SimuLab 15: The Diffusion Chamber Simulation

You need the Diffusion Chamber program to complete this SimuLab. This program is only available for Macs at this time.	Download Diffusion Chamber:
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The Diffusion Chamber program carries out the diffusing checkers model with computer speed, many more "checkers,'' and a longer chamber. With this program you can change the model in various ways in order to compare the results with experiment.

Start the Diffusion Chamber program. You will see two diffusion chambers, one above the other. Later this arrangement will allow you to compare diffusion under different circumstances. For now just use the upper chamber. In the selections on the right, choose m(green) = m(blue) which means that the two diffusing molecules have the same mass, Infinite Particles which means particles are renewed from a source at each end of the chamber, and 1-D Random which means that the model is a one-dimensional random walk. Start the run with the upper chamber.

Q6.3: At what time (displayed at the lower right of the panel) are there a "significant number'' of precipitated dust particles? You define "significant.''

Q6.4: Is the precipitated dust grouped around the center of the tube or near some other location?

Now choose a topic from the list below and carry out a brief investigation using the Diffusion Chamber program. Different groups in the class may choose different topics.

For comparison, you can use both chambers in the display, if you wish, running them simultaneously. To increase the statistical accuracy of your results, you can increase the number of particles using the Size menu items (but then the process will take longer).

1. Chamber Length: For diffusion chambers of different lengths, what are the different times for a "significant number'' of dust particles to be precipitated? Make a prediction: For a chamber half as long, do you expect this time to be cut in half? Change the length of the chamber by dragging the right hand end or by clicking on the Set box at the upper right of the control panel and entering a number. Record your data and make a graph of the "diffusion time'' as a function of chamber length. Is the result what you predicted?


2. Mass Ratio of Molecules: When two molecules of different mass have the same kinetic energy, the molecule with the larger mass moves more slowly. Here is the reason: At a given common temperature, all molecules in a gas have the same average kinetic energy (mv²/2), so one would expect the molecules with larger mass m to move slower, on the average. Perhaps slower-moving more massive molecules will diffuse more slowly than less massive molecules. Therefore, perhaps the initial precipitation disk will not occur in the middle of the tube. Is this true?

In the diffusion chamber demonstration, HCl has a molecular weight (atomic mass) of 1 + 35.5 = 36.5, while NH₃ has a molecular weight of 14 + 3 = 17, about half as much. As a result of this difference, which molecule do you expect to diffuse faster. Do you expect the initial precipitation disk to form nearer the end of the chamber from which HCl is diffusing, or nearer the end from which NH₃ is diffusing? How much nearer? Write down your predictions.

Test your predictions with the Diffusion Chamber program. You can try two different mass ratios in the two panels at the same time, determining the location of the initial precipitation "disk" in each case.

Do you obtain more accurate results with molecules of smaller size? Note: The "Length'' of the chamber is measured in the number of horizontal positions that can be occupied by the diffusing molecules. When you choose a smaller size for the molecules, more of them can line up from right to left in a given chamber. In this case, the number displayed for the "Length'' is larger for a chamber that shows the same size on the screen.


3. Motion of Particles: Our simplest model forces the molecules to execute a random walk on one dimension-each one moving back and forth along a line. What is different if the molecules can move in two dimensions, executing the so-called "2-D random walk''? Will it take longer for the initial precipitation disk to form? Will this disk appear in a different place? Write down your predictions, then test them by changing the Motion of Particles setting to 2-D Random. Write down your results and compare them with your predictions.

A third way that particles might move is ballistic. This means constant velocity in a straight line along the length of the tube. What do you expect to happen in this case? Will formation of the precipitation disk occur more quickly or more slowly than for a random walk? Will ballistic motion change the location of the initial precipitation disk? Write down your predictions.

Try the Ballistic setting for Motion of Particles and compare the results with your predictions. Predict what will happen if you carry out the ballistic experiment with the Infinite Particles setting, then the No new particles setting. Compare the results with your predictions.

Discussion - What to Expect

Before going on to a laboratory experiment with diffusion chambers made of small tubes, we try to estimate numerical values for some of the results to expect in such experiments. To simplify the analysis, we return to the one-dimensional random walk model for diffusion.

From the ManyWalkers program (SimuLab 3.5) we know that in a one-dimensional random walk the average of the square of distance traveled is proportional to the number of steps. Let Let x² stand for this average square, let N be the number of steps, and let L_step be the average length of each step, the average distance between collisions. Then the result can be written as:

x2 = NLstep2.

(6.1)

Why L_step²? This must be entered in the equation as a square to make the units correct, since N here has no units. Why L_step²? This must be entered in the equation as a square to make the units correct, since L_step here has no units.

Notice that for uniform "ballistic'' motion in a straight line, the formula would be N = x/L (without the squares).




Q6.5: How many steps N will the ammonia molecule take to be detected a distance x away? To begin to answer this question we need to know the speed of the molecules at room temperature and the average distance that a molecule moves between collisions. Here are approximate values for these two quantities. Later you will derive your own values for these quantities.






1. Speed of gas molecules. A gas molecule at room temperature moves with an average speed approximately equal to that of a rifle bullet, about 600 meters/second. Why doesn't this hurt? This speed is approximately 2200 kilometers/hour (or 1300 miles/hour). Why don't you feel a 2200 kilometer/hour wind all the time?


2. Distance between collisions. A gas molecule in the air travels a distance between collisions (the length of one step) approximately equal to L_step = 1000  x  3  x 10^-10meters = 3 x 10^-7 meters. How does this distance compare with the diameter of one hair on your head? This distance is equal to how many times the diameter of an atom?

From these quantities and earlier equations, answer the following questions:




Q6.6: Suppose that the molecule travels in a straight line, taking steps only in the forward direction. How many steps N would it take for the molecule to travel x = 10 centimeters?







Q6.7: In contrast, suppose that the molecule travels in a one-dimensional random walk. How many steps (collisions) N will it take to travel x = 10 centimeters?







Q6.8: Let the time tstep represent the average time that elapses between collisions. Find an expression for tstep as a function of the particle's average velocity and the average distance traveled in each step.







Q6.9: Therefore how many seconds t are required for the molecule to move 10 centimeters, on the average, in a one-dimensional random walk? Express your answer in an everyday unit, such as hours or days or years.







Q6.10: What if the total distance traveled is one centimeter instead of 10 centimeters. How long is required, on the average, for the molecule to cover a distance of one centimeter.