Gas Properties: Mean Free Path vs Number Density

The question asks for the relationship between the number densities ($n_A$ and $n_B$) of two ideal gases (A and B), given information about their mean free paths ($\lambda$) and molecular diameters ($d$).

Mean Free Path Formula

The mean free path ($\lambda$) of molecules in an ideal gas is inversely proportional to the square of the molecular diameter ($d$) and the number density ($n$). The formula is:

$ \lambda \approx \frac{1}{\sqrt{2} \pi d^2 n} $

Relating Gases A and B

We can write the mean free path formula for both gases:

$ \lambda_A = \frac{1}{\sqrt{2} \pi d_A^2 n_A} $

$ \lambda_B = \frac{1}{\sqrt{2} \pi d_B^2 n_B} $

Using Given Conditions

The problem states:

• The mean free path of gas A is half that of gas B: $ \lambda_A = \frac{1}{2}\lambda_B $
• The diameter of gas A molecules is twice that of gas B: $ d_A = 2d_B $

Deriving Number Density Relationship

Let's find the ratio of the mean free paths:

$ \frac{\lambda_A}{\lambda_B} = \frac{\frac{1}{\sqrt{2} \pi d_A^2 n_A}}{\frac{1}{\sqrt{2} \pi d_B^2 n_B}} = \frac{d_B^2 n_B}{d_A^2 n_A} $

Substitute the given conditions into the ratio:

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

Simplify the equation:

$ \frac{1}{2} = \frac{d_B^2 n_B}{4d_B^2 n_A} $

Cancel out $d_B^2$:

$ \frac{1}{2} = \frac{n_B}{4n_A} $

Now, solve for $n_A$:

$ 4n_A = 2n_B $

$ n_A = \frac{2}{4}n_B $

$ n_A = \frac{1}{2}n_B $

Conclusion

The correct relationship between the number densities is $n_A = \frac{1}{2}n_B$. This corresponds to Option 3.