At what temperature, the translation kinetic energy of $14 g$ of nitrogen will be the same as that of $32 g m$ of oxygen at $300 K ?$

150 K

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300 K

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600 K

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900 K

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Solution

The correct option is C $600 K$
$\frac{m_{1} T_{1}}{M_{1}} = \frac{m_{2} T_{2}}{M_{2}}$

Where m = mass of molecule , M = Molecular Wt of the molecule

$\frac{14 \times T_{1}}{28} = \frac{32 \times 300}{32}$

$T_{1} = 600 K$

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