The pressure of an ideal gas is increased by keeping the temperature constant. What happens to the mean free path of the gas molecules?

Solution

1. The mean free path $\lambda$ of an ideal gas is defined by the formula $\lambda =\frac{k_{B}T}{\sqrt{2}\pi d^{2}P}$.
2. Given that the temperature $T$ is held constant, the relationship simplifies to $\lambda \propto \frac{1}{P}$.
3. As the pressure $P$ increases, the mean free path $\lambda$ must decrease.
4. Hence the answer is (B).