusion, gas molecules move through a small opening under pressure. Both processes are governed by the same mathematical law—Graham s law of diffusion and effusion. [Pg.215]

Effusion and diffusion are substantially the same process. Diffusion is movement of a substance from an area of higher concentration to an area of lower concentration. Effusion is the movement of a gas through a small opening. Graham s Law of Effusion states the rate of effusion is inversely proportional to the square root of the molecular mass. [Pg.146]

Thomas Graham developed Graham s Law of effusion and diffusion in the 1830s. He is called the father of colloid chemistry. [Pg.228]

Graham s Law of Effusion states that at the same temperature and pressure, gases diffuse at a rate inversely proportional to the square roots of their molecular masses. What this translates to is that lighter (less dense) gases travel faster than heavier (more dense) gases. [Pg.34]

You are given the molar masses for ammonia and hydrogen chloride. To find the ratio of the diffusion rates for ammonia and hydrogen chloride, use the equation for Graham s law of effusion. [Pg.388]

Natural uranium is mostly nonfissionable it contains only about 0.7% of fissionable For uranium to be useful as a nuclear fuel, the relative amount of must be increased to about 3%. This is accomplished through a gas diffusion process. In the diffusion process, natural uranium reacts with fluorine to form a mixture of UFg(g) and UFg(g). The fluoride mixture is then enriched through a multistage diffusion process to produce a 3% nuclear fuel. The diffusion process utilizes Graham s law of effusion (see Chapter 5, Section 5.7). Explain how Graham s law of effusion allows natural uranium to be enriched by the gaseous diffusion process. [Pg.901]

MOLECULAR EFFUSION AND DIFFUSION (SECTION 10.8) It Mows from kinetic-molecular theory that the rate at which a gas undergoes effusion (escapes through a tiny hole) is inversely proportional to the square root of its molar mass (Graham s law). The diffusion of one gas through the space occupied by a second gas is another phenomenon related to the speeds at which molecules move. Because moving molecules undergo frequent collisions with one another, the mean free path—the mean distance traveled between collisions—is short. Collisions between molecules limit the rate at which a gas molecule can diffuse. [Pg.431]

Thomas Graham determined that the rates of diffusion and effusion of gases are inversely proportional to the square roots of their molecular or atomic weights. This is Graham s Law. In general, it says that the lighter the gas, the faster it will effuse (or diffuse). Mathematically, Graham s Law looks like this ... [Pg.227]

In the mid-1800s, the Scottish chemist Thomas Graham studied the effusion and diffusion of gases. The above equation is a mathematical statement of some of Graham s discoveries. It describes the rates of effusion. It can also be used to find the molar mass of an unknown gas. Graham s law of effusion states that the rates of effusion of gases at the same temperature and pressure are inversely proportional to the square roots of their molar masses. [Pg.367]

Graham s law. The rates of diffusion and effusion are inversely proporuonal to the square root of the molar mass of the gas. [Pg.978]

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.)... [Pg.207]

The correct answer is (A). There is a quick way to solve this and also a long way. The quick way is to remember that Graham s law states that the rate of effusion is inversely proportional to the square root of the molar mass. You can see pretty quickly that the molar mass of methane is 4 times that of helium. The square root of 4 is 2, meaning that helium will diffuse two times faster than methane. The longer way is to actually set up the equation — = and solve for r2. [Pg.172]

Graham s Law states that, under equal conditions of temperature and pressure, the rate of effusion or diffusion of a gas is inversely proportional to the square root of its molar mass. The most useful mathematical form of this law involves the rates of two gases ... [Pg.97]

The usual problem involving Graham s law asks for the ratio of rates of effusion or diffusion of two gases, or it gives the rate for one gas and asks the rate for another. [Pg.97]


See other pages where Graham s law of diffusion and effusion is mentioned: [Pg.110]    [Pg.166]    [Pg.214]    [Pg.162]    [Pg.110]    [Pg.166]    [Pg.214]    [Pg.162]    [Pg.53]    [Pg.455]    [Pg.129]    [Pg.209]    [Pg.159]    [Pg.164]    [Pg.82]    [Pg.387]    [Pg.394]    [Pg.404]    [Pg.206]    [Pg.87]    [Pg.110]    [Pg.183]    [Pg.341]    [Pg.360]    [Pg.361]    [Pg.361]    [Pg.197]    [Pg.26]    [Pg.165]    [Pg.226]   
See also in sourсe #XX -- [ Pg.166 ]




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