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By the end of this section, you will be able to:
  • Define and explain effusion and diffusion
  • State Graham’s law and use it to compute relevant gas properties

If you have ever been in a room when a piping hot pizza was delivered, you have been made aware of the fact that gaseous molecules can quickly spread throughout a room, as evidenced by the pleasant aroma that soon reaches your nose. Although gaseous molecules travel at tremendous speeds (hundreds of meters per second), they collide with other gaseous molecules and travel in many different directions before reaching the desired target. At room temperature, a gaseous molecule will experience billions of collisions per second. The mean free path    is the average distance a molecule travels between collisions. The mean free path increases with decreasing pressure; in general, the mean free path for a gaseous molecule will be hundreds of times the diameter of the molecule

In general, we know that when a sample of gas is introduced to one part of a closed container, its molecules very quickly disperse throughout the container; this process by which molecules disperse in space in response to differences in concentration is called diffusion    (shown in [link] ). The gaseous atoms or molecules are, of course, unaware of any concentration gradient, they simply move randomly—regions of higher concentration have more particles than regions of lower concentrations, and so a net movement of species from high to low concentration areas takes place. In a closed environment, diffusion will ultimately result in equal concentrations of gas throughout, as depicted in [link] . The gaseous atoms and molecules continue to move, but since their concentrations are the same in both bulbs, the rates of transfer between the bulbs are equal (no net transfer of molecules occurs).

In this figure, three pairs of gas filled spheres or vessels are shown connected with a stopcock between them. In a, the figure is labeled, “Stopcock closed.” Above, the left sphere is labeled, “H subscript 2.” It contains approximately 30 small, white, evenly distributed circles. The sphere to its right is labeled, “O subscript 2.” It contains approximately 30 small red evenly distributed circles. In b, the figure is labeled, “Stopcock open.” The stopcock valve handle is now parallel to the tube connecting the two spheres. On the left, approximately 9 small, white circles and 4 small, red circles are present, with the red spheres appearing slightly closer to the stopcock. On the right side, approximately 25 small, red spheres and 21 small, white spheres are present, with the concentration of white spheres slightly greater near the stopcock. In c, the figure is labeled “Some time after Stopcock open.” In this situation, the red and white spheres appear evenly mixed and uniformly distributed throughout both spheres.
(a) Two gases, H 2 and O 2 , are initially separated. (b) When the stopcock is opened, they mix together. The lighter gas, H 2 , passes through the opening faster than O 2 , so just after the stopcock is opened, more H 2 molecules move to the O 2 side than O 2 molecules move to the H 2 side. (c) After a short time, both the slower-moving O 2 molecules and the faster-moving H 2 molecules have distributed themselves evenly on both sides of the vessel.

We are often interested in the rate of diffusion    , the amount of gas passing through some area per unit time:

rate of diffusion = amount of gas passing through an area unit of time

The diffusion rate depends on several factors: the concentration gradient (the increase or decrease in concentration from one point to another); the amount of surface area available for diffusion; and the distance the gas particles must travel. Note also that the time required for diffusion to occur is inversely proportional to the rate of diffusion, as shown in the rate of diffusion equation.

A process involving movement of gaseous species similar to diffusion is effusion    , the escape of gas molecules through a tiny hole such as a pinhole in a balloon into a vacuum ( [link] ). Although diffusion and effusion rates both depend on the molar mass of the gas involved, their rates are not equal; however, the ratios of their rates are the same.

Questions & Answers

if three forces F1.f2 .f3 act at a point on a Cartesian plane in the daigram .....so if the question says write down the x and y components ..... I really don't understand
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hey , can you please explain oxidation reaction & redox ?
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hey , can you please explain oxidation reaction and redox ?
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for grade 12 or grade 11?
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a rate of change in velocity of an object whith respect to time
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t =r×f
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Source:  OpenStax, Chemistry. OpenStax CNX. May 20, 2015 Download for free at http://legacy.cnx.org/content/col11760/1.9
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