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)

Step-by-step solution

Given, U_ rms =3000 ~m / s aligned & U_ rms = 3 R T M & 9 10^6 ~m ^2 ~s ^2= 3 R T 28 ~g / mol & [ Molar mass of N ₂=28 ~g / mol ] & R T=28 3 10^3 ~J / mol & [ m ^2 ~s ^2 = J / kg , 1 ~kg =1000 ~g ] aligned As we know that, K.E for 1 mole of an ideal gas = 3 2 R T K . E = 3 2 28 10^3 3 1000 ~kJ =126 ~kJ