Grahame’s equation

Substitute the known values into Graham s equation and solve. Ratew... 

This important result relating the surface charge density to the surface potential is often referred to as Grahame s equation. A little manipulation allows us to express the potential [23] in the simpler form... 

The relation between the surface potential and charge density (i.e., Grahame s equation and the analog of Eq. [25]) is readily found by applying the boundary condition of Eq. [15] to solution [38] ... 

The surface potential is found using a generalization of Grahame s equation [25] obtained by applying Eq. [15] ... 

The integral can be solved with the help of Grahame s equation (4.32) ... 

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. 

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,... 

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 ... 

Recently Hoover 29> compared various extrapolation methods for obtaining true solution resistances concentrated aqueous salt solutions were used for the comparisons. Two Jones-type cells were employed, one with untreated electrodes and the other with palladium-blacked electrodes. The data were fitted to three theoretical and four empirical extrapolation functions by means of computer programs. It was found that the empirical equations yielded extrapolated resistances for cells with untreated electrodes which were 0.02 to 0.15 % lower than those for palladium-blacked electrodes. Equations based on Grahame s model of a conductance cell 30-7> produced values which agreed to within 0.01 %. It was proposed that a simplified equation based on this model be used for extrapolations. Similar studies of this kind are needed for dilute nonaqueous solutions. 

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. 

Have you ever walked into a room and noticed the odor of an open bottle of perfume If the open bottle of perfume is on one side of the room, why can you smell it on the other side of the room Gases travel with great speeds and they can spread out or diffuse. According to Graham s Law, at the same temperature and pressure, gases diffuse at a rate inversely proportional to the square roots of their molecular masses. This can be seen in the equation ... 

Substituting the Knudsen diffusivities from Equation 3.6 into Equation 9.14 gives the famous and debatable Graham s relationship... 

This broken-down region near the ion was the subject of mathematical discussion by Webb as early as 1926, by Conway et al., and by Booth, whosepaper also can be considered seminal. Grahame made an attempt to simplify Booth s equation for the dielectric constant as a function of field strength, and a diagram due to him is shown in Fig. 2.27. 

This equation is Graham s law, and thus the kinetic molecular model fits the experimental results for the effusion of gases. 

Solving this equation for the ratio of speeds between Va and Vb gives Graham s law of diffusion. 

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. 

Hydrogen (H2) has molecular weight dTZDISS u. Using the unknown for material 1 and hydrogen for material 2 in the equation for Graham s law, the ratio of rates is V... 

This is known as Graham s law of effusion for Knudsen diffusion of a multicomponent system at constant total pressure. For binary gas mixtures, equation (1-105) simplifies to... 

Examples.—(1) T. Graham s diffusion experiments Phil. Trans., 151,188, 1861). A cylindrical vessel 152 mm. high, and 87 mm. in diameter, contained 0 7 litre of water. Below this was placed 0T litre of a salt solution. The fluid column was then 127 mm. high. After the elapse of a certain time, successive portions of 100 c.c., or of the total volume of the fluid, were removed and the quantity of salt determined in each layer. Here x = 0 at the bottom of the vessel, and x = H at the top x — h at the surface separating the solution from the liquid when t = 0. The vessel has unit area. The limiting conditions are At the end of a certain time t, (i) when p = 0, dV/dx = 0 and (ii) when x — H, dVjdx = 0 (iii) when t = 0, V= F0 between x = 0 and x=h (iv) when t = 0, F = 0 between x — h and x — H. To adapt these results to Fourier s solution of Fick s equation, first show that (6) is a particular integral of Fick s equation. Differentiate (8) with respect to x and show that for the first condition we must have b zero, and condition (i) is satisfied. For condition (ii), sin fiH must be zero but sin mr is zero hence we can put... 

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