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A container is divided into two equal part I and II by a partition with a small hole of diameter d. The two partitions are filled with same ideal gas, but held at temperature $\mathrm{T}_{\mathrm{I}}=150 \mathrm{~K}$ and $\mathrm{T}_{\mathrm{II}}=300 \mathrm{~K}$ by connecting to heat reservoirs. Let $\lambda_{\mathrm{I}}$ and $\lambda_{\mathrm{II}}$ be the mean free paths of the gas particles in the two parts such that $\mathrm{d}>\lambda_{\mathrm{I}}$ and $\mathrm{d}>\lambda_{\mathrm{II}}$. Then $\lambda_{\mathrm{I}} / \lambda_{\mathrm{II}}$ is close to $-$
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2580 Upvotes
Verified Answer
The correct answer is:
$0.7$
As dimension of hole is very small than mean path, then at equilibrium effusion rate of gas in both direction must be equal.
For this $\quad \frac{\mathrm{P}_{1}}{\sqrt{\mathrm{T}_{1}}}=\frac{\mathrm{P}_{2}}{\sqrt{\mathrm{T}_{2}}}$
Mean free path $\propto \frac{\mathrm{T}}{\mathrm{P}}$
$$
\begin{array}{c}
\frac{\lambda_{1}}{\lambda_{2}}=\frac{T_{1}}{T_{2}} \times \frac{P_{2}}{P_{1}} \\
\frac{T_{1}}{T_{2}} \times \frac{\sqrt{T_{2}}}{\sqrt{T_{1}}} \\
\frac{\lambda_{1}}{\lambda_{2}}=\sqrt{\frac{T_{1}}{T_{2}}}=\sqrt{\frac{150}{300}}=0.7
\end{array}
$$
For this $\quad \frac{\mathrm{P}_{1}}{\sqrt{\mathrm{T}_{1}}}=\frac{\mathrm{P}_{2}}{\sqrt{\mathrm{T}_{2}}}$
Mean free path $\propto \frac{\mathrm{T}}{\mathrm{P}}$
$$
\begin{array}{c}
\frac{\lambda_{1}}{\lambda_{2}}=\frac{T_{1}}{T_{2}} \times \frac{P_{2}}{P_{1}} \\
\frac{T_{1}}{T_{2}} \times \frac{\sqrt{T_{2}}}{\sqrt{T_{1}}} \\
\frac{\lambda_{1}}{\lambda_{2}}=\sqrt{\frac{T_{1}}{T_{2}}}=\sqrt{\frac{150}{300}}=0.7
\end{array}
$$
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