 Solution

ID: past-exam-of-the-chemistry-course-of-the-year-3-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/paper-3/48/a/solution

With wall effects neglected, $a_{1} = a$ . A second hard disc cannot have its centre within distance

2 r

of the first centre. Its excluded volume in this two-dimensional setting is an excluded area

π (2 r)^{2} = 4 a_{d}

a_{2} = a - 4 a_{d} .

This is exclusion of the second centre, rather than simply subtraction of the first disc's own area.

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