Mean Free Path of Atoms:
The mean free path of atoms in a gas is the average distance that an atom travels between collisions with other atoms or molecules. It is a measure of how far an atom can move freely before it encounters another particle.

Pressure of Gas:
Pressure is defined as the force exerted per unit area. In the case of a gas, it is the force exerted by the gas molecules on the walls of the container.

Relationship between Mean Free Path and Pressure:
There is an inverse relationship between the mean free path of atoms and the pressure of the gas. When the mean free path increases, the pressure decreases and vice versa.

Explanation:
When the mean free path of atoms is doubled, it means that the atoms are traveling twice as far before they collide with other atoms or molecules. This implies that there are fewer collisions per unit time, resulting in a decrease in the force exerted by the gas molecules on the walls of the container.

Mathematical Relationship:
Let's consider the mathematical relationship between mean free path and pressure using the ideal gas law.

The ideal gas law states that:
PV = nRT

Where:
P = pressure
V = volume
n = number of moles
R = ideal gas constant
T = temperature

If we rearrange the equation, we get:
P = (n/V) * RT

In this equation, (n/V) represents the number density of the gas, which is the number of moles per unit volume.

Impact of Mean Free Path on Number Density:
When the mean free path is doubled, it implies that the distance between atoms is greater. Consequently, the number of atoms per unit volume decreases, resulting in a decrease in the number density of the gas.

Impact of Number Density on Pressure:
As the number density decreases, there are fewer gas molecules colliding with the walls of the container. This leads to a decrease in the force exerted by the gas molecules on a given area of the container walls, resulting in a decrease in pressure.

Multiplicative Factor:
To determine the multiplicative factor by which the pressure changes, we can consider the ratio of the new pressure to the initial pressure.

Let's assume the initial pressure is P1 and the final pressure is P2.

P1 = (n/V)1 * RT
P2 = (n/V)2 * RT

Since pressure is inversely proportional to the number density, we can write:
P2 = (n/V)1 * (V/n)2 * RT
P2 = P1 * (V/n)2 * (n/V)1
P2 = P1 * (V/V)2

Since V/V is equal to 1, we have:
P2 = P1 * 1
P2 = P1

Therefore, the pressure does not change when the mean free path of atoms is doubled. Hence, the correct answer is 1, not 0.5 as stated.