Gaseous State

Seeing that at one bar pressure, the mean free path $\overline{l} = \frac{100}{1} = \boxed{100 \mathring{A}}$ , the mean free path at bar pressure $5$ will therefore be $\overline{l} = \frac{100}{5} = \boxed{20 \mathring{A}}$ .

That makes $(1)$ and $(2)$ the answer.

• I don't really know what other ways there are of solving this, but this was my approach.
• The volume $V$ is constant, so I took the initial $100 \mathring{A}$ value too.