Some learners are full of hot air, but is argon?

Estimate the percentage of ‘empty space’ in a sample of gaseous argon

## Introduction

Teachers who have not used the problems before should read the section Using the problems before starting.

## Prior knowledge

Familiarity with Avogadro’s number, the characteristics of solids, liquids and gases and the concept of atomic size. A detailed knowledge is unnecessary as students are encouraged to consult textbooks and data books during the exercise.

## Resources

Scientific calculators, suitable data books should be available for reference.

## Possible solutions

The key question which underpins this problem is ‘What is meant by the volume of a gas?’ Research has shown that students have difficulty imagining an empty space or a vacuum, and perceive matter as being continuous with no empty space. The problem can be approached on at least two levels.

a. By calculating the space between the atoms. This can be done starting from atomic sizes or from the density of the substance as a solid or as a liquid, and assuming that the atoms touch one another.

b. By calculating the space within the atoms and between the atoms. This can be done starting from the sizes of the nucleus and of the electrons. The data are not easy to find and it is probably best not to introduce this method unless students think of it. It could be mentioned at the very end of the problem. The calculation is set out below.

## Suggested approach

During trialling the following instructions were given to students and proved to be extremely effective:

1. Working as a group, discuss the problem and try to work out the answer. You can divide the work amongst yourselves if you wish but keep one another informed of progress. Discussion can play a vital part in working out solutions to problems like this. Several minds working together on a problem can stimulate ideas that each one on its own could not manage. About 10 minutes should be spent on the initial discussion with further discussion as required.
2. Write a brief account of what you did.
3. Working as a group, prepare a short (ca 5-minute maximum) presentation suited to middle secondary pupils, but to be delivered to the rest of your class. If possible all group members should take part: any method of presentation (such as a blackboard, overhead projector, etc) can be used. Outline the problem at the start of the presentation and then describe what you did. After the presentation, be prepared to accept and answer questions from the rest of the class and discuss with them what you did.

## Method a. Calculating the space between atoms

There are at least two alternative approaches:

(i) From atomic sizes

The van der Waals radius of argon is 0.192 nm (Stark and Wallace 1989)

Volume of this sphere = 4/3 πr3 = 4/3 π (0.192 x 109)3 m3

= 2.96 x 1029 m3

volume of 1 mole of argon atoms = 6.02 x 1023 x 2.96 x 1029 m3

= 1.78 x 105 m3

= 18 cm3 (to two significant figures)

This figure is for the atoms alone and does not include the spaces between them

(ii) From the density of the liquid

The density of argon liquid at –186 oC (its boiling point) is 1.40 g cm3 (Stark

and Wallace); and 1 mole of argon atoms has a mass of 39.948 g (40 g to two significant figures).

Therefore 1 mole of argon atoms occupies 40/1.4 = 29 cm3 ie 6.0 x 1023 atoms of argon occupy 29 cm3 when touching one another.

The 18 cm3 figure is a minimum estimate as the spaces between the spherical argon atoms in the liquid have been ignored. The atoms are close packed: reference to a textbook gives the information that 74 % of the available space in a close packed array of spheres is occupied. It seems an unnecessary complication to introduce this figure, but 74 % of 29 cm3 is 21 cm3 – close to the 18 cm3 calculated above.

The subsequent calculation is the same for both methods using whatever value the students found. At 0 °C and normal atmospheric pressure, 1 mole of gaseous argon atoms occupies 22 400 cm3

The percentage of ‘filled’ space in argon = 18/22 400)x 100 %

(or 29/22 400 x 100 %)

= 0.08 % (0.13 %)

and the percentage of ‘empty’ space = 99.92 % (99.87 %).