Gas effusion

B—Lighter gases effuse faster. The only gas among the choices that is lighter than methane is helium. To calculate the molar mass, you would begin with the molar mass of methane and divide by the rate difference squared ... 

The correct answer is (E). In this problem, you need to remember that lighter gases effuse more quickly than heavier gases. That means that CH4 will effuse most quickly, followed by N02, and finally Cl2. That means that after the time period, more CH4 will have effused than N02. Cl2 will have effused the least, so it will be the most plentiful in the flask. Since partial pressures are proportional to mole fractions, the most plentiful gas will also exert the greatest partial pressure. 

At the same temperature and pressure, gases effuse at a rate inversely proportional to the square roots of their molecular masses. 

Two gases effuse through a hole. Gas A has nine times the molecular mass of gas B. What is the ratio of the two molecular speeds ... 

Both gases effuse through pores in balloon, but ligjiter helium gas effuses faster than heavier aigon gas... 

KINETIC-MOLECULAR THEORY OF GASES EFFUSION AND DIFFUSION (sections 10.7 and 10.8)... 

Heavier gas molecules move more slowly than ighter ones, which hefc)s explain why gases effuse (move out of a container through a tiny hole into a vacuum) or diffuse (move through one another) at rates inversely proportional to the square root of their molar mass Graham s law). 

The values in this table have usually been measured spectroscopically or by mass spectro-metric analysis of hot gases effusing from a Knudsen cell. 

Recently, Sameshima(24) has measured the rates of flow of various simple gases through a compact unglazed earthenware plate. The rates of flow definitely did not obey the Knudsen formula t = k JM, where t denotes the time required for the effusion at constant pressure of a volume F of a gas of molecular weight M, and where fc is a constant. On the other hand, the law t = k M was accurately obeyed when the gases effused through a platinum plate with a single orifice. For the earthenware plate Sameshima found a formula t = to apply. If the wall was very thin n approached zero, and the simple behaviour of the perforated platinum plate was found. If the wall was thick n approached unity and the equation became t = kij (ij denotes viscosity). 

Equation (6.24) can be used in a variety of ways. For example, it can be used to determine which of two gases effuses faster, which does so in a shorter time, which travels farther in a given time, and so on. An effective way to do this is to note that in every case, a ratio of effusion rates, times, distances, and so on, is equal to the square root of a ratio of molar masses, as indicated in equation (6.25). 

EXAMPLE 6-15 Comparing Amounts of Gases Effusing Through an Orifice... 

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