The root mean square speed of hydrogen is `sqrt(5)` times than that of nitrogen. If T is the temperature of the gas, then :

Options

A. `T_(H_(2)) = T_(N_(2))`
B. `T_(H_(2)) gt T_(N_(2))`
C. `T_(H_(2)) lt T_(N_(2))`
D. `T_(H_(2)) lt sqrt7 T_(N_(2))`

Answer

C

Explanation

`V_(rm s) = sqrt((3RT)/(M)) = ((V_(rm s))_(H_(2)))/((V_(rm s))_(N_(2))) = sqrt((T_(H_(2)))/(M_(H_(2))) xx (M_(N_(2)))/(T_(N_(2)))), (V_(rm s))_(H_(2)) = sqrt5 (V_(rm s))_(N_(2))`
`:. ((V_(rm s))_(H_(2)))/((V_(rm s))_(N_(2))) xx sqrt5 = sqrt((T_(H_(2)))/(M_(H_(2))) xx (M_(N_(2)))/(T_(N_(2)))), (sqrt5)/(1) = sqrt((T_(H_(2)))/(T_(N_(2))) xx 14) rArr 5 = (T_(H_(2)))/(T_(N_(2))) xx 14`
`T_(N_(2)) xx 5 = T_(H_(2)) xx 14 :. T_(N_(2)) gt T_(H_(2))`