Question

Three closed vessels, A, B and C at the same temperature T and contain gases which obey the Maxwellian distribution of velocities. Vessel A contains only $O_2$, B only $N_2$ and C a mixture of equal quantities of $O_2$ and $N_2$. If the average speed of the $O_2$ molecules in vessel A is $v_1$, that of the $N_2$ molecules in vessel B is $v_2$, the average speed of the $O_2$ molecules in vessel C is

• A. $\frac{(v_1+v_2)}{2}$
• B. $v_1$
• C. $(v_1{v}_2{)}^{\frac{1}{2}}$
• D. $\sqrt{\frac{3kT}{M}}$

Answer 1

The correct option is B $v_1$

The average speed of molecules of an ideal gas is given by
$<v>=\sqrt{\frac{8RT}{\pi M}}i.e.,<v>\propto \sqrt{T}$ for same gas.

Since, temperatures of A and C are same, average speed of $O_2$ molecules will be equal in A and C i.e., $v_1$.