ExamBro
ExamBro
TS EAMCET · Chemistry · States of Matter

Kinetic energy in \(\mathrm{kJ}\) of \(280 \mathrm{~g}\) of \(\mathrm{N}_2\) at \(27^{\circ} \mathrm{C}\) is approximately \(\left(R=8.314 \mathrm{~J} \mathrm{~mol}^1 \mathrm{~K}^1\right)\)

  1. A 18.7
  2. B 37.4
  3. C 56.1
  4. D 74.8
Verified Solution

Answer & Solution

Correct Answer

(B) 37.4

Step-by-step Solution

Detailed explanation

\[ \begin{aligned} \text { } 280 \mathrm{~g} \text { of } \mathrm{N}_2=\frac{280}{28}=10 \text { moles of } \mathrm{N}_2 \end{aligned} \]…