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.
Answer & Solution
Correct Answer
(C) \(6232.5 \mathrm{~J}\)
Step-by-step Solution
Detailed explanation
\( \text { Kinetic energy }=\frac{f}{2} \mathrm{nRT} \)
\( =\frac{5}{2} \times 1 \times 8.31 \times 300 \mathrm{~J} \)
\( =6232.5 \mathrm{~J}\)
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