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RMS/Mean Free Path
Properties of Matter
Mean free path
for 14-16
Simple class experiments introduce students to the idea of mean free path. This can lead to a value for the average distance the molecules travel between collisions and even to an estimate of the size of a molecule.
Class practical
Finding displacement after ‘random walk’ using triangular-grid graph paper.
Apparatus and Materials
- Dice, one for each student
- 60-degree isometric graph paper
Health & Safety and Technical Notes
If the pin-through-straw method (teaching note 3) is used, plain steel dressmaking pins would be safer than the larger optical pins
.
Read our standard health & safety guidance
For a good result you will need to take an average of as many trials as possible. It is therefore important that there is enough equipment for all students.
The dice are more likely to be thrown randomly if each one is shaken in a container whose internal dimensions are several times the width of a die.
Procedure
- Each student draws on a sheet of paper six spokes making 60 degrees with the next (the first line being vertical) and labels each direction successively 1, 2, 3, 4, 5, 6.
- The student takes a die, throws it and uses the uppermost number to tell the direction in which to move. Start in the middle of the graph paper and mark a line 1 ‘unit stride’ in that direction.
- Throw the die again and take a further stride in the new direction. Repeat the process until 25 strides have been taken.
- Measure the direct distance from the start to the finish and record this distance on the board.
- When all students have completed steps 1 to 4, preferably several times, calculate a class average of all the results.
Teaching Notes
- A simple analysis of ‘random walk’ shows that the most likely displacement R from the starting point for N steps of length _s_ is about s times √ N (here 5). That is R = s √ N .
- With only the relatively small number of trials obtained, even using the whole class several times, the average of all the results is unlikely to give good agreement with that predicted. However, it is very unlikely to be zero displacement and probably a long way from 25 times the stride length for 25 throws. It is more likely to be near enough to the predicted result to lend some support.
- (Since we really should have taken the root-mean-square of the displacement results, the ideal result should be about 0.8 times the arithmetical mean. That is about 4 times s , the stride length.)
- Students who have looked at... ...may well grasp that there are millions upon millions of bromine molecules carrying out random walks on a profuse scale. So the statistical prediction will therefore work much better.
- Indeed, if students found the average displacement in 500 seconds – the ‘half-brown distance’ – to be about 10 cm, then they are in a position to find a value for the average distance the molecules move between collisions (the mean-free-path). Appealing to measurements of density and pressure of bromine vapour gives the average speed of molecules as 200 metres/second.
- Density of bromine vapour, ρ = 7.5 kg /m 3
- Pressure of bromine vapour, P = 10 5 N/m 2 (at normal pressure)
- Using the relationship PV = 1/3 (mnv 2 ),
- Speed of bromine molecules, v = √(3 P / ρ) = √(3 x 10 5 / 7.5)= 200 m/s
- So in 500 seconds the molecules will have travelled a ‘straightened out’ path of 200 x 500 m which is equal to the path between collisions, s , multiplied by the number of collisions, N.200 x 500 = sN (1)
- The random path (the half brown distance) = s√N 0.1 = s√N (2)
- From equations (1) and (2) then s = 10 -7 m and N = 10 12
- The distance which a molecule travels between collisions is known as its ‘mean free path’.
- Some teachers may then go on to get a rough value for the size of a molecule.
- It is also possible to estimate the time for an air particle to cross a room. See the relevant guidance pages linked below.
- Another version of this investigation, which does not have any restriction on direction, uses a piece of drinking straw 4 cm long, marked at one end, and with a pin through its middle vertically into a small cork. A cross is marked on a sheet of plain paper and the centre of the cork is placed on the cross. The cork is held with one hand and the other hand is used to flick the straw to spin it. When it comes to rest, a mark is made on the paper under the marked end. The cork is moved to the new mark and the procedure repeated, until 25 steps are obtained.
This experiment was safety-tested in August 2006
Up next
Random walk experiment 2
Class practical
Finding displacement after ‘random walk’ using squared paper.
Apparatus and Materials
- Sheets of paper ruled in squares
- Logs or bags of balls (see technical note)
Health & Safety and Technical Notes
If the pin-through-straw method (teaching note 3) is used, plain steel dressmaking pins would be safer than the larger optical pins
.
Read our standard health & safety guidance
The ‘logs’ are small rectangular blocks of wood. Cheap wood strips of square cross-section, say 1 cm x 1 cm, can be cut into 5 cm lengths. Before cutting up the strip, the four faces could be painted four different colours to signal a move UP, DOWN, LEFT or RIGHT (or, after cutting, the faces could be marked U, D, L and R).
The bag of balls (or beads) may be more expensive but could get a stronger feeling of chances in a concealed lottery. The balls will need to be of four colours and the bag opaque.
