Kinetic energy in $\mathrm{kJ}$ of $280 \mathrm{g}$ of ${\mathrm{N}}_2$ at ${27}^{\circ }\mathrm{C}$ is approximately (R=8.314 Jmol^-1K^-1)

K $·\mathrm{E}$ =$\frac{3}{2}$ n R T

 ${\mathrm{n}}_{{\mathrm{N}}_2}=\frac{280}{28}=10\text{ moles }$ 

 T=27^o$\mathrm{C}=300 \mathrm{K}$ 

K.E. =$\frac{3}{2}\times 10\times 8.314\times 300$ 

=$37413 \mathrm{J}$ 

 =$37.4\mathrm{KJ}$ 

Hence, Option(B) is the answer.