The mean free path of the molecules of two gases having molecular diameters $$1Ao$$ and $$2Ao$$ is N: $$1$$. Gases are under identical conditions of temperature, pressure and volume. Find N.

Mean free path

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Question

The mean free path of the molecules of two gases having molecular diameters $1 \overset{o}{A}$ and $2 \overset{o}{A}$ is N: 1. Gases are under identical conditions of temperature, pressure and volume. Find N.

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Solution

Mean free path is given by $λ = \frac{1}{\sqrt{2} d^{2} n} α \frac{1}{d^{2}}$
Where, n = number density of molecules and d = molecular diameter.
Since, pressure, volume, temperature are same, using PV= nRT, number of moles (and hence number of molecules) is also same.
Hence, required ratio is $\frac{λ_{1}}{λ_{2}} = \frac{d_{2}^{2}}{d_{1}^{2}} = 4$ .

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Mean free path

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The mean free path of the molecules of two gases having molecular diameters 1Ao and 2Ao is N: 1. Gases are under identical conditions of temperature, pressure and volume. Find N.

Mean free path is given by λ=12d2nα1d2Where, n = number density of molecules and d = molecular diameter.Since, pressure, volume, temperature are same, using PV= nRT, number of moles (and hence number of molecules) is also same.Hence, required ratio is λ1λ2=d22d12=4.

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The mean free path of the molecules of two gases having molecular diameters $1 \overset{o}{A}$ and $2 \overset{o}{A}$ is N: 1. Gases are under identical conditions of temperature, pressure and volume. Find N.