Paper ruled in centimetre squares may be available inexpensively for primary school and maths classes.
Procedure
- Mark a starting point on the paper, roll the log (or pick a ball from the bag) and draw a mark 1 square distance. That is one ‘stride length’.
- Repeat this 25 times.
- Measure the distance from start to finish and write it on the squared paper.
- When all students have completed steps 1 to 3, preferably several times, calculate a class average of all the results.
Teaching Notes
- A simple analysis of ‘random walk’ shows that the most likely displacement R from the starting point for N steps of length _s_ is about s times √ N (here 5). That is R = s √ N .
- With only the relatively small number of trials obtained, even using the whole class several times, the average of all the results is unlikely to give good agreement with that predicted. However, it is very unlikely to be zero displacement and probably a long way from 25 times the stride length for 25 throws. It is more likely to be near enough to the predicted result to lend some support.
- (Since we really should have taken the root-mean-square of the displacement results, the ideal result should be about 0.8 times the arithmetical mean. That is about 4 times s, the stride length.)
- Students who have looked at... ...may well grasp that there are millions upon millions of bromine molecules carrying out random walks on a profuse scale. So the statistical prediction will therefore work much better.
- Indeed, if studentsfound the average displacement in 500 seconds – the ‘half-brown distance’ – to be about 10 cm, then they are in a position to find a value for the average distance the molecules move between collisions (the mean-free-path). Appealing to measurements of density and pressure of bromine vapour gives the average speed of molecules as 200 metres/second.
- Density of bromine vapour, ρ = 7.5 kg /m 3
- Pressure of bromine vapour, P = 10 5 N/m 2 (at normal pressure)
- Using the relationship PV = 1/3 (mnv 2 ),
- Speed of bromine molecules, v = √(3 P / ρ)= √(3 x 10 5 / 7.5)= 200 m/s
- So in 500 seconds the molecules will have travelled a ‘straightened out’ path of 200 x 500 m which is equal to the path between collisions, s , multiplied by the number of collisions, N. 200 x 500 = sN (1)
- The random path (the half brown distance) = s√N 0.1 = s√N (2)
- From equations (1) and (2) then s = 10 -7 m and N = 10 12
- The distance which a molecule travels between collisions is known as its ‘mean free path’.
- Some teachers may then go on to get a rough value for the size of a molecule.
- It is also possible to estimate the time for an air particle to cross a room. See the relevant guidance pages linked below.
- Another version of this investigation, which does not have any restriction on direction, uses a piece of drinking straw 4 cm long, marked at one end, and with a pin through its middle vertically into a small cork. A cross is marked on a sheet of plain paper and the centre of the cork is placed on the cross. The cork is held with one hand and the other hand is used to flick the straw to spin it. When it comes to rest, a mark is made on the paper under the marked end. The cork is moved to the new mark and the procedure repeated, until 25 steps are obtained.
This experiment was safety-tested in August 2006
Up next
Mean free path of marble in a tray
Class practical
Marbles in a tray illustrate the meaning of 'mean-free-path'.
Apparatus and Materials
- Two-dimensional kinetic model kit.
Health & Safety and Technical Notes
Students should be fully prepared for the experiment so that they are careful not to send marbles out of the tray.
Read our standard health & safety guidance
Procedure
- Keep the tray flat on the table and agitate it so the marbles move around and collide with one another.
- Concentrate your attention on the marble whose colour is different from that of the others. Watch to see the sort of path it follows.
Teaching Notes
- Carelessness can, of course, lead to chaos with marbles all over the lab. However, if students are given a short time to make observations as well as the task of writing a description of what they see the marble doing, this experiment can be fruitful.
- You will want students to see the random changes of direction made by the marble after each collision. Draw to their attention the difference between the total distance travelled by a marble and its displacement from a starting point.
This experiment was safety-tested in August 2006
Up next
Photograph of marbles in motion
Demonstration
A difficult exercise, photographing marbles to show the differences in their path lengths between collisions.
Apparatus and Materials
- Two-dimensional kinetic model kit
- Lamp
- Camera
Health & Safety and Technical Notes
This kit may be obtained from science apparatus suppliers. It consists of a set of shallow trays (preferably fitted with a thin cork sheet on the base to reduce the noise-level) and a dozen or so marbles, all bar one of the same colour.
In this experiment, the tray should be placed flat on the bench top and illuminated strongly by the lamp. Arrange the lamp in such a way that most of the light entering the lens comes from the light reflected by the marbles only and not from the lamp and other reflected light. The camera must be held rigidly over the tray. An exposure must be chosen so that it shows the motion of each marble as a blur.
Read our standard health & safety guidance
Procedure
- Agitate the tray, keeping it flat on the surface, and take a photograph.
