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 $2r$ of the first centre. Its excluded volume in this two-dimensional setting is an excluded area $\pi (2r{)}^2=4a_d$, givingThis is exclusion of the second centre, rather than simply subtraction of the first disc's own area.