If the mean free path of atoms is doubled then the pressure of gas will become

The correct option is P/2.

To understand why the pressure becomes half when the mean free path is doubled, we can look at the relationship between the mean free path and the pressure of an ideal gas at a constant temperature.

The mean free path, denoted by:

$\lambda$

is the average distance a molecule travels between successive collisions. It is mathematically expressed as:

$$\lambda =\frac{1}{\sqrt{2}\pi d^{2}n}$$

where:
• $d$ is the molecular diameter.
• $n$ is the number density of the gas (number of molecules per unit volume, or $n=\frac{N}{V}$).

According to the ideal gas law, pressure is given by:
$P=n{k}_{B}T$

This allows us to write the number density as:
$n=\frac{P}{k_{B}T}$
where $k_B$ is the Boltzmann constant and $T$ is the absolute temperature.

Substituting this density back into the mean free path formula gives:

$$\lambda =\frac{k_{B}T}{\sqrt{2}\pi d^{2}P}$$

For a given gas at a constant temperature, all terms in the formula except pressure and mean free path remain constant. Thus, the mean free path is inversely proportional to the pressure of the gas:

$$\lambda \propto \frac{1}{P}$$

This relationship can be written as a ratio between the initial state (1) and final state (2):
$\frac{{\lambda }_2}{{\lambda }_1}=\frac{P_1}{P_2}$

If the mean free path is doubled, we set:
${\lambda }_2=2{\lambda }_1$

Substituting this into our ratio equation yields:
$2=\frac{P_1}{P_2}$

Solving for the new pressure:
$P_2=\frac{P_1}{2}$

Therefore, if the mean free path of the atoms is doubled, the pressure of the gas becomes P/2.