- Keep the exposure time relatively short so that the motion of the marbles before collisions is shown as a blur. Use the length of the blur to estimate the distribution of velocities.
- If the exposure time is longer, an estimate may be made of the distribution of path-lengths between collisions.
Teaching Notes
- This experiment is not recommended except to an enthusiast, student or teacher, who wishes to experiment with exposures.
- If a number of photographs are taken, you can plot histograms of speed or path-length. This would show many marbles moving with speeds close to the average, some much slower and some much faster. (Path-lengths would show a similar pattern.)
- An enthusiast could experiment with short video sequences with a digital camera.
This experiment was safety-tested in August 2006
Up next
Proof of R = s√N
Up next
Further discussion of mean free path
√N steps gives the average mean free path for a bromine molecule in its wandering amongst air molecules. That would be similar to the stride of an air molecule amongst air molecules. You can therefore conclude that an air molecule moves about 10 -7 m between one collision and the next at atmospheric pressure.
This estimate is one of the great classical approaches to estimating the mean free path of an air molecule, first used in the nineteenth century. There are other methods, depending on measurements of viscosity, Van der Waals constants, etc. The result differs somewhat according to the method chosen. For air at room temperature, older estimates gave the mean free path of an air molecule as (0.8 – 1.0) 10 -7 m but the modern value lies between (0.6 – 0.7) 10 -7 m. In the following discussion 10 -7 m will be assumed.
The number of collisions that an air molecule makes per second will be different from the experimental estimate for bromine molecules because oxygen and nitrogen molecules move faster. Combining an estimate of molecular speed with the mean free path estimate, in one second an air molecule travels 500 m of straightened out path (a distance which contains 500/ 10 -7 mean free paths). So the molecule makes 500 x 10 7 collisions per second.
Footnote on simplifications
This is very rough calculation. Any attempt to make a correction for bromine molecules being bigger would place a very unscientific emphasis on precision in one particular place in a method that is imprecise overall. Judging from the relative densities of liquids and relative molecular masses, bromine molecules have a diameter about 1.2 times that of air molecules.
That makes the ‘average diameter’ for a bromine molecule hitting an air molecules 1.1 times on average as great as for air molecules colliding. Because any mean free path depends on the collision target area
(cross-section), a bromine molecule probably has a mean free path among air molecules about 1/(1.1)
2
or 1/1.2 or (0.83) times the mean free path of an air molecule among air molecules.
This approach makes no attempt to decide what kind of average should be used for the resultant in a random walk treatment. Does the estimate ‘half brown’ fit best with the root mean square average of the random walk of bromine molecules, or should you use the plain arithmetical average? Since progress is estimated in a vertical direction alone, should you take some component of velocity, or of a mean free path?
Unless you give up this simple experiment, in which students make a guess, and resort to colorimetry and density measurements, these questions remain unanswered. Nor would it be sensible to try to answer them here; that would miss the point of proceeding quickly in a simple story so that you do not lose your students on the way.
Up next
Estimate of molecular size: a more formal method
Imagine molecules in a gas; dots spaced far apart. Add arrows to show the random motion, not all speeds (arrows) the same but speeds around the average. You could say:
'Here is a snapshot of air molecules in this room with the camera focused at one distance. To find how one molecule would move through this vast array of moving neighbours is too difficult a business. Instead, pretend that we freeze all the molecules except one and watch that one molecule go hurtling through the crowd.'
Redraw the picture showing each molecule as one round blob without any indication of velocity. Draw the path of the chosen molecule, as it moves to collide with another, as a cylinder swept out between the two molecules. The diameter of the cylinder is equal to the diameter of the molecule and its length is equal to the mean free path. Bend the path at the collision and another cylinder is swept out as shown this diagram:
The mean free path is many times longer than the separation between molecules and so the cylinder should pass many other molecules on the way to a collision.
Now move off to a separate preparatory discussion looking at such a collision in detail. Draw a large round molecule bouncing against another molecule.
'How far apart are the molecules, centre to centre, at the collision? One diameter.'
'I am now going to show you a trick for finding out how far a molecule goes before hitting another. This trick has been invented by scientists and is not what really happens but gives good results. When two molecules collide they must be 2 radii, or 1 diameter apart. Instead of drawing the collision like that, I could pretend that the molecule flying along to make the collision is much bigger, and any other molecule that it hits is much smaller. We get the same result as long as we have the centres of the two molecules 1 diameter apart at the collision. I am now going to push this to the limit and make the flying molecule have double the radius, equal to 1 diameter, and the molecule it hits have no radius at all.'
'Now we start this story all over again. Here is the artificial molecule flying along with radius equal to one molecular diameter. It sweeps out a cylinder of 1 molecular diameter in radius and collides with the artificial point sized molecule where it bends its path.'
