Consider an ideal gas at pressure p, volume V | vedclass

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(B) The formula for the mean free path $\lambda$ of an ideal gas is given by $\lambda = \frac{k_B T}{\sqrt{2} \pi d^2 p}$, where $d$ is the diameter o

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$\frac{L}{4}$

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(B) The formula for the mean free path $\lambda$ of an ideal gas is given by $\lambda = \frac{k_B T}{\sqrt{2} \pi d^2 p}$, where $d$ is the diameter of the molecule $(d = 2r)$, $k_B$ is the Boltzmann constant, $T$ is temperature, and $p$ is pressure.Substituting $d = 2r$, we get $\lambda = \frac{k_B T}{\sqrt{2} \pi (2r)^2 p} = \frac{k_B T}{4 \sqrt{2} \pi r^2 p}$.Thus, the initial mean free path is $L = \frac{k_B T}{4 \sqrt{2} \pi r^2 p}$.According to the problem, the new radius $r' = 2r$, new pressure $p' = 2p$, and new temperature $T' = 2T$.The new mean free path $\lambda'$ is given by:$\lambda' = \frac{k_B (2T)}{4 \sqrt{2} \pi (2r)^2 (2p)}$$\lambda' = \frac{2 k_B T}{4 \sqrt{2} \pi (4r^2) (2p)}$$\lambda' = \frac{2}{8} \cdot \frac{k_B T}{4 \sqrt{2} \pi r^2 p} = \frac{1}{4} L$.Therefore, the new mean free path is $\frac{L}{4}$.

Consider an ideal gas at pressure $p$, volume $V$ and temperature $T$. The mean free path for molecules of the gas is $L$. If the radius of gas molecules, as well as pressure, volume and temperature of the gas are doubled, then the mean free path will be

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