If the average velocity of `N_(2)` molecules is `0.3 ms^(-1) at 27^(@) C` , then the velocity of `0.6 ms^(-1)` will take place at

Options

A. `273 K`
B. `927 K`
C. `1000 K`
D. `1200 K`

Answer

D

Explanation

According to the kinetic molecular theory of gases, we have
`u_(aV) = sqrt((8RT)/(pi M))`
i.e., `u_(aV) prop sqrt(T)`
`:. ((u_(aV))_(1))/((u_(aV))_(2)) = sqrt((T_(1))/(T_(2)))`
`(0.3 m s^(-1))/(0.6 m s^(-1)) = sqrt((27 + 273))/(T_(2))`
`(1)/(2) = sqrt((300)/(T_(2)))` ltbrtgt or `(1)/(4) = (300)/(T_(2))`
or `T_(2) = 1200 K`