Calculate the temperature at which the average velocity of oxygen equals that of hydrogen at 25 K.

Explanation

Average velocity, `u_("avg")=(8RT"/"pi M)^(1/2)`
Let `u_(avg_1)" at "T_(1)K(25K)" and "u_(avg_2)" at "T_(2) K` be the average velocities of `O_(2)" and "H_(2)` respectively.
When the average velocities are equal, `(u_(avg_1))/(u_(avg_2))=((T_(1)"/"M_(1))/(T_(2)"/"M_(2)))^(1/2)= 1`
Therefore, `T_(1)"/"M_(1)= T_(2)"/"M_(2)` where `M_(1)" and "M_(2)` are the molar masses of hydrogen and oxygen respectively.
`T_(1)"/"32g mol^(-1) = T_(2)"/"2g mol^(-1) implies T_(1)= (32"/"2)T_(2)= 16xx 25= 400 K`.