<< Chapter < Page Chapter >> Page >
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 ( [link] ). 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 :

rate of effusion 1

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:

rate of effusion of A rate of effusion of B = B 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)

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:

rate of effusion of hydrogen rate of effusion of oxygen = 1.43 g L −1 0.0899 g L −1 = 1.20 0.300 = 4 1

Using molar masses:

rate of effusion of hydrogen rate of effusion of oxygen = 32 g mol −1 2 g mol −1 = 16 1 = 4 1

Hydrogen effuses four times as rapidly as oxygen.

Check your learning

At a particular pressure and temperature, nitrogen gas effuses at the rate of 79 mL/s. Using the same apparatus at the same temperature and pressure, at what rate will sulfur dioxide effuse?

Answer:

52 mL/s

Got questions? Get instant answers now!

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

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:

rate of effusion = amount of gas transferred time

and combine it with Graham’s law:

rate of effusion of gas Xe rate of effusion of gas Ne = Ne Xe

To get:

amount of Xe transferred time for Xe amount of Ne transferred time for Ne = Ne Xe

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

amount of Xe time for Xe amount of Ne time for Ne = time for Ne time for Xe = Ne Xe = Ne Xe

and substitute values:

time for Ne 243 s = 20.2 g mol 131.3 g mol = 0.392

Finally, solve for the desired quantity:

time for Ne = 0.392 × 243 s = 95.3 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.

Check your learning

A party balloon filled with helium deflates to 2 3 of its original volume in 8.0 hours. How long will it take an identical balloon filled with the same number of moles of air (ℳ = 28.2 g/mol) to deflate to 1 2 of its original volume?

Answer:

32 h

Got questions? Get instant answers now!
Practice Key Terms 5

Get Jobilize Job Search Mobile App in your pocket Now!

Get it on Google Play Download on the App Store Now




Source:  OpenStax, Ut austin - principles of chemistry. OpenStax CNX. Mar 31, 2016 Download for free at http://legacy.cnx.org/content/col11830/1.13
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Ut austin - principles of chemistry' conversation and receive update notifications?

Ask