Three closed vessels A, B and C are at the same temperature T and contain gases. Vessel A contains only `O_(2)`, B contains only `N_(2)` and C contains a mixture of equal quantities of `O_(2)` and `N_(2)`. If the rms speed of `O_(2)` molecules in vessel A is `v_(1)` and that of `N_(2)` molecules in vessel B is `v_(2)` then the rms speed of `O_(2)` molecules in vessel C is

Options

A. `((v_(1) + v_(2)))/(2)`
B. `v_(1)`
C. `(v_(1)v_(2))`
D. `(v_(1))/(2)`

Answer

B

Explanation

Root mean square velocity of any gas is given by
`V_("rms") = sqrt((3RT)/(M))`
Where, M = molecular weight.
As, rms speed of any gas is depends on the temperature of gas and molecular weight of gas.
In a mixture of gases (say `N_(2)` and `O_(2)`) at a constant temperature, rms speed is independent to the quantity (moles) of gas present in the gas mixture. So, in this problems the rms speed of `O_(2)` in vessel C is `v_(1)`.
Hence, the correct option is (b).