Chemistry » Gases » Effusion and Diffusion of Gases

Effusion and Diffusion of Gases

Effusion and Diffusion of Gases

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 the figure below).

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 the figure below. 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, H2 and O2, are initially separated. (b) When the stopcock is opened, they mix together. The lighter gas, H2, passes through the opening faster than O2, so just after the stopcock is opened, more H2 molecules move to the O2 side than O2 molecules move to the H2 side. (c) After a short time, both the slower-moving O2 molecules and the faster-moving H2 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:

\(\text{rate of diffusion}=\phantom{\rule{0.2em}{0ex}}\cfrac{\text{amount of gas passing through an area}}{\text{unit of time}}\)

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 (see the figure below). 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.

This figure contains two cylindrical containers which are oriented horizontally. The first is labeled “Diffusion.” In this container, approximately 25 purple and 25 green circles are shown, evenly distributed throughout the container. “Trails” behind some of the circles indicate motion. In the second container, which is labeled “Effusion,” a boundary layer is evident across the center of the cylindrical container, dividing the cylinder into two halves. A black arrow is drawn pointing through this boundary from left to right. To the left of the boundary, approximately 16 green circles and 20 purple circles are shown again with motion indicated by “trails” behind some of the circles. To the right of the boundary, only 4 purple and 16 green circles are shown.

Diffusion occurs when gas molecules disperse throughout a container. Effusion occurs when a gas passes through an opening that is smaller than the mean free path of the particles, that is, the average distance traveled between collisions. Effectively, this means that only one particle passes through at a time.

If a mixture of gases is placed in a container with porous walls, the gases effuse through the small openings in the walls. The lighter gases pass through the small openings more rapidly (at a higher rate) than the heavier ones (see the figure below). In 1832, Thomas Graham studied the rates of effusion of different gases and formulated Graham’s law of effusion: The rate of effusion of a gas is inversely proportional to the square root of the mass of its particles:

\(\text{rate of effusion}\propto \cfrac{1}{\sqrt{\text{ℳ}}}\)

This means that if two gases A and B are at the same temperature and pressure, the ratio of their effusion rates is inversely proportional to the ratio of the square roots of the masses of their particles:

\(\cfrac{\text{rate of effusion of A}}{\text{rate of effusion of B}}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\cfrac{\sqrt{{\text{ℳ}}_{\text{B}}}}{\sqrt{{\text{ℳ}}_{\text{A}}}}\)

This figure shows two photos. The first photo shows a blue balloon which floats above a green balloon. The green balloon is resting on a surface. Both balloons are about the same size. The second photo shows the same two balloons, but the blue one is now smaller than the green one. Both are resting on a surface.

A balloon filled with air (the blue one) remains full overnight. A balloon filled with helium (the green one) partially deflates because the smaller, light helium atoms effuse through small holes in the rubber much more readily than the heavier molecules of nitrogen and oxygen found in air. (credit: modification of work by Mark Ott)

Example

Applying Graham’s Law to Rates of Effusion

Calculate the ratio of the rate of effusion of hydrogen to the rate of effusion of oxygen.

Solution

From Graham’s law, we have:

\(\cfrac{\text{rate of effusion of hydrogen}}{\text{rate of effusion of oxygen}}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\cfrac{\sqrt{1.43\phantom{\rule{0.2em}{0ex}}\require{cancel}\cancel{{\text{g L}}^{\text{−1}}}}}{\sqrt{0.0899\phantom{\rule{0.2em}{0ex}}\require{cancel}\cancel{{\text{g L}}^{\text{−1}}}}}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\cfrac{1.20}{0.300}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\cfrac{4}{1}\)

Using molar masses:

\(\cfrac{\text{rate of effusion of hydrogen}}{\text{rate of effusion of oxygen}}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\cfrac{\sqrt{32\phantom{\rule{0.2em}{0ex}}\require{cancel}\cancel{{\text{g mol}}^{\text{−1}}}}}{\sqrt{2\phantom{\rule{0.2em}{0ex}}\require{cancel}\cancel{{\text{g mol}}^{\text{−1}}}}}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\cfrac{\sqrt{16}}{\sqrt{1}}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\cfrac{4}{1}\)

Hydrogen effuses four times as rapidly as oxygen.

Here’s another example, making the point about how determining times differs from determining rates.

