The mean free path of molecules in an ideal gas A is half that of another ideal gas B. The diameter of the spherical molecules of gas A is twice the diameter of the molecules of B. If number densities of the gases A and

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Quantrex Academy · Accepted Answer

Correct answer: A. n_A=(1)/(2)n_B

The mean free path of a gas molecule is given by

\lambda = \dfrac{1}{\sqrt{2}\pi d^2 n}

Given

\lambda_A = \dfrac{1}{2}\lambda_B

and

d_A = 2d_B

Substituting the formula for mean free path:

\dfrac{1}{\sqrt{2}\pi d_A^2 n_A} = \dfrac{1}{2} \left( \dfrac{1}{\sqrt{2}\pi d_B^2 n_B} \right)

\dfrac{1}{d_A^2 n_A} = \dfrac{1}{2 d_B^2 n_B}

Substituting

d_A = 2d_B

:

\dfrac{1}{(2d_B)^2 n_A} = \dfrac{1}{2 d_B^2 n_B}

\dfrac{1}{4 d_B^2 n_A} = \dfrac{1}{2 d_B^2 n_B}

4 n_A = 2 n_B

n_A = \dfrac{1}{2} n_B

Answer:

n_A=\dfrac{1}{2}n_B

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