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(A) The mean free path $\lambda$ of a gas molecule is given by the formula $\lambda = \frac{1}{\sqrt{2} \pi d^2 n}$,where $d$ is the diameter of the m

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$4: 1$

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(A) The mean free path $\lambda$ of a gas molecule is given by the formula $\lambda = \frac{1}{\sqrt{2} \pi d^2 n}$,where $d$ is the diameter of the molecule and $n$ is the number density (concentration) of the molecules.Given that the concentration $n$ is the same for both gases,the mean free path is inversely proportional to the square of the diameter: $\lambda \propto \frac{1}{d^2}$.Therefore,the ratio of the mean free paths is $\frac{\lambda_1}{\lambda_2} = \left(\frac{d_2}{d_1}\right)^2$.Given the ratio of diameters $\frac{d_1}{d_2} = \frac{1}{2}$,we have $\frac{d_2}{d_1} = \frac{2}{1}$.Substituting this into the ratio formula: $\frac{\lambda_1}{\lambda_2} = \left(\frac{2}{1}\right)^2 = \frac{4}{1}$.Thus,the ratio of their mean free paths is $4: 1$.

For the given concentration,if the ratio of the diameters of the molecules of two gases is $1: 2$,then the ratio of their mean free paths is

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