Example

Effusion Time Calculations

It takes 243 s for 4.46 \(×\) 10−5 mol Xe to effuse through a tiny hole. Under the same conditions, how long will it take 4.46 \(×\) 10−5 mol Ne to effuse?

Solution

It is important to resist the temptation to use the times directly, and to remember how rate relates to time as well as how it relates to mass. Recall the definition of rate of effusion:

\(\text{rate of effusion}=\phantom{\rule{0.2em}{0ex}}\cfrac{\text{amount of gas transferred}}{\text{time}}\)

and combine it with Graham’s law:

\(\cfrac{\text{rate of effusion of gas Xe}}{\text{rate of effusion of gas Ne}}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\cfrac{\sqrt{{\text{ℳ}}_{\text{Ne}}}}{\sqrt{{\text{ℳ}}_{\text{Xe}}}}\)

To get:

\(\cfrac{\frac{\text{amount of Xe transferred}}{\text{time for Xe}}}{\frac{\text{amount of Ne transferred}}{\text{time for Ne}}}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\cfrac{\sqrt{{\text{ℳ}}_{\text{Ne}}}}{\sqrt{{\text{ℳ}}_{\text{Xe}}}}\)

Noting that amount of A = amount of B, and solving for time for Ne:

\(\cfrac{\frac{\require{cancel}\cancel{\text{amount of Xe}}}{\text{time for Xe}}}{\frac{\require{cancel}\cancel{\text{amount of Ne}}}{\text{time for Ne}}}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\cfrac{\text{time for Ne}}{\text{time for Xe}}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\cfrac{\sqrt{{\text{ℳ}}_{\text{Ne}}}}{\sqrt{{\text{ℳ}}_{\text{Xe}}}}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\cfrac{\sqrt{{\text{ℳ}}_{\text{Ne}}}}{\sqrt{{\text{ℳ}}_{\text{Xe}}}}\)

and substitute values:

\(\cfrac{\text{time for Ne}}{243\phantom{\rule{0.2em}{0ex}}\text{s}}\phantom{\rule{0.2em}{0ex}}=\sqrt{\cfrac{20.2\phantom{\rule{0.2em}{0ex}}\require{cancel}\cancel{\text{g mol}}}{131.3\phantom{\rule{0.2em}{0ex}}\require{cancel}\cancel{\text{g mol}}}}\phantom{\rule{0.2em}{0ex}}=0.392\)

Finally, solve for the desired quantity:

\(\text{time for Ne}=0.392\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}243\phantom{\rule{0.2em}{0ex}}\text{s}=95.3\phantom{\rule{0.2em}{0ex}}\text{s}\)

Note that this answer is reasonable: Since Ne is lighter than Xe, the effusion rate for Ne will be larger than that for Xe, which means the time of effusion for Ne will be smaller than that for Xe.

Finally, here is one more example showing how to calculate molar mass from effusion rate data.

Example

Determining Molar Mass Using Graham’s Law

An unknown gas effuses 1.66 times more rapidly than CO2. What is the molar mass of the unknown gas? Can you make a reasonable guess as to its identity?

Solution

From Graham’s law, we have:

\(\cfrac{\text{rate of effusion of Unknown}}{{\text{rate of effusion of CO}}_{2}}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\cfrac{\sqrt{{\text{ℳ}}_{{\text{CO}}_{2}}}}{\sqrt{{\text{ℳ}}_{Unknown}}}\)

Plug in known data:

\(\cfrac{1.66}{1}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\cfrac{\sqrt{44.0\phantom{\rule{0.2em}{0ex}}\text{g/mol}}}{\sqrt{{\text{ℳ}}_{Unknown}}}\)

Solve:

\({\text{ℳ}}_{Unknown}=\phantom{\rule{0.2em}{0ex}}\cfrac{44.0\phantom{\rule{0.2em}{0ex}}\text{g/mol}}{{\left(1.66\right)}^{2}}\phantom{\rule{0.2em}{0ex}}=16.0\phantom{\rule{0.2em}{0ex}}\text{g/mol}\)

The gas could well be CH4, the only gas with this molar mass.

[Attributions and Licenses]


This is a lesson from the tutorial, Gases and you are encouraged to log in or register, so that you can track your progress.

Log In

Share Thoughts