Physics · 2024
JEE Main · 27 January 2024, Shift 2 · Q39
The total kinetic energy of 1 mole of oxygen at 27^° C is: [Use universal gas constant ( R )=8.31 J / mole K ]
The total kinetic energy of $\displaystyle 1$ mole of oxygen at $\displaystyle 27^{\circ} \mathrm{C}$ is : [Use universal gas constant $\displaystyle (\mathrm{R})=8.31 \mathrm{~J} /$ mole K ]
Official answer
From NTA’s final answer key for this paper.
(4)
$\displaystyle 6232.5$ J
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