'Think about the path swept out by this flying molecule which is possessively patrolling its "share" of the volume of the box. This volume is equal to d2x 10-7m.'
For justification of mean free path being 10-7m, see Guidance note...
Further discussion of mean free path
'We need to know the volume of space that belongs to one molecule of air in this room. The volume change from liquid air to air is about 1:750. If for liquid air each molecule of diameter, d, occupies a cubical box of side d, then the volume occupied is d 3on the average.'
750d3= d2x 10-7
d = 4 x 10 -10m
'We have found the diameter of a typical molecule of air. An atom is probably about half that size. This is certainly a rough estimate because our measurements were difficult and we made all kinds of risky moves carrying out our calculations. Yet this is a very good estimate for many working purposes. It is the right order of magnitude.
All we are really measuring here is an order-of-magnitude distance of approach at which inter-molecular forces grow large enough to have a noticeable effect. Air of course is a mixture of different gases, mainly nitrogen (about 78%) and oxygen (about 21%).
Careful measurements for particular molecules give different diameters according to the experiment chosen and the method of interpretation used. After all, the diameter of a molecule is not as definite a thing as the diameter of a steel ball. Both nitrogen and oxygen are diatomic molecules. Not only are diatomic molecules oblong
but they behave as if squashy, so more violent collisions are likely to reveal a smaller effective diameter. Nitrogen molecules are very slightly larger than oxygen molecules; in their gaseous state both have effective diameters of about 3 x 10-10m.
Up next
Time for air molecules to cross a room
Once students understand the thinking behind estimates of mean free path, s , they can estimate the time for air molecules to cross a room. Assume there are no convection currents.
In time T seconds the straightened-out path of a molecule of air is 500 T metres. The number of collisions it makes in that time is 500 T / s which is 5 × 109 x T collisions. The average progress from start to finish will be s √ N which is the length of the room, say 6 m.
Hence 6 = √( 5 × 109 x T) x 10-7. And so the time for an air molecule to cross the room is 720 000 seconds, more than a week!
The same kind of story applies to neutrons diffusing from the inner regions of a nuclear reactor. Also for the particles of light (photons) cannoning their way out from the inner layers of the Sun.
Up next
Mean free path
It is interesting for students to see how far physicists can push ‘back of the envelope’ calculations.
Diffusing bromine into air can be used to estimate the mean free path of bromine molecules. See the experiment:
The clock is started as soon as the bromine is released into the tube of air. After 500 seconds the average distance which the bromine has diffused up the tube is measured (the half-brown position.). The bromine molecules are used as markers to show how one gas travels through another making many collisions on the way. This experiment shows that molecules have ‘size’ as they find targets with which to collide.
A discussion might go something like this:
Put on your super-microscopic spectacles and have another look at this bromine wandering through the air.
Can you see the bromine molecules travelling very fast for a short distance, colliding with another molecule, rebounding in a different direction, then striking another...?
A quick look at the two-dimensional model with marbles in a tray, with one distinctly coloured marble to focus on, will reinforce the idea of random motion here, if wanted.
How far can a molecule go with that sort of movement? The molecule’s progress is rather like yours if you tried to work your way through the rush-hour crowd in a busy station blindfolded so that you had no memory of direction. Scientists call that a random walk
.
Imagine you take one stride from your starting point. Then choose a new direction, any direction at random, and take one stride in that direction. Again choose a new direction at random and take one stride in that direction. Go on like that until you have taken a large number of equal strides, say 100. How far are you from your starting point, as the crow flies?
A bromine molecule makes a wandering path from collision to collision, starting out in a new direction every time. The more time a molecule has to make strides, the farther from its start it is likely to end up. But the direction of motion after each collision is random, so you cannot just add up all the steps in a straight line. All the same, it is still possible to make predictions about the distance travelled in a random walk. After a given time, one molecule will have got to a place a long way away from the start. Another will have got back near its start. Many will have got to some middling distance.
You could even make a chart of molecules at various distances. You could make the chart by imagining a random walk or sketching it on paper. If you made hundreds or even thousands of trials of that and took an average, you could find a definite average "crow-flies" distance from start to finish.
There is a definite rule for predicting that average distance. Suppose a molecule takes a 100 steps, bouncing away from each collision in a new direction (sketch the motion). Catalogue its net progress from start to finish and take the average of many trials of 100 steps. The result is not 100 steps (the maximum) or no steps (the minimum) but 10 steps. This is because 10 is the √ 100 and the general rule says that for N steps the average distance travelled is √ N steps.
You need quite a lot of algebra, adding and averaging all those wild wanderings, to predict this strange rule with the √N in it. So instead of trying the algebra, play the game yourself and see if you can test the rule.