The mean free path of molecules of a gas (radius $r$) is inversely proportional to

If the pressure of a real gas $O_{2}$ in a container is given by $P = \frac{RT}{2V - b} - \frac{a}{4b^{2}}$, then the mass of the gas in the container is: (in $\text{ g}$)

How does the mean free path of a gas depend on temperature?

There are two identical chambers,completely thermally insulated from surroundings. Both chambers have a partition wall dividing the chambers into two compartments. Compartment $1$ is filled with an ideal gas and Compartment $3$ is filled with a real gas. Compartments $2$ and $4$ are vacuum. $A$ small hole (orifice) is made in the partition walls and the gases are allowed to expand into the vacuum.
Statement $-1$: No change in the temperature of the gas takes place when an ideal gas expands in a vacuum. However,the temperature of a real gas goes down (cooling) when it expands in a vacuum.
Statement $-2$: The internal energy of an ideal gas is only kinetic. The internal energy of a real gas is kinetic as well as potential.

In the absence of intermolecular forces of attraction,the observed pressure $P$ will be

For a molecule of an ideal gas,the number density is $2 \sqrt{2} \times 10^8 \text{ cm}^{-3}$ and the mean free path is $\frac{10^{-2}}{\pi} \text{ cm}$. The diameter of the gas molecule is