If the RMS speed of nitrogen at a certain temperature is 3000 `ms^(-1)` the approximate kinetic energy of one mole of nitrogen at that temperature in Kj is ( assume nitrogen as ideal gas)

Options

A. `9.0`
B. `126.0`
C. `90.0`
D. `12.6`

Answer

B

Explanation

Given `U_("rms")= 3000 `m/s
`because U_("rms")= sqrt((3RT)/(M))`
`9xx10^(6)m^(2)//s^(2)= (3RT)/(28g//"mol")`
`[because `Molar mass of `N_(2)` = 28 g/mol]
`:. RT = 28xx3xx10^(3)` j/mol
`[because (m^(2))/(s^(2))` = J/kg , 1kg = 1000 g]
As we know that K.E for 1 mole of an ideal gas = `(3)/(2) RT`
`K.E = (3)/(2) xx(28xx10^(3))/(1000) kj = 126 kj`