For two gases A and B with molecular weights MA and MB, it is observed that at a certain temperature…

Problem : For two gases A and B with molecular weights MA and MB, it is observed that at a certain temperature T1 the mean velocity of A is equal to the root mean square velocity of B. thus the mean velocity of A can be made equal to the mean velocity of B if

(A) A is at temperature T and B at T’,  T > T’

(B) A is lowered to a temperature T2 , T2 < T while B is at T

(C) Both A and B are raised to a higher temperature

(D) Both A and B are placed at lower temperature

Ans: (B)

Solution: $\large (v_{avg})_A = \sqrt{\frac{8RT}{\pi M_A}}$

$\large (v_{rms})_B = \sqrt{\frac{3 R T}{M_B}}$

$\large \frac{8}{3\pi} = \frac{M_A}{M_B}$

For A , $\large (v_{avg})_A = \sqrt{\frac{8 R T_2}{\pi M_A}}$

For B , $\large (v_{avg})_A = \sqrt{\frac{8 R T}{\pi M_B}}$

$\large \frac{T_2}{T} = \frac{M_A}{M_B}= \frac{8}{3\pi}$

$\large T_2 = \frac{8}{3\pi} T $

T2 < T