Three closed vessels `A, B` and `C` are 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 (where `M` is the mass of an oxygen molecules)

Options

A. `((V_1 + V_2))/2`
B. `V_1`
C. `(V_1 + V_2)^(1//2)`
D. `sqrt((3KT)/M)`

Answer

B

Explanation

The average speed of molecules of an ideal gas is given by ` lt V gt = sqrt((8RT)/(pi M))`
ie., `lt V gt prop sqrtT` for same gas. Since temperature of A and C are same, average speed of `O_2` molecules will be equal A and C, i.e., `v_1`