The total kinetic energy of 1 mole of oxygen at 27 is: [Use universal gas constant (R) = 8.31 J/ mole K] (a) 6232.5 J (b) 5670.5 J (c) 6845.5 J (d) 5942.0 J

Answer: (a). Oxygen is diatomic: 3 translational + 2 rotational degrees of freedom, so total KE per mole = 5/2RT. T = 300 K KE = 2.5 × 8.31 × 300 = 6232.5 J So the answer is (a) 6232.5 J.

The total kinetic energy of 1 mole of oxygen at 27 is: [Use universal… | JEE Main 2024 (27 Jan, Shift 2)

Q19JEE Main 2024 · 27 JAN · SHIFT 2Medium

Pressure & Kinetic Energy of a Gas

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

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Answer: (a)

Oxygen is diatomic: 3 translational + 2 rotational degrees of freedom, so total KE per mole $= \frac{5}{2}RT$ .
$T = 300$ K
$KE = 2.5 \times 8.31 \times 300 = 6232.5$ J
So the answer is (a) 6232.5 J.

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