Question

A mixture of two gases A and B is in a container at a constant temperature. Gas A is diatomic and B is monoatomic. The ratio of molecular masses of A and B is 4. The ratio of the r.m.s. speed of A and B is –

• A. 1 : 1
• B. 1 : $\sqrt{2}$
• C. $\sqrt{2}$: 1
• D. 1 : 2

Answer 1

Answer: (D)

1 : 2

Answer 2

Given $\frac{(M_{w})_{A}}{(M_{w})_{B}}$ = 4

$\frac{(V_{r.m.s.})_{A}}{(V_{r.m.s.})_{B}}$ =$\sqrt{\frac{(M_{w})_{A}}{(M_{w})_{B}}}$=$\frac{1}{2}$