ExamBro
ExamBro
JEE Mains · Physics · STD 11 - Kinetic theory of gases
  • JEE Mains 2024
  • Hard
  • MCQ

The total kinetic energy of \(1\) mole of oxygen at \(27^{\circ} \mathrm{C}\) is : [Use universal gas constant \((R)=8.31 \mathrm{~J} / \mathrm{mole} \mathrm{K}\) ]

Pick an option to check your answer.
Verified Solution

Answer & Solution

Correct Answer

(C)  \(6232.5 \mathrm{~J}\)

Step-by-step Solution

Detailed explanation

\( \text { Kinetic energy }=\frac{f}{2} \mathrm{nRT} \)

\( =\frac{5}{2} \times 1 \times 8.31 \times 300 \mathrm{~J} \)

\( =6232.5 \mathrm{~J}\)

Same subject
Explore more questions on app