JEE Mains · Physics · STD 11 - 12 . kinetic theory of gases
The total kinetic energy of \(1\) mole of oxygen at \(27^{\circ} \mathrm{C}\) is : [Use universal gas constant \((R)=8.31 \mathrm{~J} / \mathrm{mole} \mathrm{K}\) ]
- A \(6845.5 \mathrm{~J}\)
- B \(5942.0 \mathrm{~J}\)
- C \(6232.5 \mathrm{~J}\)
- D \(5670.5 \mathrm{~J}\)
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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