At a given temperature, the average kinetic energy of `H_(2)` molecules is 3.742 `kJ*mol^(-1)`. Calculate root mean square velocity of `H_(2)` molecule at this temperature.

Explanation

Suppose, at TK, the average kinetic energy of `H_(2)` gas molecules=3.742 `kJ*mol^(-1)`
`therefore(3)/(2)RT=3.742xx10^(3)J*mol^(-1)`
or , `(3)/(2)xx8.314xxT=3.742xx10^(3)" "therefore T=300K`
Therefore, the root meann square velocity of `H_(2)` molecule
at `300K=sqrt((3RT)/(M))=sqrt((3xx8.314xxJ*mol^(-1)*K^(-1)xx300K)/(2g*mol^(-1))`
`=sqrt((3x8.314xx10^(3)g*m^(2)*s^(-2)xx300)/(2g))=1934.2m*s^(-1)`