The average velocity of nitrogen at `27^@C` is `0.3ms^(-1)`. At what temperature it will be `0.6 ms^(-1)`?

Explanation

Average velocity is given as `vecC = sqrt((8 RT)/(pi M))`
The ratio of velocities at two different temperatures
for the same gas is given as ` ( barC_1)/(C_2) = sqrt((T_1)/(T_2))`
Substituting the values, the ratio is` (0.3 )/( 0.6 ) = sqrt((300 )/( T_2))= 1/2`
The temperature,` T_2 = 2^2 xx 300`
`= 1200K = 927^@ C`