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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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The correct answer is:
$\frac{L}{4}$
Given, initial values of pressure, volume,
temperature, radius of molecule of gas and mean free path are $P p, V, T, r$ and $L$, respectively.
$\therefore$ Mean free path,
$\begin{aligned} \lambda & =\frac{2 \sqrt{2} R T}{\pi r^2 N p} \\ \therefore \quad L & =\frac{2 \sqrt{2} R T}{\pi r^2 N p}\end{aligned}$
According to question if the radius of gas molecules, as well as pressure, volume and temperature of gas are doubled then
$\begin{aligned} \lambda^{\prime} & =\frac{2 \sqrt{2} R \cdot 2 T}{\pi \times 4 r^2 \times N \times 2 p} \\ & =\frac{1}{4} \cdot \frac{2 \sqrt{2} R T}{\pi r^2 N p}=\frac{L}{4}\end{aligned}$
(From Eq. (i))
Hence, the mean free path will be $\frac{L}{4}$.
temperature, radius of molecule of gas and mean free path are $P p, V, T, r$ and $L$, respectively.
$\therefore$ Mean free path,
$\begin{aligned} \lambda & =\frac{2 \sqrt{2} R T}{\pi r^2 N p} \\ \therefore \quad L & =\frac{2 \sqrt{2} R T}{\pi r^2 N p}\end{aligned}$
According to question if the radius of gas molecules, as well as pressure, volume and temperature of gas are doubled then
$\begin{aligned} \lambda^{\prime} & =\frac{2 \sqrt{2} R \cdot 2 T}{\pi \times 4 r^2 \times N \times 2 p} \\ & =\frac{1}{4} \cdot \frac{2 \sqrt{2} R T}{\pi r^2 N p}=\frac{L}{4}\end{aligned}$
(From Eq. (i))
Hence, the mean free path will be $\frac{L}{4}$.
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