GCSE Chemistry - Boyle's Law

In the experiment to prove Boyle¡¦s Law we set-up the following apparatus:

{DIAGRAM}

METHOD

We pumped air into the tube until the pressure was as high as possible, which was 3.4x105 Pa. We recorded the volume of gas; it was 16.0cm3. We then let the pressure down slightly and recorded the pressure and volume again. We repeated this about 15 more times, which was when the pressure reached 1.03x105 Pa. Here are the results:

Gas Pressure (x105 Pa) Gas Volume (cm3)
3.4 16
3.23 16.9
3.09 17.8
2.91 18.8
2.78 19.7
2.62 21
2.48 22.1
2.31 23.7
2.15 25.7
1.97 28.1
1.73 31.9
1.6 34.9
1.41 39.1
1.28 43.3
1.12 50
1.03 56

EXPLANATION

Boyle¡¦s law states that the volume of a fixed mass of ideal gas at constant temperature is inversely proportional to the gas pressure:

P „f 1/v OR pv = constant

This means that if p is doubled then v is halved, or, if p is halved then v is doubled.

MOLECULAR EXPLANATION

At room temperature and pressure, there are a certain number of gas molecules in 10cm3 of air. Each molecule exerts a pressure on the walls of a container. The molecules hitting the sides of the container cause this pressure. If you decrease the volume that the air occupies from 10 cm3 to 5 cm3, then there would be twice as many molecules per cm3 than before. This means twice as much pressure will be exerted due to twice as many molecules hitting the sides of the container at a certain time.

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ANALYSIS

To analyse my results I will extend my table from before by adding 1/v and pv:

p (x105 Pa) v (cm3) 1/v (cm-3) pv (x105 Pa cm3)
3.40 16.0 0.062 54.40
3.23 16.9 0.059 54.58
3.09 17.8 0.056 55.00
2.91 18.8 0.053 54.71
2.78 19.7 0.051 54.77
2.62 21.0 0.048 55.02
2.48 22.1 0.045 54.80
2.31 23.7 0.042 54.75
2.15 25.7 0.039 55.26
1.97 28.1 0.036 55.36
1.73 31.9 0.031 55.19
1.60 34.9 0.029 55.84
1.41 39.1 0.026 55.13
1.28 43.3 0.023 55.42
1.12 50 0.020 56.00
1.03 56 0.018 57.68

Using my table, I can see that as p increases from 2.15 to 3.23, v decreases from 25.7 to 16.9. I can also see ...

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