An ideal gas is confined in a container. The initial mean free path...

The mean free path ($\lambda$) of gas molecules is given by:
$\lambda = \dfrac{1}{\sqrt{2}\pi d^2 n}$
where $n$ is the number density of the gas molecules.

The number density is defined as the number of molecules per unit volume, $n = \dfrac{N}{V}$.

When half the mass of the gas escapes, the new number of molecules becomes $N' = \dfrac{N}{2}$.
When the volume of the container is doubled, the new volume becomes $V' = 2V$.

The new number density is:
$n' = \dfrac{N'}{V'} = \dfrac{\dfrac{N}{2}}{2V} = \dfrac{N}{4V} = \dfrac{n}{4}$

Since the mean free path is inversely proportional to the number density ($\lambda \propto \dfrac{1}{n}$), the new mean free path will be:
$\lambda' = \dfrac{1}{\sqrt{2}\pi d^2 \left(\dfrac{n}{4}\right)} = 4 \lambda$

Answer: $4\lambda$