Calculate the temperature at which the average velocity of oxygen equals that of hydrogen at `20 K`.

Explanation

We know that
Average velocity `=((3RT)/(M))^(1//2)`
i.e., average velocity ` prop ((T)/(M))^(1//2)`
Let `C_(O_(2))` and `C_(H_(2))` be the average velocities of `O_(2)` and `H_(2)`, respectively, then
`(C_(O_(2)))/(C_(H_(2)))=[((T_(1))/(M_(O_(2))))/((T_(2))/(M_(H_(2))))]^(1//2)=1`
or `(T_(1))/(M_(O_(2))=(T_(2))/(M_(H_(2))`
`(T_(1))/(32 g mol^(-1))=(T_(2))/(2 g mol^(-1))`
`:. T_(1)=((32)/(2))T_(2)=16xx20 K=320 K`