... 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. 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 ...

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