Discussion Forum

 Kinetic energy in kJ of 280 g of $N_{2}$ at $27^{0} C$ is approximately (R=8.314 J mol^-1K^-1)

Answer:

Explanation:

280g  of $N_{2}$=$\frac{280}{28}=10$ moles of $N_{2}$

 $27^{0} C= (27+273)K=300K$

 K.E =   $\frac{3nRT}{2}$ =  3 x number of mol x 3 x moles of gas x gas constant x temperature /2

  = $\frac{3 \times 10 \times 8.314 \times 300}{2}=37.4 \times 10^{3} J$=37.4  kJ