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 B are n_A and n_B , respectively, then the correct option is :

Step-by-step solution

The mean free path of a gas molecule is given by = 1 2 d^2 n Given _A = 1 2 _B and d_A = 2d_B Substituting the formula for mean free path: 1 2 d_A^2 n_A = 1 2 ( 1 2 d_B^2 n_B ) 1 d_A^2 n_A = 1 2 d_B^2 n_B Substituting d_A = 2d_B : 1 (2d_B)^2 n_A = 1 2 d_B^2 n_B 1 4 d_B^2 n_A = 1 2 d_B^2 n_B 4 n_A = 2 n_B n_A = 1 2 n_B Answer: n_A= 1 2 n_B